http://www.physicsbook.gatech.edu/api.php?action=feedcontributions&feedformat=atom&user=ChazPhysics Book - User contributions [en]2026-04-10T15:29:59ZUser contributionsMediaWiki 1.42.7http://www.physicsbook.gatech.edu/index.php?title=Main_Page&diff=46863Main Page2024-12-05T14:27:45Z<p>Chaz: /* Special Relativity and the Lorentz Transformation */</p>
<hr />
<div>__NOTOC__<br />
= '''Georgia Tech Student Wiki for Introductory Physics.''' =<br />
<br />
This resource was created so that students can contribute and curate content to help those with limited or no access to a textbook. When reading this website, please correct any errors you may come across. If you read something that isn't clear, please consider revising it for future students!<br />
<br />
Looking to make a contribution?<br />
#Pick one of the topics from intro physics listed below<br />
#Add content to that topic or improve the quality of what is already there.<br />
#Need to make a new topic? Edit this page and add it to the list under the appropriate category. Then copy and paste the default [[Template]] into your new page and start editing.<br />
<br />
Please remember that this is not a textbook and you are not limited to expressing your ideas with only text and equations. Whenever possible embed: pictures, videos, diagrams, simulations, computational models (e.g. Glowscript), and whatever content you think makes learning physics easier for other students.<br />
<br />
== Source Material ==<br />
All of the content added to this resource must be in the public domain or similar free resource. If you are unsure about a source, contact the original author for permission. That said, there is a surprisingly large amount of introductory physics content scattered across the web. Here is an incomplete list of intro physics resources (please update as needed).<br />
* A physics resource written by experts for an expert audience [https://en.wikipedia.org/wiki/Portal:Physics Physics Portal]<br />
* A wiki written for students by a physics expert [http://p3server.pa.msu.edu/coursewiki/doku.php?id=183_notes MSU Physics Wiki]<br />
* A wiki book on modern physics [https://en.wikibooks.org/wiki/Modern_Physics Modern Physics Wiki]<br />
* A collection of 26 volumes of lecture notes by Prof. Wheeler of Reed College [https://rdc.reed.edu/c/wheeler/home/] <br />
* The MIT open courseware for intro physics [http://ocw.mit.edu/resources/res-8-002-a-wikitextbook-for-introductory-mechanics-fall-2009/index.htm MITOCW Wiki]<br />
* An online concept map of intro physics [http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html HyperPhysics]<br />
* Interactive physics simulations [https://phet.colorado.edu/en/simulations/category/physics PhET]<br />
* OpenStax intro physics textbooks: [https://openstax.org/details/books/university-physics-volume-1 Vol1], [https://openstax.org/details/books/university-physics-volume-2 Vol2], [https://openstax.org/details/books/university-physics-volume-3 Vol3]<br />
* The Open Source Physics project is a collection of online physics resources [http://www.opensourcephysics.org/ OSP]<br />
* A resource guide compiled by the [http://www.aapt.org/ AAPT] for educators [http://www.compadre.org/ ComPADRE]<br />
* The Feynman lectures on physics are free to read [http://www.feynmanlectures.caltech.edu/ Feynman]<br />
* Final Study Guide for Modern Physics II created by a lab TA [https://docs.google.com/document/d/1_6GktDPq5tiNFFYs_ZjgjxBAWVQYaXp_2Imha4_nSyc/edit?usp=sharing Modern Physics II Final Study Guide]<br />
<br />
== Resources ==<br />
* Commonly used wiki commands [https://en.wikipedia.org/wiki/Help:Cheatsheet Wiki Cheatsheet]<br />
* A guide to representing equations in math mode [https://en.wikipedia.org/wiki/Help:Displaying_a_formula Wiki Math Mode]<br />
* A page to keep track of all the physics [[Constants]]<br />
* A listing of [[Notable Scientist]] with links to their individual pages <br />
<br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 1==<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====GlowScript 101====<br />
<div class="mw-collapsible-content"><br />
*[[Python Syntax]]<br />
*[[GlowScript]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====VPython====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[VPython]]<br />
*[[VPython basics]]<br />
*[[VPython Common Errors and Troubleshooting]]<br />
*[[VPython Functions]]<br />
*[[VPython Lists]]<br />
*[[VPython Loops]]<br />
*[[VPython Multithreading]]<br />
*[[VPython Animation]]<br />
*[[VPython Objects]]<br />
*[[VPython 3D Objects]]<br />
*[[VPython Reference]]<br />
*[[VPython MapReduceFilter]]<br />
*[[VPython GUIs]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Vectors and Units====<br />
<div class="mw-collapsible-content"><br />
*[[Vectors]]<br />
*[[SI Units]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Interactions====<br />
<div class="mw-collapsible-content"><br />
*[[Types of Interactions and How to Detect Them]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Velocity and Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Newton's First Law of Motion]]<br />
*[[Mass]]<br />
*[[Velocity]]<br />
*[[Speed]]<br />
*[[Speed vs Velocity]]<br />
*[[Relative Velocity]]<br />
*[[Derivation of Average Velocity]]<br />
*[[2-Dimensional Motion]]<br />
*[[3-Dimensional Position and Motion]]<br />
</div><br />
</div><br />
<br />
===Week 2===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Momentum and the Momentum Principle====<br />
<div class="mw-collapsible-content"><br />
*[[Linear Momentum]]<br />
*[[Newton's Second Law: the Momentum Principle]]<br />
*[[Impulse and Momentum]]<br />
*[[Net Force]]<br />
*[[Inertia]]<br />
*[[Acceleration]]<br />
*[[Relativistic Momentum]]<br />
<!-- Kinematics and Projectile Motion relocated to Week 3 per advice of Dr. Greco --><br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Iterative Prediction with a Constant Force====<br />
<div class="mw-collapsible-content"><br />
*[[Iterative Prediction]]<br />
</div><br />
</div><br />
<br />
===Week 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Analytic Prediction with a Constant Force====<br />
<div class="mw-collapsible-content"><br />
<!-- *[[Analytical Prediction]] Deprecated --><br />
*[[Kinematics]]<br />
*[[Projectile Motion]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Iterative Prediction with a Varying Force====<br />
<div class="mw-collapsible-content"><br />
*[[Fundamentals of Iterative Prediction with Varying Force]]<br />
*[[Spring_Force]]<br />
*[[Simple Harmonic Motion]]<br />
<!--*[[Hooke's Law]] folded into simple harmonic motion--><br />
<!--*[[Spring Force]] folded into simple harmonic motion--><br />
*[[Iterative Prediction of Spring-Mass System]]<br />
*[[Terminal Speed]]<br />
*[[Predicting Change in multiple dimensions]]<br />
*[[Two Dimensional Harmonic Motion]]<br />
*[[Determinism]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Fundamental Interactions====<br />
<div class="mw-collapsible-content"><br />
*[[Gravitational Force]]<br />
*[[Gravitational Force Near Earth]]<br />
*[[Gravitational Force in Space and Other Applications]]<br />
*[[3 or More Body Interactions]]<br />
<!--[[Fluid Mechanics]]--><br />
*[[Electric Force]]<br />
*[[Introduction to Magnetic Force]]<br />
*[[Strong and Weak Force]]<br />
*[[Reciprocity]]<br />
*[[Conservation of Momentum]]<br />
</div><br />
</div><br />
<br />
===Week 5===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Properties of Matter====<br />
<div class="mw-collapsible-content"><br />
*[[Kinds of Matter]]<br />
*[[Ball and Spring Model of Matter]]<br />
*[[Density]]<br />
*[[Length and Stiffness of an Interatomic Bond]]<br />
*[[Young's Modulus]]<br />
*[[Speed of Sound in Solids]]<br />
*[[Malleability]]<br />
*[[Ductility]]<br />
*[[Weight]]<br />
*[[Hardness]]<br />
*[[Boiling Point]]<br />
*[[Melting Point]]<br />
*[[Change of State]]<br />
</div><br />
</div><br />
<br />
===Week 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Identifying Forces====<br />
<div class="mw-collapsible-content"><br />
====Isabel Hollhumer F24====<br />
*[[Free Body Diagram]]<br />
*[[Inclined Plane]]<br />
*[[Compression or Normal Force]]<br />
*[[Tension]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Curving Motion====<br />
<div class="mw-collapsible-content"><br />
*[[Curving Motion]]<br />
*[[Centripetal Force and Curving Motion]]<br />
*[[Perpetual Freefall (Orbit)]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Jeet Bhatkar====<br />
<br />
====Energy Principle====<br />
The Energy Principle is a fundamental concept in physics that describes the relationship between different forms of energy and their conservation within a system. Understanding the Energy Principle is crucial for analyzing the motion and interactions of objects in various physical scenarios.<br />
<div class="mw-collapsible-content"><br />
<br />
*[[Kinetic Energy]]<br />
Kinetic energy is the energy an object possesses due to its motion.<br />
*[[Work/Energy]]<br />
Potential energy arises from the position of an object relative to its surroundings. Common forms of potential energy include gravitational potential energy and elastic potential energy.<br />
*[[The Energy Principle]]<br />
Work and energy are closely related concepts. Work (<br />
đ) done on an object is defined as the force (<br />
đč) applied to the object multiplied by the displacement (<br />
đ) of the object in the direction of the force:<br />
The Energy Principle states that the total mechanical energy of a system remains constant if only conservative forces (forces that depend only on the positions of the objects) are acting on the system. <br />
*[[Conservation of Energy]]<br />
The principle of conservation of energy states that the total energy of an isolated system remains constant over time. In other words, energy cannot be created or destroyed, only transformed from one form to another. This principle is a fundamental concept in physics and has wide-ranging applications in mechanics, thermodynamics, and other branches of science.<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Work by Non-Constant Forces====<br />
<div class="mw-collapsible-content"><br />
*[[Work Done By A Nonconstant Force]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Energy]]<br />
*[[Potential Energy of Macroscopic Springs]]<br />
*[[Spring Potential Energy]]<br />
*[[Ball and Spring Model]]<br />
*[[Gravitational Potential Energy]]<br />
*[[Energy Graphs]]<br />
*[[Escape Velocity]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Multiparticle Systems====<br />
<div class="mw-collapsible-content"><br />
*[[Center of Mass]]<br />
*[[Multi-particle analysis of Momentum]]<br />
*[[Potential Energy of a Multiparticle System]]<br />
*[[Work and Energy for an Extended System]]<br />
*[[Internal Energy]]<br />
**[[Potential Energy of a Pair of Neutral Atoms]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Choice of System====<br />
<div class="mw-collapsible-content"><br />
*[[System & Surroundings]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Thermal Energy, Dissipation, and Transfer of Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Thermal Energy]]<br />
*[[Specific Heat]]<br />
*[[Calorific Value(Heat of combustion)]]<br />
*[[First Law of Thermodynamics]]<br />
*[[Second Law of Thermodynamics and Entropy]]<br />
*[[Temperature]]<br />
*[[Transformation of Energy]]<br />
*[[The Maxwell-Boltzmann Distribution]]<br />
*[[Air Resistance]]<br />
*[[The Third Law of Thermodynamics]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Rotational and Vibrational Energy====<br />
<div class="mw-collapsible-content"><br />
*[[Translational, Rotational and Vibrational Energy]]<br />
*[[Rolling Motion]]<br />
</div><br />
</div><br />
<br />
===Week 11===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Different Models of a System====<br />
<div class="mw-collapsible-content"><br />
*[[Point Particle Systems]]<br />
*[[Real Systems]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Friction====<br />
<div class="mw-collapsible-content"><br />
*[[Friction]]<br />
*[[Static Friction]]<br />
*[[Kinetic Friction]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Conservation of Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Conservation of Momentum]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Collisions====<br />
<div class="mw-collapsible-content"><br />
*[[Newton's Third Law of Motion]]<br />
*[[Collisions]]<br />
*[[Elastic Collisions]]<br />
*[[Inelastic Collisions]]<br />
*[[Maximally Inelastic Collision]]<br />
*[[Head-on Collision of Equal Masses]]<br />
*[[Head-on Collision of Unequal Masses]]<br />
*[[Scattering: Collisions in 2D and 3D]]<br />
*[[Rutherford Experiment and Atomic Collisions]]<br />
*[[Coefficient of Restitution]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Rotations====<br />
<div class="mw-collapsible-content"><br />
*[[Rotational Kinematics]]<br />
*[[Eulerian Angles]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Angular Momentum====<br />
<div class="mw-collapsible-content"><br />
*[[Total Angular Momentum]]<br />
*[[Translational Angular Momentum]]<br />
*[[Rotational Angular Momentum]]<br />
*[[The Angular Momentum Principle]]<br />
*[[Angular Impulse]]<br />
*[[Predicting the Position of a Rotating System]]<br />
*[[The Moments of Inertia]]<br />
*[[Right Hand Rule]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Analyzing Motion with and without Torque====<br />
<div class="mw-collapsible-content"><br />
*[[Torque]]<br />
*[[Torque 2]]<br />
*[[Systems with Zero Torque]]<br />
*[[Systems with Nonzero Torque]]<br />
*[[Torque vs Work]]<br />
*[[Gyroscopes]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Introduction to Quantum Concepts====<br />
<div class="mw-collapsible-content"><br />
*[[Bohr Model]]<br />
*[[Energy graphs and the Bohr model]]<br />
*[[Quantized energy levels]]<br />
*[[Electron transitions]]<br />
*[[Entropy]]<br />
</div><br />
</div><br />
</div><br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 2==<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====3D Vectors====<br />
<br />
<div class="mw-collapsible-content"><br />
*[[Vectors]]<br />
*[[Right-Hand Rule]]<br />
*[[Right Hand Rule]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric field====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Field]]<br />
*[[Electric Field and Electric Potential]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric force====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric field of a point particle====<br />
<div class="mw-collapsible-content"><br />
*[[Point Charge]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Superposition====<br />
<div class="mw-collapsible-content"><br />
*[[Superposition Principle]]<br />
*[[Superposition principle]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Dipoles====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Dipole]]<br />
*[[Magnetic Dipole]]<br />
</div><br />
</div><br />
<br />
===Week 2===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Interactions of charged objects====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Field]]<br />
*[[Electric Potential]]<br />
*[[Electric Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Tape experiments====<br />
<div class="mw-collapsible-content"><br />
*[[Polarization]]<br />
*[[Electric Polarization]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Polarization====<br />
<div class="mw-collapsible-content"><br />
*[[Polarization]]<br />
*[[Electric Polarization]]<br />
*[[Polarization of an Atom]]<br />
</div><br />
</div><br />
<br />
===Week 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Conductors and Insulators====<br />
<div class="mw-collapsible-content"><br />
*[[Conductivity and Resistivity]]<br />
*[[Insulators]]<br />
*[[Potential Difference in an Insulator]]<br />
*[[Conductors]]<br />
*[[Polarization of a conductor]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Charging and Discharging====<br />
<div class="mw-collapsible-content"><br />
*[[Charge Transfer]]<br />
*[[Electrostatic Discharge]]<br />
*[[Charged Conductor and Charged Insulator]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Field of a charged rod====<br />
<div class="mw-collapsible-content"><br />
*[[Field of a Charged Rod|Charged Rod]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Field of a charged ring/disk/capacitor====<br />
<div class="mw-collapsible-content"><br />
*[[Charged Ring]]<br />
*[[Charged Disk]]<br />
*[[Charged Capacitor]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Field of a charged sphere====<br />
<div class="mw-collapsible-content"><br />
*[[Charged Spherical Shell]]<br />
*[[Field of a Charged Ball]]<br />
</div><br />
</div><br />
<br />
===Week 5===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential energy====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Energy]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric potential====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential]]<br />
*[[Path Independence of Electric Potential]]<br />
*[[Potential Difference Path Independence, claimed by Aditya Mohile]] <br />
*[[Potential Difference in a Uniform Field]]<br />
*[[Potential Difference of Point Charge in a Non-Uniform Field]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Sign of a potential difference====<br />
<div class="mw-collapsible-content"><br />
*[[Sign of a Potential Difference]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Potential at a single location====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential]]<br />
*[[Potential Difference at One Location]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Path independence and round trip potential====<br />
<div class="mw-collapsible-content"><br />
*[[Path Independence of Electric Potential]]<br />
*[[Potential Difference Path Independence, claimed by Aditya Mohile]]<br />
</div><br />
</div><br />
<br />
===Week 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric field and potential in an insulator====<br />
<div class="mw-collapsible-content"><br />
*[[Potential Difference in an Insulator]]<br />
*[[Electric Field in an Insulator]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Moving charges in a magnetic field====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field]]<br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Biot-Savart Law====<br />
<div class="mw-collapsible-content"><br />
*[[Biot-Savart Law]]<br />
*[[Biot-Savart Law for Currents]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Moving charges, electron current, and conventional current====<br />
<div class="mw-collapsible-content"><br />
*[[Moving Point Charge]]<br />
*[[Current]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic field of a wire====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Long Straight Wire]]<br />
*[[Magnetic Field of a Curved Wire]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Magnetic field of a current-carrying loop====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Loop]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic field of a Charged Disk====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Field of a Disk]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic dipoles====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Dipole Moment]]<br />
*[[Bar Magnet]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Atomic structure of magnets====<br />
<div class="mw-collapsible-content"><br />
*[[Atomic Structure of Magnets]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Circuitry Basics====<br />
<div class="mw-collapsible-content"><br />
*[[Understanding Fundamentals of Current, Voltage, and Resistance]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Steady state current====<br />
<div class="mw-collapsible-content"><br />
*[[Steady State]]<br />
*[[Non Steady State]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Kirchoff's Laws====<br />
<div class="mw-collapsible-content"><br />
*[[Kirchoff's Laws]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric fields and energy in circuits====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Potential Difference]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Macroscopic analysis of circuits====<br />
<div class="mw-collapsible-content"><br />
*[[Series Circuits]]<br />
*[[Parallel Circuits]]<br />
*[[Parallel Circuits vs. Series Circuits*]]<br />
*[[Loop Rule]]<br />
*[[Node Rule]]<br />
*[[Fundamentals of Resistance]]<br />
*[[Problem Solving]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Electric field and potential in circuits with capacitors====<br />
<div class="mw-collapsible-content"><br />
*[[Charging and Discharging a Capacitor]]<br />
*[[RC Circuit]] <br />
*[[R Circuit]]<br />
*[[AC and DC]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic forces on charges and currents====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
*[[Motors and Generators]]<br />
*[[Applying Magnetic Force to Currents]]<br />
*[[Magnetic Force in a Moving Reference Frame]]<br />
*[[Right-Hand Rule]]<br />
*[[Analysis of Railgun vs Coil gun technologies]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Electric and magnetic forces====<br />
<div class="mw-collapsible-content"><br />
*[[Electric Force]]<br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
*[[VPython Modelling of Electric and Magnetic Forces]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Velocity selector====<br />
<div class="mw-collapsible-content"><br />
*[[Lorentz Force]]<br />
*[[Combining Electric and Magnetic Forces]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Hall Effect====<br />
<div class="mw-collapsible-content"><br />
*[[Hall Effect]]<br />
*[[Right-Hand Rule]]<br />
*[[Motional Emf]]<br />
*[[Magnetic Force]]<br />
*[[Magnetic Torque]]<br />
</div><br />
<br />
====Magnetic force====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Force]]<br />
*[[Lorentz Force]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Magnetic torque====<br />
<div class="mw-collapsible-content"><br />
*[[Magnetic Torque]]<br />
*[[Right-Hand Rule]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Gauss's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Gauss's Flux Theorem]]<br />
*[[Gauss's Law]]<br />
*[[Magnetic Flux]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Ampere's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Ampere's Law]]<br />
*[[Ampere-Maxwell Law]]<br />
*[[Magnetic Field of Coaxial Cable Using Ampere's Law]]<br />
*[[Magnetic Field of a Long Thick Wire Using Ampere's Law]]<br />
*[[Magnetic Field of a Toroid Using Ampere's Law]]<br />
*[[Magnetic Field of a Solenoid Using Ampere's Law]]<br />
*[[The Differential Form of Ampere's Law]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Semiconductors====<br />
<div class="mw-collapsible-content"><br />
*[[Semiconductor Devices]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Faraday's Law====<br />
<div class="mw-collapsible-content"><br />
*[[Faraday's Law]]<br />
*[[Motional Emf using Faraday's Law]]<br />
*[[Lenz's Law]]<br />
<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Maxwell's equations====<br />
<div class="mw-collapsible-content"><br />
*[[Gauss's Law]]<br />
*[[Magnetic Flux]]<br />
*[[Ampere's Law]]<br />
*[[Faraday's Law]]<br />
*[[Maxwell's Electromagnetic Theory]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Circuits revisited====<br />
<div class="mw-collapsible-content"><br />
<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
<br />
====Inductors====<br />
<div class="mw-collapsible-content"><br />
*[[Inductors]]<br />
*[[Current in an LC Circuit]]<br />
*[[Current in an RL Circuit]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
==== Electromagnetic Radiation ====<br />
<div class="mw-collapsible-content"><br />
*[[Electromagnetic Radiation]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Sparks in the air====<br />
<div class="mw-collapsible-content"><br />
*[[Sparks in Air]]<br />
*[[Spark Plugs]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Superconductors====<br />
<div class="mw-collapsible-content"><br />
*[[Superconducters]]<br />
*[[Superconductors]]<br />
*[[Meissner effect]]<br />
</div><br />
</div><br />
</div><br />
<br />
<div style="float:left; width:30%; padding:1%;"><br />
<br />
==Physics 3==<br />
<br />
===Week 1===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Classical Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Classical Physics]]<br />
</div><br />
</div><br />
<br />
[[Category:Which Category did you place this in?]]<br />
<br />
===Weeks 2 and 3===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Special Relativity and the Lorentz Transformation====<br />
<div class="mw-collapsible-content"><br />
*[[Frame of Reference]]<br />
<br />
*[[Einstein's Theory of Special Relativity]]<br />
*[[Time Dilation]]<br />
*[[Twin Paradox]]<br />
*[[Lorentz Transformations]]<br />
*[[Relativistic Doppler Effect]]<br />
*[[Einstein's Theory of General Relativity]]<br />
*[[Albert A. Micheleson & Edward W. Morley]]<br />
*[[Magnetic Force in a Moving Reference Frame]]<br />
</div><br />
</div><br />
<br />
===Week 4===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Photons and the Photoelectric Effect====<br />
<div class="mw-collapsible-content"><br />
*[[Spontaneous Photon Emission]]<br />
*[[Light Scattering]]<br />
*[[Lasers]]<br />
*[[Electronic Energy Levels and Photons]]<br />
*[[Quantum Properties of Light]]<br />
*[[The Photoelectric Effect]]<br />
</div><br />
</div><br />
<br />
===Weeks 5 and 6===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Matter Waves and Wave-Particle Duality====<br />
<div class="mw-collapsible-content"><br />
*[[Wave-Particle Duality]]<br />
*[[Particle in a 1-Dimensional box]]<br />
*[[Heisenberg Uncertainty Principle]]<br />
</div><br />
</div><br />
<br />
===Week 7===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Wave Mechanics====<br />
<div class="mw-collapsible-content"><br />
*[[Standing Waves]]<br />
*[[Wavelength]]<br />
*[[Wavelength and Frequency]]<br />
*[[Mechanical Waves]]<br />
*[[Transverse and Longitudinal Waves]]<br />
*[[Fourier Series and Transform]]<br />
</div><br />
</div><br />
<br />
===Week 8===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Schrödinger Equation====<br />
<div class="mw-collapsible-content"><br />
*[[The Born Rule]]<br />
*[[Solution for a Single Free Particle]]<br />
*[[Solution for a Single Particle in an Infinite Quantum Well - Darin]]<br />
*[[Solution for a Single Particle in a Semi-Infinite Quantum Well]]<br />
*[[Quantum Harmonic Oscillator]]<br />
*[[Solution for Simple Harmonic Oscillator]]<br />
</div><br />
</div><br />
<br />
===Week 9===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Quantum Mechanics====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Tunneling through Potential Barriers]]<br />
</div><br />
</div><br />
<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====The Hydrogen Atom====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Theory]]<br />
*[[Atomic Theory]]<br />
</div><br />
</div><br />
<br />
===Week 10===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Rutherford-Bohr Model====<br />
<div class="mw-collapsible-content"><br />
*[[Rutherford Experiment and Atomic Collisions]]<br />
*[[Bohr Model]]<br />
*[[Quantized energy levels]]<br />
*[[Energy graphs and the Bohr model]]<br />
</div><br />
</div><br />
<br />
===Week 11===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Many-Electron Atoms====<br />
<div class="mw-collapsible-content"><br />
*[[Quantum Theory]]<br />
*[[Atomic Theory]]<br />
*[[Pauli exclusion principle]]<br />
</div><br />
</div><br />
<br />
===Week 12===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====The Nucleus====<br />
<div class="mw-collapsible-content"><br />
*[[Nucleus]]<br />
</div><br />
</div><br />
<br />
===Week 13===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Molecules====<br />
<div class="mw-collapsible-content"><br />
*[[Molecules]]<br />
*[[Covalent Bonds]]<br />
</div><br />
</div><br />
<br />
===Week 14===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Statistical Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Application of Statistics in Physics]]<br />
</div><br />
</div><br />
<br />
===Week 15===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Statistical Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Temperature & Entropy]]<br />
</div><br />
</div><br />
<br />
===Additional Topics===<br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Thermodynamics====<br />
<div class="mw-collapsible-content"><br />
*[[Maxwell Relations]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Nuclear Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Nuclear Fission]]<br />
*[[Nuclear Energy from Fission and Fusion]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Particle Physics====<br />
<div class="mw-collapsible-content"><br />
*[[Elementary Particles and Particle Physics Theory]]<br />
*[[String Theory]]<br />
</div><br />
</div><br />
<div class="toccolours mw-collapsible mw-collapsed"><br />
====Solid-State/Condensed Matter Physics====<br />
<div class="mw-collapsible-content"><br />
*[[What is Condensed Matter]]<br />
*[[Crystalline Structures]]<br />
</div><br />
</div></div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46862Twin Paradox2024-12-05T14:25:18Z<p>Chaz: </p>
<hr />
<div><strong>Chaz Sporrer Fall 2024</strong><br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math><br />
Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
Here is a video showing a twin traveling a distance D in a spaceship from the perspective of the other twin on Earth: https://vimeo.com/1030116480?share=copy#t=0<br />
<br />
Here is a similar video showing the perspective of the twin in the spaceship (note the contraction of the distance the twin travels): https://vimeo.com/1030116491?share=copy#t=0<br />
<br />
<br />
<br />
==Examples==<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: Because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using the time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On the way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on the trip to the planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On the way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
<br />
== See also ==<br />
<br />
===Further reading===<br />
<br />
===External links===<br />
<br />
==References==<br />
Krane, Kenneth S. <em>Modern Physics</em>. 4th ed., Wiley, 2019.<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46861Twin Paradox2024-12-05T14:24:47Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math><br />
Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
Here is a video showing a twin traveling a distance D in a spaceship from the perspective of the other twin on Earth: https://vimeo.com/1030116480?share=copy#t=0<br />
<br />
Here is a similar video showing the perspective of the twin in the spaceship (note the contraction of the distance the twin travels): https://vimeo.com/1030116491?share=copy#t=0<br />
<br />
<br />
<br />
==Examples==<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: Because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using the time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On the way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on the trip to the planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On the way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
<br />
== See also ==<br />
<br />
===Further reading===<br />
<br />
===External links===<br />
<br />
==References==<br />
Krane, Kenneth S. <em>Modern Physics</em>. 4th ed., Wiley, 2019.<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46860Twin Paradox2024-12-05T14:03:04Z<p>Chaz: </p>
<hr />
<div>Chaz Sporrer Fall 2024<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math><br />
Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
Here is a video showing a twin traveling a distance D in a spaceship from the perspective of the other twin on Earth: https://vimeo.com/1030116480?share=copy#t=0<br />
<br />
Here is a similar video showing the perspective of the twin in the spaceship (note the contraction of the distance the twin travels): https://vimeo.com/1030116491?share=copy#t=0<br />
<br />
<br />
<br />
==Examples==<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: Because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using the time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On the way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on the trip to the planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On the way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
<br />
== See also ==<br />
<br />
===Further reading===<br />
<br />
===External links===<br />
<br />
==References==<br />
Krane, Kenneth S. <em>Modern Physics</em>. 4th ed., Wiley, 2019.<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46859Twin Paradox2024-12-05T13:57:59Z<p>Chaz: </p>
<hr />
<div>Chaz Sporrer Fall 2024<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math><br />
Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
Here is a video showing a twin traveling a distance D in spaceship from the perspective of the other twin on Earth: https://vimeo.com/1030116480?share=copy#t=0<br />
<br />
Here is a similar video showing the perspective of the twin in the spaceship (note the contraction of the distance the twin travels): https://vimeo.com/1030116491?share=copy#t=0<br />
<br />
<br />
<br />
==Examples==<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: Because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using the time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On the way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on the trip to the planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On the way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
<br />
== See also ==<br />
<br />
===Further reading===<br />
<br />
===External links===<br />
<br />
==References==<br />
Krane, Kenneth S. <em>Modern Physics</em>. 4th ed., Wiley, 2019.<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46858Twin Paradox2024-12-05T13:57:29Z<p>Chaz: </p>
<hr />
<div>Chaz Sporrer Fall 2024<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math><br />
Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
Here is a video showing a twin traveling a distance D in spaceship from the perspective of the other twin on Earth: https://vimeo.com/1030116480?share=copy#t=0<br />
Here is a similar video showing the perspective of the twin in the spaceship (note the contraction of the distance the twin travels): https://vimeo.com/1030116491?share=copy#t=0<br />
<br />
<br />
<br />
==Examples==<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: Because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using the time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On the way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on the trip to the planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On the way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
<br />
== See also ==<br />
<br />
===Further reading===<br />
<br />
===External links===<br />
<br />
==References==<br />
Krane, Kenneth S. <em>Modern Physics</em>. 4th ed., Wiley, 2019.<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46665Twin Paradox2024-12-03T00:44:30Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math><br />
Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
<br />
<br />
==Examples==<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: Because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using the time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On the way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on the trip to the planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On the way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
<br />
== See also ==<br />
<br />
===Further reading===<br />
<br />
===External links===<br />
<br />
==References==<br />
Krane, Kenneth S. <em>Modern Physics</em>. 4th ed., Wiley, 2019.<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46664Twin Paradox2024-12-03T00:42:57Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math><br />
Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
<br />
<br />
==Examples==<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: Because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using the time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On the way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on the trip to the planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On the way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
<br />
== See also ==<br />
<br />
===Further reading===<br />
<br />
===External links===<br />
<br />
==References==<br />
Krane, Kenneth S. Modern Physics. 4th ed., Wiley, 2019.<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46650Twin Paradox2024-12-02T23:20:56Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math><br />
Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
<br />
<br />
==Examples==<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: Because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using the time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On the way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on the trip to the planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On the way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
<br />
== See also ==<br />
<br />
===Further reading===<br />
<br />
===External links===<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46649Twin Paradox2024-12-02T23:12:51Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math><br />
Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
<br />
<br />
==Examples==<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46648Twin Paradox2024-12-02T23:11:57Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math><br />
Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46647Twin Paradox2024-12-02T23:10:19Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math><br />
Where <math>\Delta L_0</math> is the length measured by in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46646Twin Paradox2024-12-02T23:03:22Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46645Twin Paradox2024-12-02T22:38:59Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
<div style="padding:50% 0 0 0;position:relative;"><iframe src="https://player.vimeo.com/video/1030116480?badge=0&amp;autopause=0&amp;player_id=0&amp;app_id=58479" frameborder="0" allow="autoplay; fullscreen; picture-in-picture; clipboard-write" style="position:absolute;top:0;left:0;width:100%;height:100%;" title="Twin Paradox Visual Earth Perspective"></iframe></div><script src="https://player.vimeo.com/api/player.js"></script><br />
<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46429Twin Paradox2024-11-15T18:19:01Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46428Twin Paradox2024-11-15T18:18:19Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
<div style="padding:50% 0 0 0;position:relative;"><iframe src="https://player.vimeo.com/video/1030116480?badge=0&amp;autopause=0&amp;player_id=0&amp;app_id=58479" frameborder="0" allow="autoplay; fullscreen; picture-in-picture; clipboard-write" style="position:absolute;top:0;left:0;width:100%;height:100%;" title="Twin Paradox Visual Earth Perspective"></iframe></div><script src="https://player.vimeo.com/api/player.js"><br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46427Twin Paradox2024-11-15T18:17:06Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
<div style="padding:50% 0 0 0;position:relative;"><iframe src="https://player.vimeo.com/video/1030116480?badge=0&amp;autopause=0&amp;player_id=0&amp;app_id=58479" frameborder="0" allow="autoplay; fullscreen; picture-in-picture; clipboard-write" style="position:absolute;top:0;left:0;width:100%;height:100%;" title="Twin Paradox Visual Earth Perspective"></iframe></div><script src="https://player.vimeo.com/api/player.js"></script><br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46426Twin Paradox2024-11-15T18:13:47Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<div style="padding:50% 0 0 0;position:relative;"><iframe src="https://player.vimeo.com/video/1030116480?badge=0&amp;autopause=0&amp;player_id=0&amp;app_id=58479" frameborder="0" allow="autoplay; fullscreen; picture-in-picture; <br />
<br />
clipboard-write" style="position:absolute;top:0;left:0;width:100%;height:100%;" title="Twin Paradox Visual Earth Perspective"></iframe></div><script src="https://player.vimeo.com/api/player.js"></script><br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46425Twin Paradox2024-10-31T00:01:37Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=File:Question_2_Twin_2_Reference_Frame.png&diff=46424File:Question 2 Twin 2 Reference Frame.png2024-10-30T23:56:13Z<p>Chaz: </p>
<hr />
<div></div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=File:Question_2_Twin_1_Reference_Frame.png&diff=46423File:Question 2 Twin 1 Reference Frame.png2024-10-30T23:55:24Z<p>Chaz: </p>
<hr />
<div></div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46422Twin Paradox2024-10-30T23:42:19Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.<br />
<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
When cosmic rays collide with our upper atmosphere, particles called muons can be created. These particles are unstable and only live for about 2 microseconds. We have observed these particles reaching the surface of the Earth, which was a problem in classical physics. Because the atmosphere is around 10 km thick, a muon, which travels at around 0.9999c towards the earth, should not be observable on the ground. At a speed of 0.9999c, a muon would only travel approximately 0.6km in its lifetime of 2 microseconds. Fortunately, scientists have used special relativity to answer this question. Though a muon only lives 2 microseconds in its inertial reference frame, time dilation causes the muon to live longer in our reference frame. This longer lifespan in our reference frame is what allows us to detect muons. Just like in examples of the twin paradox, one observer aged slower relative to the other. In this case, muons aged slower than us, leading to a perceived lifespan greater than 2 microseconds.<br />
<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46421Twin Paradox2024-10-30T01:07:08Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
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<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46420Twin Paradox2024-10-30T01:05:26Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = t \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
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Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46419Twin Paradox2024-10-30T01:03:43Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 â 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46418Twin Paradox2024-10-30T00:59:53Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = 0.655/year</math>. On way back (towards Earth), <math>f_0 = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\,years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
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Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
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<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46417Twin Paradox2024-10-30T00:58:26Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = 0.655/year</math>. On way back (towards Earth), <math>f_0 = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 25\years \sqrt{1 - \frac{(0.8c)^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = 10.01 \sqrt{1 - \frac{(0.999c)^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
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==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46416Twin Paradox2024-10-30T00:53:02Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = 0.655/year</math>. On way back (towards Earth), <math>f_0 = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
Question 3: twin 1 travels to a planet <math>10</math> light years away and then back to Earth at a speed of <math>0.8c</math>. Ten Earth years after twin 1 initially leaves Earth, twin 2 travels half the distance to the planet and then travels back to Earth at a speed of <math>0.999c</math>. Which twin is older when both twins have finished their journeys and by how much?<br />
<br />
Solution: The time that passes on Earth during twin 1`s journey is <math>\frac{20\,light-years}{0.8c} = 25\,years</math>. The time that passes in twin 1`s reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 15\,years</math>. Twin 2 will leave <math>10</math> years after twin 1`s journey began, and it will take <math>\frac{10\,light-years}{0.999c} = 10.01</math> Earth years for him to complete his journey. The time that passes during his journey according to his reference frame is <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 0.45\,years</math>. In Earth years, there is still another <math>(25 â 10 â 10.01) = 4.99\,years</math> until twin 1 returns. So, during the <math>25</math> Earth years, twin 2 aged <math>10 + 0.45 + 4.99 = 15.44\,years</math>. Because twin 1 aged <math>15</math> years during the <math>25</math> Earth years, twin 1 is <math>(15.44 - 15) = 0.44</math> years younger. In this situation, even though twin 2 traveled at a faster speed, the distance twin 1 traveled was enough longer than that of twin 2 so that twin 1 aged less overall. In fact, in this example, as twin 2`s velocity approaches c, twin 2 gets closer and closer to aging the exact same amount as twin 1.<br />
<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
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Books, Articles or other print media on this topic<br />
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<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46415Twin Paradox2024-10-30T00:37:51Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years â 9.17\,years) = 0.83\,years</math>.<br />
<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.<br />
<br />
Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 â u/c}{1 + u/c}}</math><br />
<br />
While traveling to planet B, <math>f_0 = 0.655/year</math>. On way back (towards Earth), <math>f_0 = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.<br />
<br />
===Difficult===<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46414Twin Paradox2024-10-30T00:19:40Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?<br />
<br />
Solution: because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.<br />
<br />
===Middling===<br />
Question 2: If twin 1 remains on Earth and twin 2 travels to planet B and back, how many years older will twin 1 be than twin 2. Twin 2 travels at a speed <math>0.4c</math> (<math>c</math> = speed of light), and the distance between Earth and planet B is <math>2</math> light-years.<br />
<br />
Solution: First, the distance of a light-year is equal to <math>(1\,year) * c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4*c*(1\,year)/0.4c = 10 years</math>. Using time dilation formula and rearranging for to (time according to twin 2), we get to = t/gamma = 9.17 years. This means we can subtract to from t to find how much older twin 1 is than twin 2. (10 years â 9.17 years) = 0.83 years.<br />
Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 (9.17 years/2) = 4.59 years to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of 10 signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where fo is the frequency observed by twin 2 and f is the frequency observed by twin 1. fo = f((1 â u/c) / (1 + u/c))0.5. While traveling to planet B, fo = 0.655/year. On way back, f0 = 1.53/year. So on trip to planet, he receives (0.655/year)(4.59 years) = 3.01 signals. On way back, he receives (1.53/year)(4.59 years) = 7.02 signals. This means that over the entire journey, twin 2 receives 7.02 + 3.01 approx. = 10 signals.<br />
<br />
===Difficult===<br />
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[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46413Twin Paradox2024-10-30T00:08:16Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance <math>D</math> away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
===Middling===<br />
===Difficult===<br />
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[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46412Twin Paradox2024-10-30T00:07:41Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance D away at a velocity <math>v</math>. The twin traveling on the spaceship does not immediately have a velocity <math>v</math>, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
===Middling===<br />
===Difficult===<br />
<br />
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#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
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==History==<br />
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<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46411Twin Paradox2024-10-30T00:04:28Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance D away at a velocity (v). The twin traveling on the spaceship does not immediately have a velocity v, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance D and back at a velocity v. The time (t) seen by the twin on Earth equals 2D/v, where 2D equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and <math>L</math> is length measured by observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
===Middling===<br />
===Difficult===<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
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Books, Articles or other print media on this topic<br />
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<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46410Twin Paradox2024-10-28T23:11:54Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance D away at a velocity (v). The twin traveling on the spaceship does not immediately have a velocity v, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with <math>\Delta t</math> being the time for the round trip according to the twin on earth, and <math>\Delta t_0</math> being the time according to the twin on the spaceship.<br />
<br />
Time Dilation Formula: <math>\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
To more easily see how this works, consider again the situation where one twin travels a distance D and back at a velocity v. The time (t) seen by the twin on Earth equals 2D/v, where 2D equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.<br />
The formula for length contraction: <math> \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} </math><br />
Where <math>\Delta L_0</math> is length measured by inertial frame, and L is length measured by observer moving at velocity v relative to the inertial frame. Thus, time <math>t_0</math> according to twin on spaceship is 2(D/gamma)/v. Thus you can say t/to = (2D/v)/(2D/v*gamma) = gamma. Rearranging, you get t = to(gamma), which is the exact equation for time dilation.<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
===Middling===<br />
===Difficult===<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
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<br />
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<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46409Twin Paradox2024-10-28T22:54:05Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance D away at a velocity (v). The twin traveling on the spaceship does not immediately have a velocity v, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with t being the time for the round trip according to the twin on earth, and to being the time according to the twin on the spaceship.<br />
<br />
<math>\Delta t = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}}</math><br />
<br />
===A Computational Model===<br />
<br />
How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
===Middling===<br />
===Difficult===<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
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<br />
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<br />
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<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46408Twin Paradox2024-10-28T22:44:41Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance D away at a velocity (v). The twin traveling on the spaceship does not immediately have a velocity v, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with t being the time for the round trip according to the twin on earth, and to being the time according to the twin on the spaceship.<br />
<br />
<math>\Delta t = \frac{\Delta t}{2}</math><br />
<br />
===A Computational Model===<br />
<br />
How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
===Middling===<br />
===Difficult===<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
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Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46407Twin Paradox2024-10-28T22:31:22Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance D away at a velocity (v). The twin traveling on the spaceship does not immediately have a velocity v, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with t being the time for the round trip according to the twin on earth, and to being the time according to the twin on the spaceship.<br />
<br />
{{math&Deltat}}<br />
<br />
===A Computational Model===<br />
<br />
How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
===Middling===<br />
===Difficult===<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46406Twin Paradox2024-10-28T22:30:21Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance D away at a velocity (v). The twin traveling on the spaceship does not immediately have a velocity v, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with t being the time for the round trip according to the twin on earth, and to being the time according to the twin on the spaceship.<br />
<br />
<math>&Delta </math><br />
<br />
===A Computational Model===<br />
<br />
How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
===Middling===<br />
===Difficult===<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46405Twin Paradox2024-10-28T20:58:33Z<p>Chaz: /* The Main Idea */</p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
Special relativity has created the idea of time dilation, in which the clock of an observer in motion will move slower than the clock of the observer at rest relative to the object. However, in each observer`s inertial reference frame, it will appear that they are motionless and that it is really the other observer who is moving. This means that each observer will think that their clock is moving faster than that of the other observer. This creates a problem, because only one clock can truly move faster than the other. This problem is known as the twin paradox. Though it is called the twin paradox, it is not actually a paradox. Thus, one observer`s clocks does actually move faster than the other`s.<br />
<br />
===A Mathematical Model===<br />
<br />
Here is how to think about the situation. Think about two twins who are exactly the same age, and consider the situation where one twin travels by spaceship to and from a planet a distance D away at a velocity (v). The twin traveling on the spaceship does not immediately have a velocity v, but has to accelerate. We could say the twin accelerates very quickly, and thus spends almost the entirety of the trip at a constant velocity. However, the twin who stays on the planet does not need to accelerate, because he is not leaving the planet. Of course, one could argue that according to the twin on the spaceship, the twin on Earth âappearsâ to be accelerating backwards, but in reality, the only observer who had to experience a force to accelerate was the one on the spaceship. As a result of the twin on the spaceship accelerating, this twin changed inertial reference frames, and thus is the twin who is actually in motion. This means we can apply the time dilation equation to the scenario of the twins, with t being the time for the round trip according to the twin on earth, and to being the time according to the twin on the spaceship.<br />
<br />
===A Computational Model===<br />
<br />
How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
===Middling===<br />
===Difficult===<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46403Twin Paradox2024-10-05T01:21:29Z<p>Chaz: </p>
<hr />
<div>===Chaz Sporrer Fall 2024===<br />
Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
State, in your own words, the main idea for this topic<br />
Electric Field of Capacitor<br />
<br />
===A Mathematical Model===<br />
<br />
What are the mathematical equations that allow us to model this topic. For example <math>{\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net}</math> where '''p''' is the momentum of the system and '''F''' is the net force from the surroundings.<br />
<br />
===A Computational Model===<br />
<br />
How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
===Middling===<br />
===Difficult===<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chazhttp://www.physicsbook.gatech.edu/index.php?title=Twin_Paradox&diff=46399Twin Paradox2024-10-02T18:31:11Z<p>Chaz: Created page with "Short Description of Topic ==The Main Idea== State, in your own words, the main idea for this topic Electric Field of Capacitor ===A Mathematical Model=== What are the mathematical equations that allow us to model this topic. For example <math>{\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net}</math> where '''p''' is the momentum of the system and '''F''' is the net force from the surroundings. ===A Computational Model=== How do we visualize or predict using this topi..."</p>
<hr />
<div>Short Description of Topic<br />
<br />
==The Main Idea==<br />
<br />
State, in your own words, the main idea for this topic<br />
Electric Field of Capacitor<br />
<br />
===A Mathematical Model===<br />
<br />
What are the mathematical equations that allow us to model this topic. For example <math>{\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net}</math> where '''p''' is the momentum of the system and '''F''' is the net force from the surroundings.<br />
<br />
===A Computational Model===<br />
<br />
How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]<br />
<br />
==Examples==<br />
<br />
Be sure to show all steps in your solution and include diagrams whenever possible<br />
<br />
===Simple===<br />
===Middling===<br />
===Difficult===<br />
<br />
==Connectedness==<br />
#How is this topic connected to something that you are interested in?<br />
#How is it connected to your major?<br />
#Is there an interesting industrial application?<br />
<br />
==History==<br />
<br />
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.<br />
<br />
== See also ==<br />
<br />
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?<br />
<br />
===Further reading===<br />
<br />
Books, Articles or other print media on this topic<br />
<br />
===External links===<br />
<br />
Internet resources on this topic<br />
<br />
==References==<br />
<br />
This section contains the the references you used while writing this page<br />
<br />
[[Category:Which Category did you place this in?]]</div>Chaz