Electromagnetic Waves: Difference between revisions
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===A Mathematical Model=== | ===A Mathematical Model=== | ||
The mathematical equations that allow us to model this topic are Maxwell's equations; but a simpler way to figure it out is to consider a notion from General Relativity. According to General Relativity, the speed of information is <math>c</math>. Now consider an electron moving at a velocity <math>\vec{v}(t)</math>. Now consider an observation point a location <math>\vec{r}</math> from the electron. Let us say the electron is oscillating, so we have that <math>\vec{v}=\cos{t} \vec{k}</math>. Then since the information that the electron is moving takes a while to get to the position <math>\vec{r}</math>, so instead of getting the Coulomb field <math>\mathbf{E}=\frac{kq_e}{|\vec{r}|^2}\hat{r}</math> we get the electric field <math>\mathbf{E} = -\frac{q}{4\pi\epsilon_0} \left[ \frac{\mathbf{e}_{r'}}{r'^2} + \frac{r'}{c} \frac{d}{dt} \left(\frac{\mathbf{e}_{r'}}{r'^2}\right) + \frac{1}{c^2} \frac{d^2}{dt^2} \mathbf{e}_{r'} \right]</math>, where <math>\vec{r}'</math> is the direction that the electron appears to be according to the observer, but is not actually. The magnetic field can then be written as <math>\mathbf{B} = -\frac{\mathbf{e}_{r'} \times \mathbf{E}}{c}</math>. Now if you notice the first two terms in the above E-field equation can be dropped at large <math>r</math> because they drop off with the distance squared. Calculating, we have that our electric field in a wave is <math>\mathbf{E} = -\frac{q}{4\pi\epsilon_0 c^2} \frac{d^2 \mathbf{e}_{r'}}{dt^2}</math>. The second derivative term is just the derivative of our velocity, so we get that <math>\mathbf{E} = \frac{q}{4\pi\ | The mathematical equations that allow us to model this topic are Maxwell's equations; but a simpler way to figure it out is to consider a notion from General Relativity. According to General Relativity, the speed of information is <math>c</math>. Now consider an electron moving at a velocity <math>\vec{v}(t)</math>. Now consider an observation point a location <math>\vec{r}</math> from the electron. Let us say the electron is oscillating, so we have that <math>\vec{v}=\cos{t} \vec{k}</math>. Then since the information that the electron is moving takes a while to get to the position <math>\vec{r}</math>, so instead of getting the Coulomb field <math>\mathbf{E}=\frac{kq_e}{|\vec{r}|^2}\hat{r}</math> we get the electric field <math>\mathbf{E} = -\frac{q}{4\pi\epsilon_0} \left[ \frac{\mathbf{e}_{r'}}{r'^2} + \frac{r'}{c} \frac{d}{dt} \left(\frac{\mathbf{e}_{r'}}{r'^2}\right) + \frac{1}{c^2} \frac{d^2}{dt^2} \mathbf{e}_{r'} \right]</math>, where <math>\vec{r}'</math> is the direction that the electron appears to be according to the observer, but is not actually. The magnetic field can then be written as <math>\mathbf{B} = -\frac{\mathbf{e}_{r'} \times \mathbf{E}}{c}</math>. Now if you notice the first two terms in the above E-field equation can be dropped at large <math>r</math> because they drop off with the distance squared. Calculating, we have that our electric field in a wave is <math>\mathbf{E} = -\frac{q}{4\pi\epsilon_0 c^2} \frac{d^2 \mathbf{e}_{r'}}{dt^2}</math>. The second derivative term is just the derivative of our velocity, so we get that <math>\mathbf{E} = \frac{q}{4\pi\epsilon_0 c^2}\sin(t) \vec{k}</math>. Using the B formula we get that since <math>r</math> is very large, the <math>e_r</math> vector and the <math>\mathbf{E}</math> vector are perpendicular, leading us to calculate B to be <math>\mathbf{B} = -\frac{q}{4\pi\epsilon_0c^3}\sin(t) \vec{i}</math>. So, from our investigations we conclude (by way of the Lorentz formula) that an oscillating point charge will cause a similar oscillation in the same direction over large distances. This movement is how our eyes detect light! | ||
===A Computational Model=== | |||
https://www.glowscript.org/#/user/sboebel3012/folder/MyPrograms/program/EMWaveVisualization | |||
==Examples== | |||
=== | ===Simple=== | ||
Consider an electron which moves back and forth in a radio transmitter antenna with a frequency of 433 mHz. What frequency does the receiver get? Hint: Assume the velocity is 0, so there is no Lorentz contraction. | |||
Answer: The receiver gets exactly 433 mHz. Since there is no Lorentz contraction, the waves do not get compressed which implies that the frequency must remain the same w.r.t. the observer's reference frame. | |||
== | ===Middling=== | ||
Consider the same situation as before, except now we know that the receiver receives 5W of power from the transmitter. What is the magnitude of the electromagnetic wave? Assume, for simplicity, that the receiver and the transmitter both lie along the y-axis and that the oscillation on the receiver occurs exclusively in the z direction, thus there is no gain etc. This is also known as a plane wave. | |||
Answer: | |||
<math>P = \frac{1}{2}\epsilon_0|\mathbf{E}|^2</math>, so we have <math>|\mathbf{E}|=\sqrt{\frac{2*5}{8.8*10^-12*3*10^8}}=61.35</math>N/C. | |||
===Difficult=== | ===Difficult=== | ||
Latest revision as of 17:50, 13 April 2024
Spencer Boebel '24
Electromagnetic Waves
The Main Idea
An electromagnetic wave is what happens when we put all four of Maxwell's equations together. A changing electric field creates a magnetic field, which changes and creates an electric field, starting it all over again. This is the basic premise of an electric wave, and we shall derive other properties presently.
A Mathematical Model
The mathematical equations that allow us to model this topic are Maxwell's equations; but a simpler way to figure it out is to consider a notion from General Relativity. According to General Relativity, the speed of information is [math]\displaystyle{ c }[/math]. Now consider an electron moving at a velocity [math]\displaystyle{ \vec{v}(t) }[/math]. Now consider an observation point a location [math]\displaystyle{ \vec{r} }[/math] from the electron. Let us say the electron is oscillating, so we have that [math]\displaystyle{ \vec{v}=\cos{t} \vec{k} }[/math]. Then since the information that the electron is moving takes a while to get to the position [math]\displaystyle{ \vec{r} }[/math], so instead of getting the Coulomb field [math]\displaystyle{ \mathbf{E}=\frac{kq_e}{|\vec{r}|^2}\hat{r} }[/math] we get the electric field [math]\displaystyle{ \mathbf{E} = -\frac{q}{4\pi\epsilon_0} \left[ \frac{\mathbf{e}_{r'}}{r'^2} + \frac{r'}{c} \frac{d}{dt} \left(\frac{\mathbf{e}_{r'}}{r'^2}\right) + \frac{1}{c^2} \frac{d^2}{dt^2} \mathbf{e}_{r'} \right] }[/math], where [math]\displaystyle{ \vec{r}' }[/math] is the direction that the electron appears to be according to the observer, but is not actually. The magnetic field can then be written as [math]\displaystyle{ \mathbf{B} = -\frac{\mathbf{e}_{r'} \times \mathbf{E}}{c} }[/math]. Now if you notice the first two terms in the above E-field equation can be dropped at large [math]\displaystyle{ r }[/math] because they drop off with the distance squared. Calculating, we have that our electric field in a wave is [math]\displaystyle{ \mathbf{E} = -\frac{q}{4\pi\epsilon_0 c^2} \frac{d^2 \mathbf{e}_{r'}}{dt^2} }[/math]. The second derivative term is just the derivative of our velocity, so we get that [math]\displaystyle{ \mathbf{E} = \frac{q}{4\pi\epsilon_0 c^2}\sin(t) \vec{k} }[/math]. Using the B formula we get that since [math]\displaystyle{ r }[/math] is very large, the [math]\displaystyle{ e_r }[/math] vector and the [math]\displaystyle{ \mathbf{E} }[/math] vector are perpendicular, leading us to calculate B to be [math]\displaystyle{ \mathbf{B} = -\frac{q}{4\pi\epsilon_0c^3}\sin(t) \vec{i} }[/math]. So, from our investigations we conclude (by way of the Lorentz formula) that an oscillating point charge will cause a similar oscillation in the same direction over large distances. This movement is how our eyes detect light!
A Computational Model
https://www.glowscript.org/#/user/sboebel3012/folder/MyPrograms/program/EMWaveVisualization
Examples
Simple
Consider an electron which moves back and forth in a radio transmitter antenna with a frequency of 433 mHz. What frequency does the receiver get? Hint: Assume the velocity is 0, so there is no Lorentz contraction.
Answer: The receiver gets exactly 433 mHz. Since there is no Lorentz contraction, the waves do not get compressed which implies that the frequency must remain the same w.r.t. the observer's reference frame.
Middling
Consider the same situation as before, except now we know that the receiver receives 5W of power from the transmitter. What is the magnitude of the electromagnetic wave? Assume, for simplicity, that the receiver and the transmitter both lie along the y-axis and that the oscillation on the receiver occurs exclusively in the z direction, thus there is no gain etc. This is also known as a plane wave.
Answer: [math]\displaystyle{ P = \frac{1}{2}\epsilon_0|\mathbf{E}|^2 }[/math], so we have [math]\displaystyle{ |\mathbf{E}|=\sqrt{\frac{2*5}{8.8*10^-12*3*10^8}}=61.35 }[/math]N/C.
Difficult
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