Twin Paradox

Chaz Sporrer Fall 2024

Short Description of Topic

The Main Idea

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.

A Mathematical Model

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]\displaystyle{ D }[/math] away at a velocity [math]\displaystyle{ v }[/math]. The twin traveling on the spaceship does not immediately have a velocity [math]\displaystyle{ 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 “appears” 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]\displaystyle{ \Delta t }[/math] being the time for the round trip according to the twin on earth, and [math]\displaystyle{ \Delta t_0 }[/math] being the time according to the twin on the spaceship.

Time Dilation Formula: [math]\displaystyle{ \Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}} }[/math]

A Computational Model

To more easily see how this works, consider again the situation where one twin travels a distance [math]\displaystyle{ D }[/math] and back at a velocity [math]\displaystyle{ v }[/math]. The time [math]\displaystyle{ t }[/math] seen by the twin on Earth equals [math]\displaystyle{ \frac{2D}{v} }[/math], where [math]\displaystyle{ 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. The formula for length contraction: [math]\displaystyle{ \Delta L = \frac{\Delta L_0}{\sqrt{1 - \frac{v^2}{c^2}}} }[/math] Where [math]\displaystyle{ \Delta L_0 }[/math] is length measured by inertial frame, and [math]\displaystyle{ L }[/math] is length measured by observer moving at velocity [math]\displaystyle{ v }[/math] relative to the inertial frame. Thus, time [math]\displaystyle{ t_0 }[/math] according to twin on spaceship is [math]\displaystyle{ \frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v} }[/math]. Thus you can say [math]\displaystyle{ \frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}} }[/math]. Rearranging, you get [math]\displaystyle{ t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}} }[/math], which is the equation for time dilation.

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

Simple

Question 1: If there are two twins, and twin 1 travels a distance [math]\displaystyle{ d }[/math] away from twin 2 at a velocity [math]\displaystyle{ v_1 }[/math] and then travels back to twin 2 at a velocity [math]\displaystyle{ v_2 }[/math], which twin will be older?

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.

Middling

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]\displaystyle{ 0.4c }[/math] ([math]\displaystyle{ c }[/math] = speed of light), and the distance between Earth and planet B is [math]\displaystyle{ 2 }[/math] light-years.

Solution: First, the distance of a light-year is equal to [math]\displaystyle{ (1\,year) * c }[/math]. So time [math]\displaystyle{ t }[/math] to travel a round trip of [math]\displaystyle{ 4 }[/math] light-years = [math]\displaystyle{ 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 – 9.17 years) = 0.83 years. 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 – 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.

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