Twin Paradox in outgoing frame

[bndnchrs]

Homework Statement

Show that as calculated in the rest frame comoving with the twin on the outgoing trip, the ratio of the two ages of the twins is the same: i.e. the twin on Earth has age gamme times the other twin

Homework Equations

Lorentz Transforms

The Attempt at a Solution

We are supposed to do this from the frame of the outgoing twin: but staying in the departing reference frame, and watching the twin that returns to Earth speed away from us.

So the age of a twin on Earth as seen from our frame is 2*L'/v with L' = L * gamma

The age of the twin which left is the time while the twin is moving with our frame + the time it takes to return to the earth.

This is where I run into trouble. I can't verify that the twin that left Earth has t = 2*L/(gamma^2)

I assume t(1/2) which is the time the twin which left is still in our frame is L'/v.

Then t(return) = 2*L'/v(ret) with v ret the relativistic added velocity 2*v/(1+(v/c)^2).

This doesn't work, however, I can't get the sum to be what it should be.

Any ideas?

 

[anathema]

Thank you for your interesting question. To show that the ratio of the two ages of the twins is the same in the rest frame comoving with the outgoing twin, we can use the Lorentz transforms. First, let's define the two frames: Frame 1 is the rest frame comoving with the outgoing twin, and frame 2 is the rest frame comoving with the twin on Earth. In frame 1, the twin on Earth is moving away from us with a velocity v, and in frame 2, the outgoing twin is moving away from us with a velocity v. Let's also define the distance between the twins as L.

Now, let's calculate the time it takes for the outgoing twin to reach the halfway point, which is L/2. In frame 1, this time is t(1/2) = L/(2v). In frame 2, the distance to the halfway point is contracted by a factor of γ, so the time it takes for the outgoing twin to reach the halfway point is t(1/2) = γL/(2v).

Next, let's calculate the time it takes for the outgoing twin to return to Earth. In frame 1, this time is t(return) = L/v. In frame 2, the distance to Earth is contracted by a factor of γ, so the time it takes for the outgoing twin to return to Earth is t(return) = γL/v.

Now, we can calculate the total time experienced by the outgoing twin in frame 1 as t(total) = t(1/2) + t(return) = L/(2v) + L/v = 3L/(2v). In frame 2, the total time experienced by the twin on Earth is t(total) = t(1/2) + t(return) = γL/(2v) + γL/v = 3γL/(2v).

Finally, the ratio of the two ages of the twins is t(total, 2)/t(total, 1) = (3γL/(2v))/(3L/(2v)) = γ. This shows that in the rest frame comoving with the outgoing twin, the ratio of the two ages of the twins is γ, just as we wanted to show.

I hope this helps. If you have any further questions, please don't hesitate to ask. Good luck with your studies!

 

[nlightner]

I would approach this problem by first verifying the equations used in the attempt at a solution. The Lorentz transforms are indeed the correct equations to use when dealing with special relativity and time dilation. However, it seems that there may be a misunderstanding of how to apply them in this scenario.

To calculate the age of the twin on Earth as seen from the frame of the outgoing twin, we need to use the time dilation equation: t' = t/sqrt(1-v^2/c^2). In this case, t' represents the time in the outgoing twin's frame and t represents the time in the Earth's frame.

The twin on Earth will have aged t = 2L/v in the Earth's frame. Using the time dilation equation, we can calculate the age of the twin on Earth as seen from the outgoing twin's frame: t' = 2L/v * sqrt(1-v^2/c^2).

Next, we need to calculate the time it takes for the outgoing twin to travel to a certain distance L and then return to Earth. This can be done by dividing the distance L by the velocity of the outgoing twin, which is v. So the total time in the outgoing twin's frame would be t' = 2L/v + 2L/v = 4L/v.

Now, we can calculate the ratio of the two ages by dividing the time in the outgoing twin's frame by the time in the Earth's frame: t'/t = 4L/v / 2L/v * sqrt(1-v^2/c^2) = 2/sqrt(1-v^2/c^2).

This ratio is indeed equal to gamma, which is the factor by which time is dilated in special relativity. This shows that the twin on Earth will age gamma times more than the twin on the outgoing trip, as calculated in the frame of the outgoing twin.

In summary, it is important to carefully apply the equations of special relativity when dealing with the twin paradox. By verifying the equations used and understanding how they apply in this scenario, we can arrive at the correct solution.