How does the doppler effect resolve the twin paradox?

[Dennydont]

If you had twin 1 on the earth, and twin 2 fly to a star and back at a speed of v with the Earth and star separated by a distance L, twin 1 sends out flashes at intervals of t seconds (measured in his frame). Taking into consideration the numbers of redshifted and blueshifted flashes that the second twin receives, the time for T₁ (twin 1) is equal to gamma*T₂. How does one explain this using doppler effect?

 

[Nugatory]

Check out http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html and especially the section on Doppler.
If that doesn't answer your question, come back and ask about whatever part isn't making sense for you.

 

[stevendaryl]

I'm not sure if the Doppler shift explains the twin paradox, but it does illustrate the difference between the two twins.

When twin 2 blasts off, she sees the signals from twin 1 Doppler shifted lower (they arrive less often that once every t seconds). The minute she turns around, she sees the signals from twin 1 Doppler shifted higher (they arrive more often than once every t seconds).

If twin 2 sends out N signals on the way out, and N on the way back, then she sends out a total of 2N signals. Because of the Doppler shifts, she receives $N \frac{\sqrt{1-\frac{v}{c}}}{\sqrt{1+\frac{v}{c}}}$ signals on the way out from twin 1, and receives $N \frac{\sqrt{1+\frac{v}{c}}}{\sqrt{1-\frac{v}{c}}}$ on the way back. So the total number she receives from twin 1 for the whole trip is:
$N \frac{\sqrt{1-\frac{v}{c}}}{\sqrt{1+\frac{v}{c}}} + N \frac{\sqrt{1+\frac{v}{c}}}{\sqrt{1-\frac{v}{c}}} = 2N \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} > 2N$.

So the traveling twin receives more signals than she sent, so she concludes that the stay-at-home twin aged more.

If you look at it from the point of view of the stay-at-home twin, it's not exactly the same. It's still true that he sees the signals from twin 2 Doppler-shifted lower during the outward journey, and Doppler-shifted higher during the return journey. But the difference is that the stay-at-home twin doesn't see the signals from the traveling twin until long after the traveling twin turns around. That's because, it takes time for those blue-shifted signals to reach Earth, and in that time, the traveling twin is also traveling toward Earth. So by the time the signals reach Earth, the traveling twin is almost home. So the stay-at-home twin doesn't see half red-shifted signals and half blue-shifted signals; he sees mostly red-shifted signals and then a very brief time of blue-shifted signals.

 

[PeterDonis]

I'm not sure if the Doppler shift explains the twin paradox

The rest of your post is the explanation; it's the same explanation that's given in the Usenet Physics FAQ article that Nugatory linked to.

 

[jtbell]

For a numerical example of what both twins see, year-by-year, using the relativistic Doppler equation, see this old post:

https://www.physicsforums.com/threads/time-difference-cameras.69214/#post-510214