# Twin paradox, live via satellite!

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Ok I was trying to explain special relativity to a friend and I was wondering about something (he said someone explained to him why special relativity was wrong ).

Lets say you have 3 people, A, B, and C. Person A sits on earth. Person B is in a rocket ship. Person C is with person A and is observing person A while also watching person B via a 2.4ghz 30 frames per second (FPS) video linkup. Now, person B speeds up to near the speed of light. I forget what speed we had used but i remember y=4.

Now I was trying to explain exactly what person C would see on his little tv but I couldn't really figure it out. Would the receiver not pick up anything because of the doppler shift? And if it be adjusted to pick up the correct frequency, would you see everything at 7.5FPS?

That's not exactly true, because you've got to remember that apart from the timedilation of person B as observed by observer C the video signal must travel from person B to person C. The time dilation yields a factor of $$\gamma$$ in the time between to frames and the increase of the distance of B w.r.t. C yields another factor of $$1+v/c$$. Multiplying these we get the relativistic dopplershift

$$\Delta t_C = \sqrt{\frac{1+v/c}{1-v/c}} \Delta t_B$$.

So all frequencies, i.e. the frequency 2.4gHz of the signal and the frequentcy 30Hz of video frames, get shifted by a factor

$$\sqrt{\frac{1-v/c}{1+v/c}} \approx 0.127$$.

(In case $$\gamma=4$$.)

nice problem.
Have you considered the possiblity of treating the incoming signal so as to correct this Doppler shift ? It seems to be possible in principle.

Gold Member
DaTario said:
nice problem.
Have you considered the possiblity of treating the incoming signal so as to correct this Doppler shift ? It seems to be possible in principle.

Yah, that was the second part of the problem. What would happen if you adjusetd the reciever to pick up the correct signal frequency

It seems to me that the rate at which you will get information on Earth is different from the rate at which the rockett is sending. Althogh I have no direct view of a proof of this, it seems to be a consequence of space ad time modifications caused by accelerations on the rockett.

Remembering that (acceleration) equivalent to (gravitational fields) which results in (time dilatation).

Ich
Just as Timbuqtu said, the 30 Hz get shifted by a factor of ~8 when you adjust the receiver. This has nothing to do with acceleration: half of it is time dilation due to velocity, half of it is the ordinary doppler effect.

And if it be adjusted to pick up the correct frequency, would you see everything at 7.5FPS?

You'd see a slower number of frames/second, yes. Which makes this a fun thought
experiment. The thing to figure out for the twin paradox is this:

The total number of frames must be less for the traveling twin (edit: compared to
a similar video system staying with A). Why is this
when the fps goes up on the return trip?

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Antiphon said:
The total number of frames must be less for the traveling twin. Why is this
when the fps goes up on the return trip?
Why? All frames sent by B will be received by C, so they will (of course) agree on the total number of frames. But on average the C will receive the frames with less than 1/30 of a second between them. That is, B and C will NOT agree on the time elapsed during their journey.

To be precise: it will take B the same amount of time to travel to turningpoint as it will take him to travel back. So B will send as many frames on his flight away as on his flight back. Hence, C will receive the first half of the frames with a time

$$\Delta t_C = \sqrt{\frac{1+v/c}{1-v/c}} \Delta t_B$$

between them. And the second half of the frames with

$$\Delta t_C = \sqrt{\frac{1-v/c}{1+v/c}} \Delta t_B$$

between them. So the elapsed time according to C is

$$t_C = \frac{N}{2} (\sqrt{\frac{1-v/c}{1+v/c}} + \sqrt{\frac{1+v/c}{1-v/c}})\Delta t_B = \gamma N \Delta t_B =\gamma t_B ,$$
where N is total number of frames, t_C and t_B are the total ellapsed times according to C and B.

Conclusion: A/C has aged $$\gamma = 4$$ times as much as B has during the journey.

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Timbuqtu said:
Why? All frames sent by B will be received by C, so they will (of course) agree on the total number of frames.

Sorry I wasn't clear. If the same type of video system was also monitoring A's
behavior, it would have more total frames than B's transmissions even if no frames
were dropped as transmission errors.

Antiphon said:
Sorry I wasn't clear. If the same type of video system was also monitoring A's
behavior, it would have more total frames than B's transmissions even if no frames
were dropped as transmission errors.

Ok. In that case the video system is just some sort of clock, ticking 30 times per second. And indeed as I pointed out A/C observes the journey to take 4 times as long as observed by B, so A's video system will have captured in total 4 times as many frames as B's video system has.

jtbell
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See posting #3 in this thread which works out a similar situation in detail. Instead of video frames, I talk about turning the pages of a calendar, but the analysis is the same either way.