## Main Question or Discussion Point

Twin A is at rest with respect to Twin B who is traveling close to c and after taking off for a bit Twin B turns and returns to earth. Now the thing that I am wondering is what does each twin observe during the whole trip.
I might have it wrong and thats why I made this thread.

Lets say that 1 year goes by for Twin B during the whole trip and 40 years goes by for Twin A during the whole trip. I assume that everything Twin B does during his 1 year will be observed by Twin A within the 40 years. And everything Twin A does during the 40 years will be observed by Twin B in a year. When Twin B leaves Twin A they both observe each other in slow motion, as Twin B turns the light still has to reach Twin A so when Twin A finally observes Twin B's turn he should observe Twin B almost instantly travel back to earth like the doppler effect. Twin B also observes Twin A's clock run faster then his own right? I'm not sure if everything I just said was accurate but I wonder when most of Twin A's 40 years observed by Twin B takes place, during the return trip?

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You need to understand the foundations first

There are subtleties in the twin paradox that can't be avoided unless you're competent in high school algebra and can follow carefully stated mathematical reasoning, line by line. I only know of one paper that properly resolves the "twin paradox." It is carefully detailed and is written by a mathematician. Let me know if you have any questions.

http://www.everythingimportant.org/relativity/

George Jones
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I'm going to change the numbers a bit. Suppose Twin A stays at home in the (approximately) inertial reference frame of the Earth while Twin B makes a return trip between Earth and another planet (approximately at rest with respect to the Earth) that is 8 lightyears from the Earth. During the outgoing leg of the trip B travels at speed 0.8c away from the Earth, and during the incoming leg of the trip B travels at 0.8c towards the Earth. Therefore, the time dilation/Lorentz contraction factor is

$$\gamma=\left( 1-0.8^{2}\right)^{-\frac{1}{2}}=5/3$$

and the Doppler shift factor is

$$D=\sqrt{\frac{1+0.8}{1-0.8}}=3.$$

In B's frame the distance between Earth and the other planet is Lorentz contracted to $8/\gamma=4.8$ light years. Consequently, the time, according to B's clock, for the first part of the trip is 4.8/0.8 = 6 years. For the same reason, B takes another 6 years, according to his watch, to get back to the Earth, for a total elapsed time of 12 years. Twin A measures the total elapsed time to be 16/0.8 = 20 years.

As B travels, B uses a telescope to watch A's wristwatch. During the ougoing leg, B sees her second hand spin faster than A's by a factor of 3, due to the Doppler effect. Thus, at the turn around point B sees, via her telescope, a reading of 6/3 = 2 years on A's watch. During the incominging leg, B sees her second hand spin slower than A's by a factor of 3, again due to the Doppler effect. Thus, B sees the reading on A's watch increase by 6*3 = 18 years during the homecoming leg.

Bottom line: B sees the reading on A's watch increase by 2 + 18 = 20 years during the trip.

It is interesting to note that there is a point during the return trip that B sees the reading on A's watch to be exactly the same as the reading that B sees on her own watch. I calculate this to be, according to B, at the 8 year mark of the trip, i.e., 2 years after turn around. It is also interesting to note that if A and B are mouths of a ''short'' wormhole, then this is the point at which a Morris-Thorne-Yurtsever time machine forms.

Regards,
George

Through A's telescope would A observe B's clock run slower on the outgoing leg by a factor of 3 for 18 years and on the incomming leg by a factor of 3 for 2 years? Totaling 12 years of aging for B and 20 years of aging for A. I think I might understand how it works now.

Thanks.

jtbell
Mentor
George Jones said:
As B travels, B uses a telescope to watch A's wristwatch. During the ougoing leg, B sees her second hand spin faster than A's by a factor of 3, due to the Doppler effect. Thus, at the turn around point B sees, via her telescope, a reading of 6/3 = 2 years on A's watch. During the incominging leg, B sees her second hand spin slower than A's by a factor of 3, again due to the Doppler effect. Thus, B sees the reading on A's watch increase by 6*3 = 18 years during the homecoming leg.
Just to complete the picture, let's imagine that A is also using a telescope to watch B's wristwatch as she (B) travels out and back. In A's reference frame, each leg of B's trip takes 10 years. However, A does not actually see B turn around until 8 years after it happens, and 18 years after the trip begins, because the turnaround point is 8 lightyears away. During the outbound leg, A sees B's watch running slower by a factor of 1/3 becase of the Doppler effect, and therefore sees (1/3)(18) = 6 years elapse on it, up to the turnaround point.

In A's reference frame, B's round trip takes 20 years. When A sees B turn around, 18 years have elapsed, so only 2 years are left until B returns. During the homecoming leg of the trip, A sees B's watch running faster by a factor of 3 because of the Doppler effect, and therefore sees 3*2 = 6 years elapse on it, during that leg. Total elapsed time on B's watch, as seen by A: 6 years outbound plus 6 years homecoming = 12 years total, same as B observes, and in agreement with the time-dilation equation.

Thanks for completing the picture. I appreciate all the help.

Hope I'm asking for too much if I also ask for the length and mass of Twin B compared to Twin A? I'm not sure what numbers to give as an example.
Another thing I wonder about length contraction. The examples I see online and in the books I read show a 2 dimensional picture half as long as before but still the same height, so my question is that an accurate example or is the squared radius half as much as it was before in the 3 dimensional picture?

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The dimensions perpendicular to the relative velocity vector transform identically - the lengths in those directions don't change.

So x and t get shorter and slower while y and z remain? So the length contraction can be solved using the principle of equalivance right? So the acceleration of Twin B reaching the speed of light would be similar to the gravitional field of a black hole? It would be like a flat 2 dimensional disc?

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Enos said:
So x and t get shorter and slower while y and z remain? So the length contraction can be solved using the principle of equalivance right? So the acceleration of Twin B reaching the speed of light would be similar to the gravitional field of a black hole? It would be like a flat 2 dimensional disc?
When the realtive velocity between the twins is close to c, they will each look flattened to the other. As long as neither twin turns around (= accelerates = has a curved worldline), the effects are symmetrical between the twins. Each twin observing herself sees only a normal rest frame situation.

Perspicacious said:
There are subtleties in the twin paradox that can't be avoided unless you're competent in high school algebra and can follow carefully stated mathematical reasoning, line by line. I only know of one paper that properly resolves the "twin paradox." It is carefully detailed and is written by a mathematician. Let me know if you have any questions.

http://www.everythingimportant.org/relativity/
I draw everyone's attention to the host site of this paper. Should physicists begin to worry about "evolutionisation" of physics?

Note also that in at least four of Perspicacious' posts he, Mr Shubert, I can safely presume, refers to the same paper. The comment above therefore borders on dishonesty, representing as it does the author's own work as a reliable reference piece. Is this permissible behaviour in physicsforums? It would surely be academic misconduct in the real world.

When reading things on the internet, I suggest it is sensible to check the context in which what you are reading was written.

I didn't read his paper because I noticed it's a copy and paste reply to any subject on the twins paradox.

pervect
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