Understanding Relativity from A and B's Point of View

[EternitysEnd]

I've just started studying relativity and I am having some slight problems.

Say there are 2 persons, a and b. A is on Earth and B is accelerating away from earth.

While B gets faster and faster he will gain mass and his time will get slower from A's point of view.
How does this look from B's point of view? I suppose he has normal mass, but how does he see the time relative to earth. Is the time on Earth faster relative to his, and if so, why? (since moving away from Earth is the same as the Earth moving away from you, right?)
And if not, then why is the time on Earth so much later when B turns around and accelerates back to earth?

Hope you understand my question.

 

[cesiumfrog]

Yes, it looks the same from B's point of view (after the acceleration). No, there's no paradox on B's return. When B turns around (ie. experiences measurable acceleration) he also changes his mind about where (in space, and in time) A is "now" located (simultaneity isn't what you used to think it was).

This is in countless FAQ's somewhere..

 

[EternitysEnd]

I couldn't find a FAQ, not sticky on the forum at least. I've read a book about it saying that if B returns the time on Earth is later, but this is not the case?

I just don't understand how it would look if B had a telescope and was looking at the Earth on the way back.

 

[George Jones]

A Doppler time-rate change is observed with a telescope. In the end, this agrees with time dilation.

Are you studying a mathematical introduction to relativity? If not, then the following might not make sense.

Two twins, Alfred and Betty, are together on the planet Omicron 7. They synchronize their watches to zero, and Betty then sets off on a return trip from Omicron 7 to Earth and back to Omicron 7. The distance between Omicron 7 and Earth is $L$ lightyears in the (approximately) inertial reference frame of Omicron 7 and the Earth. Betty takes the most direct route and moves at a constant speed $v$ during both the outgoing and incoming segments of the trip. Therefore, the time dilation/Lorentz contraction factor is

$$\gamma =\left( 1-v^{2}\right) ^{-\frac{1}{2}}>1$$

and the Doppler shift factor is

$$k=\sqrt{\frac{1+v}{1-v}}>1.$$

In Betty's frame the distance between Earth and the other planet is Lorentz contracted to $L^{\prime}=L/\gamma$. Consequently, the time taken, according to Betty's clock, for the first part of the trip is $t_{1}^{\prime}=L^{\prime }/v$ years. For the same reason, Betty takes another $t_{2}^{\prime}=L^{\prime}/v$ years, according to her watch, to get back to the Earth, for a total elapsed time of $t^{\prime}=t_{1}^{\prime}+t_{2}^{\prime}=2L^{\prime}/v$ years. Alfred measures the total elapsed time to be $t=2L/v=\gamma t^{\prime}$ years.

Alfred and Betty can directly read both watches at the start and end of the trip. As Betty travels, Betty uses a telescope to watch Alfred's wristwatch. As Betty watches Alfred's watch, the frequency of revolution that she sees for Alfred's second hand is related to the frequency of revolution of her own second had by the Doppler effect factor $k$. During the ougoing leg, Betty sees Alfred's second hand spin slower than hers by a factor of $k$. Thus, at the turn around point Betty sees, via her telescope, a reading of $T_{1}=t_{1}^{\prime}/k$ years on Alfred's watch. During the incominging leg, Betty sees Alfred's second hand spin fasterer than hers by a factor of $k$. Thus, Betty sees the reading on Alfreds's watch increase by $T_{2}=t_{2}^{\prime}k$ years during the homecoming leg. The total time that Betty sees elapse on Alfred's watch, $T_{1}+T_{2}=t^{\prime}\left( 1/k+k\right) /2=\gamma t^{\prime}=t$, agrees with the time dilation calculation above, as it should.

 

[EternitysEnd]

Thanks!

Took a few reads but I think I got it now.

 

[yogi]

Try to find a used copy of Paul Hewetts "Conceptual Physics" Little or no math - a good visualizable treatment of these kinds of problems