(2nd Repost) Twin Paradox Problem?

Question

IThis was posted before, here's the slightly corrected version).

At 12:00 noon, Twin A with clock stays at rest while Twin B with telescope jumps to 4/5ths speed of light. Twin B, using telescope, observes Twin A's clock going more slowly because of relativistic doppler. After some time, Twin B suddenly reverses course and heads home at 4/5ths speed of light. Still using the telescope, Twin B observes Twin A's clock going faster, again because of relativistic doppler. At 1:00 pm, he notices that Twin A's clock now also shows 1:00 pm. When Twin B arrives back home with Twin A, how much older is Twin A than Twin B?

Bonus question: Why does Twin A, using telescope, never sees Twin B's clock showing the same time as Twin A's clock?

This is an idealized problem in Special Relativity, so any acceleration is instantaneous. In fact, Special Relativity handles acceleration poorly, which was why Einstein went on to General Relativity.

Remo Aviron, I had very carefully chosen 4/5 as the beta value to make this problem as easy as possible to solve. You finally came up with the right answer (by different means than the way I did), but I'm going to grant the best answer to Prof Zikzak only because he beat you to it. Congrats, both of you.

Answer 1

I get exactly 1 hr. older, again assuming that by "observing" you mean "viewing through the telescope and not correcting for light travel times." The key is to find how long B travels before turning around, and this can almost all be done using phys 101 kinematics in A's frame.

Twin A never sees B's clock showing the same time as his own because in the reference frame of A, B's clock is monotonically slower than A's and A experiences no shift of reference frame.

Answer 2

Ah:

1. Get the time Lorentz factor = sqr(1/(1-(4/5)^2)
=sqr(1/(9/25))
=5/3

2. Let's use A as a reference. When B has travelled one hour (i.e., 1 p.m), in A reference the time will be 1:40. This means that light from when A's clock read 1 pm. will have travelled 40 minutes or 2/3 light - hour.

3. In order to travel the 2/3 light hour back home at 4/5 c, B will take the following time t=2/3 light hours / 4/5 c = 5/6 hour

4. Therefore, according to A, B's trip will take 5/3 hour + 5/6 hour or 15/6 hour = 5/2 hours or 2.5 hours.

5. Since B's Lorentz factor is 5/3, B's total time will be 5/2 *3/5 or 1.5 hours. Therefore the time difference will be 1 hour.

[Note: B will travel out 1 light year (A's reference) and return 1/3rd of a light year at the time the two clocks synch].

Yea, I finally did it right!

The bonus question: A sees a redshifted B on the way out so B's clock will always be behind, and then there is a frame shift at the turn around so that B's clock will remain behind A's at all points.

Answer 3

Ill try and answer the bonus question first, but i might be wrong, relativity is something i need to work on.

Is the reason A never sees B's clock showing the same time because the light that A is receiving is shifted due to B's speed, hence he is seeing what has already been, or seeing what is happening at a faster rate than it actually is, due to B's speed.

In regards to the age of the twin, i know that we have to use the lorentz contraction to calculate time dilation l'=l*sqr(1-v²/c²)
however B's velocity changes after an unknown amount of time ( i was going to assume half an hour for simplicity), does the direction in which B is travelling affect what happens to Bs measure of time, assuming no deceleration, and instant reversal in direction?.....

Apologies for answering a question with a question, but this paradox raises more questions than answers for me.

Answer 4

Time is absolute and speed of light is infinite.
At all instants both clocks are at sync.

(a)
When Twin B arrives back home with Twin A, he is as old as his brother.

(Bonus question)
Because per specification twin B is with telescope, and twin A is without. It is hard to observe distant clock without a telescope.