7
$\begingroup$

Alice and Bob are moving in opposite direction around a circular ring of Radius $R$, which is at rest in an inertial frame. Both move with constant speed $V$ as measured in that frame. Each carries a clock, which they synchronize to zero time at a moment when they are at the same position on the ring. Bob predicts that when they next meet, Alice's clock will read less than his because of the time dilation arising because she is moving relative to him. Alice predicts that Bob's clock will read less with the same reason. They both can't be right. What's wrong with their arguments? What will the clocks really read?

I have try to answer it that their all wrong, Since the they are all moving at the same speed, and they will all cover the same distance ($\pi R$), so they will be at same place with the same time? but I am not sure about my reasoning, Then the clock reading will be $\Delta t=\sqrt{1-\frac{V^2}{c^2}}\Delta t_B.$

Can any one give me a clear reasons on What's wrong with their arguments?

$\endgroup$
  • 5
    $\begingroup$ So I think part of the issue here may be that neither Alice nor Bob is an inertial observer. They are both accelerating radially inward with acceleration V^2/R, so regular old special relativity isn't going to be valid in this problem. It seems clear to me by symmetry that they will read the same time when they pass next. What that time will be compared to a stationary observer, I have no clue (though obviously it ought to be less than the outside observer due to time dilation effects). $\endgroup$ – mhodel Mar 28 '14 at 2:27
9
$\begingroup$

Both move with constant speed V as measured in that frame.

Constant speed but not constant velocity; both Alice and Bob are accelerated, i.e., each observes the other's accelerometer to read non-zero so standard SR reasoning doesn't apply.

However, their worldlines, between the initial event and the event they next meet, are congruent thus, the proper time along each worldline is identical so their clocks will agree when they next meet.

$\endgroup$
  • $\begingroup$ Is it possible to measure the time dilation of Alice with respect to Bob? or Bob with respect to Alice? without considering the rest frame? $\endgroup$ – Lembris Mar 29 '14 at 10:09
  • $\begingroup$ How would one go about actually calculating the time elapsed in the two different frames. I.e. calculate how much time elapsed on Alice's Clock from Bob's point of view. They are the same but I think it would be interesting to compute. @AlfredCentauri $\endgroup$ – ClassicStyle Jun 17 '14 at 4:42
  • $\begingroup$ Should I post a new question for this? $\endgroup$ – ClassicStyle Jun 17 '14 at 4:43
0
$\begingroup$

Each carries a clock, which they synchronize to zero time at a moment when they are at the same position on the ring. Bob predicts that when they next meet, Alice's clock will read less than his because of the time dilation arising because she is moving relative to him. What's wrong [...] ?

There are two things wrong with Bob's argument as presented in the question:

  1. There was no explicit assumption or expectation given how the proper "rates" of Alice's clock readings and of Bob's clock readings, at least on average between their meetings,

$$\frac{(t_A[ \text{reunion} ] - t_A[ \text{departure} ])}{\Delta \tau_A[ \small{ \text{from departure until reunion}} ]}$$

and

$$\frac{(t_B[ \text{reunion} ] - t_B[ \text{departure} ])}{\Delta \tau_B[ \small{ \text{from departure until reunion}} ]}$$,

ought to be related to each other; e.g. whether these two rates were assumed/expected to be equal to each other, or what else.

(And given nothing more than the initial synchronization of the two clocks, there is of course no reason at all to hold any particular expectation about how these two rates might be related to each other.)

And 2.:
Based on the setup description of the movements of Alice and Bob relative to each other, accounting for "time dilation" as usual, Bob (and Alice, and everyone) would determine

$\Delta \tau_A[ \small{ \text{from departure until reunion}} ] == \Delta \tau_B[ \small{ \text{from departure until reunion}} ] == \sqrt{ 1 - \frac{V^2}{c^2} } \pi \frac{R}{V}$.

So that's certainly no reason to predict/expect

$t_A[ \text{reunion} ] < t_B[ \text{reunion} ]$

(where according to the initial synchronization $t_A[ \text{departure} ] = t_B[ \text{departure} ] = 0$ ).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.