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Consider Bob moving toward Alice at 86.7% speed of light. Alice sends two light pulses with an interval of one hour (in her frame of reference). Bob starts his clock when he sees first pulse and stops immediately after the second one.

Alice point of view: Bob is moving toward her so his clock is counting time two times slower and when it stops it displays half-hour.

Bob point of view: Alice is the one who is moving so she sends light pulses with an interval of 2 hours and that is time that is measured by his clock.

What time will be displayed on the clock when Bob will arrive to Alice and he will stop moving? How can it be fitted to commom frame?

K Cz
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    Why do you think there is a common frame? – Jon Custer Sep 01 '21 at 18:13
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    Have you tried drawing a spacetime diagram with your situation? That might clear some confusion. – jng224 Sep 01 '21 at 18:25
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    Your question does not make sense. – Marco Ocram Sep 01 '21 at 21:12
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    Sometimes when people have imperfect understanding, they can’t ask perfect questions because they don’t know exactly what they don’t know. I thought this was a reasonable attempt to communicate one’s thinking and identify points of confusion for which clarity is sought. – Al Brown Sep 03 '21 at 03:40

2 Answers2

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In this kind of problem it's always vital to be explicit about exactly when and where things happen. Assign spacetime coordinates to all events, and use one consistent frame of reference throughout; or, alternatively, draw a spacetime diagram. Note that relativity of simultaneity, time dilation, and length contraction all can play a role. It's safest to use explicit Lorentz transformations. In particular, Alice and Bob will have different notions of how far apart they are when Alice sends the signals.

Alice's view

Bob is moving towards Alice, so in the hour that Alice waits between signals Bob has moved 0.87 light hours closer to Alice. In effect, the combined speed of Bob and the light means that the light will get to him faster than if he were stationary. As measured by Alice, the first pulse is sent at (0,0), when Bob is at $(0, x)$. That pulse will be received by Bob at spacetime event $(\frac{x}{1+v}, \frac{x}{1+v})$. The second pulse, sent at by Alice at (1,0), when Bob is at $(1, x-v)$, is received at $(1+\frac{x-v}{1+v}, (x-v)-\frac{x-v}{1+v}) = (\frac{1+x}{1+v}, (x-v)(1-\frac{1}{1+v})$ (again, for clarity, all of this is in Alice coordinates).

Thus, in Alice's frame Bob receives the pulses with an elapsed time of $\frac{1}{1+v}$ = $\frac{1}{1.87}$ hours. Since Bob's clock is time dilated (according to Alice), she will expect it to show half of this time, or $\frac{1}{3.74}$ hours elapsed.

Bob's view

Alice's clock is ticking slower relative to Bob, so for Bob the light pulses are sent out at an interval of 2 hours. In that time they've come even closer together (the light has even less distance to travel) so Bob expects to receive the pulses at a longer interval apart. In Bob's frame the emission events are at $(0, x')$ and $(2, x'-2v)$ (Note that $x$ and $x'$ are probably not equal!) Doing the math again, we see that in Bob's frame the time elapsed is $\frac{1}{2(1+v)} = \frac{1}{3.74}$ hours.

Note that both Alice and Bob believe the other's clock is ticking slowly. This is not a contradiction; it just means that they're measuring different things (time in "different directions" through spacetime).

Eric Smith
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Alice point of view: Bob is moving toward her so his clock is counting time two times slower and when it stops it displays half-hour.

Their clocks can be compared that way when both are momentarily at the same position. The Lorentz tranformation for time (setting $c = 1$): $\Delta t' = \gamma(\Delta t - v\Delta x)$ reduces to $\Delta t' = \gamma \Delta t $ when $\Delta x = 0$. It is not the case in your question.

Claudio Saspinski
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