Apparent paradox in special relativity

Question

I was recently set a simple homework problem. Bob sees Alice come past him in the positive $x$ direction. They have identical spaceships, each with three clocks spaced evenly along the ships. In the rest frame, the length of the ships is $L_0 = 90$m. Bob measures Alice's to be $L'=30$m.

Now, the first part is trivial, we are asked to find $\gamma$. Afterwards, we are asked what the clocks in Alice's frame say if:

1. Bob's clocks all read 0
2. Bob observers Alice's front clock as reading 0.

Now, my line of logic is as follows. In Alice's frame, the clocks must be synchronized. If they aren't, then their asynchronicity makes this problem nonsensical, the clocks could be off by hours from one another and then not even Einstein could predict what they'd say. Thus, if we know one of Alice's clocks, we know them all. At the front of the ship, Alice and Bob pass one another and Bob measure's Alice's clock at 0. Now, if this is an event at $x^\mu = (c(0),0) = (0,0)$, then Lorentz transformation is trivial and gives $x_{Alice}^\mu = \Lambda^\mu_\nu x^\nu$ = (0,0). Thus Alice's front clock reads 0 in her frame and so too do all of her other clocks.

Now, this relies on them passing each other by overlapping, with zero physical distance at the event, but its an idealization, and no information was given about the passing distance. The solutions argue that because you know the position of the clocks in both frames, and the time in Bob's frame, you can do a Lorentz transformation of the form

$x_{clock 1} = \gamma(x'_{clock_1}+vt'_{clock_1})$ Where $x$ is the measurement in Bob's frame and $x'$ is the measurement in Alice's Frame. Because we can solve for $t'$, we should be able to find the measurement on $t'$.

Now, I cannot find a way to either refute or confirm the professor's approach. On one side, it seems to suggest that the interaction of Bob's frame with Alice's causes asynchronicity to occur, which is a direct violation of the postulates of SR. My other thought is that there is some confusion between frame time (the set of synchronized clocks at rest in Alice's frame at all points in space), and the time actually "seen" by Alice. After all, at 45 and 90 meters, there would be some 300 ns delay between her and the furthest clock, which is remarkably close to the reported answer. Even so, this seems very much no in the spirit of relativity, were we assume Alice can successfully account for the time delay. Am I missing a clear resolution here? Is the professor just wrong? Am I just wrong?

Answer 1

My assumption for this answer is that the front of Alice's rocket passes by the front of Bob's rocket at the moment Bob's clocks all read $0$. In Alice's frame, we can define the event coordinates for Alice's clocks as $(c(0),-90),(c(0),-45),(c(0),0)$.

Let the coordinates for the same events in Bob's frame be $(ct_1, x_1), (ct_2,x_2), (ct_3,x_3)$ respectively. Since the fronts of rockets coincide when both Alice and Bob's front clocks read $0$ in Bob's frame, $x_3=0$ and $t_3=0$.

Next, you can solve for $t_1,t_2$: $t_1=\gamma(0+\frac{\beta}{c}(-90))$ and $t_2=\gamma(0+\frac{\beta}{c}(-45))$, where $\beta=2\sqrt 2/3$ and $\gamma=3$.

Here's Bob's spacetime diagram in Desmos. I hope that gives you further intuition of the situation. Note that all points falling on the dashed line represent simultaneous events for Alice, but clearly they aren't simultaneous for Bob since they correspond to different $ct$ coordinates in Bob's spacetime diagram.

Answer 2

Your analysis seems reasonable to me. The problem (at least as you've paraphrased it) seems poorly worded and ambiguous. It should explicitly state that the clocks are synchronized, as you said.

My guess is that the clocks are all synchronized in the usual way, and you're supposed to find the readings on all of Alice's clocks when the nearest clock of Bob's reads 0. What you need for that is the other half of the Lorentz transformation, $0=γ(t'+vx'/c^2)$. If you want the reading as a function of position in Bob's rest frame, rather than Alice's, then you also need $x=γ(x'+vt')$.

Answer 3

Your mistake is to assume that all of Alice's clocks show the same time when Bob's do.

This is a question about the relativity of simultaneity. Because Bob and Alice are moving relative to each other, their respective planes of simultaneity become tilted. A fixed time for Bob is a flat slice through his spacetime, whereas it is a sloping slice through Alice's, so at a fixed time everywhere in Bob's frame, Alice's time varies at every point along the direction in which she is moving relative to Bob.

ADDENDUM

You are confused about the meaning of the question and about the significance of the relativity of simultaneity.

If you have two frames A and B moving relative to each other, and you take in one a plane of constant time, say t=0 in plane B, then that plane is angled through time in the other frame. At every point in space it is t=0 in Bob's frame; in Alice's frame there will be a varying value of t' all along the plane.

The question is not saying that Alice's clocks have become unsynchronised. The question is recognising that the various points along the plane of t=0 in Bob's frame exist at different times in Alice's.

To put it the converse way- in Alice's frame, Bob's clocks will appear out of synch. She will 'see' each of them read t=0 at different times. Can you see that?