Lorentz Jr

B and C are signal receptions at t = 0 and x = 1 or -1.

Forget about them, martillo! They're worthless! You can't analyze relativity problems in terms of objects. You have to focus on events, and there are no events involving A after it sends the signals. The only thing you can do is calculate the time and positions when the signals arrive in A's reference frame, and then transform them to get x' and t' for the signal's arrival at C in B's frame.

You can't do that. The only way A can be placed midway between B and C is if their positions are all measured at the same value of t (i.e. at the same time in A's frame), but then they would correspond to different values of t' (i.e. different times in B's frame).

In B's frame: xi' = 2γ.

This is the distance from A in A's reference frame:

You can't do this, martillo. You're assuming distB(A,C) = L, but it's not.

Good point. Anyway, the problem is that @martillo keeps trying to apply the simultaneity of the signals in A's frame to the analysis in B's frame. Or maybe he's gotten beyond that now, but he hasn't decided when C received its signal.

Maybe you're right. They're equidistant in A's frame. So how do you know when C received its signal?

That's true, martillo. We need to know what assumptions you're making. In all frames. The positions of B and C are basically synchronized ahead of time in A's frame. (They're equidistant from A.) A sends the signal, and B and C are moving toward A at speed v from opposite directions during the experiment.

The observer at the midpoint.

Relative to A. The relative speed between B and C is v' = 2v/(1 + v2).

C received its signal at t' = -2γv (t'' = 0). B receives its signal at t' = 0. C and B meet at t' = 1/γv.

But you said B and C change their clocks to t = 0 when they receive their signals. The signal is sent at t = 0 in A's reference frame, but that has nothing to do with Genady's diagram, because it's in B's reference frame. No, that line doesn't mean anything important for this problem. The diagram is confusing. It should show when the signals are received, not when t' = 0 on C's world line.

As @Genady said, there are two signals, so those are two separate events. They're not simultaneous in B's reference frame. B receives its signal at t'=0, but C already received its signal at t' = -2γv. So, in B's reference frame, C's clock is slower but there's more time between C's signal and the meeting. Those two effects exactly cancel each other out. That's okay, martillo. Everyone is confused about relativity. It doesn't make sense in the simple way that old-fashioned physics does. Can you show the events? C's signal received at -2γv, B's signal received at 0, and the crossing at 1/γv?

But there are no fairies in it. 🤔

A

Stress is the external influence being exerted on the system. Strain is the system's passive response to the stress, giving in or succumbing to it.

Sure, as long as it doesn't get too cold. It's similar to modeling electric current as a flow of "holes", which are gaps in the electron density in a conductor. It works as long as there's something left to remove.

Thank you, martillo. I appreciate your candor. Right! Except guess what, martillo: Using your gamma formula is a simpler ending for the solution! I was thinking about what you said, that the simple gamma formula works for events on the primed world line, and it turns out that finishing the solution in B's frame is simpler than that huge mess at the end of my first solution (in C's frame). We need to start with the parameters in B's frame: ti' = -2γv v' = 2v/(1 + v2) tf' = 1/γv Then calculate the elapsed time from C's signal, still in B's frame: Δ' = tf' - ti' = 1/γv + 2γv vΔ'/γ = 1 - v2 + 2v2 = 1 + v2 And finally, just divide that by Γ: Δ = Δ'/Γ (vΔ/γ)2 = (1 - v'2)(1 + v2)2 = (1 + v2)2 - 4v2 = (1 - v2)2 = 1/γ4 Δ = 1/γv MUCH easier! 😋

That applies when the initial and final events are the same for both intervals and the two events occur at the same location in the primed frame. In the current problem, B and C receive two different signals, so the simple gamma formula doesn't work. More time passes for B if you start counting from when C receives a signal, but the elapsed time for B starts when B receives a signal, not at the earlier time when C does.

This is incorrect, martillo. You can't just use gamma for your calculations. You have to use the Lorentz transformations.

So x = -ct for B's signal and x = vt - L for B. The signal arrives when -ct = vt - L. (c+v)t = L t = L/(c+v) x = -L/(1 + v/c) for B and x = +L/(1 + v/c) for C. It's exactly the same problem, martillo. The only difference is B and C receive their signals at a distance of L/(1 + v/c) from A instead of L, and the signals arrive at t = L/(c+v) instead of zero. Same problem with different numbers. That doesn't work. You need to start in A's frame, because that's the frame in which B and C's positions are symmetric and they receive their signals simultaneously. Right. That's -2γv. I come with the correct solution to your problem, martillo. A slight variation with different numerical values, but the same math and same result. Any solution without the velocity-addition formula will be incorrect, unless you solve the problem in A's frame. Otherwise, you need to know the speed of C relative to B in order to calculate the correct gamma factor between them.

Okay, let's start with an observer at rest with respect to A and at B's location x = -L when the signals arrive, and let's set c and L equal to 1 for simplicity, so we don't have to keep writing them in every step. Then we have the following: B receives a signal at t = x = 0. C receives a signal at t = 0 and x = 2. Now let's switch to B's reference frame, which moves at speed +v: B receives a signal at x' = t' = 0. C received a signal at xi' = γ(2 - 0v) = 2γ and ti' = γ(0 - v(2)) = -2γv. Now we need the speed of C relative to B. We get that from the velocity-addition formula: u' = (u-v) / (1 - uv) = ((-v) - v) / (1 - (-v)v) = -2v/(1 + v2) So the (positive) relative speed between B and C is v' = 2v/(1 + v2), and C's motion in B's reference frame is x' = xi' - v'(t' - ti') = 2γ - v'(t' - (-2γv)) = 2γ - v'(t' + 2γv)) In B's reference frame, B and C will meet when C gets to x' = 0: 0 = 2γ - v'(tf' + 2γv) v'(tf' + 2γv) = 2γ tf' + 2γv = 2γ/v' tf' = 2γ/v' - 2γv = 2γ((1 + v2)/2v - v) = (γ/v) ((1 + v2) - 2v2) = 1/γv L/γv is just the time-dilated time for B to return to the midpoint at A. Now, what happens in C's frame, as calculated from B's frame? We need the time when C received a signal (at xi' = 2γ and ti' = -2γv in B's frame). Defining Γ to be the gamma factor for v' (i.e. between B and C), we have ti'' = Γ(t' + v'x') = Γ(- 2γv + 2γv') = 2Γγ(v' - v) (1 + v2) ti'' = 2Γγ( 2v - v(1 + v2) = 2Γγv(1 - v2) = 2Γv/γ Of course, C's clock was set to zero, but that just means it's not synchronized correctly in B's frame. The important thing is that it passed through the desired point in spacetime. And the time when C meets B is tf'' = Γ(t' + v'x') = Γ(1/γv + 0v') = Γ/γv. So the elapsed time in C's frame is Δ = tf'' - ti'' = Γ/γv + 2Γγ(v - v'), and now the math gets a little grungy, so we'll define w = v2 and w' = v'2. γvΔ/Γ = 1 + 2γ2v(v - v') (1+w)γvΔ/Γ = 1 + w + 2γ2(w(1+w) - 2w) = 1 + w + 2γ2(w2 - w) (1-w)(1+w)γvΔ/Γ = (1-w)(1+w) + 2w(w-1) (1+w)γvΔ/Γ = (1+w) - 2w = 1-w (1+w)2 wΔ2 (1-w') = (1-w)3 = wΔ2 ((1+w)2 - 4w) = wΔ2(1-w)2 (1-w) = wΔ2 Δ = 1/γv So the same amount of time elapses for C as does for B.

And what do B and C do while the signals are traveling toward them? How do we know where they will be when they receive their signals?

Good job! This is the right way to start. 🙂 What do you mean by "initially", martillo? A's and B's clocks can't be synchronized if A is at x=0 and B is at x=-L.