Special Relativity - Relativity of Simultaneity?

[Luxucs]

Homework Statement

A thin rod of proper length 4a is traveling along the x-axis of a frame S with a speed ${\frac {\sqrt 3} 2}c$ in the positive x-direction. A hollow cylinder CD of proper length 2a is placed with its axis along the x-axis, so that when the ends of the cylinder are open the rod will pass through the cylinder. Then, end C of the cylinder is located at x = -2a and the end D at x = 0, and both ends are equipped with hypothetical devices capable of closing them off with impenetrable and immovable walls.

a) Show that in the frame S in which the cylinder is at rest, both the rod and the cylinder have length 2a.

b) Show that in the frame S' in which the rod is at rest, the length of the rod is 4a, whereas the length of the cylinder is only a.

c) Suppose at time t = 0 in frame S the front end B of the rod is located at x = 0, and both ends of the cylinder are suddenly closed. Is the rod trapped in the cylinder as might be inferred from the answer to part a), or is the rod cut into two parts as might be inferred from the answer to part b)? Reconcile this paradox.

Homework Equations

We have the Lorentz transforms,

$x' = \gamma(x - vt)$ (1)
$t' = \gamma(t - \frac {vx} {c^2})$ (2)

And the length of an object can be related to its proper length by,

$L = L_{o}\sqrt {1 - \frac {v^2} {c^2}}$ (3)

where v is the velocity of the frame S' with respect to S. S' is attached to the moving rod in this case.

The Attempt at a Solution

Part a) and part b) are relatively straightforward (just plug-and-chug with (3)). Part c) is where the action of the problem lies, and it is the part I seek confirmation on. I considered the front and back ends of the cylinder using (2) to analyze the times at which both ends of the cylinder close with respect to S'.

The Lorentz factor, $\gamma$, for the given velocity of the rod (and thus of the frame S'), is,

$\gamma = 2$

From (2), we'll first look at the time at which the front end of the cylinder closes with respect to the rod in S',

$t'_{front} = 2(0 - 0) = 0$ (4)

(4) basically means the front end of the cylinder closes at the same time in both S and S'. We now consider the back end of the rod, again from (2),

$t'_{end} = 2(0 - {\frac {{\frac {\sqrt 3} 2}c} {c^2}(-2a)}) = {\frac {2\sqrt{3}a} {c}}$ (5)

So, (5) means that the back end of the cylinder won't close at the same time as the front end in S', unlike in S. However, I'm unsure as to the resolution of the paradox from here. Did I mess up somewhere along the way here? How am I to interpret this result in terms of the rod either being cut in half or fitting inside the cylinder?

 

[Nathanael]

There's actually a simple resolution that requires no calculation. If two events happen at the same time and the same place, then that is universally true in every frame. Just by that you can reason that, since the blade slices at the same time and place as the front end of the rod reaches it, it will also be at the same time in the other frame. The same can be said for the back end and back knife. Thus, in any frame, the front slice and back slice must happen whenever the front and back ends reach them.

Anyway, you are on the correct path and almost done, so let's continue your method. (It's a good check of the Lorentz transformation.) Now as you said the front blade comes down right when the front end crosses, then the back end comes down a little time later (later because it's positive) given by your equation (5). Just a couple more questions to see how it ties up; How far does the rod travel in that time? How far behind the back knife is the back end when t=0?

 

[Luxucs]

There's actually a simple resolution that requires no calculation. If two events happen at the same time and the same place, then that is universally true in every frame. Just by that you can reason that, since the blade slices at the same time and place as the front end of the rod reaches it, it will also be at the same time in the other frame. The same can be said for the back end and back knife. Thus, in any frame, the front slice and back slice must happen whenever the front and back ends reach them.

Is this due to the spacetime interval between these two events being invariant, or is it something else?

Anyway, you are on the correct path and almost done, so let's continue your method. (It's a good check of the Lorentz transformation.) Now as you said the front blade comes down right when the front end crosses, then the back end comes down a little time later (later because it's positive) given by your equation (5). Just a couple more questions to see how it ties up; How far does the rod travel in that time? How far behind the back knife is the back end when t=0?

Alright, I think there's some clarification that I need here. Are we assuming the front end of the cylinder wall retracts after coming down (as in, it snaps down and comes back up immediately, so the rod can continue moving)? It's probably a silly clarification, but that might be part of my confusion here.

As for your questions (assuming the retracting-wall case), we know that in S', the back end of the rod is at x = -4a, and the cylinder, from part b) of the problem, is at x = -a. Thus, we require the back end of the cylinder to move a distance of 3a (in S') in the time interval that I calculated above in order for it to not become cut off, or to go from x = -a to x = -4a, as we consider the rod stationary in this frame.

Here, we can apply simple kinematics to determine if this will be the case,

$v' = \frac {\Delta{x'}} {\Delta{t'}}$

Notably, since we are considering the frame in which the rod is at rest, v' will simply be the negative of the velocity of the rod in S, and the back end of the cylinder starts at x = -a, or,

$-\frac {{\sqrt{3}}} 2c = \frac {{x_f'} + a} {\frac {2{{\sqrt{3}}}a} c}$

$x_f' = -4a$

This should is the expected result. Does this look good to you?

 

[Nathanael]

Alright, I think there's some clarification that I need here. Are we assuming the front end of the cylinder wall retracts after coming down (as in, it snaps down and comes back up immediately, so the rod can continue moving)? It's probably a silly clarification, but that might be part of my confusion

Yes that’s right, and it’s not silly, it’s an absolutely necessary clarification! The problem is misleading for saying “is the rod trapped or cut?” ... the answer is neither! We can’t trap the rod or else it would stop and no longer be contracted. All we can do is momentarily contain the rod and instantly release it. It’s very necessary that we instantly retract the walls.

Does this look good to you?

Yep that’s good. As you said we need the cylinder to move 3a in that time interval and 3a is exactly what you get when you take (t’_back - t’_front)v

Is this due to the spacetime interval between these two events being invariant, or is it something else?

Well the spacetime interval is always invariant for any two events, so it’s not that.

The idea is that if you have a difference of coordinates (Δx, Δy, Δz, Δt) = (0,0,0,0) then a Lorentz transformation (in any direction) will give (Δx’, Δy’, Δz’, Δt’) = (0,0,0,0) as well.

Loosely speaking, two events at the same point in spacetime are basically the same event and can’t be separated. Changes in space get mixed up with changes in time (and vice versa) when changing frames, but if there is no change in time or space then there’s nothing to get mixed up.

I very well may be making that sound more complicated than it is; if so, sorry.

 

[Luxucs]

Well the spacetime interval is always invariant for any two events, so it’s not that.

The idea is that if you have a difference of coordinates (Δx, Δy, Δz, Δt) = (0,0,0,0) then a Lorentz transformation (in any direction) will give (Δx’, Δy’, Δz’, Δt’) = (0,0,0,0) as well.

Loosely speaking, two events at the same point in spacetime are basically the same event and can’t be separated. Changes in space get mixed up with changes in time (and vice versa) when changing frames, but if there is no change in time or space then there’s nothing to get mixed up.

I very well may be making that sound more complicated than it is; if so, sorry.

No worries. I think I grasp the fundamentals of what you're getting at anyway. The point is that space and time separately are not invariant (due to length contraction/time dilation), but when put together into spacetime are.

Thanks for taking the time to help with this problem! I'm good to go from here, and I definitely did learn something!