Reconciling Length Contraction and Lorentz Transformation Interpretations of a Special Relativity Problem [closed]

Question

I am stuck on understanding how to interpret the following findings. I was working on a special relativity problem in which two observers standing next to a conveyor belt (which happens to be moving at a relativistic speed $v$ relative to the ground) decide to paint marks on the belt simultaneously. I know in the painters' separation as measured in the ground frame, call it $L$. They will say the paint marks are that same distance apart on the conveyor belt: $L$. Since that length $L$ is measured in a frame which is moving relative to the belt, the proper separation of the paint spots, $L_0$, (in the frame of the belt) will end up being longer than the separation observed by the painters, because of the length contraction relationship: $L_0 = \gamma L$. That's fine.

Then the issue arises: how can an observer on the belt say the paint marks are separated by $L_0 > L$, when the painters' separation should be contracted to less than $L$ as observed from the belt. That is resolved by realizing that events that are simultaneous in one frame (in this case the ground's frame) are not necessarily simultaneous in a different inertial frame. You can use the Lorentz transformation equations to find that the front painter (the one way ahead of an observer at the origin in the belt's frame) paints a mark before the back painter (i.e. the painter that's closer to the origin in the belt's frame), according to the belt's frame.

However, here's where I run into a problem: I know $L=10$ ft, and $v=0.5c$ corresponding to $\gamma=1.15$. I know $L_0 = 11.5$ ft. I know (or at least I thought I knew) that the painters apparent separation according to the belt frame is $L_\mathrm{app} = 8.7$ ft.

Now for the Lorentz transformations: calling the painting events A and B, with $x_A = 0$, $t_A = 0$, $x_B = 10$ ft, and $t_B = 0$, I find $x'_A = 11.5$ ft, $x'_B = 0$, $t'_A = -5.85$ ns, and $t'_B = 0$. That looks good. $L_0 = x'_A-x'_B = 11.5$ ft, and the front painter paints his spot $5.85$ ns before the back painter paints his spot.

Now the issue arises: at $t' = -5.85$ ns, the origin of the belt frame is $\left|0.5c\cdot -5.85 ns\right| = 5.76$ ft to the left of the origin in the ground frame (assuming the positive $x$-axis points to the right). But, since $x'_A = 11.5$ ft, that must mean that the front painter is only $11.5$ ft $- 5.76$ ft $= 5.74$ ft from the back painter (who is standing at the origin in the ground's frame). But that means the painters appear to be $5.74$ ft apart, not the $L_\mathrm{app} = 8.7$ ft apart that I found using the regular old length contraction expression.

So, finally the question: As measured in the belt's frame, are the painters $8.7$ ft apart, or are they $5.74$ ft apart?

Answer 1

I think I see how you managed to confuse yourself; the way you wrote it up had me briefly confused too.

Here is the correct analysis, with both calculations coming out the same way. I'll leave it to you to pinpoint the spot where your calculations went astray:

The thick black lines are worldlines of the painters and the thick blue lines are the worldlines of the marks. The unthick lines are lines of simultaneity in the two frames.

First do everything in the black frame:

$B$ is at $(t=0,x=L)$.

$C$ is at $(t=vL,x=L)$

$A$ is on the line $vx=t$ and on the line $x=vt+L$, hence at $\big(t={vL/(1-v^2)},x=L/({1-v^2})\big)$

The marks are painted at $O$ and $B$, which in the black frame are simultaneous and a distance $L$ apart.

Now Lorentz transform to the blue frame:

$B$ is at $(t'=-Lv/\sqrt{1-v^2},x'= L/\sqrt{1-v^2})$.

$C$ is at $(t'=0,x'=L\sqrt{1-v^2})$

$A$ is at $(t'=0,x'=L/\sqrt{1-v^2})$.

Now how far apart are the marks in the blue frame?

First calculation: Just as one mark reaches $A$, the other mark is being created at $O$. The spatial distance from $A$ to $O$ is $L/\sqrt{1-v^2}$. So that's the distance between the marks.

Second calculation: The distance between the painters is the distance between their locations at time $t'=0$, i.e. the distance from $O$ to $C$, or $M=L\sqrt{1-v^2}$.

The time between the placing of the marks is the time that passes between events $0$ and $B$, or $N=Lv/\sqrt{1-v^2}$. During this time, the painters move left a distance $vN$.

Therefore the distance between the marks is $M+vN=L/\sqrt{1-v^2}$.

The two calculations give the same result.