I Solving Length Contraction Paradox

[Jonsson]

Hello there,

Suppose there are two inertial frames of reference $S$ and $S'$ with coordinates $(x,ct)$ and $(x',ct')$ such that $S'$ is moving relative to $S$ with velocity $v$. Suppose $v>0$, that implies $\gamma >1$.

We know that a Lorentz boost is given by:
$$
x' = \gamma (x -vt), \qquad t' = \gamma (t - \frac{v}{c^2}x ) \qquad (1)
$$
And the inverse transforms are given by
$$
x = \gamma (x' +vt'), \qquad t = \gamma (t' + \frac{v}{c^2}x' ). \qquad (2)
$$
Suppose $|x_A - x_B| = L$ and $|x'_A - x'_B| = L'$ is the length of a rod in the $S$ and $S'$ frame of reference respectively, then we know that there is a length contraction:
$$
L' = |x'_A - x'_B| \stackrel{(1)}{=} |\gamma (x_A - vt) - \gamma(x_B - vt)| = \gamma L \qquad (3)
$$
However using equation (2) a paradox arises:
$$
L = |x_A - x_B| \stackrel{(2)}{=} |\gamma (x'_A +vt') - \gamma (x'_B +vt')| = \gamma |x'_A - x'_B| = \gamma L' \qquad (4)
$$
That means
$$
L' \stackrel{(3)}{=} \gamma L \stackrel{(4)}{=} \gamma \gamma L' \stackrel{(3)}{=}\gamma^3 L \iff L = \gamma^3 L \implies 1 = \gamma^3 \implies \gamma = 1,
$$
a contradiction.

What is going wrong? I really cannot find it.

 

[Orodruin]

In (3) you are assuming that the events are simultaneous in S by assuming they occur at the same time (you are using the same t in both their transformations). In (4) you are assuming that they are simultaneous in S' instead (you are using the same t'). This is inconsistent withe the time Lorentz transformations.

 

[Dale]

Suppose $|x_A - x_B| = L$ and $|x'_A - x'_B| = L'$ is the length of a rod in the $S$ and $S'$ frame of reference

This is the problem. This can only be true if the rod is at rest in both $S$ and $S'$. In which case $v=0$ and $\gamma=1$

The reason is, as @Orodruin mentions, the relativity of simultaneity. Length requires that the ends of the rod be measured simultaneously. So the difference in x coordinates between events A and B can only represent a length if they are simultaneous, and they can only be simultaneous in both frames if $v=0$

 

[stevendaryl]

Here's a little javascript animation I made showing how length contraction (moving objects are shorter), time dilation (moving clocks run slower) and the relativity of simultaneity (the clocks of a moving rest frame are out of synch with one another) work together consistently.

http://dee-mccullough.com/relativity/

In the demo, there are two very long rocket ships traveling side-by-side in opposite directions, the Red Rocket and the Green Rocket. The relative speed of the rockets is 0.866 c, leading to a value of \gamma = 2. Each rocket is 52 light-minutes long in its own rest frame, meaning it takes 52 minutes for light to travel from one end to the other, in the rocket's rest frame; it takes the other rocket 60 minutes to cover the same distance. Each rocket has a clock at each end (the time is shown in the demo). The two clocks for a rocket are synchronized in the rocket's rest frame, but not in the other rocket's rest frame.

In the demo, the Red Rocket is moving left-to-right relative to the Green Rocket, and the right end of the Red Rocket passes the left end of the Green Rocket at a time when the corresponding clocks show 12:00.

In the rest frame of the Green Rocket:

1. The Red Rocket is shrunk to half its normal size: 26 light-minutes, rather than 52.
2. The clocks on the Red Rocket run slow---they advance at half the rate of the Green Rocket's clocks.
3. The clocks on the Red Rocket are out of sych: the left clock of the Red Rocket is 45 minutes ahead of the right clock.

In the rest frame of the Red Rocket:

1. The Green Rocket is shrunk to half its normal size.
2. The clocks on the Green Rocket advance at half the rate of the Red Rocket's clocks.
3. The clocks on the Green Rocket are out of sych: the right clock of the Green Rocket is 45 minutes ahead of the left clock.

Even though the coordinate-dependent facts are completely different in the two frames, there are a few invariant facts, that they both agree on:

1. The right end of the Red Rocket and the left end of the Green Rocket pass when they both show time 12:00.
2. The right end of the Red Rocket passes the right end of the Green Rocket when the closest Red clock shows 12:30 and the closest Green clock shows 1:00 (which confirms to the Green Rocket passengers that the Red clock is running slow--it only advanced 30 minutes while the Green clocks advanced 1 hour)
3. The left end of the Red Rocket passes the left end of the Green Rocket when the closest Red clock shows 1:00, while the closest Green clock shows 12:30 (which confirms to the Red Rocket passengers that the Green clock is running slow).

You can click the appropriate buttons to switch between the two frames, or click on one of the clocks to switch to the frame in which that clock is at rest (without changing the time or location of that clock).