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I am trying to prove the contraction. I know there are several ways one can do it. The way I am trying to solve is said to be complicated and that there is an easier way to do so. Because I don't have a fundamental understanding of the LTs, I can't tell when a situation is easier or more difficult than the other. This is what I am doing:

Rest frame of the rod F.

Moving frame F'.

In frame F:

$x_1=0$ (the beginning of the rod is in the origin) $t_1=t$ (some arbitrary time).

$x_2=l$(end point of the rod), $t_2=t$(same arbitrary time).

In frame F':

$x_1'=-vt_1'-l'$ (beginning of the rod) $t_1'=t'$ (some arbitrary time in the moving frame).

$x_2'=-vt_2'$ (end of the rod), $t_2'=t'$ (same arbitrary time in the moving frame).

Now we want to detect contraction, which occurs in the moving frame:

Using the direct coordinates of the observer in F':

$\Delta x'= x_2'-x_1'=l'$.

Using the LTs :

$x_1'=\gamma(x_1-vt)=-\gamma vt$

$x_2'=\gamma(x_2-vt)=\gamma (l-vt)$

Than $\Delta x'=x_2'-x_1'=\gamma(l)$.

From here $l'=\gamma l$. The intepretation of this last equation, if I understand correctly says that the length measured by the observer in F' is bigger than the one measured by the one in F, which is totally wrong, since is the opposite of what Lorentz contraction describes, I believe

Qmechanic
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imbAF
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  • the origins of $F$ and $F'$ don't seem to be the same, which they should be for a Lorentz transformation – JEB Aug 23 '23 at 22:48
  • See the linked question. The LT into S' puts the ends of the rod at different times, and that's why you are finding it to be longer. When you account for the time difference you will find the rod is indeed shorter. – John Rennie Aug 24 '23 at 06:08

2 Answers2

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You need to label your events as $(t, x)$, just for clarity. (This will make solving SR problems a lot easier.)

For example:

$$E_0=(0, 0)$$ $$E_1=(0, L)$$

are the end points of the rod in its rest-frame at $t=0$. The proper length is $L$.

Then you can transform these to a moving frame to get:

$$ E'_0=(0, 0) $$

and, here's the rub,

$$ E'_1=(t_1'\ne 0, \gamma L) $$

So in the moving frame, you are not measuring the endpoints at the same time, which is no way to measure a moving object.

So you need to find a third event that is the far end point of the rod at $t' = 0$:

$$ E'_2=(0, x'_2) $$

When you do that, you will get:

$$x'_2 = L/\gamma $$

JEB
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  • What do you mean with this: "So in the moving frame, you are not measuring the endpoints at the same time, which is no way to measure a moving object." ? I don't understand it – imbAF Aug 23 '23 at 20:07
  • It means you need to measure the endpoints of the rod at the same time. If that's confusing, replace the rod with a train, and put synchronized clocks on each end. What do people on the platform see? – JEB Aug 23 '23 at 20:10
  • But can't the same be argued for the rod in it's rest frame? The observer there measures one end of it and then the other. So he conducts two measurements in two different moment in time. Unless I am missing something in what you say about the inability to measure the rob simultaneously in the movin frame – imbAF Aug 24 '23 at 06:03
  • SR stuff are thought experiments, so you don’t have someone taking two consecutive measurements. And observer means an infinite lattice of clocks and rulers. You just measure the end points at the same time. In the OP you are measuring the endpoints at different times, while they are moving. – JEB Aug 24 '23 at 11:37
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In the S frame, in which the rod is moving at speed $v$ in the $x$ direction, we must mark the positions, $x_1$ and $x_2$, of its ends simultaneously (at time $T$, say) if we are to find the rod's length, $L$, as $x_2-x_1$. So we have two events, whose space-time co-ordinates are ($x_1, T$) and ($x_2, T$)

Therefore in the S' frame, in which the rod is stationary, the $x$-co-ordinates of these end-of-rod events are $$x'_2=\gamma(x_2-vT),\ \ \ \ x'_1=\gamma(x_1-vT), \ \ \ \ \text{so}\ \ \ \ x'_2-x'_1=\gamma(x_2-x_1)=\gamma L$$

The length, $L_0$, of the rod in its own rest frame is therefore $L_0=x'_2- x'_1$, so $L=\frac 1\gamma L_0$.

BUT, you may say, the events of marking the ends of the rod were simultaneous in the S frame, but they won't be simultaneous in S'. Surely that invalidates $L_0=x'_2- x'_1$ ? No it doesn't, because in the rod's own rest frame we don't have to make simultaneous note of the positions of its ends; it wouldn't matter if we made them days apart!

Philip Wood
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