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I´m trying to solve a special relativity problem, and I think I need some help. There are two inertial frames of reference, $O$ and $O'$, the last one moving with relative velocity $v$ in the $x$ direction. There's a rod with length $L'$ fixed to frame $O'$, such that extreme A is at $x'=0$, and extreme B is at $x'=L'$. Clocks are synchronized at time $t=t'=0$, when position is $x=x'=0$. An observer from $O$ measures the rod, and the result is $L$. Now, from Lorentz transformations, we know that

$$x'=\gamma(x-vt)=\gamma(L-v.0)=\gamma L$$

And, as $x'=L'$, we have $L'=\gamma L$, with

$$\gamma=\frac{1}{\sqrt(1-(v/c)^2)}$$

Now I need to find the result $L'=\gamma L$, but with another method. First, the problem requests to find the $\Delta t'$ (from the point of view of O') since extreme A of the rod is measured by O, until extreme B is measured by O. Of course, both events are simultaneous for O, then $\Delta t=t_B-t_A=0$, so $t_B=t_A=t$. Here, I did (I'm not sure if it's ok)

$$t'_A=\gamma(t_A-vx_A/c^2)=\gamma(t-v.0/c^2)=\gamma t$$ $$t'_B=\gamma(t_B-vx_B/c^2)=\gamma(t-vL/c^2)=\gamma t-\gamma vL/c^2$$

Therefore,

$$\Delta t'=t'_B-t'_A=-\gamma vL/c^2$$

Which means that extreme B is measured before extreme A.

After that, the problem requests to find the position of coordinate origin at O, at the moment when extreme B is measured by O, as seen by O'; and also the position of coordinate origin at O, at the moment when extreme A is measured by O, as seen by O'. Finally, with this and $\Delta t'=-\gamma vL/c^2$, I should find again the difference of length between $L$ and $L'$.

Please let me know if I wasn't clear. Thanks.

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I can give you a hint by explaining how length contraction comes about.

Imagine you and I have to measure the length of a train as it passes a platform. One way we might try is to position ourselves along the platform so that I will be standing by the front of the train as it passes and you will be standing by the back. If we make marks on the platform to show exactly where the front and back of the train are at a given instant, then we can measure the gap between the marks to give us the length of the train. However, that only works if we each make a mark at exactly the same time, otherwise, if I make my mark by the front of the train first, the train will have a chance to move further down the platform before you make your mark by the rear- as a result, the two marks will be closer together than the true length of the train.

That is how length contraction arises. From the perspective of the object in its rest frame, the coordinates of its two ends in the moving frame are at two different times, the front coordinate being earlier than the rear. Any measurement of the length of the object in the moving frame is effectively pinning down the position of the front of the object before the rear, which allows the object to move forward in between, so it seems shorter.

The question you are wrestling with is asking you to find the degree of length contraction by considering the time difference between the measurements in frame A from the object's perspective. All you need to do is to consider how far the rear of the object moves forward between the two measurements, and that will tell you how much shorter it appears.

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  • $\begingroup$ Thanks for the answer. What comes to my mind, is to calculate the position of the origin of O, when O' is at time $t'_A$: $x=0=\gamma(0+v.0)$. And then, to calculate the origin of O, when O' is at time $t'_B$: $x=0=\gamma\left(L'+v\left(-\frac{v}{c^2}\right)\gamma L\right)$, but from this, I get $L'=-\frac{v^2}{c^2}\gamma L$, and I'm not sure of what I'm doing wrong. $\endgroup$
    – Question
    Jan 28, 2022 at 0:04

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