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i have struggled to understand how length contraction is derived. My lecture book says: length in frame S is defined as $L=x(b)-x(a)$ and in S' as: $L'=x'(b)-x'(a)$ transforming $x'(b)$ and $x'(a)$ we get: $L'=\gamma\ L$ . This is how pole of length L' (which does not move in S') is seen in frame S... my question is: Why do we transformed $x'(b)$ and $x'(a)$ with lorentz transforms instead of $x(b)$ and $x(a)$ with inverse lorentz transforms (which gives apparently opposite result)? ... Similar problem is with derivation of time dilation. Sorry for my English.

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The subtlety here is that the lengths are defined at a given time. And in special relativity, "simultaneity is relative". We'll see what this means by working it out.

We assume that the object is at rest in the frame $S'$, and has length $L' = x'(b)-x'(a)$ in this frame. We take $L = x(b) - x(a)$, with say $t(b) = t(a) = 0$. We then use the Lorentz transformations, assuming the speed of $S$ with respect to $S'$ is $v$, to get: $$L = x(b) - x(a) = \gamma\left(\ x'(b) - vt'(b)\ \right) - \gamma\left(\ x'(a) - vt'(a)\ \right)$$ where $\gamma^{-1} = \sqrt{1-\frac{v^2}{c^2}}$. But we also have: $$t(b) = \gamma t'(b)-\gamma\frac{v}{c^2}x'(b) = 0$$ $$\implies t'(b) = \frac{v}{c^2}x'(b)$$ and similarly, $$t'(a) = \frac{v}{c^2}x'(a)$$

This means that we get $$L = \gamma \left(1-\frac{v^2}{c^2}\right) (x'(b)-x'(a))$$ or $$L = \gamma^{-1} L'$$

Now let's try to do it the other way. We get: $$L' = \gamma\left(\ x(b) + vt(b)\ \right) - \gamma\left(\ x(a) + vt(a)\ \right)$$ But this time, we are once again concerned about the length measured in $S$, which is done when $t(b) = t(a) = 0$. The simple result is: $$L' = \gamma\left(x(b) - x(a)\right)$$ $$\implies L' = \gamma L$$ which is no different from what we got earlier. The length measured in the moving frame $S$ is always shorter than the 'rest length' of the object. And it turns out to be the same, because when you are measuring the position difference at a given time in one frame, that actually corresponds to two different times at two different positions in a frame moving relative to it.

Something of length $L'$ in $S'$ has length $\gamma^{-1}L'$ in $S$. But something of length $L_2$ at rest in $S$ has a length of $\gamma^{-1}L_2$ in $S'$. The only thing determining length contraction is the motion of your frame relative to the rest frame of the object. That's how relativity works.

The situation is analogous for time dilation. The same position at different times in one frame is at different positions at the corresponding times in a moving frame.

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  • $\begingroup$ Great explanation! I was wondering why the gamma factor appears as a length dilation in the Lorentz transformation for the x, I had a hunch that it was connected to the different measuring times, and you made it very clear. $\endgroup$ May 25, 2020 at 8:55

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