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I've been working through the Feynman Lectures on Physics. I'm currently on lecture 15: The Special Theory of Relativity, specifically 15-5, the section on the deriving the Lorentz Transformation from the idea of length contraction.

I'm confident this question has been asked/answered before, or that my understanding is flawed, but I've been up to this point unable to find an answer to this specific question, so I'd rather just ask.

Feynman considers Joe, with coordinate system $(x,y,z,t)$, and Moe, with coordinate system $(x',y',z',t')$. Moe and Joe are said to be in relative motion, with Moe having some velocity u relative to Joe in the x direction. Both are trying to measure the x coordinate of a point P.

Moe measures the distance to be x'. However, from Joe's perspective, Moe's ruler has been shortened due to length contraction; thus Moe's measurement will be greater than Joe's, by a factor of the Lorentz factor (not yet accounting for the fact that the origin of Moe's coordinate system will be constantly moving).

This assertion makes sense, given the explanation, but I would have tackled it in a different manner, apparently yielding a different result: Point P is in relative motion to Moe, so, from Moe's perspective, the distance between him and P contracts. However, P is at rest from Joe's frame of reference, so Joe will be measuring the proper length. Given this explanation, it would appear that Moe's measurement would be less than Joe's.

Is it that it matters what frame of reference we take to be rest? I'm not naive enough to believe that I've found some "flaw" or paradox in special relativity, but I am ignorant enough to be unable to resolve this issue on my own. Any help will be appreciated, Thanks!

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  • $\begingroup$ This sounds like an awful, awful way of teaching the Lorenz Transform. You just can't measure length contraction with a ruler! I don't think Feynman really got relativity . . . If you have the option, try Bondi, he seems to me to be a better teacher - archive.org/details/RelativityCommonSense/page/n0 $\endgroup$ – m4r35n357 Oct 2 '18 at 15:47
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No, it doesn't matter in special relativity which frame of reference is taken to be at rest. In this case, Feynman never explicitly says that P is at rest in Joe's reference frame, but it's implied from the reasoning. He correctly infers that $x' = \dfrac{x-ut}{\sqrt{1-u^2/c^2}}$ is the distance Moe measures to P at time $t$, using a ruler that is length contracted as seen from Joe's reference frame.

If we change the story and say that P is at rest in Moe's reference frame instead, then obviously $x = x_0 + ut$, where $x_0$ is Joe's distance to P at time $t=0$, and $x'$ is now no longer time-dependent: $x' = \dfrac{x_0}{\sqrt{1-u^2/c^2}}$ or alternatively $x = x'\sqrt{1-u^2/c^2}+ut$, which becomes the inverse Lorenz transformation once you take relativity of simultaneity into account and express $t$ in terms of $t'$ and $x'$. The result is: $x = \dfrac{x'+ut'}{\sqrt{1-u^2/c^2}}$.

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It does not strictly matter which frame is considered the rest frame. But for length contraction, you want to identify which frame is at rest relative to the thing whose length you are measuring. Here is a simple example that reaches the same result (of course) as Feynmann and Cuspy Code, and illustrates a possible mistake that you might be making in your attempted rework of Feynmann's example.

Let's assume two reference frames in the standard orientation, and let one frame be attached to a moving train car ($S^{'}$) and the other to the ground beside the railroad tracks ($S$). An observer in $S^{'}$ defines a duration $\Delta t^{'}$ as the time needed for a lamppost attached to the ground to move from the front of the car to the rear of it. An observer in $K$ defines a duration $\Delta t$ as the time needed for the car to move past the lamppost from front to back.

Each observer is calculating a speed: the observer in $S^{'}$ the speed $\Delta x^{'}/\Delta t{'}$ of the lamppost relative to his frame, and the observer in $S$ the speed $\Delta x/\Delta t$ of the car relative to his frame. These speeds will necessarily be equal:

$$\frac{\Delta x^{'}}{\Delta t^{'}} = \frac{\Delta x}{\Delta t} \implies \Delta x^{'} = \Delta x \frac{\Delta t^{'}}{\Delta t}.$$

$\Delta t^{'}$ and $\Delta t$ are related by the time dilation formula, but we need to identify which frame is moving relative to the train car since that is the object whose length is contracted. A mistake here will put $\gamma$ in the wrong spot.

Because $S$ is the 'moving frame' (the train car is at rest in $S^{'}$), the time dilation formula needs to be written as

$$\Delta t^{'} = \gamma \Delta t,$$

which says the moving clocks in the $S$ frame are running slow by a factor of $\gamma$ as seen from the $S^{'}$ frame. Substitute this into the expression for $\Delta x^{'}$ to get

$$\Delta x^{'} = \gamma \Delta x \implies \Delta x = \frac{\Delta x^{'}}{\gamma}.$$

The length of the train car is contracted by a factor of $\gamma$ in the $S$ frame. An observer in $S^{'}$ also claims (correctly) to observe a length contraction. She claims that meter sticks in the $S$ frame are shortened so that more of those meter sticks fit between the rear of the car and the front of it than do meter sticks in her frame. But of course, if she reasons from this that an observer in $S$ claims a length for the car that is too long, then she is making a mistake. I suspect you might be making essentially the same mistake when you attempt to rework Feynman's example, but I can't be sure. Perhaps you can elaborate?

From here, note that the front of the train continues to move down the track at speed $v$ in the $S$ frame, so its $x-$coordinate in that frame is given by

$$x = vt + \frac{x^{'}}{\gamma}.$$

Rearrangement gives the Lorentz transformation for $x$ to $x^{'}$:

$$x^{'} = \gamma(x - vt).$$

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