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Suppose an inertial frame $S^\prime$ is moving with a relative velocity $\textbf{v}=v\hat{n}$ w.r.t another intertial frame S with their axes parallel and $\hat{n}$ is an arbitrary direction. In that case the coordinates of the two frames are related by boost only and no rotation.

Now consider three such frames $S_1$, $S_2$ and $S_3$. Let the relative velocities $\textbf{v}_{21}$ and $\textbf{v}_{23}$ are not parallel but the axis of all the three frames are. In that case, it is inferred that the coordinates of $S_1$ abd $S_3$, are related not only by a boost but a boost as well as rotation. Now, here is my problem.

There is a relative velocity between $S_1$ and $S_3$. Let us call it $\textbf{v}_{31}$. As the axes of all the frames are parallel, so for the frames $S_1$ and $S_3$, we are back to the situation mentioned in the first paragraph. So why shouldn't the coordinates of 1 and 3 are only related by boost and no rotation? Where am I making the mistake?

Qmechanic
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SRS
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  • "why shouldn't" is not the right question. If you claim that something is true (in this case the statement that $S_1$ and $S_3$ are related only by boost) you have to show it is the case, not claim it to be true by default until disproven. In this case, it's simply a fact that the composition of two boosts is not a boost, as you can directly check by multiplying the corresponding matrices of the boosts and observing the result is not of the form of a boost. – ACuriousMind Feb 18 '16 at 01:14
  • But why should I have to relate the frames $S_1$ and $S_3$ through $S_2$? Let us forget about $S_2$ for a minute. I do know that $S_3$ has a relative velocity $\textbf{v}_{31}$ w.r.t $S_1$, and I also know in such cases what is the relation between $S_1$ coordinates and $S_3$. It is just a simple boost. Where am I wrong here? Two arbitrary frames, whose axes are parallel is related by a Lorentz boost given by $ct^\prime=\gamma(ct-\boldsymbol{\beta}\cdot\textbf{r})$ and $\textbf{r}^\prime=\textbf{r}+\frac{\beta\cdot \textbf{r})\boldsymbol{\beta}(\gamma-1)}{\beta^2}-\boldsymbol{\beta}\gamma ct$. – SRS Feb 18 '16 at 07:40
  • You didn't get my point of confusion. I understand that when I try to relate $S_1$ and $S_3$ through $S_2$, the result turns out to be a rotation plus a boost. But where is the error if I directly use the fact there is a relative velocity between $S_1$ and $S_3$? – SRS Feb 18 '16 at 07:43

3 Answers3

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Your statement

the relative velocities $v_{21}$ and $v_{23}$ are not parallel but the axis of all the three frames are

is not precise enough. You are lacking the information from which point of view they are parallel, since "being parallel" is frame-dependent in Minkowski space (hyperbolic rotations do not preserve "being parallel").

What is true is that $S_1$ and $S_2$ are parallel as viewed from either of them, and that $S_2$ and $S_3$ are parallel as viewed from either of them. Only from the viewpoint of $S_2$ are all three frames parallel, but $S_1$ does not perceive $S_3$ as parallel, so your reasoning why they should be related by a simple boost fails.

In a way, this is the whole point of Wigner rotation/Thomas precession: Just boosting a frame twice in different directions produces a frame that is not parallel to the original frame as seen from it or the original frame. But the intermediate frame of course sees all three as parallel, since it is related through a simple boost to them. We just have to let go of the Euclidean intuition that changes of frame preserve the property of being parallel.

ACuriousMind
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Minkowski space is affine and therefore it preserves parallelism. The 3+1 dimensions case is special in that this is not apparent when dealing with boosts but a better choice of (complexified) basis makes it clear in the Lie álgebra level.

Rufus
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    Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. – Community May 22 '22 at 12:53
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    @Frobenius: general boosts indeed behave as you say. The generators in the complexified basis are not boosts. – Rufus May 23 '22 at 10:27
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In Special Relativity we couldn't say in general that the axes of two inertial frames S and S′ in relative translational motion (boost) are parallel, see Figure-02, except of special cases, see Figure-01.

In case of a general boost with velocity $\:\boldsymbol\upsilon$, see Figure-02, we could not talk about parallel axes. For example, the points of the $\:x'_2-$axis in frame $\:\rm S'\:$ at a given moment $\:t'\:$ are simultaneous events in Minkowski space, so the $\:x'_2-$axis is a well-defined straight line in $\:\rm S'$. But these events are not simultaneous in $\:\rm S\:$ so there doesn't exist such a thing or curve or whatever else in $\:\rm S\:$ to be parallel to the $\:x_2-$axis.

Reference : My answer here How to represent a pair of inertial frames in relativity?.

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Frobenius
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