Parallel axis is special relativity?

Question

Landau vol.2 first chapter explain special relativity, and in particular it is considered the case of two frame of references $K$ and $K’$ such that the axis $x$ and $x’$ are coincident while $y$ and $z$ are parallel to $y’$ and $z’$.

I’m not sure about what Landau means with parallel because in the contest of special relativity it’s not so clear. A doubt relative to the definition of parallel axis is the following: can i say that the plane $y=0$ and $y’=0$ are coincident only as a consequence of the fact that the two frames are parallel or do i need further hypothesis to say it (for example to show that $y=y’$ i need to assume the invariance of physical law in every frame of reference)?

Answer 1

Landau's arguments are usually based on very general considerations, symmetries, and such. They are very briefly stated. Landau thought the arguments below were so obvious they were only worth a sentence or two. So it can take the reader some work to insert the details.

But it does make his arguments elegant and to the point.

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Special relativity considers inertial frames. Two inertial frames, $K$ and $K'$, may be moving with respect to each other at a constant velocity. If they are, each frame has a direction of motion of the other frame. You can set up the coordinate systems of each frame so that these directions are parallel to the $x$ and $x'$ axes. From each frame, you need to consider what the other frames' axes look like.

You can pick some event as the origin of spacetime for both frames. Then the $x$ and $x'$ axes both contain that event.

The $K$ frame sees the $x'$ axis moving in the $x$ direction. At $t = 0$, the point on the $x'$ axis labelled $x' = 0$ passes through the origin. Because it is moving in the $x$ direction, thereafter the point $x' = 0$ stays on the $x$ axis.

You can make the same argument from the $K'$ frame to show that the point $x = 0$ stays on the $x'$ axis. In the $K$ frame, that means the $x$ point $x = 0$ always contains some point of the $K'$ axis. If two points of the $x'$ axis always lie on the $x$ axis, the two axes coincide. That is, the two axes slide along each other.

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The other axes are perpendicular to $x$ and $x'$. Make some choice for the $y$ and $z$ axes. In $K'$ at $t' = 0$, they define a plane in spacetime. That plane also contains the $y'$ and $z'$ axes.

Choose $y'$ and $z'$ so they lie on top of $y$ and $z$. At that time, $y'$ and $y$ are parallel. So are $z'$ and $z$.

Since the $y$ and $z$ axes are moving in the $x'$ direction, they stay parallel to $y'$ and $z'$.

Answer 2

Just from a coordinate standpoint, without regard to application.. if the $x-$ and $x’-$axes are coincident and the other axes are parallel, then $y=y’$, $z=z’$. It’s true that the infinite $xy-$ and $x’y’-$planes, as well as $xz$ and $x’z’$, would cover the same points, but the points would have different values for their X coordinate. The set-up is: any point $(a, b, c)$ in $XYZ$ would equal $(a+k, b, c)$ in $X’Y’Z’$, where before movement $k$ is constant.

So yes it is generally true even before looking at the application that $y=y’$.

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If it is moving, we allow $k$ to change. Even before any physics, we know they are not moving relatively in the $Y$ or $Z$ directions. From a special relativity standpoint this means that the coordinate systems would be moving relatively $v=\frac{d k}{dt}$ according to both of them.

They will not measure objects as having the same length in the $X$ direction and will have time dilation if one is faster from a still frame.

But length contraction is directional, and parallel to the relative motion. The planes will remain parallel. We know this because to change that fact, two points in the (wlog) $XY$ plane would have to move different amounts in the $Z$ direction. This cannot happen as there is no length contraction in that direction.