In special relativity, how do we know that distance doesn't change in the direction perpendicular to velocity? In the theory of special relativity it is said that the distance in the direction of the speed changes by a factor of
$$\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$
How do we know that the distance perpendicular to the velocity doesn't change?
 A: Consider two thin cylinders, each of radius $r$ in their own rest frame, moving towards each other with their axes aligned.
Suppose that distances in the perpendicular direction shrink. Then each cylinder observes the other as shrunken and has a radius less than $r.$ Each cylinder finds that the other cylinder will pass directly through it.
Two cylinders cannot both pass inside each other, but the situation is symmetric. There is no way to choose which cylinder should shrink, and so if the universe obeys the symmetries of special relativity, neither cylinder can shrink.
The same argument applies to the cylinders expanding, so they can neither shrink nor expand, showing that distances in the perpendicular direction are the same between frames.
This argument appears in Spacetime Physics by Taylor and Wheeler, but I don't know if it's original to that book.
A: The answer is straightforward. The formula shows that there is no dilation when your speed is zero, which is what you would expect, otherwise length dilation would be observable in connection with stationary objects. The component of your speed perpendicular to your direction of motion is always zero, hence dilation must be zero in that direction.
A: We can simply derive this from the Lorentz transform. With the $x$ direction parallel and the $y$, $z$ directions perpendicular the Lorentz transform is: $$t'=\gamma \left(t-\frac{vx}{c^2} \right)$$ $$x'=\gamma (x - vt)$$ $$ y' = y$$ $$z' = z$$ For an object with a length $L$ in the $y$ direction at rest in the unprimed frame we have the worldlines of the ends as $y_0 = y$ and $y_L= y+L$ so the distance between the ends is $y_L-y_0=L$.
To obtain the distance in the primed frame we simply transform $y'_0=y=y'$ and $y'_L=y+L=y'+L$ so the distance is $y'_L-y'_0=L$.
Therefore the distance in the perpendicular direction does not contract.
A: I think the best way to understand this is to figure out why we needed special relativity in the first place.
One of the fundamental principles of Physics is that no matter which frame of reference you go to, the laws of Physics should remain same (Also known as lorentz invariance). Relativity was introduced by Galileo and Newton creating what is traditionally known as classical mechanics. The problem that arose was that when classical electromagnetic theory was applied to Maxwell's equations (more specifically to moving charges) things went haywire. There were asymmetries that started to rise up. Classical relativity and electromagnetism were not working together and thus something was wrong.
So we needed some new formulation of relativity which can have a transformation (not galilean) that makes all physical laws looking the same + the speed of light is same in all frames. While the second condition may look arbitrary, it comes out from the electrodynamics issue.
Lets start the derivation. We have two inertial frames F and F' in which a light front is moving and they have a relative velocity of v (in the x direction). The equation for this would be
$ x^2+y^2+z^2=c^2t^2 $ and in the second frame it would be $x'^2+y'^2+z'^2=c^2t'^2$. Once you solve these equations and try to find relationships between x,x' y,y' z,z' t,t' you end up with the so called lorentz transformation. These transformation laws show us that length contraction happens in whichever direction the relative velocity is between the two frames. ( $x'=\gamma (x-vt) \mathrm{\: and\: }y'=y\: z'=z$)
