# Deriving Length contraction without time dilation? [duplicate]

I was wondering if there is a way to prove the length contraction without using the time dilation? because every time I see a derivation to length contraction it comes with the time dilation and start based on it.

It depends on your postulates: on what premisses are you going to build your theory. If you take the Lorentz transforms as your premisses then you don't really use time dilation to establish length contraction. But you still need to consider time in order to understand length contraction. In particular you need to realise that in a frame, S, in which a body is moving (in the +$$x$$ direction, you must make simultaneous measurements of the positions of $$x_A$$ and $$x_B$$ of A and B on the body in order to measure the distance ($$x_B-x_A$$) in your frame. In the S' frame, in which the body is stationary there is no need for simultaneous measurement of $$x'_A$$ and $$x'_B$$. Using the Lorentz transform for displacements parallel to the relative velocity between frames, and the simultaneity of measuring $$x_A$$ and $$x_B$$ we have: $$x'_A = \gamma(x_A-vt)\ \ \ \ \text{and}\ \ \ \ x'_B = \gamma(x_B-vt)\ \ \ \ \text{so}\ \ \ \ x'_B-x'_A=\gamma(x_A-x_B)$$ Since $$\gamma > 1$$ we have $$x_A-x_B
• What I want to do is prove the $\gamma$ in the length contraction without using the time dilation. As if I didn't know about Lorentz transformation Nov 26 '20 at 23:01
• The assumptions are: Let there be a photon emitting from A (which has a mirror) to B (Also has a mirror) and both of them are in a parallel line with the velocity $v$ (which is all the system velocity), for the stationary observer the distance between A and B is $X$. How can I prove that it will have a length contraction for the outside frame? Nov 26 '20 at 23:15