# Length contraction via Lorentz transformation matrix

1. Sep 23, 2015

### fluidistic

1,2,3. The problem statement, all variables and given/known data
I tried to derive the length contraction using the Lorentz transformation matrix and considering 2 events. I reached the correct result but there's a step that I had to assume that I don't understand.

Consider a ruler of length L along the x-axis for an observer at rest with respect to the ruler. The inertial reference frame of that observer is K.

Consider an observer moving alongside the x-axis with speed v. The inertial reference frame of that system is K'.

I considered the events (omitting y and z): $(t_1,x_1)$ and $(t_2,x_2)$. They are worth $(t_1,0)$ and $(t_2,L)$, respectively. (*)

Then $\begin{bmatrix} t_1'\\x_1' \end{bmatrix}=\begin{bmatrix} \gamma & - \beta \gamma \\ -\beta \gamma & \gamma \end{bmatrix}\begin{bmatrix} t_1\\x_1 \end{bmatrix}$. This yields the two equations $t_1'=\gamma t_1$ and $x_1'=-\beta\gamma t_1$.

While $\begin{bmatrix} t_2'\\x_2' \end{bmatrix}=\begin{bmatrix} \gamma & - \beta \gamma \\ -\beta \gamma & \gamma \end{bmatrix}\begin{bmatrix} t_2\\L \end{bmatrix}$ yields $t_2'=\gamma t_2 -\beta \gamma L$ and $x_2'=-\beta\gamma t_2 +\gamma L$.

Now since I want to obtain a distance measurement in K', I set $t_1'=t_2'$ and then I solve for $x_2'-x_1'$ and I indeed reach that it's worth $L/\gamma$.

What didn't like/understand about my own derivation is that for it to work, I had to assume that even though $t_1 \neq t_2$, $x_1=0$ and $x_2=L$ (see (*)). To me, this is equivalent to say that the length of the ruler is NOT worth L for an observer at rest with respect to it.

Oh wait, actually it could well be... since the observer is at rest with respect to the ruler, it doesn't matter when it measures the spatial distance between the 2 extrema of the ruler, it will always be L no matter when each measurement was performed. Is this reasoning correct?

2. Sep 23, 2015

### TSny

Yes, sounds good!

Note, you might try using the inverse transformation matrix for going from primed to unprimed coordinates and seeing if it's any easier. You can let the observation time in the primed frame be $t_1' = t_2' = 0$.

Last edited: Sep 23, 2015