# Special relativity kinematics problem - measure of a rod and simultaneity

I´m trying to solve a special relativity problem, and I think I need some help. There are two inertial frames of reference, $$O$$ and $$O'$$, the last one moving with relative velocity $$v$$ in the $$x$$ direction. There's a rod with length $$L'$$ fixed to frame $$O'$$, such that extreme A is at $$x'=0$$, and extreme B is at $$x'=L'$$. Clocks are synchronized at time $$t=t'=0$$, when position is $$x=x'=0$$. An observer from $$O$$ measures the rod, and the result is $$L$$. Now, from Lorentz transformations, we know that

$$x'=\gamma(x-vt)=\gamma(L-v.0)=\gamma L$$

And, as $$x'=L'$$, we have $$L'=\gamma L$$, with

$$\gamma=\frac{1}{\sqrt(1-(v/c)^2)}$$

Now I need to find the result $$L'=\gamma L$$, but with another method. First, the problem requests to find the $$\Delta t'$$ (from the point of view of O') since extreme A of the rod is measured by O, until extreme B is measured by O. Of course, both events are simultaneous for O, then $$\Delta t=t_B-t_A=0$$, so $$t_B=t_A=t$$. Here, I did (I'm not sure if it's ok)

$$t'_A=\gamma(t_A-vx_A/c^2)=\gamma(t-v.0/c^2)=\gamma t$$ $$t'_B=\gamma(t_B-vx_B/c^2)=\gamma(t-vL/c^2)=\gamma t-\gamma vL/c^2$$

Therefore,

$$\Delta t'=t'_B-t'_A=-\gamma vL/c^2$$

Which means that extreme B is measured before extreme A.

After that, the problem requests to find the position of coordinate origin at O, at the moment when extreme B is measured by O, as seen by O'; and also the position of coordinate origin at O, at the moment when extreme A is measured by O, as seen by O'. Finally, with this and $$\Delta t'=-\gamma vL/c^2$$, I should find again the difference of length between $$L$$ and $$L'$$.

Please let me know if I wasn't clear. Thanks.

• Thanks for the answer. What comes to my mind, is to calculate the position of the origin of O, when O' is at time $t'_A$: $x=0=\gamma(0+v.0)$. And then, to calculate the origin of O, when O' is at time $t'_B$: $x=0=\gamma\left(L'+v\left(-\frac{v}{c^2}\right)\gamma L\right)$, but from this, I get $L'=-\frac{v^2}{c^2}\gamma L$, and I'm not sure of what I'm doing wrong. Jan 28, 2022 at 0:04