# How to derive length contraction from Lorentz transformations? [closed]

I am trying to work out the length contraction using the Lorentz transformations. Here is how I stated the problem:

Suppose a bar (whose proper length is $$L$$) is moving at speed $$u$$ (to the right) with respect to a reference frame $$R$$. Let $$x_2$$ and $$x_1$$ be the two coordinates of the rightmost and the leftmost points of the bar respectively. Measured in the moving frame $$R'$$, their difference is: $$x_2'-x_1'= \gamma(x_2-ut_2)-\gamma(x_1-ut_1) = \gamma(x_2-x_1)-\gamma u(t_1-t_2)$$ But measuring a length means knowing both $$x_1$$ and $$x_2$$ at simultaneity in a frame. Therefore, $$t_2=t_1$$. It follows that: $$\Delta x'=\gamma \Delta x$$ or $$L'=\gamma L$$ Then $$\gamma = \frac{L'}{L}>1 \Rightarrow L'>L$$ Which implies that length have dilated not contracted, as I was trying to solve. What is the problem with this reasoning?

I think I have understood the problem. Since I want to compute the length in the moving reference frame $R'$, I must assume that $t_2' = t_1'$ but not $t_1=t_2$, because I want to measure the position of the two ends of the bar at the same time in the moving frame. The calculation is as follows: $$x_2-x_1 = \gamma(x_2'+ut_2')-\gamma(x_1+ut_1) = \gamma(x_2'-x_1')+\gamma u(t_2'-t_1')$$ Now, because $t_2' = t_1'$: $$\Delta x = \gamma \Delta x'$$ or $$L = \gamma L' \Rightarrow L'=\frac{L}{\gamma}$$
• This derivation and result is actually for the length of a rod stationary to frame $R$, as seen from frame $R'$, in contrast to what is stated in the question. The result is (of course) the same, but I hope that you are aware that. Jan 22, 2022 at 16:34
• How could an observer measures length of a moving object in direction of motion. If an object in motion, then it took time to pass from an observational point, so $t_1'\ne t_2'$. Nov 11, 2022 at 6:20
The relation $L'=\gamma L$ is correct. $L$ is the length of the moving object from $R$ frame, which is shorter than $L'$, which is the length of the object measured in a frame which is moving with the object, i.e. the object is stationary in that frame. So, the stationary length of the object is $L'$, which is greater than $L$, the length of the moving object.
Since $L' > L$, the length of a moving object contracts.
• $L$ is the proper length means, an object is placed in a frame at rest or measured by a frame moving with that object. Nov 11, 2022 at 6:17