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

Question

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?

Answer 1

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 $$ L = \gamma L' \Rightarrow L'=\frac{L}{\gamma} $$

Answer 2

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.