# A question about length contraction

I'm new to special reltivity theory and length contraction. I can't figure out the logic or algorithm of calculating lengths in length contraction problems. Let me explain where I am stuck.

There is a simple and first given example. There are two inertial frames of reference, one is $$S$$ and another one is $$S^{'}$$. $$S^{'}$$ is moving along the $$\hat{x^{'}}$$ direction with speed $$V$$. $$S$$ is stationary frame of reference. There is a bar or stick layed down $$x$$-axis (stationary in $$S$$ frame of reference). The observer in $$S$$ frame of reference, calculates the ends of the bar at the same time and finds the length of the bar (proper length) as $$\Delta{x}=x_2-x_1=L_0$$. The question is, what is the length of the bar calculated by the observer in $$S^{'}$$ frame of reference.

In my opinion, we know the proper length of the bar $$L_0=\Delta{x}=x_2-x_1$$ that is calculated in frame $$S$$. This calculation process happened simultaneously in frame $$S$$, so $$\Delta{t}=0$$. With this information, we need to find out $$x_2^{'}$$ and $$x_1^{'}$$ to measure the length of the bar by the eyes of the observer in frame $$S^{'}$$.

$$x_2^{'}=\gamma(x_2-Vt_2), x_1^{'}=\gamma(x_1-Vt_1)$$

$$L=\Delta{x^{'}}=x_2^{'}-x_1^{'}=\gamma(x_2-x_1-V[t_2-t_1])$$

$$L=\gamma(\Delta{x}-V\Delta{t})=\gamma(L_0-V.0)=\gamma{L_0}$$

$$\gamma\ge1$$ so I find $$L\ge L_0$$, I should have found $$L\leq L_0$$.

Where do I made mistake in my logic? If you can explain, I would be happy. Thanks!

• Roger's answer is absolutely correct. If you'd like a little more explanation, you could take a look at my answer here to a very similar question. Jun 6, 2020 at 9:28
• Also see my answer to “Reality” of length contraction in SR Jun 6, 2020 at 10:25

You need to make $$\Delta t'=0$$, not $$\Delta t=0$$.
Your bar is at rest in $$S$$ and hence moving in $$S'$$, so if the two comparisons in $$S'$$ are to be reckoned as a valid measurement of the length they have to be done at the same time: $$\Delta t'=0$$
• Oh, thanks for explanation @RogerJBarlow. I understand where I made mistake. We already know the proper length which is measured in $S$. Why would I measure the bar in $S^{'}$ at the same time with $\Delta{t}$, it must be $\Delta{t^{'}$. Thank you so much! Jun 6, 2020 at 9:39