# 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 $$\Delta t=0$$.

You make a measurement by putting a ruler against the bar and reading the values at the two ends.

If the ruler and the bar are in relative motion and you take the readings at different times then this is clearly not a good measurement!

To be valid you have to have either the two values taken simultaneously (as you reckon) or have the ruler and the bar at rest relative to each other (in which case you can take your time).

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
• The nice thing about special relativity is that when you think there's a puzzle or paradox there's ALWAYS a solution. Jun 6, 2020 at 9:44