# Length Contraction: is $t'$ or $t = 0$?

To demonstrate my confusion - let's say there is a rod traveling with velocity +v relative to S, and in S, the length of the rod is measured to be $$L$$.

If I want to go from S to S', the frame where a rod is motionless, and I apply the Lorentz transformation for x to find the proper length of the rod in S', do I put that $$\Delta t = 0$$ or that $$\Delta t' = 0$$? David Morin's textbook says to use the latter since we want to make the measurements of the end of the rod simultaneous. But intuition (and the answers to other problems) tell me that the first should be true, since the initial measurement of $$\Delta x = L$$ assumed that $$\Delta t = 0$$ (so that it would be simultaneous). For example, in I.E. Ordov's Problems in General Physics, problem 1.352 (I think) is following the latter.

I feel like I have some fundamental misunderstanding of what is going on, and would appreciate any help.

• But doesn't the event of the measurement of $\Delta x = L$ assume t = 0, so when we put it in the lorentz transformation, why wouldn't this be the case? Put in another way, aren't both vertices of the rod measured at time t in S-frame to produce $\Delta x = L$? Commented Jul 3 at 12:53
• I'm not sure I understand you completely, but there is a difference as to whether you are (1) measuring the length of the rod in frame S' or (2) describing a length measurement in S from the point of view of S'. For (1), you must ensure that $\Delta t' = 0$. For (2), the correct thing to do would be to set $\Delta t = 0$. However, note that the result of the measurement in case (2) will not be the length of the rod in S', but in S. Commented Jul 3 at 13:06
• Got it. As for the IE Ordov question, I'm confused why they multiplied by $\gamma$ instead of dividing by $\gamma$.... Since $\Delta t' = 0$, shouldn't you get $\gamma * l_0 = l$ so $l_0 = l / \gamma$ Commented Jul 3 at 13:13
• I don't know the problem you're discussing, or the meanings of $l_0$ or $l$, but the general rule of thumb is that the rod will always be longest in its rest frame. If it is at rest in $S'$ and its length in $S'$ is $L'$, then ordinary length contraction would have that the length in another frame $S$ would be $L = \frac{L'}{\gamma}$. Commented Jul 3 at 13:38