Consider the following experimental setup:

  1. 2 parallel lines close to each other. The direction of +ve x-axis is same for both the lines.
  2. There is a standard 1-meter rod on each of these lines. Let they be called S and S'. The end-point towards the +ve x-axis for S is P and S' is P'. The other end-point is O and O' respectively.
  3. O and O' are origins of reference frame S and S'. And O' moves +V with respect to O
  4. Event E1 is when O and O' coincide (i.e. are right next to each other. Assume please: the lines are very close to each other) - By convention, this is also considered as the origin for both the reference frame S and S'
  5. Event E2 is when P and P' coincide.

Now a few questions:

  1. What is the velocity of P' with respect to P?
  2. In each (or any) of the reference frames S and S' - Are events E1 and E2 simultaneous? If not which precedes the other?

This is not a homework question - I have been thinking about this to better get an intuitive grasp of special relativity. I think I know the answers (I have used Lorentz Transformation to get the answers above) - but I lack a good physical understanding, intuition or argument (I don't exactly know what I lack here :)) to arrive at those conclusions.

So for e.g. even an answer like E1 precedes E2 in the reference frame S because <some physical argument> will be very helpful.

(I don't really want to use math to answer the above)

  • $\begingroup$ How can those points coincide if the rods are moving on parallel tracks? $\endgroup$
    – PM 2Ring
    Commented Jul 16, 2020 at 4:56
  • $\begingroup$ Assume the tracks are right next to each other. Hence I can bring them arbitrarily closer. You can also assume there are "pointers" perpendicular to the direction on the motion. And I define that when the "pointers" collide the points coincide. I will edit to Q to make this more clear. Thanks $\endgroup$
    – aman_cc
    Commented Jul 16, 2020 at 5:02

2 Answers 2


The velocity of $P'$ w.r.t $P$ is same as the velocity of that frame, i.e. $+V$. The events $E1$ and $E2$ cannot be simultaneous in either frame due to length contraction. At the instant of event $E1$, looking from the unprimed frame, $P'$ is always shorter (lengthwise) than $P$.

  • $\begingroup$ Yes it is the Q2 which I need a better grip on. thnks $\endgroup$
    – aman_cc
    Commented Jul 16, 2020 at 4:57
  • $\begingroup$ Thanks True - but is there a more intuitive physical argument? That is what I'm after. $\endgroup$
    – aman_cc
    Commented Jul 16, 2020 at 5:06
  • $\begingroup$ @aman_cc If you're confident (or may I say, intuitive) about the length contraction, then I think it is obvious to see the events won't be simultaneous. $\endgroup$ Commented Jul 16, 2020 at 5:26
  • $\begingroup$ yes indeed - i guess i lack that intuition. and maybe need to think more clearly $\endgroup$
    – aman_cc
    Commented Jul 16, 2020 at 5:44

Answer to question 1: Since we are talking about inertial coordinates systems, If origin O' (an endpoint of the rod in S') moves with constant velocity +V relative to O , so the entire rod in S' will move with velocity +V with respect to S (in particular, in the origin O)

Answer to question 2: So, assuming that the hole play begins with the event E1 (say, the rods start to move relatively to each other) at the instant t= t' =0 , for t be measured in S and t' measures in S', and using your assumption that both rods have 1 meter in their own reference frames, the only possibility that E1 and E2 are simultaneous is when +V = 0, e.g. the reference frames are in rest relative to each other. This happens because in your example, in the S frame point of view, the S' rod will be moving with velocity +V, so by length contraction phenomenon, it's length seen by S frame will be $1* \sqrt{1- v²/c²} \le 1$ (for a arbitrary length L, it would be $ L * \sqrt{1- v²/c²}$). By this, there's no way that the rods start to move relatively at E1 and the their endpoints meet, E2, at the same time for any of the frames (since the rod always tend to be smaller and smaller as +V goes to C)

PS: sorry for any english mistakes, I trying the improve my Grammat


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