Suppose we have two simultaneous events, $A$ and $B$, separated by a distance $L$ (the simultaneity is in frame of reference $S$). Now suppose we have a second frame of reference $S'$ moving with speed $V$ relative to $S$.

According to the invariance of intervals, the interval between events $A$ and $B$ in frame $S'$ is also $L$, but since there is also a temporal separation between events (simultaneity isn't conserved), one finds that the spatial distance between events in $S'$ is larger than $L$. In contrast, according to the length contraction prediction of special relativity, the maximum length is in the rest frame, $S$. Why the inconsistency?

  • $\begingroup$ please give a link for "invariance of intervals" in special relativity $\endgroup$ – anna v Mar 25 '16 at 8:24
  • 1
    $\begingroup$ I think he is referring to the following in special relativity $$ds^2=-(dt)^2+dx^2$$ ds^2 is independent of the frame of reference en.wikipedia.org/wiki/… $\endgroup$ – Secret Mar 25 '16 at 8:30

You need to review the derivation of Lorentz contraction. The result and your reasoning are correct, but wrongly interpreted.

The observer in the primed frame indeed measures $$ dx' > L $$ I've used a different notation for reasons that will become apparent. In fact, one can show the primed observer sees $$ dx' = \gamma(v) L $$ where $\gamma(v) \ge 1$ is a Lorentz factor.

However, to find the length of an object, he would need the simulatneous position of its endpoints (the events are not simultaneous in his frame). He needs to account for the fact that he knows the second endpoint at a later time, and during that time, it moved! He accounts for this by subtracting the amount it moved, $vdt'$, $$ L' = dx' - v dt' = \cdots= L / \gamma < L. $$ With that correction, he indeed sees a contracted length that is less than the proper length.

  • $\begingroup$ Thank you very much innisfree. I understood your excellent explanation. It seems i still need to work a lot in order to understand special relativity in a satisfactorily way. $\endgroup$ – user2554 Mar 25 '16 at 10:28
  • $\begingroup$ No problem! You won't really understand it unless you look at the maths. Fortunately the math isn't that bad and there are lots of good resources. Your question wasn't a silly one at all, I think with work you'll get the hang of it. $\endgroup$ – innisfree Mar 25 '16 at 10:33
  • $\begingroup$ @innisfree Isn't the extension in your answer, the portion after 'However,...', artificial? I mean would there be any physical meaning to the 'length' between two events that are not simultaneous? $\endgroup$ – Dvij Mankad Mar 27 '16 at 7:52
  • $\begingroup$ I thought a bit and I figured out that this can be percieved in the following way: The observer imagines an invisible moving stick that has bulbs at its ends and they blink simultaneously in that stick's frame - but only once in a lifetime. Assuming that the two events that he is observing are actually the two events of blinking - happening simultaneously in stick's frame, the observer finds the what would be the spatial spread of the stick in his frame at a given instant of time in his frame. Is my interpretation correct? $\endgroup$ – Dvij Mankad Mar 27 '16 at 7:54
  • $\begingroup$ Clearly, If we employ this procedure of interpretation then we can associate a 'length' with any two causally disconnected events - no matter how far separated in time. But this all seems far stretched to me. Is there any reference to such use of the word 'length' in some books or so? I like the concept per se and would love to read more about it. $\endgroup$ – Dvij Mankad Mar 27 '16 at 7:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.