Imagine we have two reference frames, $S$ and $S'$. Imagine a point $x_0'$, moving with velocity $v$ in the $S$, and with $v' = 0$, in the $S'$ frame. From the $S$ frame, the time dilatation experienced by the moving point is: $$ t = \frac{t'}{\sqrt{1 - \frac{v^2}{c^2}}} $$ I read this as:

  • $t$, time taken by a process in the $S$ frame.
  • $t'$, time taken by the same process in the $S'$ frame.

Then, the length contraction as seen from the $S$ frame is: $$ \ell = \ell' \sqrt{1 - \frac{v^2}{c^2}} $$ Maybe this is a silly thing, but, the velocity is: $$ v = \frac{\ell}{t} $$ But because, no reference frame is preferred, then also: $$ v= \frac{\ell'}{t'} $$ applies, no? But substituting $t'$ and $\ell'$ in the second equation, we found: $$ v = \gamma^2 \frac{\ell}{t} $$ Is this correct or you should always have $v = \ell/t$?

  • $\begingroup$ You have to use general Lorentz transformations to derive velocity using $v=\frac {l'}{t'}$. For why the equations you wrote above cannot be used, refer to Kleppner and Kolenkow, Irodov, or Morin if you are relatively new to the subject, if not then refer to Tai Chow. $\endgroup$ Aug 23, 2022 at 21:04
  • $\begingroup$ What is $l$ the length of? $\endgroup$
    – WillO
    Aug 23, 2022 at 21:30

1 Answer 1


You are confusing length with distance traveled.

Velocity is distance traveled divided by time, not the length of anything divided by time.

Suppose you start at the origin, and finish your journey (in my frame) at time $t$ and location $x$. Then I will compute your velocity as $x/t$. You will say you finished at time $t'$ and location $x'=0$. You clearly know the formulas for $t'$ and $x'$ in terms of $t$ and $x$ and vice versa.

Now here is a completely different problem: Suppose that you are traveling at speed $v$, holding a rod out in front of you. I say that at time $0$, one end of the rod is at location $0$ and the other end is at location $L$. (Therefore the length of the rod is $L$). You say that one end is at location $0$ and the other end is at location $L'$. You can work out a formula for converting back and forth between $L$ and $L'$. (Working this formula out for yourself is a really really good exercise if you're just getting your feet wet with relativity.) It is not the same formula that gets you back and forth between $x$ and $x'$. Your argument incorrectly assumes otherwise.

Here is how to see that the formula for transforming distance cannot possibly be the same as the formula for transforming length: Suppose that in my frame your rod has length $L$ and you travel a distance $L$ (ending up right where the far end of your rod began). Coincindentally, there's a green cone at the start of your journey and a red cone at the end.

Then I define the length of your rod to be the distance from the green cone at time $0$ to the red cone at time $0$. I define the distance you've traveled to be the distance from the green cone at time $0$ to the red cone at some later time. When we transfer all that to your coordinates, we have to correct for the fact that you and I measure distances differently AND for the fact that you and I measure times differently. So the formula for length contraction cannot involve only distances.


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