Consider measuring the length of an object in another frame of reference.
Of course this should happen at the same instance in the frame of reference the measurement takes place in.
but using Lorentz transformation the time in the other frame of reference would be different and there would be a period
\begin{equation}\Delta t^\prime = \gamma v/c^2(x1 - x2)\end{equation} so we have $\Delta t = 0$ but $\Delta t^\prime$ has a value
doesn't this contradict with the time dilation basic law $$\Delta t^\prime = \gamma \Delta t$$

  • $\begingroup$ What do you mean by "measuring the length of an object in another frame of reference"? Measurements performed in one frame of reference can only ever yield the measurement results in their own frame of reference, how could they not? $\endgroup$
    – ACuriousMind
    Jul 3, 2014 at 13:05

1 Answer 1


The transformation law for time intervals is given by

$$c\Delta t'=\gamma(c\Delta t-\beta\Delta x),$$

with $\gamma=1/\sqrt{1-\beta^2}$ and $\beta=v/c.$ This coincides with your second formula only if $\Delta x=0$.

  • $\begingroup$ you mean if $\Delta t = 0& !! $\endgroup$ Jul 3, 2014 at 13:39
  • $\begingroup$ @AbdoSaeedAnwar: In this case, your first formula holds. $\endgroup$ Jul 3, 2014 at 13:40
  • 1
    $\begingroup$ @AbdoSaeedAnwar The point is that you forgot that 'measuring an object' is actually a set of two events, that are separated in space as well as (potentially) in time. $\endgroup$
    – Danu
    Jul 3, 2014 at 14:05
  • $\begingroup$ it seems that there is something wrong in my understanding, but what I realized now is that the second formula is just a special case and holds only when Delta x = 0 and that the general rule is from Lorentz $\endgroup$ Jul 3, 2014 at 14:09
  • $\begingroup$ @AbdoSaeedAnwar: That is exactly what I wanted to say in my answer. $\endgroup$ Jul 3, 2014 at 15:21

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