# what if the time is zero in one reference frame

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 $$\Delta t^\prime = \gamma \Delta t$$

• 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? – ACuriousMind Jul 3 '14 at 13:05

$$c\Delta t'=\gamma(c\Delta t-\beta\Delta x),$$
with $\gamma=1/\sqrt{1-\beta^2}$ and $\beta=v/c.$ This coincides with your second formula only if $\Delta x=0$.