# Lorentz Transformation at t=0

Suppose I have two reference frames $S$ and $S'$, where $S'$ is moving with velocity $v$ with respect to $S$.

The Lorentz transformation equation for time in reference frame $S$ is given by: $$t'=\gamma\left(t-x\frac{v}{c^2}\right)$$ where $\gamma$ is the Lorentz factor.

Now for an event happening at $t=0$ and at some position $x$ in $S$ frame, then $t'<0$ for $S'$ frame. But if $x=0$ then $t'=0$.

How should I interpret this result?

The interpretation is that two events being simultaneous as measured in frame $S$ doesn't imply that the events are simultaneous in frame $S'$. Which events count as being "simultaneous" depends on the frame of reference. This is known as the relativity of simultaneity.

The coinciding of the origins is an event, call it event $A$. The coordinates of $A$ as measured in $S$ are $t_A=0$, $x_A=0$, and the coordinates of $A$ as measured in $S'$ are $t'_A=0$, $x'_A=0$.

You're also considering another event, call it $B$, whose coordinates as measured in $S$ are $t_B=0$, $x_B=x$. As measured in $S'$, the $t'$ coordinate of $B$ is $t'_{B}=-\gamma x v/c^2$.

$t'_B$ being negative (if $x>0$) means that according to $S'$, event $B$ occurs before event $A$. However, if $x=0$, then $B$ is the same event as $A$, so $t'_B=0$ and $x'_B=0$.

There is no physical discontinuity between the situations where $x=0$ and $x>0$, i.e., between the situations where $t'_B=0$ and $t'_B<0$. It's just that the further away event $B$ is from event $A$ spatially, the bigger the difference will be between $t'_B$ and $t_B$.