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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?

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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.

Added clarification due to comment:

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 $t_B$.

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