Confusion about Lorentz Coordinate Transformation A normal Lorentz coordinate problem might say:
At $t=t'=0$, two coordinate systems $S$ and $S'$ have their origins coincide with the $S'$ system moving with speed $v$ in the $+x$ direction relative to $S$. If event 1 happens at $x=a$, $t=0$ in the $S$ system then when/where does this event happen in the $S'$ system. I know that $x'=\gamma(x-vt)$ and I have no confusion with that equation, but in problems like these, I am confused specifically about the time. Since $t=t'=0$ and $t=0$, I would imagine that the time in the prime system would also be 0. But when one uses the Lorentz transformation $t'=\gamma(t-vx/c^2)$, $t'$ comes out as a non-zero negative number. How do we reconcile this?
 A: You have discovered the relativity of simultaneity. In the basic Minkowski diagram:

the $x$ axis marks the all the coordinates for which $t=0$, so that for the $S$ frame's observer (living at $x=0$), at the time $t=0$, all events on the $x$-axis occur simultaneously. That is, it is "now" in his version of space + time.
Meanwhile, for the observe in $S'$, who sits at $x'=0$, the $x'$ axis marks his definition of "now" at the $t'=0$. His definition of space + time is different.
This is very non intuitive: both observers are in the same location, at the same time, but their definitions of "now" at distant points in space are different.
You will find that most (but not all) special relativity paradoxes are resolved by the relativity of simultaneity: If there is any spatial separation between events that are simultaneous in one frame, then there are other frames where their time-ordering can go either way.
A: Consider Einstein's train thought experiment that demonstrates non-simultaneity. Let's say the train moves left to right. The lightning strikes at the front and back of the train occur simultaneously for the platform observer (S frame) but for the train observer (S' frame) the strike at the front of the train occurred first. If S and S' are set up so that the two frames coincide with t = t' = 0 at the back of the train when the strike there occurs, we have t = 0 in the S frame for both strikes but in the S' frame the strike at the front would have occurred at t' < 0. In your example, the two two events in the S frame are (x,t) = (0,0) and (a,0). These are simultaneous in the S frame so we know they are not simultaneous in the S' frame and the Lorentz transformation give us the t' values in S'. The (0,0) event also occurs at (0,0) in S', but the (a,0) event does not occur at t' = 0.
