I am somewhat confused about how to interpret negative time in Lorentz transformation. In the usual case of two reference systems S and S' where the distance X (the one that measures S) to an event, is very large with respect to the distance between S and S' (also measured by S) the time t' of S' gives a negative result.
I don't understand why if at $t = t '= 0$, S and S' were together, the event has been observed by S' before this synchronization..., so it could not be in front of S, and also, in this synchronization, S' could warn S about the event that is future for S ... which seems to me paradoxical.
Just in case my general interpretation of the Lorentz transformation is incorrect, I clarify that I assume that S and S' are in positive X, they are together at $t = t' = 0$ and after a certain time t, S observes S' at a certain distance and an event X; with X and t I can calculate that for S' the event has occurred at X' and at t '.
Then, using natural units for simplicity, if $V = 0.8$, and at $t = 7.5$, S observes an event at $X = 11$, then S' which at that time is $V * t$ away from S, observes ,applying Lorentz, $X’= 8.3333$ and $T’ = - 2.16666$
How should we interpret this result? Did S' really saw the event before synchronization?