# Special Relativity-Lorentz Transformation

Can you guys please verify/help me with some questions.

Consider 2 events, A & B. In frame S they are seperated by Δx and Δt. It is reasonable to say that the events are causally connected if it is possible for a signal (such as a light pulse) to travel between them. We mean that one might have caused the other. Write down a relation (an inequality) between Δx and Δt that expresses this. Make sure your relation is still valid when Δx and/or Δt are negative (hint: use absolute values)

I was thinking, since I = -(cΔt)² + (Δx)² + (Δy)² + (Δz)², with Δy and Δz = 0. We have I = -(cΔt)² + (Δx)²

Since this is a time like interval or at most light like interval, I <= 0.
so -(cΔt)² + (Δx)² <= 0.

I belive this should be correct but i didn't use any absolute value signs... or should i express it in terms of Δt & Δx instead of its square?

Show that the concept of being causally connected is invarient, i.e, that all inertial observers will agree on whether A & B are causally connected.
We have -(cΔt)² + (Δx)²
I define a new frame S', with interval (cΔt', Δx', Δy', Δz').
Δy' & Δz' is 0.
so using lorentz transformations:
x = γ(x'+vt), t = γ(t'+vx'/c²)
-(cγt' + γvx'/c)² + (γx' + γvt')² <= 0
some expanding & simplifying gives
γ²t'²(v²-c²) + γ²x'²(1-v²/c²) <= 0
subbing in γ² for c²/(c²-v²) for the first product and 1/(1-v²/c²) for the second gives
-(ct')² + (x')² <= 0.
Therefore all inertial observers will agree on whether A & B are causally connected.

Does this make sense? Or am i doing the question wrong?

Suppose that A & B are causally connected, and that A occurs before B in frame S, Show that all observers will agree that A occurs before B.

Suppose there is a time t > 0 between events A and B (and a distance x),
i have to show that t' in any reference frame is > 0
that is to say, the time between the events is always positive, meaning A happened first.

Something along those lines, if someone could help me and check my answers, that would be great.
Thanks everyone for helping.