Understand result of Lorentz transform

Question

Assume an event that happens at P=(ct:2,x:4) in some inertial frame of reference S. Assume a second inertial frame S' in standard configuration and $\beta=4/5$ ($\gamma=5/3$). The Lorentz transform is:

$$ P'=\frac{5}{3}\begin{pmatrix} 1 & -4/5\\ -4/5 & 1\end{pmatrix}\begin{pmatrix} 2\\ 4\end{pmatrix}=\begin{pmatrix} -2\\ 4\end{pmatrix}$$.

That means that the event is perceived in S' before t=0. When S' pass by (ct:0,x:0) it could "say" to S the information about the event. In this way, S knows its future.

An error in my calculations or an apparent paradox ?

Answer 1

The mistake in your reasoning is to assume that because an event happens somewhere in S' before t'=0, it means that someone at the origin of S' would know about it at t'=0, and could pass that information to someone in S when the origins of the two frames pass. In both S and S', the event is so far from the origin that no information about the event could reach the origin in 2 seconds.

Answer 2

Information can never travel faster than light. You can draw a light cone eminating from the origin and the point $P$, both before and after the transform. Since $P$ is outside the lightcone of the origin it can't communicate anything with the origin.

In relativity causality means that if a point $P$ is outside a light cone it is impossible to get it inside the light cone with a Lorentz transformation.

Answer 3

Note that OP is a spacelike vector, where O=(0,0).

The temporal ordering of events O and P are not preserved by the boost.

Answer 4

This is how causality gets preserved in special relativity, even when you mess with space and time. Timelike seperated events have a well-preserved time ordering. Spacelike-seperated events do not. Since your event is spacelike seperated from the origin, whether it happens before or after $t=0$ is not preserved by a boost.

This is also the heart of the "you can't travel faster than the speed of light" thing -- faster than light travel is moving along a spacelike line, and there are observers that would parse this travel as moving into the past. The line is not time-orderable. Travelling faster than light isn't just "hard" or "forbidden", it is meaningless in special relativity -- "superluminal motion" is the same thing in the theory as "a spatial seperation in some reference frame".

Answer 5

An event being simultaneous to observer S' doesn't mean that the observer perceives it at that instant in time; for S' to become aware of it, a signal of some sort has to travel from the event to S', and this can only happen at the speed of light, or slower.