Space-time points

An event can be fully localized by providing a space-time point that consists of the event's spatial and temporal coordinates, relative to any one inertial frame. Let A = (xA, yA, zA, tA) and B = (xB, yB, zB, tB) be two such points relative to K, and let A' = (x'A', y'A', z'A', t'A') and B' = (x'B', y'B', z'B', t'B') denote the corresponding points relative to K', respectively. If Δx ≔ xB – xA, and similar notation is used for the other deltas too, we can write:

Δx' = (Δx – v · Δt) / √1 – v2   / c2   

Δt' = (Δt – Δx · v / c2) / √1 – v2   / c2   

Δy' = Δy

Δz' = Δz

Now, multiply both sides of the second equation by c, then square the first two equations, and after that subtract the second from the first to get:

Δx'2 – (c · Δt')2 = Δx2 – (c · Δt)2

Finally, adding the squared third and fourth equations to both sides leads to an invariant measure between space-time points:

d2(A, B) ≔ Δx2 + Δy2 + Δz2 – (c · Δt)2

Apart from the fact that it can be negative, it resembles a (squared) distance measure. It is absolute like classical distance in that its value does not change when switching from one inertial frame to the other. That is, d2(A, B) = d2(A', B') always holds, for arbitrary K'. This suggests that space-time points constitute, at least in a mathematical sense, a four-dimensional absolute space, and that very same space is observed from all inertial frames. (In classical mechanics, the spatial and the temporal points form two absolute spaces, a three- and a one-dimensional one, respectively, which are independent of each other.)

As for the interpretation of d2(A, B): if it's positive, it means that two events at A and B, respectively, cannot have any causal relationship. If e.g. tB > tA, even a light signal that is sent from A would be too slow to influence the event at B.

« Previous | View full article | Next »

 

Leave a Reply

Your email address will not be published.