Two events in S are separated by a distance D = x₂ - x₁ and a time T = t₂ - t₁.

(a) Use the Lorentz transformation to show that in frame S’, which is moving with speed V relative to S, the time separation is t₂’ - t₁’ = γ(T -VD/c²).

(b) Show that the events can be simultaneous in frame S’ only if D is greater than cT.

(c) If one of the events is the cause of the other, the separation D must be less than cT, since D/c is the smallest time that a signal can take to travel from x1 to x2 in frame S. Show that if D is less than cT, t₂’ is greater than t₁’ in all reference frames. This shows that if the cause precedes the effect in one frame, it must precede it in all reference frames.

(d) Suppose that a signal could be sent with speed c’ > c so that in frame S the cause precedes the effect by the time T = D/c’. Show that there is then a reference frame moving with speed V less than c in which the effect precedes the cause.