Two events in S are separated by a distance D=x2 - x1 and a time T = t2 - t1 . a) Use the Lorenrz transformation to show that in frame S' , which is moving with speed v relative to S, the time separation is t2 - t1 = ã(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 refence 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.
a) To show the Lorentz transformation for the time separation in frame S', let's start with the Lorentz transformation equations:
t' = γ(t - vx/c^2)
x' = γ(x - vt)
where γ = 1 / sqrt(1 - v^2/c^2) is the Lorentz factor.
In frame S, the time separation is given by T = t2 - t1 and the distance separation is given by D = x2 - x1.
In frame S', the time separation is t2' - t1'. Using the Lorentz transformation for time, we have:
t2' - t1' = γ(t2 - vx2/c^2) - γ(t1 - vx1/c^2)
Multiply and divide the second term of the equation by c2 to get a common denominator:
t2' - t1' = γ[(t2 - t1) - v(x2 - x1)/c^2]
Since D = x2 - x1, we can substitute it into the equation:
t2' - t1' = γ(T - vD/c^2)
b) In order for the events to be simultaneous in frame S', we need the time separation, t2' - t1', to be zero. Therefore, we have:
0 = γ(T - vD/c^2)
This can be rearranged to:
vD/c^2 = γT
But γT = ãT, where ã is the Lorentz factor for time separation. So we have:
vD/c^2 = ãT
If we rearrange this equation, we get:
D/cT > 1/v
Since v is the velocity of frame S' relative to frame S, it must be less than the speed of light, c. Therefore, 1/v is greater than c.
So, we conclude that D/cT > c, i.e., D > cT, for the events to be simultaneous in frame S'.
c) If one event is the cause of the other, the separation between them 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.
Let's assume that D < cT. Then, we have D/c < T, which means that the smallest time it takes for a signal to travel from x1 to x2, D/c, is less than the time separation T between the events.
Using the Lorentz transformation for time, we showed in part a) that t2' - t1' = ã(T - vD/c^2).
Since D/c < T, the term -vD/c^2 in the equation is negative. Therefore, T - vD/c^2 is positive.
Considering that ã > 1, we can say that t2' - t1' > 0 in all reference frames. Thus, t2' is greater than t1' in all reference frames.
d) Suppose we have a signal that can be sent with speed c' greater than the speed of light, c. In frame S, the cause precedes the effect by the time T = D/c'.
To find a reference frame where the effect precedes the cause, we need to find a velocity v of frame S' relative to frame S such that t1' - t2' < 0.
Using the Lorentz transformation equations, we have:
t1' - t2' = γ(t1 - vx1/c^2) - γ(t2 - vx2/c^2)
Substituting D = x2 - x1, we get:
t1' - t2' = γ(t1 - t2) + γ(vx1/c^2 - vx2/c^2)
Since t1 - t2 = -T (the cause precedes the effect), we have:
t1' - t2' = -γT + γ(vx1/c^2 - vx2/c^2)
Substitute T = D/c' into the equation:
t1' - t2' = -γ(D/c') + γ(vx1/c^2 - vx2/c^2)
If we choose v = D / (c'T), then vx1/c^2 - vx2/c^2 = x1 / T - x2 / T = 0, as D / cT = 1.
Therefore, we have:
t1' - t2' = -γ(D/c') + γ(D/c'T)
We can factor out a γ, and simplify the expression:
t1' - t2' = γ(D/c') * (1 - 1/T)
Since T > c' therefore 1/T < 1/c', we have:
t1' - t2' = γ(D/c') * (1 - 1/T) < 0
Thus, in the reference frame S' with velocity v = D / (c'T), the effect precedes the cause.