Page 160 - Intro to Tensor Calculus

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That is, the velocity of P with respect to R equals the velocity of P with respect to Q plus the velocity of Q
with respect to R. For example, a person (P) running north at 3 km/hr on a train (Q) moving north at 60
km/hr with respect to the ground (R) has a velocity of 63 km/hr with respect to the ground. What happens
when (P) is a light wave moving on a train (Q) which is moving with velocity V relative to the ground? Are
the velocities still additive? This type of question led to the famous Michelson-Morley experiment which
has been labeled as the starting point for relativity. Einstein’s answer to the above question was ”NO” and
required that V P/R = V P/Q = c =speed of light be a universal constant.
In contrast to the Newtonian equations, Einstein considered the motion of light from the origins 0 and
0 of the systems S and S.If the S system moves with velocity v relative to the S system and at time t =0
a light signal is sent from the S system to the S system, then this light signal will move out in a spherical
wave front and lie on the sphere
2
2 2
2
2
x + y + z = c t (1.5.146)
where c is the speed of light. Conversely, if a light signal is sent out from the S system at time t = 0, it will
lie on the spherical wave front
2 2
2
2
2
x + y + z = c t . (1.5.147)
Observe that the Newtonian equations (1.5.144) do not satisfy the equations (1.5.146) and (1.5.147) identi-
cally. If y = y and z = z then the space variables (x, x) and time variables (t, t) must somehow be related.
Einstein suggested the following transformation equations between these variables

x = γ(x − vt)and x = γ(x + vt) (1.5.148)

where γ is a constant to be determined. The differentials of equations (1.5.148) produce

dx = γ(dx − vdt) and dx = γ(dx + vdt) (1.5.149)

from which we obtain the ratios
dx γ(dx − vdt) 1 v
= or v = γ(1 − ). (1.5.150)
γ(dx + vdt) dx γ(1 + dx ) dx
dt
dt
dx dx
When = = c, the speed of light, the equation (1.5.150) requires that
dt dt
v 2 v 2
2 −1 −1/2
γ =(1 − 2 ) or γ =(1 − 2 ) . (1.5.151)
c c
From the equations (1.5.148) we eliminate x and find
v
t = γ(t − x). (1.5.152)
c 2
We can now replace the Newtonian equations (1.5.144) by the relativistic transformation equations

x =γ(x + vt) x =γ(x − vt)
y =y y =y
or (1.5.153)
z =z z =z
v v
t =γ(t + x) t =γ(t − x)
c 2 c 2