11. A Little More Relativity: Comoving Frames and Proper Time

First, we recall some terminology.

By Proposition 9.2, if M is Minkowskian and m M, then one can find a locally inertial frame at m such that the metric at m has the form diag(1, 1, 1, -1). We actually have some flexibility: we can, if we like, adjust the scaling of the x⁴-coordinate to make the metric look like diag(1, 1, 1, -c²). In that case, the last coordinate is the local time coordinate. Later, we shall convert to units of time to make c = 1, but for now, let us use this latter kind of inertial frame.

Note If M is Minkowski space E₄, then inertial frames are nothing more than Lorentz frames. (We saw in Theorem 6.3 that Lorentz frames were characterized by the fact that the metric had the form diag(1, 1, 1, -c²) at every point, so they are automatically inertial everywhere.)

Now let C be a timelike curve in the Minkowskian 4-manifold M.

Proof Fix p₀ C and a Lorentz frame W(1), W(2), W(3), W(4) of M_p0 (so that g_** = diag(1, 1, 1, -c²).) We want to change this set to a new Lorentz frame V(1), V(2), V(3), V(4) with

V(4) =

dxi d

Recall that = s/c

So let us take V(4) as above. Then it is tangent to C at p₀. Further,

||V(4)||2 =

dxi d

2

=

dxi ds

2

ds d

2

= (-1)c2 = -c2.

V(i) = W(i) +

2 c2

W(i), V(4) V(4)

V(i),V(j)	=	W(i),W(j) +	4 c2	W(i),W(4) W(j),W(4) +	4 c2	W(i),V(4)W(j),V(4) ||V(4)||2
	=	0

V(i),V(i)	=	W(i),W(i) +	4 c2	W(i),W(4)2 +	4 c2	W(i),V(4)2 ||V(4)||2
	=	||W(i)||2 = 1

At another point p along the curve, proceed as follows. For V(4), again use dxⁱ/d (evaluated at p). For the other axes, start by talking W(1), W(2), and W(3) to be the parallel translates of the V(i) along C. These may not be orthogonal to V(4), although they are orthogonal to each other (since parallel translation preserves orthogonality). To fix this, use the same time shearing trick as above to obtain the V(i) at p. Note that the spatial coordinates have not changed in passing from W(i) to V(i)-all that is changed are the time-coordinates. Now again use 9.2 to flesh this out to an inertial frame.

"Proof"
We are assuming starting with some coordinate system x, and then switching to the MCRF . Notice that, at the point m,

d4 d	=	4 xi dxi d
	=	4 xi V(4)i   (by definition of V(4))
	=	4 V(4) = 1.     (since V(4) has coordinates (0,0,0,1) in the barred system)

In other words, the time coordinate ⁴ is moving at a rate of one unit per unit of proper time . Therefore, they must agree.

A particular (and interesting) case of this is the following, for special relativity.

Proof

=

s c

=

1 2

-gij

dxi dt

dxj dt

1/2

dt.

The curve C has parametrization (vt, 0, 0, t) (we are assuming here movement in the x¹-direction), and g_** = diag (1, 1, 1, -c²). Therefore, the above integral boils down to

	=	1 2 -(v2 - c2) 1/2     dt
	=	1 2 c 1 - v2/c2 1/2     dt
	=	t (1 - v2/c2)1/2.

But, by the (inverse)Lorentz transformations:

t	=	t + v/c2 (1-v2/c2)1/2
	=	t (1-v2/c2)1/2   since = 0 for the particle.

Thus,

t

=

t(1-v2/c2)1/2

=

,

as required.

Note By the proof of Proposition 11.3, we have

u, u

=

dxi d

2

=

-c2.

In other words, four velocity is timelike and of constant magnitude.

Example 11.7 Four Velocity in SR
Let us calculate the four-velocity of a particle moving with uniform velocity v with respect to some (Lorentz) coordinate system in Minkowski space M = E₄. Thus, xⁱ are the coordinates of the particle at proper time . We need to calculate the partial derivatives dx^I/d, and we use the chain rule:

dxi d	=	dxi dx4 dx4 d
	=	vi dx4 d     for i = 1, 2, 3

since x⁴ is time in the unbarred system. Thus, we need to know dx⁴/d. (In the barred system, this is just 1, but this is the unbarred system...) Since 4 = , we use the (inverse) Lorentz transformation:

x4

=

4 + v1/c2 (1 - v2/c2)1/2

,

assuming for the moment that v = (v, 0, 0). However, in the frame of the particle, ¹ = 0, and ⁴ = , giving

x4

=

(1 - v2/c2)1/2

,

and hence

dx4 d

=

1 (1 - v2/c2)1/2

Now, using the more general boost transformations, we can show that this is true regardless of the direction of v if we replace v² in the formula by (v¹)² + (v²)² + (v³)² (the square magnitude of v). Thus we find

ui

=

dxi d

=

vi

dx4 d

=

vi (1 - v2/c2)1/2

(i = 1, 2, 3)

and

u4

=

dx4 d

1 (1 - v2/c2)1/2

.

Hence the coordinates of four velocity in the unbarred system are given as follows.

We can now calculate u, u directly as

u, u	=	u* 1 0 0 1 0 1 0 0 0 0 1 0 0 0 0 -c2 uT
	=	v2-c2 (1-v2/c2)1/2 = -c2.

If a contravariant "force" field F (such as an electromagnetic force) acts on a particle, then its motion behaves in accordance with

m0

=

du d

=

F,

where m₀ is a scalar, the rest mass, corresponding to the mass of the particle as measured in its own frame.

We use the four velocity to get four momentum, defined by

pⁱ = m₀uⁱ,

Its energy is given by the fourth coordinate, and is defined as

E

=

c2p4

=

m0c2 (1-v2/c2)1/2

.

Note that, for small v,

E

=

m0(1-v2/c2)-1/2

m0c2

+

1 2

m0v2.

In the eyes of a the comoving frame, v = 0, so that

E = m₀c².

This is called the rest energy of the particle, since it is the energy in a comoving frame.

1. What are the coordinates of four velocity in a comoving frame? Use the result to check that u, u = -c² directly in an MCRF.

2. What can you say about p, p, where p is the 4-momentum?

3. Is energy a scalar? Explain

4. Look up and obtain the classical Lorentz transformations for velocity. (We have kind of done it already.)

5. Look up and obtain the classical Lorentz transformations for mass.