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11. A Little More Relativity: Comoving Frames and Proper Time
First, we recall some terminology.
Definition 11.1 A Minkowskian 4manifold is a 4manifold in which the metric has signature (1, 1, 1, 1) (eg., the world according to Einstein). 
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 x^{4}coordinate to make the metric look like diag(1, 1, 1, c^{2}). 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 E_{4}, 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, c^{2}) at every point, so they are automatically inertial everywhere.)
Now let C be a timelike curve in the Minkowskian 4manifold M.
Definition 11.2 A momentary comoving reference frame for C (MCRF) associates to each point m C a locally inertial frame whose last basis vector is parallel to the curve and in the direction of increasing parameter s. Further, we require the frame coordinates to vary smoothly with the parameter of the curve. 
Proposition 11.3 (Existence of MCRF's)
If C is any timelike curve in the Minkowskian 4manifold M, then there exists an MCRF for C. 
Proof Fix p_{0} C and a Lorentz frame W(1), W(2), W(3), W(4) of M_{p0} (so that g_{**} = diag(1, 1, 1, c^{2}).) We want to change this set to a new Lorentz frame V(1), V(2), V(3), V(4) with
V(4) = dx^{i}
dRecall that = s/c
So let us take V(4) as above. Then it is tangent to C at p_{0}. Further,
using Proposition 6.5. Intuitively, V(4) is the time axis for the observer at p_{0}: it points in the direction of increasing proper time . We can now invoke Proposition 9.2 to flesh out this orthonormal set to obtain an inertial frame at p_{0}. For the other vectors, take
V(4)^{2} = dx^{i}
d^{2} = dx^{i}
ds^{2} ds
d^{2} = (1)c^{2} = c^{2}.
for i = 1, 2, 3. Then
V(i) = W(i) + 2
c^{2}W(i), V(4) V(4)
by orthogonality of the W's and the calculation of V(4)^{2} above. Also,
V(i),V(j) = W(i),W(j) + 4
c^{2}W(i),W(4) W(j),W(4) + 4
c^{2}W(i),V(4)W(j),V(4) V(4)^{2} = 0
so there is no need to adjust the lengths of the other axes. Call this adjustment a time shear. Since we now have our inertial frame at p_{0}, we can use 9.2 to flesh this out to an inertial frame there.
V(i),V(i) = W(i),W(i) + 4
c^{2}W(i),W(4)^{2} + 4
c^{2}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 dx^{i}/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 timecoordinates. Now again use 9.2 to flesh this out to an inertial frame.
Proposition 11.4 (Proper Time is Time in a MCRF)
In a MCRF , the x^{4}coordinate (time) is proper time . 
"Proof"
We are assuming starting with some coordinate system x, and then switching to the MCRF . Notice that, at the point m,
d 
= 


= 


= 

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.
Proposition 11.5 (In SR, Proper Time = Time in the Moving Frame)
In SR, the proper time of a particle moving with a constant velocity v is the tcoordinate of the Lorentz frame moving with the particle. 
Proof
=  c 
=  1 2 
g_{ij}  dt 
dt 
1/2 
dt. 
The curve C has parametrization (vt, 0, 0, t) (we are assuming here movement in the x^{1}direction), and g_{**} = diag (1, 1, 1, c^{2}). Therefore, the above integral boils down to
= 


= 


= 

But, by the (inverse)Lorentz transformations:
t  = 


= 

Thus,
t 
=  t(1v^{2}/c^{2})^{1/2}  =  , 
as required.
Definition 11.6 Let C be the world line of a particle in a Minkowskian manifold M. Its four velocity is defined by

Note By the proof of Proposition 11.3, we have
u, u  =  d 
2  =  c^{2}. 
In other words, four velocity is timelike and of constant magnitude.
Example 11.7 Four Velocity in SR
Let us calculate the fourvelocity of a particle moving with uniform velocity v with respect to some (Lorentz) coordinate system in Minkowski space M = E_{4}. Thus, x^{i} 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:
d 
= 


= 

since x^{4} is time in the unbarred system. Thus, we need to know dx^{4}/d. (In the barred system, this is just 1, but this is the unbarred system...) Since 4 = , we use the (inverse) Lorentz transformation:
x^{4}  =  (1  v^{2}/c^{2})^{1/2} 
, 
assuming for the moment that v = (v, 0, 0). However, in the frame of the particle, ^{1} = 0, and ^{4} = , giving
x^{4}  =  (1  v^{2}/c^{2})^{1/2} 
, 
and hence
d 
=  (1  v^{2}/c^{2})^{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 v^{2} in the formula by (v^{1})^{2} + (v^{2})^{2} + (v^{3})^{2} (the square magnitude of v). Thus we find
u^{i}  =  d 
=  v^{i }  d 
=  (1  v^{2}/c^{2})^{1/2} 
(i = 1, 2, 3) 
and
u^{4}  =  d 
(1  v^{2}/c^{2})^{1/2} 
. 
Hence the coordinates of four velocity in the unbarred system are given as follows.
Four Velocity in SR

We can now calculate u, u directly as
u, u  = 


= 

If a contravariant "force" field F (such as an electromagnetic force) acts on a particle, then its motion behaves in accordance with
m_{0}  =  d 
=  F, 
where m_{0} 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
Its energy is given by the fourth coordinate, and is defined as
E  =  c^{2}p^{4}  =  (1v^{2}/c^{2})^{1/2} 
. 
Note that, for small v,
E  =  m_{0}(1v^{2}/c^{2})^{1/2}  m_{0}c^{2}  +  1 2 
m_{0}v^{2}. 
In the eyes of a the comoving frame, v = 0, so that
This is called the rest energy of the particle, since it is the energy in a comoving frame.
Definitions 11.7 If M is any locally Minkowskian 4manifold and C is a timelike path or spacelike (thought of as the world line of a particle), we can define its four momentum as its four velocity times its rest mass, where the rest mass is the mass as measured in any MCRF. 
1. What are the coordinates of four velocity in a comoving frame? Use the result to check that u, u = c^{2} directly in an MCRF.
2. What can you say about p, p, where p is the 4momentum?
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.
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