The energy-momentum four-vector
REFERENCE: Hartle
Recall the definition of the four-momentum
given by
p = m0u.
of a particle of mass m0
[Eq. 1]
We will call m0 the rest mass for reasons that will follow.
definition:
Here is a
"The rest mass m0 of a particle is that mass which would be
measured in the rest frame of the particle."

Recalling that the four-velocity u was related to the three velocity V
by the equation

[Eq. 2]
u = ( c, V ),
we can write the four-momentum as

p = (cm0, m0V ).
[Eq. 3]
pt = cm0,
[Eq. 4]
From Eq. 3 we see that
and

[Eq. 5]
pi = m0V.
We now wish to analyze the components of our four-momentum vector.
Recall the binomial expansion of (1  x)n:
(1  x)n = 1  nx
1!
+
n(n-1)x2
+ ... (x2 < 1).
2!
[Eq. 6]

We may thus apply Eq. 6 to  = [1 - V 2/c2]-1/2 to obtain

V2
 = 1 + 2c2 + ... =

1 2
V2
c +
+ ...
c2
2
.
[Eq. 7]
Using this result in Eq. 4 we have
pt = m0c =
1
m c2 +
c 0

1m V
2 0
2
+ ...
.
[Eq. 8]
Note the leading two terms in the parentheses of Eq. 8.
The second
term is the familiar three-space kinetic energy of the particle in its
rest frame:
1 2
[Eq. 9]
K =
mV .
2 0
The first term is an unfamiliar term (in the classical sense) and
somehow represents an energy component of the particle that has nothing
to do with the particle's motion:
E0 = m0c2.
[Eq. 10]
The two terms together constitute (to first order) the energy component
of p. Thus
pt 
1E =
c
1 [ K + E + (higher order terms) ]
0
c
[Eq. 11]
Expanding the spatial component of the four-momentum using Eq. 7 and
Eq. 5, we obtain
pi

V
= m0V = 2 m0c2 +
c

so that

1m V
2 0

pi = m0V + ...
2
+ ...
,
.
[Eq. 12]
The first term is recognizable as the three-momentum of the particle in
its rest frame.
We thus rewrite Eq. 3 in a more suggestive form as

p = ( m0c, m0V )  (
where

1
E , p ),
c
1
t
c E = p = m0c,
and


p = pi = m0V.
[Eq. 13]
[Eq. 14]
[Eq. 15]
In light of Eq. 13, we call p the energy-momentum four vector.
Recall that
p  p = -m02c2.
[Eq. 16]
Substituting Eq. 13 into Eq. 16 we obtain

E = (m02c4 + c2p 2)1/2.
[Eq. 17]
Eq. 17 gives the relationship between the energy and the momentum.

Note that if the particle is at rest, p = 0 and Eq. 17 reduces to
E = m0c2,
[Eq. 18]
as previously seen in Eq. 10. Eq. 18 is no doubt recognizable as one
of the most famous equations in physics. It represents the rest energy
of the particle. This is an energy that the particle contains that is
not kinetic.
It is a new form of energy predicted famously by
Einstein's special theory of relativity, and proposes that mass and
energy are, in some sense, equivalent.
We will look at one more surprising feature of relativity.
we have


p = m0V.
From Eq. 15
[Eq. 19]

The quantity p is the spatial portion of the four-momentum.
We can
rewrite Eq. 19 as


[Eq. 20]
p = mV,
where
m0
.
[Eq. 21]
m = m0 =
.
[1 - V 2/c2]1/2
We call m the particle's relativistic mass.
The relativistic mass is
the mass we would
measure
in
the
coordinate
frame
in which the particle

is moving at V. Eq. 21 predicts that the relativistic mass is always
greater than the rest mass.
Thus we see that not only are time and
length relative, but so is mass.