Scott A.

Hughes Introduction to relativity and spacetime physics

Massachusetts Institute of Technology
Department of Physics
8.033 Fall 2021

Lecture 7
4-momentum and 4-velocity

7.1 Transforming energy and momentum between reference frames

The requirement that all observers measure the speed of light to be c has led us to rather
different formulations of energy and momentum: a body of rest mass m (i.e., the mass that
we measure it to have when it is at rest with respect to us) moving with velocity u has an
energy E and a momentum p given by
E = γ(u)mc2 , p = γ(u)mu . (7.1)
These quantities respect conservation laws: a system’s total E and p are conserved as its
constituents interact with one another. In the limit u/c  1, these formulas reduce to
1
E = mc2 + mu2 , p = mu + O(u3 ) . (7.2)
2
This makes it clear that Newtonian momentum agrees with relativistic momentum for speeds
much smaller than c. The energies likewise agree in this limit, provided we account for the
body’s rest energy mc2 . In the vast majority of circumstances a body’s rest energy is bound
up in the body, and cannot be “used” for anything in their interaction, so it can be ignored.
The relativistic quantities and the Newtonian quantities thus agree perfectly when u  c.
Suppose we measure a body to have energy EL and momentum pL in our laboratory.
What energy ET and momentum pT will an observer moving past our lab in a train with
velocity v = vex measure the body to have? To figure this out, follow this recipe:
1. Deduce the 3-velocity uL of the body in the lab from the values of EL and pL .
2. Use the velocity addition formulas to compute the 3-velocity of uT of the body as
measured by observers on the train.
3. From uT , compute ET and pT .
You will work through these steps on a problem set. The result you find is
ET = γ(EL − vpxL ) , pxT = γ(pxL − vEL /c2 ) ,
pyT = pyL , pzT = pzL . (7.3)
Tweaking notation slightly, we rewrite this
    
ET /c γ −γβ 0 0 EL /c
 pxT  −γβ γ 0 0  x 
 y =   pLy  . (7.4)
 pT   0 0 1 0  pL 
z
pT 0 0 0 1 pzL
In other words, the relativistic formulations of energy and momentum form a set of quantities
that transform under a Lorentz transformation.

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7.2 An invariant for energy and momentum
Recall that we found ∆s2 = −c2 ∆t2 + ∆x2 + ∆y 2 + ∆z 2 is a Lorentz invariant: all Lorentz
frames agree on the value of ∆s2 between two events. Can we do something similar with
energy and momentum?
Looking at how E and p behave under a Lorentz transformation, let’s think of energy
as the “timelike” component of momentum (E/c actually — which hopefully makes sense
since we need our quantities to have the right dimensions). Let’s see what happens when we
examine “negative time bit squared” plus “space bit squared”:

E2 x 2 y 2 z 2 E2
− + (p ) + (p ) + (p ) = − + |p|2 . (7.5)
c2 c2
Plug into this
m 2 c4
E 2 = γ 2 m2 c4 = , (7.6)
1 − u2 /c2
m2 u2
|p|2 = γ 2 m2 u2 = . (7.7)
1 − u2 /c2
Putting these together, we have

E2 2 m2 u2 − m2 c2
− + |p| =
c2 1 − u2 /c2
= −m2 c2 . (7.8)

Let us reorganize this expression slightly:

E 2 − |p|2 c2 = m2 c4 or E 2 = |p|2 c2 + m2 c4 . (7.9)

In other words, although different Lorentz frames will measure E and p differently, all frames
agree that E 2 and |p|2 are related by the expressions given in Eq. (7.9).
Notice that if m = 0, then |p| = E/c: massless bodies carry non-zero momentum. This
relationship corresponds perfectly to the energy and momentum carried by electromagnetic
radiation; compare with the Poynting vector if you need a refresher in this concept.

7.3 The 4-momentum

By virtue of the way in which E/c and px,y,z transform, we can see that they behave exactly
like the components of the displacement 4-vector. This tells us that we really should define
a 4-vector whose components all have the dimensions of momentum:
3
X
p~ = pµ~eµ , (7.10)
µ=0

with
p0 = E/c , p 1 = px , p1 = py , p 3 = pz . (7.11)
This p~ is then a geometric object: observers in all Lorentz frames use this 4-vector to
describe the system’s energy and momentum, but break it up into components and unit

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vectors differently. If the components and unit vectors according to O are pµ and ~eµ , then
an observer O0 constructs p~ using
0 0
pµ = Λµ α pα , ~eµ0 = Λα µ0 ~eα (7.12)
0
(switching to the Einstein summation convention). The matrix elements Λµ α perform the
Lorentz transformation of events from the frame of O to the frame of O0 ; the matrix elements
Λα µ0 perform the inverse transformation.
The reason why this is useful for us is that conservation of energy and conservation of
momentum are now combined into a single law: the conservation of 4-momentum. Suppose
Ni bodies interact, resulting in Nf bodies afterwards. Then,
Ni Nf
X X
p~jinit = p~jfinal , (7.13)
j=1 j=1

where p~jinit is the initial 4-momentum of particle j, and p~jfinal is the final 4-momentum of
particle j.

7.4 4-vectors in general; scalar products of 4-vectors

Let’s pause a momentum to reflect on the logic by which we assembled the 4-momentum.
We essentially followed the following recipe:

1. We found that a grouping of 4 quantities plays a meaningful role in physics: p0 = E/c,

p1,2,3 = px,y,z .
2. We found that when we change reference frames, these 4 quantities are transformed
to the new frame by the Lorentz transformation exactly as the components of the
0 0
4-displacement are: pµ = Λµ α pα .
3. Since it behaves under the transformation law exactly like the 4-vector we discussed
previously, we define pµ as the components of a new 4-vector, p~, and use this 4-vector
as a tool in our physics moving forward.

We can do this for any set of 4 quantities that turns out to be meaningful for our analysis.
In other words,
If any set bµ with µ ∈ [0, 1, 2, 3] has the property that when we change reference
0 0
frames their values are related by a Lorentz transformation, bµ = Λµ α bα , then
bµ represent the components of a 4-vector: ~b = bµ~eµ .
Once we have identified these quantities as the components of a 4-vector, we can start
identifying invariants. Whatever the vector ~b represents, we are guaranteed that all Lorentz
frames agree on the value of −(b0 )2 + (b1 )2 + (b2 )2 + (b3 )2 . In fact, it is not hard to show that
we can define a more general notion of an invariant. Suppose ~a = aµ~eµ and ~b = bµ~eµ . Then,

~a · ~b ≡ −a0 b0 + a1 b1 + a2 b2 + a3 b3 (7.14)

is a Lorentz invariant: all Lorentz frames agree on the value of ~a · ~b. This is simply proven
by transforming the components of ~a and ~b to another frame and then showing that the
right-hand side of (7.14) in the new frame is unchanged from its value in the original frame.

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 Equation (7.14) defines what we call the “scalar product” between two 4-vectors. A
(rather obvious) corollary is that the scalar product of any 4-vector with itself is a Lorentz
invariant. Two quantities we’ve recently examined can be rephrased using this definition:

∆~x · ∆~x = ∆s2 , (7.15)

p~ · p~ = −m2 c2 . (7.16)

The resemblance to the invariant interval ∆s2 gives us a convention for describing 4-
vectors. For any 4-vector ~a, if
~a · ~a < 0 (7.17)
then we say that ~a is timelike. This means that we can find a Lorentz frame in which only
the time component of ~a is non-zero: ~a has no spatial components in that frame. If

~a · ~a > 0 (7.18)

then we say that ~a is spacelike. There exists a1 Lorentz frame in which ~a has no component
in the time direction; it points purely in a spatial direction. Finally, if

~a · ~a = 0 (7.19)

then ~a is lightlike or null. In all Lorentz frames, ~a points along light cones.
Notice that p~ is either timelike or lightlike, and is only lightlike for m = 0.

7.5 4-velocity
In Newtonian mechanics, velocity and momentum were related by a factor of the body’s
mass. Let’s do the same thing using the 4-momentum; we will define the quantity that
results as the 4-velocity:
1
~u = p~ . (7.20)
m
What does this quantity mean? Let’s look at its components:

p0 E
u0 = = =γc, (7.21)
m mc
1
p
u1 = = γ (u)x , (7.22)
m
p2
u2 = = γ (u)y , (7.23)
m
3
p
u3 = = γ (u)z . (7.24)
m
(Note the somewhat unusual notation on the spatial components: (u)x means the x compo-
nent of the 3-vector u, and likewise for the y and z components.) The spatial components of
~u look just like “normal” 3-velocity, but multiplying by a factor of γ. How do we interpret
this factor?
1
Actually, many such Lorentz frames: once we find one, any Lorentz frame that is related to the first by
a rotation will do the trick.

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 Consider someone passing by with 3-velocity u. That person’s clocks run slow according
to you: as an interval dτ passes on their clock, an interval dt passes on your clock, with

dt = γ dτ . (7.25)

Notice that their clock is running slow — if γ = 2, then we measure 2 seconds passing for
every 1 second interval that they record.
We define the interval dτ as the proper time: it is an interval of time according to the
clock of the observer (or object) who we say is moving. Note that the word “proper” in
this case comes from an older meaning denoting “belonging to oneself.” Hence an observer’s
proper time is the time which that observer measures.
Proper time is a useful quantity because it is a Lorentz invariant: all Lorentz frames agree
that the observer in motion measures a time interval dτ . That won’t be the time interval
we measure as observer O whizzes by us at 90% of the speed of light; it won’t be what our
friend measures as they whizz by at 90% of the speed of light in another direction; but we
all agree that it is what O measures. Hence it is a useful benchmark which we can agree on.
With this in mind, let’s re-examine the spatial components of the 4-velocity:
dx dx
ux = γ (u)x = γ = , (7.26)
dt dτ
dy dy
uy = γ (u)y = γ = , (7.27)
dt dτ
dz dz
uz = γ (u)z = γ = , (7.28)
dt dτ
Let’s also look at the timelike component:
dt dt
ut = γ c = γ c =c . (7.29)
dt dτ
Comparing with how we defined the displacement 4-vector, we see that
d~x
~u = . (7.30)
dτ
The 4-velocity is the rate at which something moves through spacetime per unit proper time.
It’s worth computing the invariant associated with the 4-velocity:

1 m2 c2
~u · ~u = 2
p
~ · p
~ = − 2
= −c2 . (7.31)
m m
Notice that the 4-velocity of a body which is at rest in some Lorentz frame has the same
~u · ~u as a body which is moving 0.99999999999c in that frame.

7.6 4-velocity contrasted with 3-velocity

We now have two important ways to characterize a moving body’s motion:

• 3-velocity u = dx/dt describes motion through space per unit time. Both “space”
and “time” are frame-dependent concepts, and so u depends on the frame in which it
is measured.

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 • 4-velocity ~u = d~x/dτ describes motion through spacetime per unit proper time. It
is a frame-independent, geometric object; the same ~u is used by all observers.

A major conceptual difference between these two quantities is how we regard them when
observed in different Lorentz frames:

• As a frame-independent geometric object, all observers agree on an object’s 4-velocity

~u. They assign it different components, however, and use different unit vectors in
expanding ~u into components:
0
~u = uµ~eµ = uα ~eα0 , (7.32)

where
0 0
uα = Λα µ uµ , ~eα0 = Λµ α0 ~eµ . (7.33)

• The 3-vector is actually different in the two frames. Given u, we find the components of
u0 which describe the body’s motion in a new frame by applying the velocity addition
formulas: if the relative motion of the two frames is given by v = vex , then

x0 (u)x + v
(u) = , (7.34)
1 + (u)x v/c2
y0 (u)y
(u) = , (7.35)
γ(v)(1 + (u)x v/c2 )
0 (u)z
(u)z = . (7.36)
γ(v)(1 + (u)x v/c2 )

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