Hassani S. Mathematical Methods: For Students of Physics and Related Fields
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244 Vectors in Relativity
or
a ·b =
8
aηb where η =
⎛
⎜
⎜
⎝
1000
0 −10 0
00−10
00 0−1
⎞
⎟
⎟
⎠
, (8.8)
and
8
a and b are the row and column vectors in Equation (8.7).
A linear transformation that leaves the inner product of Equation (8.8)—
general Lorentz
transformation
and therefore the spacetime length Δs—invariant is called a Lorentz trans-
formation. By Equation (7.11), such a transformation Λ—which is a 4 × 4
matrix—satisfies
8
ΛηΛ = η. (8.9)
The study of the general structure of Lorentz transformations is beyond
the scope of this book. Here we shall confine ourselves to the Lorentz trans-
formations in two dimensions, in which the third and fourth components of
vectors are ignored. This means that vectors are of the form a =(a
0
,a
1
),
b =(b
0
,b
1
), the inner product is of the form a · b ≡ a
0
b
0
− a
1
b
1
,andthe
matrix η reduces to
η =
10
0 −1
.
In addition, the Lorentz transformations become 2 ×2 matrices.
Let Λ =
a
11
a
12
a
21
a
22
be a two-dimensional Lorentz transformation that
acts on 2-vectors in O to give the corresponding 2-vectors in O
.ThenΛmust
satisfy Equation (8.9) or
a
11
a
21
a
12
a
22
10
0 −1
a
11
a
12
a
21
a
22
=
10
0 −1
, (8.10)
which is equivalent to the following three equations [see (6.21) for a guide]:
a
2
11
− a
2
21
=1,a
11
a
12
− a
21
a
22
=0,a
2
12
− a
2
22
= −1. (8.11)
As in the case of rotations (see Section 6.1.3), we can conclude that
a
2
22
= a
2
11
,a
2
12
= a
2
21
,a
2
12
= a
2
11
− 1. (8.12)
So, all parameters are once again given in terms of a
11
.
To determine a
11
, consider the 2-vector (cΔt, Δx), the difference between
the time and position of two events in O. This 2-vector is represented by
(cΔt
, Δx
)inO
, and, by the definition of the Lorentz transformations,
cΔt
Δx
=
a
11
a
12
a
21
a
22
cΔt
Δx
. (8.13)
Now suppose that Δx = 0, i.e., that the two events occur at the same location.
Then O is measuring the proper time,sothatΔt =Δτ. From Equation (8.13),
8.3 Lorentz Transformation 245
we also have cΔt
= a
11
cΔt or Δt
= a
11
Δτ. Comparison with Equation (8.3)
yields
a
11
=
1
1 − (v/c)
2
.
Introducing the two symbols β ≡ v/c and γ =1/
1 − (v/c)
2
,weobtain
a
11
=
1
1 − β
2
≡ γ. (8.14)
The rest of the matrix elements can now be found. The first equation in
(8.12) gives a
22
= ±γ. To choose the correct sign for a
22
,notethatifO and
O
are not moving relative to one another, the coordinates do not change.
Therefore Λ must be the unit matrix. So, a
22
=1whenv =0. Thiscan
happen only if a
22
=+γ. The second equation in (8.12) now gives a
12
= a
21
;
and the third equation yields
a
2
12
= γ
2
− 1=
1
1 − β
2
− 1=
β
2
1 − β
2
= β
2
γ
2
⇒ a
12
= ±βγ.
The ambiguity in the sign comes from the choice we have for the direction of
motion. We absorb this choice of sign in β,andwrite
Λ=
γγβ
γβ γ
. (8.15)
For the important case of spacetime “position” vector (ct, x), this yields
Lorentz
transformation in
two spacetime
dimensions
ct
= γ (ct + βx) ,
x
= γ(x + βct). (8.16)
β is positive (negative) when observer O—who uses (ct, x) for events—travels
in the positive (negative) direction of O
—who uses primed coordinates. Equa-
tion (8.16) displays the celebrated Lorentz transformations in two spacetime
dimensions.
Example 8.3.1.
Emmy (observer O) is riding a train and she is standing in the
middle of one of the cars of length L at the two ends of which are two firecrackers that
explode simultaneously. Karl (observer O
) is standing on the platform watching
Emmy go by with speed β.Timezerofor both coincides with the moment that Emmy
passes by Karl. Suppose that the simultaneous explosion of the two forecrackers
(according to Emmy) also takes place at t = 0. We want to see how all this appears
to Karl.
Assume that Emmy and Karl are located at their respective origins. Let the front
firecracker be labeled as 1 and the back as 2. Then the front and back firecrackers
have coordinates (0,L/2) and (0, −L/2), respectively, in Emmy’s RF. Karl, on the
other hand, measures the coordinates of the firecrackers as
ct
1
= γβL/2,x
1
= γL/2 ct
2
= γ(−βL/2),x
2
= γ(−L/2)
246 Vectors in Relativity
from Equation (8.16). This shows that, for Karl, the back firecracker occurs first.
In fact, it occurs before Emmy reaches him (at time t
=0). Thetimedifference
between the two events is
Δt
= t
1
− t
2
= γβL/c.
Take L to be 30 m. Then, for the time difference to be a mere one second, we must
have
30γβ =3×10
8
or
β
1 − β
2
=10
7
,
giving β =0.999999999999995, awfully close to the speed of light!
On the other hand, if L is a typical interstellar distance of say 10 light years,
then
γβ =
Δt
10
with Δt
measured in years. For a time difference of one hour, we have γβ =
1.14 × 10
−5
, yielding β =1.14 × 10
−5
,orv = 3425 m/s, an easily attainable
speed.
Example 8.3.2. Observer O moves in the positive space direction of observer O
at speed v (or β = v/c). A particle moves at speed β
p
in the positive space direction
of O.Whatisβ
p
, the speed of the particle relative to O
?
The definition of speed is distance between two events divided by time interval
between those events: spotting of the particle at a point in space and an instant in
time (first event), and spotting the particle at a nearby point a little later (second
event). For example, observer O assigns the coordinates (ct, x) to the first event
and (ct + cΔt, x +Δx) to the second event, and concludes that the (dimensionless)
speed of the particle is β
p
=Δx/(cΔt).
Similarly, observer O
assigns the coordinates (ct
,x
) to the first event and
(ct
+ cΔt
,x
+Δx
) to the second event, and concludes that the speed of the
particle is β
p
=Δx
/(cΔt
), where Δx
and cΔt
are related to Δx and cΔt via the
Lorentz transformation. Using Equation (8.16), we find
β
p
=
Δx
cΔt
=
γ(Δx + βcΔt)
γ (cΔt + βΔx)
,
dividing the numerator and denominator by cΔt,wegetrelativistic law of
addition of
velocities
β
p
=
β
p
+ β
1+ββ
p
, (8.17)
which is called the relativistic law of addition of velocities.
One can show that if 0 <β
p
< 1and0<β<1, then 0 <β
p
< 1. So, it is
impossible to add two velocities close to light speed and get a velocity larger than
light speed. Furthermore, if the particle happens to be a photon (or a light beam),
then β
p
=1and
β
p
=
1+β
1+β
=1,
verifying the universality of the speed of light, the starting point of relativity
theory!
8.4 Four-Velocity and Four-Momentum 247
In many situations, an observer in three dimensions moves along the x-
axis. Then, the y and z coordinates of events—being perpendicular to the
direction of motion—do not change. This suggests a slightly more general
Lorentz transformation than (8.16):
ct
= γ (ct + βx) ,
x
= γ(x + βct),
y
= y, (8.18)
z
= z.
If an object moves in the xy-plane of an observer O with a velocity whose
components are (v
x
,v
y
), then the same object moves in the x
y
-plane of
another observer O
with a velocity whose components are
v
x
=
dx
dt
=
γ(dx + βcdt)
γ (dt + βdx/c)
=
v
x
+ βc
1+βv
x
/c
,
v
y
=
dy
dt
=
dy
γ (dt + βdx/c)
=
v
y
γ(1 + βv
x
/c)
, (8.19)
where β is the velocity of O relative to O
. In particular, if the object is light
and the angle it makes with the x-axis is α,thenv
x
= c cos α, v
y
= c sin α,
v
x
= c cos α
and v
y
= c sin α
, and the equations above yield
cos α
=
cos α + β
1+β cos α
,
sin α
=
sin α
γ(1 + β cos α)
. (8.20)
Now suppose that an observer O carries an EM radiation source which
an ultrarelativistic
source radiates
only in the
forward direction.
radiates uniformly in all directions. If β is very close to 1, then (8.20) im-
plies that cos α
→ 1 (and of course, sin α
→ 0), regardless of α.Thus,
an ultrarelativistic source of EM wave radiates (almost) only in the forward
direction.
8.4 Four-Velocity and Four-Momentum
In Newtonian mechanics velocity is defined as the derivative of the position
vector with respect to time. In terms of (Cartesian) coordinates, an observer
O locates the object in motion by assigning it the coordinates (x, y, z), and
differentiates these coordinates with respect to (the universal) time t to get
the velocity of the object: v =(˙x, ˙y, ˙z).
In relativity, the “position vector” is r =(ct,x,y,z) ≡ (ct, r), and there is
no universal time. However, each moving object has a proper time (measured
by a clock carried by the object), which is universal in the sense that all
observers measure it to be the same [see Equation (8.4) and the comments
248 Vectors in Relativity
after it]. Therefore, it is natural to define the dimensionless four-velocity as
u ≡
dr
ds
=
1
c
dr
dτ
=
dt
dτ
,
1
c
dx
dτ
,
1
c
dy
dτ
,
1
c
dz
dτ
= γ
1,
˙x
c
,
˙y
c
,
˙z
c
= γ (1,v/c) ,
(8.21)
where a dot represents differentiation with respect to the coordinate time t,
and we used dt = γdτ [see Equation (8.3)].
An interesting property of the four-velocity is that its spacetime length is
one:
4-velocity has
constant length
u ·u = u
2
0
−u
2
1
−u
2
2
−u
2
3
= γ
2
[1 − (v/c) · (v/c)] = γ
2
1 − v
2
/c
2
!
=1, (8.22)
from the definition of γ in (8.14). The four-velocity of an object in the object’s
rest frame is (1, 0, 0, 0), i.e., it is a unit vector in the time direction. If we define
the four-acceleration as the rate of change of the four-velocity with respect
to proper time, then the inner product of the 4-velocity and the 4-acceleration
of any object is zero, i.e., because of (8.22), the 4-acceleration is η-orthogonal
to the 4-velocity. Summarizing these two properties of the 4-velocity, we get
4-velocity is
perpendicular to
4-acceleration
u ·u =1, u · a =0. (8.23)
Example 8.4.1.
A particle is moving in the two-dimentional spacetime of an
inertial frame on a path given parametrically as
t(σ)=b sinh(σ),x(σ)=cb cosh(σ),
where σ is a dimensionless parameter. The differential of the particle’s proper time
is
(cdτ )
2
=(cdt)
2
− (dx)
2
=(cb)
2
cosh
2
(σ)(dσ)
2
− (cb)
2
sinh
2
(σ)(dσ)
2
=(cb)
2
(dσ)
2
⇒ dσ =
1
b
dτ,
and σ = τ/b. Thus, as a function of the proper time, the path becomes
t(τ )=b sinh(τ/b),x(τ )=cb cosh(τ/b).
The components of the (dimensionless) 4-velocity are
u
0
=
dt
dτ
=cosh(τ/b),u
1
=
dx
cdτ
= sinh(τ/b),
which satisfy u
2
0
− u
2
1
= 1 as they should.
The acceleration of the particle has components
a
0
=
du
0
dτ
=
1
b
sinh(τ/b),a
1
=
du
1
dτ
=
1
b
cosh(τ/b).
It is easily verified that a · u =0andthat
a · a = a
2
0
− a
2
1
= −
1
b
2
.
So, the particle has a uniform acceleration of 1/b. The negative sign in the last
equation is due to the fact that the magnitude of the acceleration has to be defined
as −a · a = a
2
− a
2
0
, with the space part appearing as positive (so that when a
0
is
absent, we get back the Newtonian acceleration).
8.4 Four-Velocity and Four-Momentum 249
The (kinematic) 4-velocity leads to the (dynamic) 4-momentum: just mul- 4-momentum
defined
tiply u by mc—the c is to give dimension to the 4-velocity. In a reference
frame in which an object of mass m moves with velocity v,the4-momentum
p is given by
p ≡ (p
0
,p
1
,p
2
,p
3
) ≡ (p
0
,p)=mcu = γmc (1,v/c)=(γmc,γmv) . (8.24)
The space part of the 4-momentum is
relativistic
momentum
p = γmv =
mv
1 − (v/c)
2
, (8.25)
and gives ordinary Newtonian momentum when |v| << c, because in that
limit, γ ≈ 1. Therefore, we call p the relativistic momentum.
What about p
0
? How are we to interpret that? If we set γ ≈ 1, we get
p
0
≈ mc which does not correspond to any Newtonian quantity.
3
However, if
we make the next best approximation to γ (see Example 10.2.1 and Problem
10.8), i.e.,
γ =
1
1 − (v/c)
2
≈ 1+
1
2
(v/c)
2
,
then
p
0
= mcγ ≈ mc
1+
1
2
v
2
/c
2
!
⇒ p
0
c ≈ mc
2
+
1
2
mv
2
.
The second term gives us the clue that p
0
c must be the relativistic energy
E.Sowewrite
relativistic energy
p ≡ (p
0
,p)=(E/c, p)=(γmc, γmv),E= γmc
2
=
mc
2
1 − (v/c)
2
. (8.26)
An important special case of this is the 4-momentum p of a particle in its rest
frame:
p =(mc,
0) = (mc, 0, 0, 0). (8.27)
The definition of the relativistic energy allows objects to have rest energy:
when v =0,weget
the most famous
equation in
physics!
E = mc
2
, (8.28)
which states the equivalence of mass and energy and allows their conversion
into one another.
The invariance of the length of a 4-vector tells us that p · p is a quantity
that is independent of observers. From Equation (8.26), we get
p ·p =(E/c)
2
−|p|
2
= γ
2
m
2
c
2
− γ
2
m
2
v
2
= γ
2
m
2
c
2
(1 −v
2
/c
2
)=m
2
c
2
,
which we rewrite for future reference
p · p = m
2
c
2
or E
2
−|p|
2
c
2
= m
2
c
4
. (8.29)
3
One may interpret mc as the momentum of an object moving at the speed of light.
However, while objects moving at light speed are possible in Newtonian physics, relativity
does not allow a massive object to go at the speed of light [see (8.25)].
250 Vectors in Relativity
Thus, although different observers measure different values for the energy
and 3-momentum of an object, when they subtract the square of their value
of momentum (times c) from their corresponding value of energy squared, all
get the same numerical value, namely the square of the mass of the object
(time c
4
).
Equation (8.29) allows particles with zero mass to have energy and mo-
mentum. For such particles,
E
2
−|p|
2
c
2
=0 or E = |p|c. (8.30)
Since p/E = v/c
2
[see Equation (8.26)], we conclude from (8.29) and (8.30)
that
Box 8.4.1. A particle is massless if and only if it moves at light speed.
The particle (quantum) of electromagnetic waves is photon.Ittravelsatthephoton is
massless!
speed of light (obviously!). Therefore, it must be massless.
Example 8.4.2.
A particle has 4-momentum p relative to an observer O
whose 4-
velocity is u
. In the rest frame of this observer u
=(1, 0, 0, 0), and if p =(E
/c, p
)
in this frame, then
p · u
= E
/c.
Now consider another observer O with respect to whom the 4-momentum of the
particle is p =(E/c, p )andthe4-velocityofO
is u
=(γ,γv/c). In the frame of O,
p · u
= γE/c − γp ·v/c.
The invariance of the inner product now gives
E
= γ(E − p ·v). (8.31)
In the special case in which the particle is at rest with respect to O, p =0and
E = mc
2
. This leads to
E
= γmc
2
=
mc
2
1 − (v/c)
2
,
which is the expected expression for the relativistic energy of a particle moving with
velocity v relative to O
.
8.4.1 Relativistic Collisions
Conservation of energy and momentum in relativistic collisions is stated suc-
cinctly in terms of the total four-momenta before and after: p
bef
tot
= p
aft
tot
,
where in each case, p
tot
is the sum of the 4-momenta of all particles involved.
As a first example, consider two particles that collide and form a single
third particle. Let the masses of the first two particles be m
1
and m
2
.Wecan
immediately find the mass M of the third particle. Before doing so, we set
8.4 Four-Velocity and Four-Momentum 251
c = 1 to avoid the cluttering of calculations. This is common in high energy
physics, in which energy, momentum, and mass are all measured in the same
unit (usually electron volt, eV). If desired, we can easily restore the factors of
c at the end by a simple dimensional analysis. With this convention, Equation
(8.29) becomes p · p = m
2
.
The conservation of 4-momentum in the present situation is p
1
+ p
2
= P,
where P is the four-momentum of the final particle. Since this is a vector
equation, all components must equal. In particulare, separating the time and
the space parts, we get
p
01
+ p
02
= P
0
, or E
1
+ E
2
= E,
p
1
+ p
2
=
P, (8.32)
which are the conservation of energy and momentum.
Squaring both sides of p
1
+ p
2
= P gives
(p
1
+ p
2
) ·(p
1
+ p
2
)=P ·P,
or
p
1
· p
1
+ p
2
· p
2
+2p
1
·p
2
= P · P,
or
m
2
1
+ m
2
2
+2p
1
· p
2
= M
2
. (8.33)
Because of the invariance of the dot product, this equation holds in any in-
ertial frame.
Let us evaluate (8.33) in the rest frame of the second particle, where
p
2
=(m
2
,
0) by (8.27), and the energy of the first particle is assumed to be
E
1
.Then
p
1
·p
2
=(E
1
,p
1
) ·(m
2
,
0) = E
1
m
2
,
and Equation (8.33) immediately gives the mass of the final particle:
M
2
= m
2
1
+ m
2
2
+2m
2
E
1
, or M
2
= m
2
1
+ m
2
2
+2m
2
E
1
/c
2
, (8.34)
where the second equation restores the necessary powers of c.Notehowthe
initial energy E
1
on the right-hand side has turned into (part of) the final mass
M on the left-hand side. This is how large accelerators create new particles
out of the energy of collision.
We can also find the momentum of the final particle from the second
equation in (8.32). This easily gives
P = p
1
, indicating that, in the rest frame
of particle 2, the final particle moves in the initial direction of particle 1. The
magnitude of
P can be calculated in terms of energies and masses:
|
P | = |p
1
| =
E
2
1
− m
2
1
. (8.35)
The first equation in (8.32) gives the energy of the final particle
E = E
1
+ m
2
. (8.36)
252 Vectors in Relativity
Combining Equations (8.35) and (8.36), we can obtain the speed of the final
particle:
V =
|
P |
E
=
E
2
1
− m
2
1
E
1
+ m
2
. (8.37)
A more common collision has two particles initially and two finally. So
the conservation of 4-momentum becomes p
1
+ p
2
= p
3
+ p
4
. Separating the
time and the space parts yields the conservation of energy and momentum:
E
1
+ E
2
= E
3
+ E
4
,
p
1
+ p
2
= p
3
+ p
4
. (8.38)
Squaring both sides of p
1
+ p
2
= p
3
+ p
4
gives
m
2
1
+ m
2
2
+2p
1
·p
2
= m
2
3
+ m
2
4
+2p
3
·p
4
, (8.39)
which holds in any inertial frame. Evaluating this equation in the rest frame
of the second particle, yields
m
2
1
+ m
2
2
+2m
2
E
1
= m
2
3
+ m
2
4
+2(E
3
E
4
− p
3
· p
4
). (8.40)
In this frame, Equation (8.38) becomes E
1
+ m
2
= E
3
+ E
4
and p
1
= p
3
+ p
4
.
Solving for E
4
and p
4
from these equations and substituting the results in
(8.40) yields (after some algebra and using E
2
3
−|p
3
|
2
= m
2
3
)
m
2
1
+ m
2
2
+2m
2
E
1
= m
2
4
− m
2
3
+2E
3
(E
1
+ m
2
) −2p
1
· p
3
,
or
m
2
1
+ m
2
2
+2m
2
E
1
= m
2
4
− m
2
3
+2E
3
(E
1
+ m
2
) −2|p
1
||p
3
|cos θ
13
, (8.41)
where θ
13
is the scattering angle of the third particle. Once the energy E
1
of the initial incident particle is known, Equation (8.41) gives the scattering
angle as a function of the energy of the third particle (|p
1
| and |p
3
| are related
to E
1
and E
3
, respectively).
Example 8.4.3.
The particle nature of light, which had been proposed by EinsteinCompton
scattering in his explanation of the photoelectric effect, was demonstrated by Compton in what
is now called the Compton scattering. In this scattering, a photon of energy E
is scattered off a stationary electron of mass m
e
. The scattered photon is detected
at an angle θ from the direction of the incident photon. What is the change in the
wavelength of the photon as a function of θ?
In (8.41), let 1 denote the incident photon, 2 the stationary electron, 3 the
scattered photon, and 4 the scattered electron. Let E
denote the energy of the
scatterd photon, then, with m
1
= m
3
= 0, Equation (8.41) becomes
m
2
e
+2m
e
E = m
2
e
+2E
(E + m
e
) − 2EE
cos θ,
or
m
e
E = E
(E + m
e
) − EE
cos θ ⇒ m
e
(E − E
)=EE
(1 − cos θ).
8.4 Four-Velocity and Four-Momentum 253
Restoring the factors of c and noting that E = hc/λ,weobtain
m
e
c
2
hc
λ
−
hc
λ
=
hc
λ
hc
λ
(1 − cos θ),
which can be simplified to
Δλ ≡ λ
− λ =
h
m
e
c
(1 − cos θ) ≡ λ
c
(1 − cos θ), (8.42)
where λ
c
= h/m
e
c is called the Compton wavelength of the electron. By mea-
suring the difference between the wavelengths of scattered and incident photons,
Compton could verify Equation (8.42) and demonstrate that light had particle
property.
8.4.2 Second Law of Motion
The Newtonian mechanics defines force as the rate of change of momentum.
We generalize this to relativity and define
f =
dp
dτ
= m
du
dτ
= ma, (8.43)
where τ is the proper time of the moving object with mass m,four-velocityu,
and four-momentum p. Let us explore the meaning of the components of f.
In a particular inertial frame, we assume that Newton’s second law holds:
dp
dt
=
F, (8.44)
where p is the space part of the 4-momentum. The space part of f can now
be written as
f =
dp
dτ
=
dp
dt
dt
dτ
= γ
F. (8.45)
Thetimepartoff is a little trickier. First note that
f
0
=
dp
0
dτ
=
1
c
dE
dτ
.
Next differentiate (8.29) with respect to τ to obtain E(dE/dτ)=c
2
p·(dp/dτ).
Finally use p/E = v/c
2
to arrive at
f
0
=
1
c
dE
dτ
=
1
c
c
2
p
E
·
dp
dτ
=
1
c
γv ·
F = γ
β ·
F,
where
β ≡ v/c.Thus,
f ≡ (f
0
,
f)=(γ
β ·
F,γ
F ). (8.46)
The fact that f
0
= γ
β ·
F could also be obtained by using f · u =0,which
is a result of Equation (8.43) and the orthogonality of the 4-velocity and
4-acceleration (see Problem 8.13).