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Particle Notes
Ryan D. Reece
July 9, 2007
Chapter 1
Preliminaries
1.1
1.1.1
Overview of Special Relativity
Lorentz Boosts
Searches in the later part 19th century for the coordinate transformation that
left the form of Maxwell’s equations and the wave equation invariant lead to
the discovery of the Lorentz Transformations. The “boost” transformation
from one (unprimed) inertial frame to another (primed) inertial frame moving
with dimensionless velocity β = v/c respect to the former frame, is given by
0 ct
γ
−γβ
ct
=
(1.1)
x0
−γβ
γ
x
Because a boost along one of the spacial dimensions leaves the other two unchanged, we can suppress the those two spacial dimensions. γ is the Lorentz
Factor, defined by
1
(1.2)
γ=p
1 − β2
γ ranges from 1 to ∞ monotonicly in the nonrelativistic and relativistic
limits, respectively. It is useful to remember that γ ≥ 1.
The inverse transformation is given by
0 ct
γ γβ
ct
(1.3)
=
x
γβ γ
x0
1
The differences betweeen two points in spacetime follow from the transformations:
c ∆t0
∆x0
c ∆t
∆x
1.1.2
=
=
=
=
γ c ∆t − γβ ∆x
−γβ c ∆t + γ ∆x
γ c ∆t0 + γβ ∆x0
γβ c ∆t0 + γ ∆x0
(1.4)
(1.5)
(1.6)
(1.7)
Length Contraction and Time Dilation
Consider a clock sitting at rest in the unprimed frame (∆x = 0). Equation
(1.4) and the fact that γ ≥ 1 implies that the time interval is dilated in the
primed frame.
∆t0 = γ ∆t
(1.8)
Now consider a rod of length ∆x in the unprimed frame. A measurement
of the length in the primmed frame corresponds to determining the coordinates of the endpoints simultaneously in the unprimed frame (∆t0 = 0).
Then (1.7) implies that length is contracted in the primed frame.
∆x
(1.9)
∆x0 =
γ
We call time intervals and lengths “proper” if they are measured in the
frame where the subject is at rest (in this case, the unprimed frame). In
summary, proper times and lengths are the shortest and longest possible,
respectively.
1.1.3
Four-vectors
Knowing that lengths and times transform from one reference frame to another, we wonder if there is anything is invariant. Consider the following,
using equations (1.6) and (1.7).
2
2
(c ∆t)2 − (∆x)2 = (γ c ∆t0 + γ β ∆x0 ) − (γβ c ∆t0 + γ ∆x0 ) (1.10)
(
(
(0(
2β(c(∆t
∆x0 + β 2 (∆x0 )2
= γ 2 (c ∆t0 )2 + (
(
(
(0(
2β(c(∆t
∆x0 − (∆x0 )2
−β 2 (c ∆t0 )2 − (
(1.11)
2
2
0 2
0 2
= γ (1 − β ) (c ∆t ) − (∆x )
(1.12)
| {z }
γ −2
0 2
= (c ∆t ) − (∆x0 )2 ≡ (∆τ )2
2
(1.13)
Which shows that ∆τ has the same value in any frames related by Lorentz
Transformations. ∆τ is called the “invariant length.” Note that it is equal
to the proper time interval.
This motivates us to think of (t, ~x ) as a four-vector that transforms according to the Lorentz transformations, in a “spacetime vector space,” and
there should be some kind of “inner product,” or contraction, of these vectors
that leaves ∆τ a scalar. This can be done by defining the Minkowski metric
tensor as follows.


1 0
0
0
 0 −1 0
0 

(1.14)
g=
 0 0 −1 0 
0 0
0 −1
Four-vectors are indexed by a Greek index, xµ = (c t, ~x )µ , ranging from 0
to 3 (x0 = c t, x1 = x, x2 = y, x3 = z). The contraction of a spacetime
four-vector with itself, its square, is give by
xµ gµν xν ≡ xµ xµ = (c t)2 − ~x · ~x = (∆τ )2
(1.15)
giving the square of the invariant length between xµ and the origin. In
equation (1.15), we have defined that the lowering of a four-vector index is
done by multiplication by the metric tensor. Explicit matrix multiplication
will show that the Minkowski metric has the same components regaurdless
of the orientation of its indices.
g µν ≡ gνλ g µλ = g µν
(1.16)
gµν ≡ gµσ gνλ g σλ = g µν
(1.17)
Anything that transforms according to the Lorentz Transformations like
(ct, ~x ) is four-vector Another example of a four-vector is four-velocity, defined
by
η µ ≡ γ (c, ~v )µ
(1.18)
One can show that the square of η µ is invariant as required.
η µ ηµ = γ 2 (c2 − v 2 )
1
=
(1 − β 2 ) c2
1 − β2
= c2
3
(1.19)
(1.20)
(1.21)
which is obviously invariant. Any equation where all of the factors are scalars
(with no indices or contracting indices), or are four-vectors/tensors, with
matching indices on the other side of the equal sign, is called “manifestly
invariant.”
1.1.4
Momentum and Energy
The Classically conserved definitions of momentum and energy, being dependent on the coordinate frame, will not be conserved in other frames. We are
motivated to consider the effect of defining momentum with the four-velocity
instead of the classical velocity. The rest (invariant) mass of a particle, m, being an intrinsic property of the particle, must be a Lorentz scalar. Therefore,
the following definition of the four-momentum is manifestly a four-vector.
pµ ≡ m η µ = γ m (c, ~v )µ
(1.22)
p µ p µ = m 2 η µ ηµ = m 2 c 2
(1.23)
The square of which is
Now let’s give some interpretation to the components of the four-momentum.
To consider the nonrelativistic limit, let us expand γ in the β → 0 limit.
γ '1+
1 2 3 4
β + β + ···
2
8
(1.24)
Then the leading order term of the space-like components of the four-momentum
is just the Classical momentum.
p~ = m ~v + · · ·
(1.25)
We therefore, interpret the space-like components of the four-momentum as
the relativistic momentum.
p~ = γ m ~v
(1.26)
The expansion of the time-like term gives
m c2 +
1
m v2 + · · ·
2
4
(1.27)
We can now recognize the second term as the Classical kinetic energy. The
first term is evidently the “rest mass energy,” energy present even when
v = 0. Higher order terms give relativistic corrections.
E = γ m c2
(1.28)
We can therefore write the four-momentum in terms of the relativistic energy,
E, and relativistic momentum, p.
pµ = (E, p~ )µ
(1.29)
The four-momentum is the combination of momentum and energy necessary to transform according to Lorentz Transformations. Both E and p~ are
conserved quantities in any given frame, but they are not invariant; they
transform when going to another frame. Scalar quantities, like mass, are invariant but are not necessarily conserved. Mass can be exchanged for kinetic
energy and vice versa. Charge is an example of a scalar quantity that is also
conserved.
Taking the ratio of equations (1.26) and (1.28) gives the following interesting relation.
p~
~v
(1.30)
= 2
E
c
which leads to
|~p | c
=β
E
(1.31)
Note that in order for energy and momentum to remain finite, velocity
has an upper bound at c, because γ diverges as v → c in equations (1.26)
and (1.28). The only way around the prohibition against light speed is for
a particle to have zero mass. Conversely, any massless particle must move
at speed c, since otherwise it would have no energy, no momentum — no
existence at all. For massless particles, (1.31) reduces to
E = |~p | c
(1.32)
Also note the relationship to the equation from quantum mechanics for
the energy of a photon, E = ~ ω ⇒ |~p | = ~ ωc = ~ k, consistent with
de Broglie’s relation.
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1.1.5
Relationship Between Energy, Momentum, and
Mass
We now can use equations (1.28) and (1.31) to eliminate the velocity dependence to get a frame independent relationship.
E = γ m c2
(1.33)
2 − 12
2
= (1 − β ) m c
2 !− 12
|~p | c
=
1−
m c2
E
E 2 = |~p |2 c2 + m2 c4
⇒
(1.34)
(1.35)
(1.36)
which is consistent with the zero mass case (1.32).
Another way to arrive at equation (1.36) is to apply our interpretation of
the components of pµ to equation (1.23).
pµ pµ = E 2 − |~p |2 c2 = m2 c4
1.2
1.2.1
(1.37)
Units
Natural Units
Factors of c were explict in the above review of special relativity. From now
on, we will use a form of natural units, where certain natural constants are
set to one by using units derived from the God-given scales in Nature.
~ = c = ε0 = 1
(1.38)
From ~ = 6.58 × 10−25 GeV · s = 1, it follows that if we choose to measure
energy in units GeV, then time can be measured in units GeV−1 .
1 GeV−1 = 6.58 × 10−25 s
From c = 3 × 108 m/s = 1, it follows that
1 = 3 × 108 m 6.58 × 10−25 GeV = 1.97 × 10−16 m · GeV
⇒
1 GeV−1 = 1.97 × 10−16 m
(1.39)
(1.40)
(1.41)
Summarizing the dimensionality:
time = length =
6
1
energy
(1.42)
1.2.2
Barns
We will later see later that when calculating cross sections, the conventional
unit of area in particle physics is a barn.
1 barn ≡ (10 fm)2 = 10−24 cm2
(1.43)
1 mb ≡ 10−3 barns = 10−27 cm2
(1.44)
From (1.41), it can be shown that
1 GeV−2 = 0.389 mb
1.2.3
(1.45)
Electromagnetism
Finally, from ε0 = 1 and c = 1
c= √
1
µ0 ε 0
⇒
µ0 = 1
giving Maxwell’s equations the following form.
Field Tensor:
F µν ≡ ∂ µ Aν − ∂ ν Aµ
(1.46)
(1.47)
Homogeneous:
∂µ Fνλ + ∂ν Fλµ + ∂λ Fµν = 0
(1.48)
∂ν F µν = J µ
(1.49)
Inhomogeneous:
1.3
Relativistic Kinematics
1.3.1
Lorentz Invariant Phase Space
1.3.2
Mandelstam Variables
1.4
Perturbation Theory
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Chapter 2
Quantum Electrodynamics
2.1
The Dirac Equation
The Dirac Equation is the equation of motion for fermion (spin 1/2) fields.
It is written
(iγ µ ∂µ − m)ψ = 0
(2.1)
It is actually four coupled equations for the four components of the fermion
field, ψ. The four components are that of a “spinor space,” describing the
spin, like the usual two component spinor from non-relativistic quantum
mechanics, and also describing the particle/antiparticle nature of particles.
Each γ, indexed by a four-vector index, µ, is a 4 × 4 matrix, operating in
the spinor space. The mass of the fermion, m, is multiplied by an assumed
identity matrix in the spinor space.
2.1.1
The γ-matrices
The γ-matrices satisfy the following defining anticommutation relation
{γ µ , γ ν } = 2g µν
(2.2)
where again, multiplied by g µν , there is an assumed identity matrix in the
spinor space. Be careful to distinguish between spacetime and the spinor
space. Unfortunately, they both have four components. Note that g µν , a
component of the Minkowski metric, is just a number (+1, −1, or 0).
From the anticommutation relation, one can derive the following properties:
†
γ0 = γ0
(2.3)
8
and
†
γ k = −γ k
(2.4)
γ µ† = γ 0 γ µ γ 0
(2.5)
which generalize as
Like four-vectors, the space-index parts of γ µ change sign when lowering
the index.
γ 0 = γ0 ,
γ k = −γk
(2.6)
There are several representations (Pauli-Dirac, Weyl, Majorana. . . ) of
the γ-matrices, where each γ µ is represented by a specific 4 × 4 matrix, but it
is often not necessary to assume a specific representation to do a calculation.
Often, all that is needed is the anticommutation relation and its resulting
properties.
2.1.2
Bar and Slash Notation
We will see that it is convenient to define some compact notation. The
adjoint (row) spinor, said “psi-bar,” is defined
ψ̄ ≡ ψ † γ 0
(2.7)
The Feynman Slash abreviates the contraction of γ-matrices with a fourvector. It is defined by
a
(2.8)
/ ≡ γ µ aµ ≡ γµ aµ
where a is any four-vector.
2.1.3
The Adjoint Equation and Conserved Current
Written in a more explicit notation notation, the Dirac Equation is
∂ψ
∂ψ
+ iγ k k − mψ = 0
∂t
∂x
(2.9)
∂ψ †
∂ψ † 0
γ − i k (−γ k ) − mψ † = 0
∂t
∂x
(2.10)
iγ 0
Taking the adjoint gives
−i
9
To restore the covariant form, we need to remove the minus sign of −γ k while
leaving the first term unchanged. Since γ 0 γ k = −γ k γ 0 , this can be done by
multiplying on the right by −γ 0 .
i
∂ψ †
∂ψ † 0 0
γ γ + i k γ 0 γ k + mψ † γ 0 = 0
∂t
∂x
(2.11)
Using the bar notation, we have
i∂µ ψ̄γ µ + mψ̄ = 0
(2.12)
We can now show that the fermion field conserves charge current, as it
should in order to describe particles like electrons. Multiplying the Dirac
Equation (2.1) from the left by ψ̄ and adding it to the adjoint equation
(2.12) multiplied by ψ from the right gives
iψ̄γ µ ∂µ ψ − m
ψ̄ψ
+ i∂µ ψ̄γ µ ψ + m
ψ̄ψ
=0
(2.13)
ψ̄γ µ ∂µ ψ + ∂µ ψ̄γ µ ψ = ∂µ (ψ̄γ µ ψ) = 0
(2.14)
Multiplying by the charge of an electron (−e), we interpret
j µ = −eψ̄γ µ ψ
(2.15)
as the charge current density four-vector of electrodynamics, satisfying the
continuity equation:
∂µ j µ = 0
(2.16)
2.1.4
Antiparticles and Free Particle Spinors
The Dirac Equation has plane-wave solutions of the form
ψ = u(p) eip·x
(2.17)
where u(p) is a four-spinor. Plugging equation (2.17) into the Dirac Equation
(2.1) gives
(γ µ pµ − m) u(p) = 0
or
(p/ − m) u = 0
(2.18)
where we have used the Feynman slash notation.
10
We will now assume the Pauli-Dirac Representation of the γ-Matrices in
order to study the components of solutions to the Dirac Equation in some
detail. It is given by
I 0
0 σi
0
i
γ =
,
γ =
(2.19)
0 −I
−σi 0
where σi are the Pauli Matricies and I is the 2 × 2 identity matrix.
Let
 (1) 
u
(1) (3)  u(2) 
u
u
u
A

u=
,
uA =
, uB =
(2)
 u(3)  = uB
u
u(4)
u(4)
Plugging in the γ-matrices into (2.18), we have
E−m
~σ · p~
uA (p)
=0
−~σ · p~ −E − m
uB (p)
(2.20)
(2.21)
We can see right away that in the limit |~p | → 0, solutions have E = ±m.
Does this mean that the Dirac Equation has negative energy solutions?! No.
We interpret the “negative energy solutions” as antiparticles with positive
energy. Antiparticles are like their corresponding particles with all signs
of charge-like quantities (electric charge, baryon number, lepton number. . . )
flipped. Therefore, the conserved currents from antiparticles can be described
as that of particles going backward in time (t → −t). The phase factor of
the field, then is unchanged because the negative sign from the time cancels
the negatives sign of the energy.
e−i(−E)(−t) = e−iEt
(2.22)
Solutions u(1) and u(2) correspond to particles while solutions u(3) and u(4)
correspond to antiparticles.
2.2
Scattering
2.2.1
Interaction with the Electromagnetic Field
2.2.2
e− µ− → e− µ−
11