Rotations, radiation and polarization patterns
Magnus Johan Axelsson
13th April 2007
Contents
1 Classical and quantum mechanical rotations in two and three dimensions
1.1 Independent coordinates of a rigid body . . . . . . . . . . . . . . . . . . . . . .
1.2 Orthogonal transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 Rotation in two dimensions . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.2 The Euler angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.3 Second rotation about the η axis . . . . . . . . . . . . . . . . . . . . . .
1.3 Rigid body Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Rate of change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.2 Kinetic energy for rotation about a point . . . . . . . . . . . . . . . . .
1.3.3 Diagonalization of the Inertia tensor . . . . . . . . . . . . . . . . . . . .
1.3.4 Angular velocity vector expressed through the Euler angles . . . . . . .
1.4 Lagrangian expressed through the Euler angles . . . . . . . . . . . . . . . . . .
1.5 Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.1 Axial symmetric body not subjected to any foreces . . . . . . . . . . . .
1.6 Rotations in quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 Algebraic approach to the angular momentum eigenvalue problem . . . . . . . .
1.8 Matrix representation of Jx , Jy and Jz . . . . . . . . . . . . . . . . . . . . . . .
1.8.1 The case j = 1/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8.2 The case j = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.9 Orbital angular momentum and the spherical harmonics . . . . . . . . . . . . .
1.10 Representation of the rotation operator . . . . . . . . . . . . . . . . . . . . . . .
1.10.1 Quantum mechanical rotation of a rigid body - Euler angles revisited .
1.10.2 The case l = 1/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.10.3 the case l = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.11 Angular momentum components in the laboratory and body-fixed coordinate
frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(l)∗
(l)∗
1.11.1 Proof of J2 Dmk = h̄2 l(l + 1) Dmk . . . . . . . . . . . . . . . . . . . . .
1.11.2 Further properties of J+ and J− . . . . . . . . . . . . . . . . . . . . . .
1.11.3 Energy in the body-fixed system . . . . . . . . . . . . . . . . . . . . . .
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2 Radiation from scalar and electromagnetic fields
2.1 Scalar field radiation . . . . . . . . . . . . . . . . . . . . . .
2.1.1 Scalar wave equation . . . . . . . . . . . . . . . . . .
2.1.2 Classical angular distribution . . . . . . . . . . . . .
2.2 Determination of the constants Alm . . . . . . . . . . . . . .
2.3 Quantum angular distribution of radiation from scalar fields
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2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
The Maxwell equations . . . . . . . . . . . . . . . . . . . . . . . .
2.4.1 Static fields . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.2 Time dependent fields . . . . . . . . . . . . . . . . . . . .
Poynting’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . .
Multipole radiation . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.1 Electric dipole terms . . . . . . . . . . . . . . . . . . . . .
2.6.2 Magnetic dipole terms . . . . . . . . . . . . . . . . . . . .
2.6.3 Electric quadrupole terms . . . . . . . . . . . . . . . . . .
2.6.4 Linear antenna . . . . . . . . . . . . . . . . . . . . . . . .
2.6.5 Circular loop antenna . . . . . . . . . . . . . . . . . . . .
Multipole expansion and the vector spherical harmonics . . . . .
2.7.1 Vector wave equation . . . . . . . . . . . . . . . . . . . . .
Determination of the multipole coefficients aE (l, m) and aM (l, m)
2.8.1 The radiation zone limit . . . . . . . . . . . . . . . . . . .
2.8.2 Angular distribution in the radiation zone . . . . . . . . .
Antennas revisited . . . . . . . . . . . . . . . . . . . . . . . . . .
2.9.1 Linear antenna . . . . . . . . . . . . . . . . . . . . . . . .
2.9.2 Circular loop antenna . . . . . . . . . . . . . . . . . . . .
Quantum angular distribution of radiated photons . . . . . . . .
2.10.1 Plane wave expansion . . . . . . . . . . . . . . . . . . . .
2.10.2 Angular distribution in terms of spin-±1 harmonics . . . .
Polarization and the Stoke’s parameters . . . . . . . . . . . . . .
2.11.1 The polarization ellipse . . . . . . . . . . . . . . . . . . .
2.11.2 The Stoke’s parameters . . . . . . . . . . . . . . . . . . .
Stoke’s parameters for a linear antenna . . . . . . . . . . . . . . .
Stoke’s parameters from a circular loop antenna . . . . . . . . . .
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3 Spin weighted harmonics and their application to electromagnetism
3.1 The photon wave function and parity . . . . . . . . . . . . . . . . . . . .
3.2 Spherical helicity states . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Spin-weighted spherical harmonics . . . . . . . . . . . . . . . . . . . . .
3.4 Solution of the vector Helmholtz equation . . . . . . . . . . . . . . . . .
3.4.1 Divergenceless solutions of the vector Helmholtz equation . . . .
l
to the multipole expansion . . . . . . . . . . . . .
3.5 Application of Ym,±1
3.5.1 The radiation zone fields . . . . . . . . . . . . . . . . . . . . . . .
3.5.2 Determination of a± . . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Angular distribution of photon emission . . . . . . . . . . . . . . . . . .
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4 Gravitational radiation and spin-2 fields
4.1 Linearized field equations . . . . . . . . . . . . . . .
4.2 Gravitational radiation in the weak-field limit . . . .
4.2.1 The TT-gauge . . . . . . . . . . . . . . . . .
4.2.2 Matter in the prescence of gravitational waves
4.3 Constructing spin-weighted tensors . . . . . . . . . .
4.4 Solution of the linearized Einstein equation . . . . .
4.4.1 The method of Green functions . . . . . . . .
4.5 Angular distribution of gravitational radiation . . .
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1
4.6
4.7
4.8
4.9
4.5.1 Calculating the energy-momentum tensor . . . . . . . . .
Angular distribution of tensor radiation . . . . . . . . . . . . . .
Binary star system . . . . . . . . . . . . . . . . . . . . . . . . . .
Classical angular distribution of tensor radiation . . . . . . . . .
4.8.1 Divergenceless solutions of the spin-2 Helmholtz equation
4.8.2 The radiation zone fields . . . . . . . . . . . . . . . . . . .
4.8.3 Angular distribution in the radiation zone . . . . . . . . .
Quantum angular distribution of tensor radiation . . . . . . . . .
4.9.1 Expansion of the graviton field . . . . . . . . . . . . . . .
A Calculation of ∇2 F
A.1 The gradient . . . .
A.2 The divergence . . .
A.3 The curl . . . . . . .
A.4 The Laplace operator
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∇2
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Chapter 1
Classical and quantum mechanical
rotations in two and three dimensions
We start by considering classical rotations, and at the same time we reviewsome of the basics
of the underlying theory. For that reason we first define what we mean when we talk about
a rotation of a rigid body in space. First, a rigid body is defined as a system of mass points
subjected to the holonomic constraint that the distance between all pairs of points remain
constant throughout the motion. This of course is somewhat of an idealization, but in many
cases, and especially in this, the definition above is more than adequate - and very useful.
1.1
Independent coordinates of a rigid body
Intuitively we guess that a rigid body has six degrees of freedom, three external coordinates
to specify the position of the center of mass relative to a reference frame and three to specify
the orientation with respect to the external coordinates. In general, unless further constraints
than rigidity are imposed on the body, one needs six generalized coordinates qi to specify it’s
configuration. This result is independent of how many particles the body contains, even for a
continuous body.
Next we need to consider how these coordinates should be assigned. The configuration of
a rigid body is completely specified if we introduce a Cartesian set of coordinates fixed in the
body relative to the external coordinate axes. For an illustration see figure (1.1). Here we
clearly see that to specify it’s orientation in space relative to the external reference frame we
need six independent parameteres. If in addition, the two coordinate frames have a common
origin, then it will suffice with three to completely specify it’s configuration.
There are many ways to specify the orientation of a Cartesian set of axes relative to another
set of axes with common origin. One way is to specify the direction cosines for the primed
axes relative to the unprimed ones. Let {ei }, i ∈ (x, y, z) be set of unit vectors corresponding
to the xyz-axes and {e′i }, i ∈ (x′ , y ′ , z ′ ) be the set of unit vectors along the x′ y ′ z ′ -axes. Then
2
1.1 Independent coordinates of a rigid body
3
Figure 1.1: Unprimed axes represent the external reference frame, while the primed ones are
fixed in the body, taken from [8])
the direction cosines are defined as
cos θ11 = cos e′1 · e1 = e′1 · e1
cos θ12 =
cos θ31 =
cos e′1
cos e′3
· e2 =
· e1 =
e′1
e′3
(1.1)
· e2
· e1
and so on. For θij the first index refers to the primed system while the second one refers to
the unprimed system. There are a total of nine direction cosines which completely specify the
orientation of the x′ y ′ z ′ system relative to the xyz set. The unit vectors in the primed system
can be written with the aid of the direction cosines as
e′i
=
3
X
j=1
ej cos θij ≡ aij ej
(1.2)
where noe where we have defined aij = cos θij and the summation convention is used. This
means that any vector r can be written in the unprimed and primed system as
r = x′1 e′1 + x′2 e′2 + x′3 e′3 = x′i e′i = x′i aij ej = xj ej
= x1 e1 + x2 e2 + x3 e3
(1.3)
where x′i = ãij xj and ãij is the inverse matrix element of aij . Hence we see that the primed
coordinates are linear combinations of the unprimed ones. If primed axes are fixed in the
body, then the direction cosines are functions of time as the body changes it’s orientation.
There are nine direction cosines but we know that only three of them are independent. The
connection between them can be written compactly as
3
X
l=1
cos θlm′ cos θlm = δm′ m ⇔ alm′ alm = δm′ m
(1.4)
and arises because of the fact that the unit vectors in both the primed and unprimed coordinate
system are orthogonal to each other and have unit magnitude: ei · ej = δij = e′i · e′j .
1.2 Orthogonal transformations
1.2
4
Orthogonal transformations
We now move on to consider rotations in space. Equation (1.2) is an example of a linear or
vector transformation, and together with the constraint eq. (1.3) the transformation is called
an orthogonal transformation. We see that eq. (1.2) can be written in matrix form as


 ′  
a11 a12 a13
x1
x1
 x′2  =  a21 a22 a23   x2 
(1.5)
x′3
a31 a32 a33
x3
|
{z
}
R
where the 3 × 3 matrix R is called the matrix of transformation. We will simply call it the
rotation matrix. It can be viewed upon as an operator acting on the unprimed coordinates
and transforming them into the primed system through a rotation. The vector r itself remains
unchanged, we just specify it’s components in two different reference frames. Since a rotation
does not change the norm of a vector, any real rotational matrix must satisfy the orthogonality
relation
RT R = RRT = 1.
(1.6)
where 1 is the unit matrix. For a proper rotation, the determinant of the matrix R is equal
to +1. However, one can also have improper rotations, for which the determinant is -1.
It is important to notice that these matrices do not represent a physical rotation, since it
includes the inversion operation. This operation cannot be performed with a rigid change in
the coordinate axes and this means that we must disregard any orthogonal transformation
matrix whose determinant is -1.
1.2.1
Rotation in two dimensions
As a simple example of an orthogonal transformation, we restrict ourselves to two dimensions
i.e. motion in a plane. This means that we rotate through the x3 -axis, and therefore x3 = x′3 .
The transformation matrix reduces to


a11 a12 0
R =  a21 a22 0 
(1.7)
0
0 1
Here, the off-diagonal matrix elements correspond to a mixing of the x1 and x2 coordinates.
The four matrix elements are connected by three orthogonality conditions eq. (1.4) which
means that we only need one independent parameter to specify the transformation. After a
little algebra we obtain
a11 = a22 = cos θ
(1.8)
a12 = −a21 = sin θ
where θ is the angle between the x1 and x′1 axes. We will now consider the orthogonality
conditions. Written out explicitly we have
a211 + a221 = 1
a212
+ a222
=1
a11 a12 + a22 a21 = 0
(1.9)
1.2 Orthogonal transformations
5
which is just the wellknown trigonometric identities
cos2 θ + sin2 θ = 1
2
(1.10)
2
sin θ + cos θ = 1
cos θ sin θ − sin θ cos θ = 0
The matrix transformation corresponds to a counterclockwise rotation of the coordinate axes
(passive transformation) by an angle θ in the plane. There are two ways of viewing the rotation;
one interpretation is that R acts on the coordinate system itself, a passive transformation, or
it can act on the vector r itself, an active transformation. This distinction is purely formal as
the algebra remains the same no matter what view we adopt.
1.2.2
The Euler angles
In the three dimensional case, we shall employ the so called Euler angles to describe the rotation of a rigid body. These are three independent parameters that specify the orientation
of the body in such a way that the corresponding orthogonal matrix of transformation has
determinant equal to +1 (proper transf.). We start by carrying out the transformation of the
xyz-system into the x′ y ′ z ′ -system through three succesive rotations in a specific sequence. The
Euler angles are then defined as the rotation angles. First we rotate an angle α counterclock-
Figure 1.2: Definition of the Euler angles, [8]
wise about the z-axis. This corresponds to a two dimensional rotation which we discussed in
the previous section. The rotational matrix is


cos α sin α 0
C =  − sin α cos α 0 
(1.11)
0
0
1
It takes the xyz-coordinates and turn them into the ξηζ set, see the figure. Next we rotate
an angle β counterclockwise about the intermediate ξ axis. The set ξηζ goes to ξ ′ η ′ ζ ′ and the
1.2 Orthogonal transformations
matrix which corresponds to this transformation is


1
0
0
B =  0 cos β sin β 
0 − sin β cos β
6
(1.12)
Finally we rotate an angle γ counterclockwise about the ζ ′ axis. The set ξ ′ η ′ ζ ′ goes to x′ y ′ z ′
which is the fully rotated system. The rotation matrix A in this case has the same form as C:


cos γ sin γ 0
A =  − sin γ cos γ 0 
(1.13)
0
0
1
The full rotation matrix R is the product of the three individual matrices
R(α, β, γ) = A(γ)B(β)C(α)
(1.14)
and when we perform the matrix multiplication we obtain


cos γ cos α − cos β sin α sin γ
cos γ sin α + cos β cos α sin γ sin γ sin β
R =  − sin γ cos α − cos β sin α cos γ − sin γ sin α + cos β cos α cos γ cos γ sin β 
sin β sin α
− sin β cos α
cos β
We have R−1 = RT since it is orthogonal. The convention we have used is one of twelve
conventions used to define the Euler angles. The transformation matrix elements in general
depend on which order we perform the rotations, since they do not commute with each other.
The rotation matrices we obtain in this section are based purely on classical considerations
and in the quantum mechanical case of rotations we will in general obtain different matrices.
We also mention that in numerical calculations it is common to replace the Euler angles with
the so called Cayley-Klein parameters. This is because of the large number of trigonometric
functions involved in the computations.
1.2.3
Second rotation about the η axis
Since the rotation matrix R is dependent on which axis we rotate about, we will, instead of
performing a rotation about the intermediate ξ axis, rotate the system through the η axis.
This means that B → B ′ , which has different matrix elements. The main reason for doing
this is that when obtaining the quantum mechanical rotational matrix in three dimensions,
which also is a product of three rotational matrices, the second rotation is performed about
the intermediate η axis. One then obtains real matrix elements, since the y-component of the
angular momentum operator is purely imaginary. More on this later.
For a counterclockwise rotation about the y-axis the matrix of transformation is


cos β 0 − sin β

B′ =  0
1
0
sin β 0 cos β
(1.15)
It is trivial to calculate this so we drop the explicit calculation. Now, the new rotational
matrix is R′ = A(γ)B ′ (β)C(α) which we calculate to be


cos α cos β cos γ − sin α sin γ
sin α cos β cos γ + cos α sin γ − cos γ sin β
R′ =  − sin γ cos α cos β − sin α cos γ − sin α cos β sin γ + cos α cos γ sin γ sin β 
sin β cos α
sin β sin α
cos β
1.3 Rigid body Lagrangian
7
We see that as predicted R′ 6= R. We will use these expressions for the rotation matrices R, R′
to derive the Lagrangian of a rigid rotating body. It will be a function of the time-derivatives of
the Euler angels; α̇, β̇, γ̇ and the so-called moments of inertia. We will in the following sections
see how this is done and we also derive the equations of motion for an axially symmetric body
not subjected to any constraints.
1.3
Rigid body Lagrangian
As we mentioned above we will in this section make use of the Euler angles to describe rigid
body motion. The rotation angles α, β, γ are the three generalized coordinates needed to
completely specify it’s orientation in space, and we now wish to find a Lagrangian scalar
function expressed through these. Later on we will use the Lagrangian to find the equations
of motion, but first we need to consider the relation between an arbitrary vector in the rigid
body system and a vector in an inertial frame as the body rotates and the vector changes.
1.3.1
Rate of change
In this short section we will follow the derivation made in [8]. First, consider some arbitrary
vector r pointing in the direction of a point in the body. As the body moves the vector will
change in time, and the change is dependent on which reference frame we observe from. Assume that this vector is the radius vector from the body origin. The change of the components
of the vector r in an infinitesimal time dt in the space coordinate system and the change of
the same vector in the rigid body system are related through
(dr)s = (dr)b + (dr)rotation
(1.16)
where the subscript s stands for the space coordinate system, and b is the body coordinate
system. As the body rotates the change in the vector r as seen from the s system is given by
(dr)rotation = dΩ × r
(1.17)
where (dΩ)s is the angle of rotation. If we divide by the differential time element dt we obtain
the time rate of change:
dr
dr
=
+ω×r⇔v =V+ω×r
(1.18)
dt s
dt b
where ω is the angular velocity of the body defined through ω dt = dΩ and lies along the
axis of rotation. Here V is the center of mass velocity of the body, and v is the velocity in
the space coordinate system.
1.3.2
Kinetic energy for rotation about a point
The total kinetic energy of a rigid body can be separated into two contributions - one term
from the translation of the center of mass and the other from the rotation about it:
T =
1
M (ẋ2 + ẏ 2 + ż 2 ) + TR (α, β, γ)
2
(1.19)
where M is the total mass and ˙ = d/dt. Often the potential energy of the system can be
divided in a similar way. We will now calculate the kinetic energy in a coordinate frame K ′
1.3 Rigid body Lagrangian
8
fixed in the body’s center of mass. The definition of the center of mass system implies that if
the body we are considering is made up of N particles, then
N
X
(1.20)
mn rn = 0
n=1
in the discrete case. The kinetic energy becomes
T
=
=
N
N
1X
1X
2
mn vn =
mn (V + ω × rn )2
2
2
n=1
n=1
!
N
N
N
X
1X
1 X
mn (ω × rn ) +
mn (ω × rn )2
mn V 2 + V ·
2
2
n=1
n=1
n=1
where we have used (1.18). The second term in this expression can be rearranged using cyclic
permutation:
N
N
X
X
mn rn = 0
(1.21)
mn (ω × rn ) = (V × ω) ·
V·
n=1
n=1
when we use the definition of the center of mass system eq. (1.20). The third term involves
the square of the vector ω × rn . We can write it in the following way (omitting for a second
the particle index for clearer notation);
(ω × r)2 = ω 2 r 2 sin2 χ = ω 2 r 2 (1 − cos2 χ) = ω 2 r 2 − (ω · r)2
3
3 X
X
ω j r 2 δjk − xj xk ω k
=
(1.22)
j=1 k=1
If the insert these expressions into the formula for the kinetic energy we obtain
3
3
1
1 XX j
1
1
T = M V2 +
ω Ijk ω k = M V2 + ω T · I · ω
2
2
2
2
(1.23)
j=1 k=1
The last term on the right is called a dyad-product, and is really a multiplication of matrices,
which in the end results in a scalar. We have also defined the quantity
Ijk ≡
N
X
n=1
mn rn2 δjk − xnj xnk
(1.24)
which we recognize as the inertia tensor. Here we again use the particle index explicitly. The
kinetic energy of a rigid body is separated into two terms, one that contains the center of mass
velocity squared and a term which is purely rotational. The second term involves the inertia
tensor, which in a general situation can have a complex time dependence. Let us see if we
can derive a relation between the angular momentum vector L and this tensor. The angular
momentum with respect to the center of mass of a rigid body is
L =
=
N
X
n=1
N
X
n=1
mn rn × vn =
N
X
n=1
mn rn × (ω × rn )
mn ωrn2 − rn (rn · ω)
(1.25)
(1.26)
1.3 Rigid body Lagrangian
9
using a standard vector identity. Now, if we write out the components of L we find that it
can be written in matrix form as

PN
PN


  PN
2 − x2 ) −
m
(r
m
x
y
−
m
x
z
ωx
Lx
n n
n n n
n n n
n
n=1
n=1
n=1
P
PN
PN

2
2
 Ly  = 
ωy 
− N

 − n=1 mn xn yn
n=1 mn yn zn
n=1 mn (rn − yn ) P
P
PN
N
N
2
2
ωz
Lz
− n=1 mn yn zn
− n=1 mn xn zn
n=1 mn (rn − zn )
One can easily see that the matrix element (...)jk can be written exactly as (1.24) which is
the inertia tensor. Thus the angular momentum vector can be expressed as L = Iω (matrix
product) and the final expression for the kinetic energy in a coordinate system fixed in the
body is
1
1
T = M V2 + ω · L
(1.27)
2
2
where V is the center of mass velocity relative a fixed coordinate system in space.
In the limit of a continuous body one can obtain analogous results making the substitution
N
X
n=1
mn →
Ijk
M
d3 r ρ(r)
R
d3 r r ρ(r) = 0 for the center of mass are:


Z
Z
3
3 X
X
1 2
1
=
V
ω j r 2 δjk − xj xk ω k 
d3 r ρ(r) +
d3 r ρ(r) 
2
2
j=1 k=1
Z
≡
d3 r ρ(r) r 2 δjk − xj xk
Z
=
d3 r ρ(r)
The results we obtain using the condition
T
Z
(1.28)
which means that the kinetic energy for a continuous body is T = 21 M V2 + 12 ω T · I · ω even
in this case. If we let n be a unit vector in the direction of ω then we can write ω = ωn, and
we obtain a different form for the kinetic energy:
1
1
n · I · n = I ω2
2
2
N
X
mn rn2 − (rn · n)2
I = n·I·n =
T
=
(1.29)
n=1
The last quantity is called the moment of inertia about the axis of rotation. It depends upon
the direction of the axis of rotation and is in general a function of time, except when the body
is constrained to rotate about a fixed axis, then the moment of inertia is constant. To obtain
an especially simple expression for the kinetic energy one can choose to specify a coordinate
system in which the inertia tensor I is diagonal, and we will discuss how to do this in the
following section.
1.3 Rigid body Lagrangian
10
Figure 1.3: Visualisation of the axis of rotation, courtesy of [8]
1.3.3
Diagonalization of the Inertia tensor
From the definition of the inertia tensor we see that the off diagonal components are symmetric,
that is Ijk = Ikj . This reduces the number of independent coordinates from nine to six (three
diagonal, and three off-diagonal elements). The components will depend on the location of
the body’s coordinate system and the relative orientation with respect to these axes. This
means that it is possible to find a coordinate system in which the inertia tensor is diagonal. A
proof of this fact exists, but will not be written down here. In this special coordinate system
the matrix elements of the inertia tensor can be written as Iij δij . The angular momentum
components are then
3
X
Iij ωi δij = Ijj ωj .
Lj =
i=1
The mixing of the angular velocity vector components disappear and we obtain a particularly
simple expression for the energy.
Consider the moment of inertia of a body passing through it’s center of mass. To make
the inertia tensor diagonal, we perform a similiarity transformation using the rotation matrix
so that ID = R I RT . In this system, the inertia tensor is diagonal if we chose α, β, γ with
care. The coordinate axes in this system are called the principal axes. To find it, let us return
to the rigid body coordinate axes, and re-label them as x,y, and z. We label this coordinate
system K. In K the inertia tensor is in general not diagonal. If we rotate K using the rotation
matrix R(α, β, γ) derived in section 1.2.2, the rotated system K ′ will have coordinate axes
x′ , y ′ , z ′ . Then in the new system the angular velocity vector is
ω ′ = R ω ⇒ ω = RT ω ′
(1.30)
since R−1 = RT (orthogonality). We find for the angular momentum in K ′
L′ = R L = R I ω = R IRT ω ′ = ID ω ′
(1.31)
which means that in K ′ the inertia tensor is diagonal. To find the principal axes in K ′ assume
that n1 , n2 and n3 are the unit vectors written as column vectors, along these axes. Then the
1.3 Rigid body Lagrangian
11
matrix product Ini must yield the same vector ni times a constant:
Ini = λi ni ,
i ∈ (1, 2, 3)
(1.32)
which is an eigenvalue equation with the eigenvalues λi . This set of equations has a solution
provided that the determinant vanishes:
I11 − λ
I
I
12
31
I12
I22 − λ
I23 = 0
(1.33)
I31
I23
I33 − λ Here we display the symmetry of I explicitly. This secular equation has three roots λ ∈ (I1 ,
I2 , I3 ) which are the principal moments we seek. Any kind of symmetry in the rigid body (e.g.
symmetri around the z-axis) leads to a degeneration in the eigenvalues which makes it easier to
solve the equations of motion. The eigenvectors we obtain from the different eigenvalues point
out the principal axes which makes the inertia tensor diagonal. When we obtain these, we
use them to redefine the body’s coordinate system and calculate all the important quantities
there. We then transform them back to the original inertial reference frame using the inverse
rotation matrix.
1.3.4
Angular velocity vector expressed through the Euler angles
In the previous section we’ve outlined a method for simplifying the problem of finding a
Lagrangian scalar function as a function of the Euler angles. We chose a coordinate system
in which the moment of inertia tensor is diagonal by performing a similarity transformation
using the rotation matrix and it’s transpose. This matrix can be expressed through the Euler
angles α, β and γ. The kinetic energy in the transformed coordinate system is
3
3
X X
1
1
1
1
1
Iij ωi = M V2 + I1 ω12 + I2 ω22 + I3 ω32
ωj
T = M V2 +
2
2
2
2
2
j=1
(1.34)
i=1
Now, we have to express the angular velocity vector ω through the Euler angles and their time
derivatives. A general angular velocity vector is a sum over three separate angular velocity
vectors ω α , ω β and ω γ which are not symmetrically placed.
The x-convention
In the case of the rotations performed in section 1.2.2, which we will now call the x-convention
since the second rotation is performed about the intermediate x-axis (ξ). The vector ω α is
directed along the space coordinate frame’s z-axis, ω β is along the line of nodes while ω γ
is directed along the body’s z ′ -axis. If we wish to find the total vector ω in the body’s
coordinate system K ′ then we need to transform the components of the vectors using the
three orthogonal transformation matrices A, B and C defined in section 1.2.2. For the first
rotation counterclockwise around the space coordinate system K’s z-axis, we have to use the
full rotation matrix R:
 ′ 




ωx
α̇ sin β sin γ
0
 ωy′  = R ·  0  =  α̇ sin β cos γ 
(1.35)
α̇
cos
β
ω
ωz′
z
α
| {z α}
| {z α}
in K′
in K
1.3 Rigid body Lagrangian
12
where we have used (ωz )α = α̇. R is the rotation matrix. The subscript α indicates that these
are the α-components of the angular velocity vector. Now, for the second rotation around the
line of nodes in the ξ ′ η ′ ζ ′ (P ) system we just use the transformation matrix A(γ) to find the
angular velocity vector components in the K ′ system:




 ′ 
ωx
β̇ cos γ
β̇
 ωy′  = A ·  0  =  −β̇ sin γ 
(1.36)
′
ωz β
0 β
0
β
| {z }
| {z }
in K′
in P
Finally, no transformatin is needed for the components of ω γ since it is already directed along
the z ′ -axis in the K ′ system:
 ′ 
 
ωx
0
 ωy′  =  0 
(1.37)
′
γ̇ γ
ωz γ
The total angular velocity vector in the K ′ system (ω ′ ) is the sum over all rotations which we
find to be


α̇ sin β sin γ + β̇ cos γ
ω′ =  α̇ sin β cos γ − β̇ sin γ 
(1.38)
α̇ cos β + γ̇
where the unit vectors are along the body set of axes. This method allows us to find the
components of the angular velocity vector in any coordinate frame. We can for instance
find the components of this vector in the inertial frame K using the inverse transformation
ω = RT ω ′ . A rather lengthy calculation gives


β̇ cos α + γ̇ sin α sin β
(1.39)
ω =  β̇ sin α − γ̇ cos α sin β 
α̇ + γ̇ cos β
where the unit vectors are the usual Cartesian unit vectors. We see that we have a large
degree of symmetry here. Making the substitutions α → γ and γ → α nearly transforms the
vectors from K to K ′ and vice versa, the only thing standing in the way from a complete
transformation is an overall minus sign in the second component. This reveals a certain
symmetry in the transformation matrix R which is very interesting.
The y-convention
We now turn our attention to the convention used in section 1.2.3, which we label as the
y-convention, since the second rotation is performed through the intermediate y-axis (η). The
calculation of the angular velocity vector is using this convention slightly different. We will be
employing the rotation matrix R′ when transforming the different quantities from one system
to another. Now, the first rotation around the space coordinate system z-axis, K, transforms
to K ′ as:




 ′ 
ωx
−α̇ sin β cos γ
0
 ωy′  = R′ ·  0  =  α̇ sin β sin γ 
(1.40)
α̇
cos
β
α̇
ωz′
α
| {z α}
| {z α}
in K′
in K
1.4 Lagrangian expressed through the Euler angles
13
Now, in the y-convention the angular velocity vector ω β is directed along the intermediate
y-axis, so it transforms as




 ′ 
0
ωx
β̇ sin γ
 ωy′  = A ·  β̇  =  −β̇ cos γ 
(1.41)
′
ωz β
0 β
0
β
| {z }
| {z }
in K′
in P
where P is the intermediate coordinate system. The third rotation is identical to the one in
the x-convention so
 ′ 
 
ωx
0
 ωy′  =  0 
(1.42)
′
γ̇ γ
ωz γ
as before. Once again the total angular velocity vector in the K ′ system is the sum of the
three vectors. Performing the summation one obtains


−α̇ sin β cos γ + β̇ sin γ
ω ′ =  α̇ sin β sin γ + β̇ cos γ 
(1.43)
α̇ cos β + γ̇
where the unit vectors are directed along the body principal axes. The connection between
eqs. (1.38) and (1.43) is revealed if one makes the substitutions sin γ → − cos γ, cos γ → sin γ
in eq. (1.43). This corresponds to the transformation γ → γ + π/2 when we change from
the y- to the x-convention. The transcription from the x- to the y-convention is particularly
simple because β retains its meaning in both conventions and the changes for the other angles
are easily obtained. As in the previous section we calculate the angular velocity vector along
the space axes with the result


−β̇ sin α + γ̇ sin β cos α
ω =  β̇ sin α + γ̇ sin β cos α 
(1.44)
α̇ + γ̇ cos β
The results obtained in this section will be useful later in this thesis as we wish to make
comparisons between the classical and quantum mechanical case of rigid body rotation.
1.4
Lagrangian expressed through the Euler angles
We’ve developed all the tools necessary to completely describe the rigid body, and we now
write down a Lagrangian scalar function of the center of mass coordinates and the rotation
angles in the principal axes system K ′ , since the inertia tensor is diagonal in this system. For
the rotation convention used in section 1.2.2 we have
2 1 2
1 1
M V2 + I1 α̇ sin β sin γ + β̇ cos γ + I2 α̇ sin β cos γ − β̇ sin γ
L =
2
2
2
1
I3 (α̇ cos β + γ̇)2 − V (α, β, γ)
(1.45)
+
2
where V is some potential derivable from the forces acting on the body. In general it will only
be a function of the position coordinates or the Euler angles, not their derivatives.
1.5 Equations of motion
1.5
14
Equations of motion
Now that we have written the general Lagrangian down, we can use it on holonomic conservative systems to find the equations of motion. We work in the K ′ system where equation
(1.45) is valid. In general the generalized coordinates are the Euler angles, but if the motion
is confined effectively to two dimensions, then the axis of rotation is fixed and we need only
one angle of rotation to specify the system. This was demonstrated in section 1.2.1. To derive
the equations of motion in the general case we use the following form of Lagrange’s equations:
∂T
d ∂T
= Qi , i = 1, 2, 3 . . . D
(1.46)
−
dt ∂ q̇i
∂qi
where D is the number of degrees of freedom. This form is valid for conservative systems
where the forces are derived from a scalar potential function: Fi = −∇i V . In this equation,
the generalized force Qi = −∂V /∂qi . If V is not an explicit function of q̇i then
∂L
d ∂L
=0
−
dt ∂ q̇i
∂qi
which is the ordinary form of Lagrange’s equations. For our rigid body the generalized forces
are the torques τi corresponding to the rotation angles. If one applies eq. (1.46) to the kinetic
energy term in the Lagrangian then one obtains the equation for the i′ th component of the
angular velocity vector:
Ii ω̇i + ǫijk ωj ωk Ik = τi
(1.47)
where we have suppressed the primes. Only one of the Euler angles has it’s torque along
one of the body principal axes, namely the γ coordinate. The remaining Euler equations are
obtained by cyclic permutation.
1.5.1
Axial symmetric body not subjected to any foreces
In this section we supress the prime notation, but it is important to remember that all quantities in this section refer to the K ′ system, and to obtain the corresponding quantities in the
K inertial system one uses the inverse transformation matrix RT .
As a concrete example, let us consider a rigid body not subjected to any net forces or torques.
According to Newton’s first law the center of mass is either at rest or moving uniformly and we
will therefore consider rotational motion in a reference frame where it is at rest. The angular
momentum then solely arises from the rotation about the center of mass, and in the abscense
of torques we have
I1 ω̇1 + (I3 − I2 ) ω2 ω3 = 0
I2 ω̇2 + (I1 − I3 ) ω1 ω3 = 0
I3 ω̇3 + (I2 − I1 ) ω1 ω2 = 0
(1.48)
These equations are normally referred to as the Euler equations of motion. Since there are no
net forces acting on the system the kinetic energy and the angular momentum are conserved
quantities. We use a coordinate frame oriented along the body’s principal axes and let the
angular velocity vector ω components be measured along the instantaneous axis of rotation.
1.6 Rotations in quantum mechanics
15
We will now demonstrate that if the body is axial symmetric with respect to some axis, then
ω will precess about this axis when we consider force free motion. Let the symmetry axis be
the z ′ -axis in the K ′ system. This means that I1 = I2 < I3 . Eqs. (1.48) reduce to
I1 ω̇1 + (I3 − I1 ) ω2 ω3 = 0
I1 ω̇2 − (I3 − I1 ) ω1 ω3 = 0
I3 ω̇3 = 0
which means that ω3 = constant. We can rewrite the two first equations as
I1 − I3
ω3 ω2
ω̇1 =
I1
I3 − I1
ω3 ω1
ω̇2 =
I1
Now define Ω =
I3 −I1
I1
(1.49)
(1.50)
· ω3 = constant. This gives us
ω̇1 = −Ω ω2
(1.51)
ω̇2 = Ω ω1
This set of equations have solution ω1 = A cos Ωt and ω2 = A sin Ωt. The vector ω1 e1 +ω2 e2 =
A (cos Ωt e1 + sin Ωt e2 ) has constant magnitude A and rotates uniformly counterclockwise
about the z ′ -axis with angular frequency Ω. The total angular velocity vector ω = ω1 e1 +
ω2 e2 + ω3 e3 precesses with time about the same axis with the same frequency. The closer I3
is to I1 the slower the precession frequency will be.
1
· ω 3 e3
Another way of demonstrating the precession of ω is to define a vector Ω = I3I−I
3
and notice that eqs. (1.49) are equivalent to ω̇ = ω × Ω, which demonstrates that the vector
ω precesses with frequency Ω about the e3 -axis.
1.6
Rotations in quantum mechanics
Having exhausted the classical case, we now turn our attention to the theory of rotations in
quantum mechanics. As we have seen in the previous sections, rotation of a system brings
in it’s angular momentum. Although we haven’t mentioned it the rotational invariance of
the equations of motion is reflected in the fact that the angular momentum is conserved. In
quantum mechanics this translates into the commutation relation [H, L] = 0 where H is the
Hamiltonian operator of the system and L its angular momentum operator.
Differences between classical mechanics arises because the angular momentum is a vector
operator and not an ordinary vector. The components of this vector don’t commute but satisfy
the wellknown relation
[Li , Lj ] = ih̄ ǫijk Lk
(1.52)
where ǫijk is the ordinary Levi-Civita antisymmetric tensor defined by ǫ123 = +1. Our goal is
to establish the connection between rotations and angular momentum, and to derive rotational
(l)
matrices (which we call Dmk (α, β, γ)) as in the classical case. In order to do this though, we
need to establish the basic theory concerning angular momentum in quantum mechanics.
1.7 Algebraic approach to the angular momentum eigenvalue problem
1.7
16
Algebraic approach to the angular momentum eigenvalue
problem
The orbital angular momentum is a physical quantity of great importance, which justifies a
derivation of the properties of L using some basic commutation relations. In the classical case,
the angular momentum is defined for a single particle as L = r × p. Quantization means that
p → −ih̄∇ so in wave mechanics we have
L ≡ −ih̄r × ∇
which is a vector operator, satisfying commutation relations eq. (1.52). Generalization to a
system on N particles is straightforward. The same relations hold but this time
L=
N
X
Li
i=1
is the sum of all individual angular momentum vectors. We therefore adopt the following
definition of an angular momentum operator:
A vector operator J is an angular momentum operator if it’s components are
observables satisfying commuation relations [Ji , Jj ] = ih̄ǫijk Jk .
We have replaced L by J to emphasize that the eigenvalue problem, which will be solved
in this section by the algebraic method, has the posibility to represent a larger class of physical situations than orbital angular momentum.
The square of the angular momentum J2 = Jx 2 + Jy 2 + Jz 2 commutes with Jx , Jy and Jz .
This is a consequence of commutation relations (1.52). This is written as
[J, J2 ] = 0
(1.53)
We define also for future convenience the two Hermitean conjugate operators
J+ = Jx + iJy
(1.54)
J− = Jx − iJy
(1.55)
The three operators J+ , J− and Jz completely define J and it is more convenient to use them
for algebraic manipulations. The following relations are easily deduced:
[Jz , J+ ]
= h̄J+
(1.56)
[Jz , J− ]
= −h̄J−
(1.57)
[J+ , J− ] = 2h̄Jz
(1.58)
With the aid of eq. (1.53) we further deduce that
[J2 , J+ ] = [J2 , J− ] = [J2 , Jz ] = 0
So J2 is given by the expression
J2 =
1
1
(J+ J− + J− J+ ) + Jz 2 = {J+ , J− } + Jz 2
2
2
(1.59)
1.7 Algebraic approach to the angular momentum eigenvalue problem
17
where {A, B} ≡ AB + BA is the anticommutator between the operators A and B. Let us
now consider the eigenvalue problem. From theory we know that we cannot simultaneously
assign definite values for all angular momentum components. From (1.53) we can obtain
simultaneous eigenvectors for say J2 and Jz . We denote the eigenvalues of Jz by mh̄ and
those of J2 by λh̄2 , so the eigenvalue problem can be stated as
Jz |λ, mi = mh̄|λ, mi
2
2
J |λ, mi = λh̄ |λ, mi
(1.60)
(1.61)
where |λ, mi are the eigenkets or eigenvectors. Next we develop a ladder procedure by acting
on eq. (1.60) with J+ and J− we obtain with the help of eqs. (1.56), (1.57) and (1.58):
Jz J+ |λ, mi = (m + 1)h̄J+ |λ, mi
(1.62)
Jz J− |λ, mi = (m − 1)h̄J− |λ, mi
J2 J± |λ, mi = λh̄2 J± |λ, mi
Hence, if |λ, mi is an eigenket of Jz and J2 with eigenvalues mh̄ and λh̄2 , then J± |λ, mi is also
an eigenket of the same operator and has eigenvalues (m ± 1)h̄ and λh̄2 respectively. From
this fact we infer that we must have
J± |λ, mi = A± · |λ, m ± 1i
(1.63)
where A± are complex numbers which we shall determine later. To see this note that if
Jz |λ, mi = mh̄|λ, mi and J+ |λ, mi = const. · |λ, m + 1i we obtain
beqJz J+ |λ, mi = Jz const. · |λ, m + 1i = const. · (m + 1)h̄|λ, m + 1i = (m + 1)h̄J+ |λ, mi
The case of Jz J− follows an identical argument. We will state without proof (this proof can be
found in most introductory textbooks on the subject) that the eigenvalues λ and m belonging
to the same eigenket satisfy the inequality
λ ≥ m2
(1.64)
which limits the magnitude of m. Therefore it must exist a maximum value mmax = j
for any λ. To take the argument one step further we note that if we act on the eigenket
|λ, mmax i = |λji with the raising operator J+ we will obtain 0! Therefore:
(1.65)
J− J+ |λ, ji = J2 − Jz 2 − h̄Jz |λ, ji = λ − j 2 − j h̄2 |λ, ji = 0
which translates into λ = j(j + 1), and we will from now on use the parameter j instead of
λ to specify a state. For similar reasons there must be a minimum value mmin = p. This
condition leads to λ = p(p − 1) so to have consistency one must have
j(j + 1) = p(p − 1) ⇔ p = −j or j + 1
(1.66)
The second solution is meaningless because it violates the assumptions we’ve made. Hence
we must have p = −j. The eigenvalues of Jz are limited both upward and downward so by
repeated application of the lowering operator one can for a given j move from the state |j, ji
to the state |j, −ji. In each step m is decreased by unity. We then have −j ≤ m ≤ j. This
1.8 Matrix representation of Jx , Jy and Jz
18
means that for a given j we have 2j + 1 ortogonal states or eigenkets. Also j − p = 2j must
be a nonnegative integer which implies that j is a nonnegative half-integer or integer, thus
(1.67)
j = 0, 1/2, 1, 3/2, 2 . . .
So for every j it is possible to construct a (2j + 1)-dimensional vector space which is closed
under algebra of the operators Jx , Jy and Jz and the commutations relations between them.
This is the foundation for the irreducible representation of the rotation group which is an idea
we will return to later in this chapter.
To obtain the constants A± we begin by noting that from eq. (1.59) and eqs. (1.54), (1.55)
one obtains
J2 − Jz 2 = Jx 2 + Jy 2
(1.68)
2
2
2
2
J± J∓ = (Jx ± iJy )(Jx ∓ iJy ) = Jx + Jy ± i[Jy , Jx ] = Jx + Jy ± h̄Jz
which can be combined to the identity
J2 − Jz 2 ± h̄Jz = J± J∓
(1.69)
To derive expressions for the coefficients we note that hj, m|J− = hj, m + 1|A∗+ h̄ so
hj, m|J− J+ |j, mi = |A+ |2 h̄2 hj, m + 1|j, m + 1i
(1.70)
If we assume that all eigenkets are normalized to unity then if we use our newly derived
identity we obtain
hj, m|J− J+ |j, mi = hj, m|J2 − Jz 2 − h̄Jz |j, mi = j(j + 1) − m2 − m h̄2 hj, m|j, mi (1.71)
Comparing the two expressions reveals that
|A+ |2 = j(j + 1) − m2 − m = j(j + 1) − m(m + 1)
(1.72)
The phase of A+ is not determined, and we choose it to be equal to zero. Using A− (j, m) =
A∗+ (j, m − 1) one can finally write down the final expressions for the raising and lowering
operators acting on eigenkets
p
(1.73)
J+ |j, mi = h̄ j(j + 1) − m(m + 1)|j, m + 1i
p
(1.74)
J− |j, mi = h̄ j(j + 1) − m(m − 1)|j, m − 1i
1.8
Matrix representation of Jx , Jy and Jz
The expressions for the raising and lowering operators can be used to derive matrices representing the angular momentum components in a basis that consist of the common eigenkets of
Jz and J2 . The matrix elements are (Ji )mn = hm|Ji |ni since we always have the same j.
We begin with Jz which satisfies Jz |j, mi = h̄m|j, mi, which means that
(Jz )mn = hm|Jz |ni = h̄mδmn
1.8 Matrix representation of Jx , Jy and Jz
19
since the eigenkets satisfy hj, m|j ′ , ni = δj,j ′ δmn . The operators Jx and Jy are harder to
determine since we don’t know how they transform the kets. But we can express them as
Jx =
Jy =
1
(J+ + J− )
2
1
(J+ − J− )
2i
(1.75)
(1.76)
and use eqs. (1.73) and (1.74). This helps us to express them as
p
h̄ p
(Jx )mn = hm|Jx |ni =
j(j + 1) − n(n + 1)δm,n+1 + j(j + 1) − n(n − 1)δm,n−1
2
(1.77)
and the calculation of the matrix elements of Jy results in
p
−ih̄ p
j(j + 1) − n(n + 1)δm,n+1 − j(j + 1) − n(n − 1)δm,n−1
(Jy )mn = hm|Jy |ni =
2
(1.78)
The matrices all have dimension (2j +1) for any given j and we note that since the operator Jz
is an eigen-operator of any ket in this (2j + 1)-dimensional subspace it’s matrix representation
is diagonal. The matrices representing Jx and Jy both have off-diagonal contributions. The
matrix elements of Jx are real, while those of Jy are purely imaginary.
1.8.1
The case j = 1/2
In this case the quantum number m ∈ [−1/2, 1/2] in steps of one. The matrix representation
of the operators are
h̄ 0 −i
h̄ 1 0
h̄ 0 1
, Jy =
, Jz =
(1.79)
Jx =
1 0
i 0
0 1
2
2
2
From this expression one defines the famous Pauli σ-matrices:
0 1
0 −i
1 0
σx =
, σy =
, σz =
1 0
i 0
0 −1
(1.80)
We will state some of the basic relations between them without proof. All matrices are unitary
and
σx 2 = σy 2 = σz 2 = 1.
(1.81)
The basic commutation and anti-commutation relations are
[σi , σj ] = 2i ǫijk σk ,
{σi , σj } = 2 δij
(1.82)
The traces also vanish for each matrix which is easy to see from the definition. It is possible
to form from the σ-matrices and the unit matrix a basis for all 2×2 matrices. Then it follows
that if U is a unitary 2×2 matrix it can always be expressed as
U = eiγ (1 · cos ω + in · σ sin ω)
(1.83)
where γ, ω are real angles and n is a real unit vector. The following identity is sometimes
useful. For two arbitrary vector A and B one can show that
(σ · A)(σ · B) = A · B + iσ · (A × B)
In quantum mechanics the Pauli matrices describe spin-1/2 fermions.
(1.84)
1.9 Orbital angular momentum and the spherical harmonics
1.8.2
20
The case j = 1
In this case m ∈ [−1, 0, 1], so we obtain 3×3 matrices. They are easily calculated to be






0 1 0
0 −i 0
1 0 0
h̄
h̄
(1.85)
Jx = √  1 0 1  , Jy = √  i 0 −i  , Jz = h̄  0 0 0 
2
2
0 1 0
0 0 −1
0 i
0
These matrices act on states describing spin-1 particles which are bosons. These matrices are
not widely used so we will content ourselves with calculating them.
We have solved the angular momentum eigenvalue problem algebraicly in Hilbert space but
there are still some important questions which remain unanswered. In our case we would like
to know which of the integral and half-integral values that make up the spectrum of j. So
far we know that j = 0, ±1, ±2, . . . A priori it will depend on the problem considered. Of
course, as we will show in the following section, in the case of the angular momentum of a
single particle the spectrum of j will consist of all integers from 0 to ∞, all half-integral values
being excluded.
1.9
Orbital angular momentum and the spherical harmonics
In this section we consider the angular momentum L of a single particle. For this purpose
we work in coordinate space using spherical coordinates. We chose the z-axis as the polar
direction. Since the angular momentum operators L and L2 are functions of the polar angles
and their derivatives we ignore the radial coordinate. In coordinate space the eigenvalue
problem is stated as
Lz Ylm (θ, φ) = h̄m Ylm (θ, φ)
2
2
L Ylm (θ, φ) = h̄ l(l + 1) Ylm (θ, φ)
(1.86)
(1.87)
The function Ylm (θ, φ) can be separated into two parts Ylm (θ, φ) = Θ(θ)Φ(φ). When we insert
this and Lz = −ih̄∂/∂φ into the above equations we end up with a simple equation for Φ(φ):
h̄ ∂Φ
= mh̄Φ(φ)
i ∂φ
which has solution Φ(φ) = eimφ . Since the wave function must be single valued at every point
in space we must impose the restriction that Φ(φ) = Φ(φ + 2π) which has the consequence
that m = 0, ±1, ±2, . . . i.e an integer. Since m is integral so is j = l ⇒ there is no half-integral
orbital angular momentum.
We’ve answered the question we posed earlier. Now we again wish to determine the
eigenvalues of L and their degeneracy. In order to do this we need an expression for L2 in
coordinate space. We will write it down here as
1 ∂
∂
1 ∂2
2
2
+
sin θ
(1.88)
L = −h̄
sin 2 θ ∂φ2 sin θ ∂θ
∂θ
1.9 Orbital angular momentum and the spherical harmonics
21
For future convenience we also write down the expressions for the raising and lowering operators of angular momentum as differential operators in coordinate space. They are
∂
∂
iφ
+ i cot θ
(1.89)
L+ = h̄e
∂θ
∂φ
∂
∂
L− = −h̄e−iφ
− i cot θ
(1.90)
∂θ
∂φ
Let us also relate the angular momentum operator to the Laplacian, which represents the
kinetic energy term. One approach is to write the angular momentum components in the
form
∂
Lk = −ih̄ǫijk xi
∂xj
Formulate the simple identity ǫijk ǫpqk = δip δjq − δiq δjp . Then it follows that (using the
summation convention)
∂
∂
∂
∂
∂
∂
2
2
2
L = Lk Lk = −h̄ ǫijk ǫpqk xi
xp
= −h̄ xi
xi
− xi
xj
∂xj ∂xq
∂xj ∂xj
∂xj ∂xj
∂xj
∂2
∂2
∂
∂
2
+ xi xi
−
− xi xj
xi
= −h̄ xi
∂xi
∂xj ∂xj
∂xj
∂xi
∂xi ∂xj
2 ∂2
∂
2
2 2
2 ∂
2
2 2
= −h̄ r ∇ − 2r · ∇ − xi xj
−r
= −h̄ r ∇ − 2r
∂xi ∂xj
∂r
∂r 2
which can be rearranged to give
∂
L2
= −r 2 ∇2 +
2
h̄
∂r
∂
r
∂r
2
(1.91)
Now if we insert expression (1.88) into the eigenvalue equation for L2 we obtain a differential
equation for Θ
1 d
d
m2
+ l(l + 1) Θ(θ) = 0
(1.92)
sin θ
−
sin θ dθ
dθ
sin 2 θ
The physically allowed solutions of this equation are the wellknown Legendre functions with
m ≤ l:
(−1)m (l + m)!
dl−m
Plm (ξ) = l
(1 − ξ 2 )−m/2 l−m (ξ 2 − 1)l
(1.93)
2 · l! (l − m)!
dξ
with ξ = cos θ. This function is proportional to the Legendre polynomial which is a power
series which terminate at l(l + 1) for any given l which must be a nonnegative integer. The
associated Legendre functions are orthogonal in the sense that
Z 1
(1.94)
Plm (ξ) Pl′ m (ξ) dξ = δl,l′
−1
Note that the two subscripts are the same. In general Legendre functions with different values
of m are not orthogonal.
1.9 Orbital angular momentum and the spherical harmonics
22
Spherical harmonics
One usually defines the sperical harmonics as the simultaneous solution of the eigenvalue
problems for L and L2 . This means that when we move between Hilbert and coordinate
spaces we have
hθ, φ|l, mi ↔ Ylm (θ, φ)
(1.95)
The functions are normalized with respect to integration over a total solid angle
s
2l + 1 (l − m)! imφ
e
Plm (cos θ)
Ylm (θ, φ) =
4π (l + m)!
(1.96)
where Plm (cos θ) is the Legendre polynomial. They have parity (−1)l i.e. definite parity,
Table 1.1: The first few normalized Spherical Harmonics
l
m
Ylm (θ, φ)
0
0
√1
4π
1
±1
∓
q
q
0
±2
2
±1
0
q
q
3
8π
15
32π
e±iφ sin θ
3
4π
cos θ
e±2iφ sin 2 θ
15 ±iφ
∓ 8π
e
sin θ cos θ
q
5
2
16π (3 cos θ − 1)
which is required since L commutes with the parity operator. The spherical harmonics form
a complete set and satisfy the orthonormality conditions
Z
(∗)
(1.97)
Ylm (θ, φ) Yl′ m′ (θ, φ) dΩ = δl,l′ δm,m′
and
X
(∗)
sin θ Ylm (θ, φ) Yl′ m′ (θ, φ) = δl,l′ δm,m′
(1.98)
l,m
Since the spherical harmonics form a complete and orthonormal set of functions, any function
of θ, φ can be expanded in terms of the spherical harmonics. For example: one can connect
l (R) to the spherical harmonics by noting that the application of the
the rotation matrices Dmk
rotation operator R to an eigenvector of Lz transforms it to a new eigenvector of L′z (where
the new z‘-axis is obtained by rotating the z-axis):
X
l
′
R|l, mi =
Dm
(1.99)
′ m (R)|l, m i
m′
1.10 Representation of the rotation operator
Multiply by hθ, φ| and use hθ, φ|l, mi = Ylm (θ, φ) to obtain
X
l
Ylm (θ ′ , φ′ ) =
Ylm′ (θ, φ)Dm
′ m (R)
23
(1.100)
m′
As a final exercise we demonstrate our normalization convention: As a solution to eqs. (1.86)
and (1.87) Ylm (θ, φ) is undefined up to the extent of an arbitrary constant. We wish to
normalize and adopt a phase convention, and since for a given l the relative phases of the
2l + 1 spherical harmonics are fixed we only need to determine one of them. We implement
this by requiring that Yl0 (0, 0) is real and positive.
We know that acting with the raising operator L+ on the ket |l, mmax i should produce zero,
i.e.
∂
∂
+ i cot θ
Yll (θ, φ) = 0
(1.101)
L+ |l, li = 0 ⇔
∂θ
∂φ
The solution to the above partial differential equation is easily found to be Yll (θ, φ) =
cl sinl θ eilφ where cl is a constant depending on l. One can obtain the modulus of the normalization constant by the normalization condition
Z π
Z 2π
2
|cl |
dθ (sin θ)2l+1 = 1
(1.102)
dφ
0
0
The integration is easily performed with the result
r
(2l + 1)!
1
|cl | = l
2 l!
4π
(1.103)
To determine the phase one takes advantage of the following representation:
r
(2l + 1)!
l cl
Plm (cos θ) ⇒ Yl0 (0, 0) ≥ 0
Yl0 (θ, φ) = (−1)
|cl |
4π
cl
⇒
= (−1)l
|cl |
And the desired result follows, since
s
Ylm (θ, φ) = cl
1.10
(l − m)!
(−1)m eimφ Plm (cos θ)
(2l)!(l + m)!
(1.104)
Representation of the rotation operator
Having established most of the relevant theory regarding angular momentum we now turn our
attention to the rotation of a rigid body in quantum mechanics. For this purpuose we begin
with a discussion of rotations and symmetry.
One of the fundamental assumptions in the application of quantum mechanics is that
ordinary space is homogeneous and isotropic, meaning that there is no preferred direction in
space. Gravity seems to violate this assumption, but for all purpuoses this force is entirely
negligeable when we consider the scale in which we operate. If this truly is so then it follows
logically that all translation and rotation are symmetry operations. We mean by symmetry
operation a transformation that leaves the mutual physical relations of the system unaltered.
1.10 Representation of the rotation operator
24
This is the content of what is called the Euclidean principle of relativity. One deduces
that if this principle holds, then all rotations in quantum mechanics are represented by unitary
transformations. To see this consider the following. If we rotate a quantum system in space
the state vector transforms into a new state vector
|Ψi → |Ψ′ i = R |Ψi
where R is the rotation operator acting on |Ψi. The Euclidean principle of relativity requires
that all probabilities be invariant under rotations, or said in another way: all inner products
remain invariant in absolute value. Mathematically it is written
hΨ|Ψi = hΨ′ |Ψ′ i = hΨ|R† R|Ψi → R† R = 1
which means that R must be unitary. To determine the rotation operator one takes advantage
of the fact that any finite rotation can be viewed upon as a succession of infinitesimal rotations,
and the rotation operator is then the product of these infinitesimal rotation operators. It can
be expressed in terms of the angular momentum and one can show that
R(θ) = e−in·L θ/h̄
for a rotation through an axis with unit vector n. In our representation we have chosen the
common eigenvectors of L2 and Lz as our basis vectors which we’ve labeled as |l, mi. The
matrix elements corresponding to the operator R is then
(l)
Dmk = hl, m| R |l, ki = hl, m| e−in·L θ/h̄ |l, ki
and the total representation matrix is then
D (l) (R) = e−in·Lθ/h̄ ,
where L is ’a vector of matrices’. All components except Lz are non-diagonal matrices which
makes life a little difficult. In general the dependence on the rotation parameters n and θ is
quite complicated. However; there are special cases where one can obtain simple expressions.
For small values of l we construct the rotation matrices in terms of the first 2l powers of n · L.
We first state the general case then consider our case explicitly:
Let f (z) be an analytic function. The function f (A) of the normal operator A can then
be expanded in finite powers of A. From the completeness relation
f (A) =
N
X
i=1
f (A′i )|A′i ihA′i |
Provided that the singularities of f don’t coincide with the eigenvalues of A, and if p of these
eigenvalues are distinct we label these non-repeating eigenvalues by i = 1, . . . , p and write
f (A) =
p
X
f (A′j )
j=1
p
Y
A′i − A
A′i − Aj
l
Y
h̄k′ − n · L
h̄(k′ − k)
′
i=1,i6=j
So for small l we have
D (l) (R) =
l
X
k=−l
e−ikθ
k ′ =−l,k6=k
1.10 Representation of the rotation operator
25
Figure 1.4: Definition of the Euler angles in quantum mechanics, taken from [11]
Another case where the representation matrices are easy to handle is in the case of infinitesimal
rotations, i.e. θ << 1. Let ǫ = n θ and expand the exponential to first order:
e−in·Lθ/h̄ ≈ 1 −
i X
i
ǫi Li
ǫ·L=1−
h̄
h̄ x,y,z
Then one use the ladder operators L± to obtain
iǫx − iǫy p
(l)
Dmk = δmk (1 − iǫz ) −
l(l + 1) − k(k + 1)δm,k+1
2
iǫx − iǫy p
l(l + 1) − k(k − 1)δm,k−1
−
2
Further diagonalization is impossible since diagonalizing for example Lx would mean undiagonalizing Lz .
1.10.1
Quantum mechanical rotation of a rigid body - Euler angles revisited
We finally turn our attention to the task of representing the rotation of a rigid body using
(l)
quantum mechanics. We parametrize the expression for the rotation matrix Dmk (R) in terms
of the three Euler angles α, β, γ rather than in terms of n, φ. As indicated in figure (1.10.1)
we will use the y-convention here (see section 1.3.4). The reason for this is apparent: it makes
the matrix elements real.
The rotation operator in this case is the product of three successive rotations
i
′
i
R(α, β, γ) = e− h̄ γz ·L e− h̄ βy
′′ ·L
i
e− h̄ αz·L
(1.105)
We transform the right hand side into a product of of rotations around the space set of axes
(x, y, z). To see how this is done consider the following: Since all inner products must be
1.10 Representation of the rotation operator
26
invariant under rotations (remember the Euclidean principle of relativity) the observable Q in
the state |Ψi and it’s transform Q′ = RQ in the rotated state |Ψ′ i are related through
Q = R† Q′ R or equivalently Q′ = RQR†
which means that the observable undergoes the same unitary transformation as the states.
Using this fact and the property
†
eSAS = S eA S †
valid for all similarity transformations involving the two arbitrary operators S and A one
obtains
i
e− h̄ αLz′
i
e− h̄ βLy′′
i
i
i
= e− h̄ βLy′′ e− h̄ γLz e h̄ βLy′′
i
i
i
= e− h̄ αLz e− h̄ βLy e h̄ αLz
which can be combined to give a total rotation operator
i
i
i
R(α, β, γ) = e− h̄ αLz e− h̄ βLy e− h̄ γLz
(1.106)
which is a fairly simple result. The matrix representation which is different for different l and
has dimension 2l + 1 is then
(l)
(l)
Dmk (α, β, γ) = hl, m| R(α, β, γ) |l, ki = e−i(mα+kγ) dmk
− h̄i βLy
(l)
dmk ≡ hl, m| e
(1.107)
(1.108)
|l, ki
(l)
since Lz is represented by a diagonal matrix. The matrix (dmk ) is called the reduced matrix.
Our next task is to calculated these matrices for different values of l. The matrices are called
Wigner D functions after the physicist Wigner. A general formula for reduced matrix elements
exist, and it is derived in [7] using combinatorical arguments. The result is
(l)
dmk (β) =
×
p
(j + k)!(j − k)!(j + m)!(j − m)!
X (−1)n (cos θ/2)2j+m−k−2n (sin θ/2)k−m+2n
n
(k − m + n)!(j + m − n)!(j − k − n)!n!
(1.109)
where n runs over the values which make the factorial terms ≥ 0. This formula ′′ looks a bit
messy′′ (in the words of Feynman) but it produces the wellknown result
s
(l − m)!
(l)
m
Plm (cos β)
(1.110)
d0m (β) = (−1)
(l + m)!
so we are sure that it works.
1.10.2
The case l = 1/2
In the case of l = 1/2 we have m ∈ [−1/2, 1/2] in steps of one. We use expression (1.108) to
calculate the matrix elements. We write
i
hl, m| e−iβLy /h̄ |l, ki = e− h̄ β·L/h̄ = e−iβ·σ/2
1.10 Representation of the rotation operator
27
where β = (0, β, 0) and σ = (σx , σy , σz ). We’ve also used the result found in section 1.8.1;
Ly = h̄2 σy . We expand the exponential:
i
e− h̄ βLy
= e−iσy β/2 =
∞
X
(−iβ/2)n
n=0
= 1+
= 1+
σyn = 1 +
n!
∞
X
(−iβ/2)n
n=1
n!
σyn
∞
∞
X
(−iβ/2)2m 2m X (−iβ/2)2m+1 2m+1
σy +
σy
(2m)!
(2m + 1)!
m=1
m=0
∞
∞
X
X
(−1)m (β/2)2m+1
(−1)m (β/2)2m
−i
σy
(2m)!
(2m + 1)!
m=0
m=1
= cos(β/2) − i sin(β/2) σy
where we have used the Taylor expansions of the trigonometric functions sin x, cos x and the
properties σy2m = 1 and σy2m+1 = σy . This is actually a special case of the more general
wellknown formula
e−iθn·σ = cos θ − in · σ sin θ
(1.111)
In terms of a matrix one obtains
(1/2)
(dmk )
=
cos(β/2) − sin(β/2)
sin(β/2) cos(β/2)
Now we insert this result into (1.108) and obtain the final expression for the rotation matrix
in terms of the Euler angles:
−i(α+γ)/2
e
cos(β/2) −e−i(α−γ)/2 sin(β/2)
(l=1/2)
(1.112)
D
(α, β, γ) =
ei(α−γ)/2 sin(β/2)
ei(α+γ)/2 cos(β/2)
1.10.3
the case l = 1
As usual we have m = [−1, 0, 1]. Again, we want to use (1.108) to calculate the matrix
elements. This time we have to take into consideration the fact that the matrix representation
of Ly derived in section 1.8.2 only has non-diagonal elements and therefore it is somewhat
harder to calculate multiples of. We have


0 −i 0
h̄
Ly = √  i 0 −i 
2
0 i
0
As we observe, we must calculate (Ly /h̄)n which is a truly challenging task. A little thought
makes one realize that if infer a simliarity transformation Ly = SAS −1 where A is a diagonal
matrix and S is a matrix containing eigenvectors of Ly one obtains
(Ly /h̄)n = S (A/h̄)n S −1
which is easier to calculate. We know the eigenvalues of Ly (−1, 0, 1 in units of h̄), and one
obtains
(Ly /h̄)n = Ly /h̄
odd n
2
= (Ly /h̄)
even n
1.11 Angular momentum components in the laboratory and body-fixed coordinate
frames
28
Inserting this result into the expansion of the exponential one obtains
− h̄i βLy
e
=
∞
X
(−iβ)n
n!
n=0
= 1+
= 1+
n
(Ly /h̄) = 1 +
∞
X
(−iβ)n
n=1
n!
(Ly /h̄)n
∞
∞
X
X
(−iβ)2m
(−iβ)2m+1
(Ly /h̄)2m +
(Ly /h̄)2m+1
(2m)!
(2m
+
1)!
m=1
m=0
∞
∞
X
X
(−1)m β 2m
(−1)m β 2m+1
(Ly /h̄)2 − i
(Ly /h̄)
(2m)!
(2m + 1)!
m=1
m=0
= 1 + (cos β − 1)(Ly /h̄)2 − i sin β (Ly /h̄)
Inserting the various matrices one obtains
 1
√1
2 (1 + cos β) − 2 sin β

(1)
cos β
(dmk ) =  √12 sin β
1
1
√ sin β
2 (1 − cos β)
2
1
2 (1 − cos β)
− √12 sin β
1
2 (1 + cos β)
Now, the total rotation matrix becomes
 1 −i(α+γ)
(1 + cos β) − √12 e−iα sin β
2e

√1 e−iγ sin β
cos β
D (l=1) (α, β, γ) = 
2
1 iα
1 i(α−γ)
√
(1 − cos β)
e sin β
2e
2
1.11



1 −i(α−γ)
(1 − cos β)
2e
1 iγ
√
− 2 e sin β
1 i(α+γ)
(1 + cos β)
2e
(1.113)


 (1.114)
Angular momentum components in the laboratory and
body-fixed coordinate frames
The expression for the operator of rotation is given by
R(Ω) = e−iΩ·J/h̄
(1.115)
where J is the total angular momentum operator and Ω = (Ω1 , Ω2 , Ω3 ). We can write Ω = nΩ
where n is the axis we rotate about. Thus
Rn (Ω) = e−iΩn·J/h̄
(1.116)
From this expression we see that the rotation is generated by the differential operator
∂
(1.117)
∂Ω
where Ω is the angle of rotation. For the rotation convention used in section 1.10, the yconvention we obtain the total angular momentum components along the z, y ′′ and z ′ -axes:
n · J = −ih̄
∂
∂
∂
Jz = nα · J = −ih̄ ∂α
, Jy′′ = nβ · J = −ih̄ ∂β
, Jz′ = nγ · J = −ih̄ ∂γ
(1.118)
Here, Jz′ is the z-component of the total angular momentum operator in the body-fixed system
K ′ . We are seeking expressions for the operators in the inertial reference frame, the K system,
and the K ′ system, given by
J = Jx ex + Jy ey + Jz ez
J
′
=
Jx′ e′x
+
Jy′ e′y
+ Jz′ e′z
in K
(1.119)
in K’
(1.120)
1.11 Angular momentum components in the laboratory and body-fixed coordinate
frames
29
we can use the left-hand side of (1.117) to obtain equations for the angular momentum components expressed as differential operators in the Euler angles α, β, γ. We first need to decompose
the axes of rotation into K and K ′ . This requires some geometrical thinking and using figure
?? we decompose the second axis of rotation (the y ′′ -axis) in K:
nβ = cos αey − sin αex
(1.121)
Inserting this into the generator one obtains
−ih̄
∂
= cos αJy − sin αJx
∂β
(1.122)
The third rotation axis is a little bit harder to determine. Using the fact that nγ is the image
of ez after two successive rotations one obtains
nγ = sin β cos αex + sin β sin αey + cos βez
(1.123)
which gives our second equation:
−ih̄
∂
= sin β cos αJx + sin β sin αJy + cos βJz
∂γ
(1.124)
The first axis of rotation is already decomposed into the K system. So, collecting all results
we obtain three equations from which we can solve for Jx , Jy and Jz in terms of the Euler
angles:
∂
∂α
∂
−ih̄
∂β
∂
−ih̄
∂γ
−ih̄
= Jz
(1.125)
= cos α Jy − sin α Jx
(1.126)
= cos β Jz + sin β(cos α Jx + sin α Jy )
(1.127)
After some straight forward algebra we obtain
∂
cos α ∂
∂
− sin α
+
Jx = −ih̄ − cos α cot β
∂α
∂β
sin β ∂γ
∂
∂
sin α ∂
Jy = −ih̄ − sin α cot β
− cos α
−
∂α
∂β
sin β ∂γ
∂
Jz = −ih̄
∂α
(1.128)
(1.129)
(1.130)
These are the components of the angular momentum operator J with respect to the laboratory
system (the xyz-axes). To calculate the components with respect to the body-fixed axes x′ y ′ z ′
we use the fact that the spherical components Jµ where
1
−1
(Jµ ) = (J−1 , J0 , J1 ) ≡ √ (Jx − iJy ), Jz , √ (Jx + iJy ) ,
2
2
is a spherical tensor of rank one. This definition assumes that
J = J− e(+) + J0 e(0) + J+ e(−)
(1.131)
1.11 Angular momentum components in the laboratory and body-fixed coordinate
frames
30
where the spherical unit vectors
e(±) = ± √12 (ex ± iey ), e(0) = ez
(1.132)
We will be using this convention throughout this thesis. Now, the spherical components of
the angular momentum with respect to the body-fixed axes are given though the relation
X (1)
′
Jm
=
(1.133)
Dm′ m (α, β, γ) Jm′
m′
We now have three equations for the angular momentum components in the body-fixed system:
Jx′ − iJy′ = e−iγ (cos α cos β − i sin α)Jx − e−iγ (sin α cos β + i cos α)Jy − e−iγ sin βJz
Jx′ + iJy′ = eiγ (cos α cos β + i sin α)Jx + eiγ (− sin α cos β + i cos α)Jy − eiγ sin βJz
Jz′ = cos α sin βJx − sin α sin βJy + cos βJz
Some straight forward but tedious algebra then gives
∂
∂
cos γ ∂
′
+ sin γ
+ cos γ cot β
Jx = −ih̄ −
sin β ∂α
∂β
∂γ
sin γ ∂
∂
∂
′
Jy = −ih̄
− cos γ
− sin γ cot β
sin β ∂α
∂β
∂γ
Jz′ = −ih̄
∂
∂γ
(1.134)
(1.135)
(1.136)
The angular momentum vector operator J′ = (Jx′ , Jy′ , Jz′ ) in the body fixed system has components which are related to the angular momentum components in the fixed inertial reference
frame by the relation J′i = J · e′i where e′i is any unit vector along the axes in the body frame.
Our next task is to compute the squared total angular momentum operator. Since it is
given through J2 = Jx 2 + Jy 2 + Jz 2 the derivation involves squaring the different operators
representing the angular momentum components. When doing this one finds that the result
is independent of which coordinate system the calculation is performed, which is completely
in agreement with the fact that the scalar product J · J is invariant under rotations. The full
expression is in both coordinate systems found to be
2
∂
∂2
∂
∂
∂2
1
2
′2
2
+
− 2 cos β
+ cot β
+
(1.137)
J = J = −h̄
sin 2 β ∂α2 ∂γ 2
∂γ
∂β 2
∂β
As I said, it is independent of the choice of coordinate system, but under the requirement that
they have a common origin, the same axis of quantization (in our case the z−, z ′ -axis) and
obey a consistent set of quantization rules. A consequence of these rules are the commutation
relations between the angular momentum operators in the body-fixed system. One can by
direct calculation, using eqs. (1.134) - (1.136), show that
[Jx′ , Jy′ ] = −ih̄Jz′
in the body-fixed system. We see that the commutation relations in this system differ from
the commutation relations in the laboratory system by a minus sign.
1.11 Angular momentum components in the laboratory and body-fixed coordinate
frames
31
Since we have quantized along the z−, z ′ −axes we search for the common eigenfunctions
of the Jz , Jz′ and the J2 operators. In this case one quickly realizes that the matrix elements
l (∗)
Dmk satisfy
(l)∗
(l)∗
Jz Dmk (Ω) = h̄m Dmk (Ω)
(l)∗
(l)∗
Jz′ Dmk (Ω) = h̄k Dmk (Ω)
(l)∗
(l)∗
J2 Dmk (Ω) = h̄2 l(l + 1) Dmk (Ω)
The first two relations are trivial to obtain by direct calculation, while the calculation leading
to the last expression is certainly non-trivial, the calculation being based on some subtleties,
and these subtleties justify a special treatment in this next section.
1.11.1
(l)∗
(l)∗
Proof of J2 Dmk = h̄2 l(l + 1) Dmk
The derivation of the above relation is not based on brute force evaluation of the partial
derivatives in the expression for the total angular momentum operator. Instead we use another
tactic which lets us derive a few bonus-relations in addition. Let us consider a particle placed
in a fixed Cartesian coordinate system (the laboratory system) with axes labeled x, y, z. In
spherical coordinates the particle position can be described using the two angles (θ, φ) relative
to the axes. We ignore the r coordinate since we will be keeping it fixed through the entire
derivation.
Now rotate the coordinate axes into a set of new axes x′ , y ′ z ′ and attach a rigid body to
these. The particle position can now be described in this system using the new set of angles
(θ ′ , φ′ ) relative to the rotated axes. The position of the rigid body relative to the lab system is
given through the Euler angles α, β, γ and the D-functions describe rotations in these angles.
Let the operator L be the total angular momentum operator for the particle, and analogously
let J be the operator of total angular momentum (expressed as a differential operator in the
Euler angles) acting on the D-functions. The eigenfunctions of the particle operators L2 , Li
are denoted hθ, φ|l, mi in the lab system and hθ ′ , φ′ |l, mi in the body system. We will now
(l)∗
show that if we know the eigenvalues of L then we can find the action of J on Dmk (Ω).
The eigenfunctions in the two systems are related through a rotation which can be put in
the mathematical form
X (l)∗
hθ, φ|l, mi =
(1.138)
Dmk (Ω) hθ ′ , φ′ |l, ki
k
where we’ve used the passive transformation description. We rotate again the body-fixed coordinate axes an infinitesimal angle δψ through the unit vector n. The D-functions transform
as
i
(l)∗
(l)∗
2
′
−i J·n δψ/h̄ (l)∗
Dmk (Ω) = e
Dmk (Ω) = 1 − (J · n) δψ + O(δψ ) Dmk (Ω)
h̄
where we have expanded to first order in δψ. The wave functions defined with respect to the
lab system remain unchanged
hθ, φ|l, mi′ = hθ, φ|l, mi
while the wave functions defined in the body-fixed system transform as
i
hθ ′ , φ′ |l, mi′ = ei L·n δψ/h̄ hθ ′ , φ′ |l, mi = 1 + (L · n) δψ + O(δψ 2 ) hθ ′ , φ′ |l, mi
h̄
1.11 Angular momentum components in the laboratory and body-fixed coordinate
frames
32
The altered form of equation (1.138) is
X (l)∗
hθ, φ|l, mi′ =
Dmk (Ω)′ hθ ′ , φ′ |l, ki′
k
hθ, φ|l, mi =
≈
−
0 =
i
i
(l)∗
1 − (J · n) δψ Dmk (Ω) 1 + (L · n) δψ hθ ′ , φ′ |l, mi
h̄
h̄
k
X (l)∗
X (l)∗
i
Dmk (Ω) hθ ′ , φ′ |l, ki + δψ
Dmk (Ω)(L · n)hθ ′ , φ′ |l, ki
h̄
k
k
X
i
(l)∗
′ ′
δψ
hθ , φ |l, ki(J · n)Dmk (Ω)
h̄
k
X (l)∗
(l)∗
Dmk (Ω)(L · n)hθ ′ , φ′ |l, ki − hθ ′ , φ′ |l, ki(J · n)Dmk (Ω)
X
k
where we have neglected a term of second order in δψ and used eq. (1.138). We can remove
the angle dependence and thus we arrive at our ’master equation’
X (l)∗
X
(l)∗
Dmk (Ω)(L · n)|l, ki =
(1.139)
|l, ki(J · n)Dmk (Ω)
k
k
If we now let n = e′z - the unit vector along the z ′ -axis then we obtain L · n = Lz , J · n = Jz′
which gives
X (l)∗
X
X
(l)∗
(l)∗
(l)∗
Dmk Lz |l, ki =
|l, kiJz′ Dmk ⇐⇒
|l, ki Jz′ Dmk − h̄kDmk = 0
k
k
k
where we have used that Lz is an eigenfunction of |l, ki giving Lz |l, ki = h̄k|l, ki. The sum
must be identically zero for all k which means that
(l)∗
(l)∗
Jz′ Dmk = h̄kDmk
(1.140)
for all k. We see that the Jz′ operator in the body fixed system acts on the second index of
(l)∗
Dmk which means that k is the quantum number in the body system which determines the
spin projection. We have reproduced one of the above promised relations. To advance further
we introduce new ladder operators which act on the D-functions:
J+ ≡ Jx′ − iJy′
J− ≡
Jx′
+
(1.141)
iJy′
(1.142)
Notice the difference in the sign convention as to that one adopted earlier which still holds
for the particle angular momentum operator J. Furthermore one can easily show (using eq.
(1.139) )that
X (l)∗
X
(l)∗
Dmk L+ |l, ki =
|l, kiJ− Dmk
k
X
k
k
(l)∗
Dmk L− |l, ki
=
X
k
(l)∗
|l, kiJ+ Dmk
1.11 Angular momentum components in the laboratory and body-fixed coordinate
frames
33
Proceeding in the same way as before we find the following results:
p
(l)∗
(l)∗
J+ Dmk = h̄ (l − k)(l + k + 1)Dm k+1
p
(l)∗
(l)∗
J− Dmk = h̄ (l + k)(l − k + 1)Dm k−1
(1.143)
(1.144)
′ , J ′ } = (J ′ )2 + (J ′ )2 so that (J ′ )2 = (J ′ )2 + (J ′ )2 + (J ′ )2 = 1 {J , J } +
Also note that 12 {J+
+
−
−
x
y
x
y
z
2
′
2
(Jz ) . We know the eigenfunctions and their eigenvalues. Insertion gives after some simple
algebra
(l)∗
(l)∗
(l)∗
(J ′ )2 Dmk = J 2 Dmk = h̄2 l(l + 1)Dmk
(1.145)
where we have used that the squared total angular momentum is a scalar operator which
means that it has the same form regardless of which coordinate system we use. We see
(l)
(l)∗
that as expected it is the complex conjugated matrix elements Dmk and not Dmk that are
eigenfunctions of the angular momentum operators in the body-fixed system.
In the derivation of this result we obtained the action of the angular momentum ladder
operators in the body-fixed system on the D-functions which is a standard result. However
the definition of the new raising and lowering operators acting on the D-functions is not the
same.
1.11.2
Further properties of J+ and J−
The angular momentum ladder operators defined in eqs. (1.89) and (1.90) raise and lower
m. In the previous section we saw that J+ and J− did the same for k. Let us write out the
operators explicitly. By using the definitions of Jx′ and Jy′ one obtains
J+
J−
∂
∂
1 ∂
−i
− cot β
= −ih̄e
sin β ∂α
∂β
∂γ
∂
∂
1 ∂
iγ
+i
− cot β
= ih̄e
sin β ∂α
∂β
∂γ
−iγ
(1.146)
(1.147)
Let us switch to spherical coordinates through the substitutions γ = 0, α = φ and β = θ.
This gives immediately
∂
i ∂
+
J+ = −h̄
(1.148)
∂θ sin θ ∂φ
∂
i ∂
J− = −h̄
−
(1.149)
∂θ sin θ ∂φ
1.11.3
Energy in the body-fixed system
With these relations in place we can derive an expression for the quantum mechanical energy
in the body-fixed system.
Consider a rotating rigid body with principal moments I1 , I2 and I3 . We transform from
the laboratory to the body-fixed system in which the principal moments coinside with the
coordinate axes x′ y ′ z ′ . In this case the angular momentum components Li are related to the
the principal moments of inertia through Li = Ii ωi′ where ωi′ is the angular velocity seen from
1.11 Angular momentum components in the laboratory and body-fixed coordinate
frames
34
the body-fixed system. Now, consider an axial-symmetric body with the z ′ -axis as the axis of
symmetry. Then I3 > I2 = I1 . The Hamiltonian is (we drop the primes for ease of notation)
1
1
1
1
1
1 ′2
1
(Jx 2 + Jy 2 ) +
Jz 2 =
(J2 − Jz′ 2 ) +
J
H = I1 ω1 2 + I2 ω2 2 + I3 ω3 2 =
2
2
2
2I1
2I3
2I1
2I3 z
where we in the last expression restored the primes and used J′2 = J2 . The eigenfunctions
(l)∗
are simply the Wigner D-functions Dmk with eigenvalues
h̄2
I3 − I1 2
Elk =
k
l(l + 1) −
2I1
I3
(1.150)
independent of m. This expression for the energy is of course valid in the laboratory system as
well and we see that the energy only depends on quantum numbers obtainable in this system.
This is a very convenient result indeed.
We have now explored a great portion of the classical and quantum mechanical theories
regarding rigid-body rotations. We have shown that the energy of the rotating system is
related to derivatives of the Euler angles α, β, γ defined in section (1.2.2) (classical case) and
section (1.10.1) (quantum mechanical case). The next step is to implement all these relations,
but this will be done later in this thesis.
The next chapter deals with a different subject; radiation from simple charge distributions.
The gap between chapter one and two will be bridged when we expand the electromagnetic
fields in multipoles. These multipoles can be put in a closed form with the aid of the already
introduced spherical harmonic functions.
Chapter 2
Radiation from scalar and
electromagnetic fields
In this chapter we shall deal with different forms of radiation. We start by considering the
simplest type of radiation, i.e. from scalar fields. Since the quanta are spinless Klein-Gordon
particles we should expect the radiation pattern to be invariant under rotations which in turn
mean that the angle dependence should be given by the ordinary spherical harmonics. When
we are finished with the classical case we move on to canonical quantization of the scalar field
by introducing operators. More on this later.
The quanta of the electromagnetic field are the massless photons which are spin-1 particles.
We will develop the formalism for electromagnetic fields and show how to find the angular
distribution of radiation. Then we will calculate the patterns arising from simple charge
distributions, and we will consider linear and circular antennas explicitly. We will also consider
quantization of the electromagnetic vector potential and show the correlation between the
classical case and the quantum mechanichal case when it comes to the angular distribution
patterns.
In the end of the chapter we consider polarization and the Stokes paramters.
2.1
Scalar field radiation
Let’s consider the simplest case of radiation; the case where the field quantum does not carry
any spin. This is the case for the Klein-Gordon particles, which are massless spin-0 bosons.
We will start by considering the classical case using classical field theory, of course, scalar fields
have yet to be discovered in nature, but this does not prevent us from constructing theories
in which they occur.
Let φ(x) = φ(r, t) be a real massless scalar field (φ∗ = φ) defined everywhere in space. We
will assume that it varies harmonicly in time, i.e.
φ(r, t) = φ(r) e−iωt
(2.1)
We couple the field to a charge-density ρ(r, t) through the following Lagrangian density:
L=
1 µν
η ∂µ φ∂ν φ + ρφ
2
35
(2.2)
2.1 Scalar field radiation
36
To obtain the equation of motion we utilize the covariant Lagrange equations
∂L
∂L
− ∂µ
=0
∂φ
∂(∂µ φ)
(2.3)
The reader can easily verify that the resulting equation is the wave-equation for φ with the
charge-density as a source term:
∂2
2
(2.4)
− 2 + ∇ φ = −ρ
∂t
When one takes into account (2.1) one obtains the inhomogenous scalar Helmholtz equation
∇2 + k2 φ = −ρ
(2.5)
since ω = k = |k|, where k is the scalar particle momentum vector. We start by setting
ρ = 0, which means we are solving for the field outside the source. We will in the following
assume that the particles are inside some normalized volume V = 1. Then we can write φ as
a superposition of waves with different momentum:
o
X 1 n
√
φ(x) =
(2.6)
ak eik· x e−iωt + a∗k e−ik· x eiωt
2ω
k
This form guarantees that φ∗ = φ. The ak ’s are complex constants. The exponentials are
solutions of eq. (2.5) with ρ = 0 provided that ω = |k|. Since k varies continuously we can
convert the sum into an integral over momentum:
Z
o
d3 k 1 n
ik· x −iωt
∗ −ik· x iωt
√
(2.7)
e
e
a
e
e
+
a
φ(x) =
k
k
(2π)3 2ω
The scalar field is here expressed through Cartesian coordinates {x, y, z}. Now assume instead
that we replace the momentum dependence with a dependece on the quantum numbers l, m.
We then obtain a sum over the quantum numbers l, m and energy ω which gives
X Z dω φ(x) =
(2.8)
alm ulm (r)e−iωt + a∗lm u∗lm (r)eiωt
2π
l,m
where ulm (r) is called a mode-function. We always demand that the mode-function satisfies
(2.5) with ρ = 0 but we are free to choose which set of coordinates to use. Radiation problems
usually have spherical symmetry so it’s advantageous to use spherical coordinates.
2.1.1
Scalar wave equation
When k 6= 0 we can use the method of separation of variables to write the solution ulm (r) to
the scalar Helmholtz differential equation (eq. (2.5) with ρ = 0) as a radial function times the
ordinary scalar harmonics:
ulm (r) = Rlm (r)Ylm (θ, φ)
When we insert this ansatz into the Helmholtz equation we find that the radial functions
satisfy
2
2 d
l(l + 1)
d
2
+
+
k
−
Rlm (r) = 0
(2.9)
dr 2 r dr
r2
2.1 Scalar field radiation
37
independent of m. We will anyhow keep the m subscript since the constants in R theoreticly
can depend on l and m. Here we have used relation (1.91) to expand the squared gradient
operator. It can be shown that the general solution of the radial equation is a linear combin√
ation of spherical Bessel and Hankel functions by making the substitution Rl (r) = fl (r)/ r.
The new function fl (r) then satisfies
2
d
1 d
(l + 1/2)2
+
+1−
fl (x) = 0
(2.10)
dx2 x dx
x2
where the dimensionless parameter x = kr. This is the well-known Bessel differential equation
for ν = l + 1/2. Therefore we write the solution as
Blm
Alm
flm (kr) = √ Jl+1/2 (kr) + √ Nl+1/2 (kr)
kr
kr
(2.11)
where the formal expression for the spherical Bessel function jl (kr) of the first kind is
jl (x) =
r
∞
X (−1)n (n + l)!
π
Jl+1/2 (x) = 2l xl
x2n = (−1)l xl
2x
n!(2n + 2l + 1)!
n=0
d
x dx
l
sin x
x
(2.12)
The function Nl+1/2 (kr) is a Bessel function of the second kind, which for our purpuoses later
on will be irrelevant. We therefore spare the reader from the mathematical details on this
function. The general solution to the differential equation can be put in the form
i
h
(2) (2)
(1) (1)
(2.13)
ulm (r, θ, φ) = Alm hl (kr) + Alm hl (kr) Ylm (θ, φ)
where we have defined the linear combinations
r
π
(1)
hl (kr) =
Jl+1/2 (kr) + iNl+1/2 (kr) = jl (kr) + inl (kr)
kr
r
π
(2)
hl (kr) =
Jl+1/2 (kr) − iNl+1/2 (kr) = jl (kr) − inl (kr)
kr
(2.14)
(2.15)
which are called the spherical Hankel functions of the first and second kind respectively. The
constants A(1) , A(2) are determined by the boundary conditions.
One can also choose the solution as
ulm (r, ω) = N jl (kr)Ylm (θ, φ)
(2.16)
(1)
where N is some normalization constant. From eq. (2.14) we see that jl (ωr) = Re hl (ωr).
This solution has the correct boundary condititions when r → 0. We will work with this
representation both classically and when we quantize the scalar field. Let us now carry on
with a calculation of the angular distribution of radiated power.
2.1.2
Classical angular distribution
From field theory the angular distribution of radiated power into a solid angle element dΩ per
time intervall dt is given by
dP
= r 2 ei T 0i
(2.17)
dΩ dt
2.1 Scalar field radiation
38
where r is the distance out to the observer, and T 0i is the 0i’th component of the energy
momentum tensor for the scalar field φ. We are interested in finding the spectral angular
intensity per frequency which can be found by slight manipulation:
Z
Z
dP
dP
2
0i
=r
(2.18)
dt ei T = dω
dΩ
dΩ dω
where dP/(dΩdω) is the spectral angular intensity we are looking for. This is equivalent to
working with Fourier transformed fields. Inserting our solution for the mode-function eq.
(2.16) into eq. (2.8) we obtain
X Z dω ∗
∗ iωt
Rlm (ωr)Ylm e−iωt + Rlm
(ωr)Ylm
e
(2.19)
φ(x) = N
2π
l,m
The constants have been absorbed into the radial functions. Since we are interested in the
angular distribution of radiated power in the radiation zone where the fields vary as 1/r we
now take the limit r → ∞ and only keep the lowest order terms. For ωr ≫ 1 we have
(1)
hl (ωr) → (−i)l+1
eiωr
(2)∗
= hl (ωr)
ωr
(2.20)
At infinity we only have outgoing waves proportional to eikr . The first term of eq. (2.13)
survives, but not the second. Then Blm = 0 and
Rlm (kr) → (−i)l+1 Alm
eiωr
ωr
The scalar field becomes in the radiation zone:
−iωr
iωr
X Z dω −iωt
∗ l+1 e
∗ iωt
l+1 e
Ylm e
+ Alm i
Y e
φ(r, t) = N
Alm (−i)
2π
ωr
ωr lm
(2.21)
(2.22)
l,m
To obtain the angular distribution of radiated power we first need to derive the energymomentum tensor for the scalar field. We now from Noether’s theorem that the energymomentum tensor, which is obtained from the Lagrangian density
T µν =
∂L
∂ ν φi − η µν L
∂(∂µ φi )
(2.23)
is conserved in the sense that ∂µ T µν = 0. With (2.2) as the Lagrangian density, and φi = φ
we obtain after a quick calculation that outside the source where the density is zero the
energy-momentum tensor is
T µν = ∂ µ φ∂ ν φ
(2.24)
We need to find ei T 0i which is a scalar associated with energy-flow through a surface:
ei T 0i = ∂ 0 φei ∂ i φ = −φ̇ (er , eθ , eφ ) · ∇φ
(2.25)
Here, the dot means differentiation with respect to time. Using eq. (2.22) we easily obtain
(to lowest order in 1/r)
Z
o
N X dω n
∗ −iωr iωt
(2.26)
Alm (−i)l Ylm eiωr e−iωt + A∗lm il Ylm
e
e
φ̇ =
r
2π
l,m
Z
o
N X dω n
∗ −iωr iωt
∇φ = −er
(2.27)
Alm (−i)l Ylm eiωr e−iωt + A∗lm il Ylm
e
e
r
2π
l,m
2.1 Scalar field radiation
39
so
ei T 0i
#
"
o 2
X Z dω n
1
∗ −iωr iωt
Alm (−i)l Ylm eiωr e−iωt + il A∗lm Ylm
e
e
(2.28)
N
=
r2
2π
l,m
2 X X Z
Z
o2
dω ′ n
N
dω
∗ −iωr iωt
e
e
Alm (−i)l Ylm eiωr e−iωt + il A∗lm Ylm
=
r
2π
2π
′
′
l,m l ,m
When squaring the quantity inside the brackets we obtain four terms:
h
′
′
′
′
∗
Yl∗′ m′ ei(ω+ω )tr
il+l (−1)l+l Alm Al′ m′ Ylm Yl′ m′ e−i(ω+ω )tr + A∗lm A∗l′ m′ Ylm
i
′
′
∗ i(ω−ω ′ )tr
+ (−1)l Alm A∗l′ m′ Ylm Yl∗′ m′ e−i(ω−ω )tr
e
+(−1)l A∗lm Al′ m′ Yl′ m′ Ylm
(2.29)
(2.30)
where tr = t − r is called the reduced time. More on this later. Take a look at the first and
second term on the first line. When we integrate over t we obtain delta functions of the form
δ(ω ′ + ω) which would require ω ′ = −ω. This is not possible however since the energy must
be positive! Therefore the first two terms do not contribute. Note also that the two last terms
are completely symmetric under summation over l, l′ and m, m′ . Since δ(ω − ω ′ ) = δ(ω ′ − ω)
they give equal contributions. We will as a consequence multiply by two and only keep one
term in what follows. When inserting our results back into eq. (2.18) we obtain
Z
X X ′ Z dω Z dω ′
dP
′
l
2
l
l
∗
∗
i
=N
i
× 2(−1) Alm Al′ m′ Ylm Yl′ m′ dt e−i(ω−ω )tr
(2.31)
dΩ
2π
2π
′
′
l,m
l ,m
since the integration over t is infinite we can substitute dt with dtr without affecting the
integration range. Using
Z ∞
′
(2.32)
dtr e−i(ω−ω )tr = 2πδ(ω ′ − ω)
−∞
we obtain
dP
dΩ
Z

N2 X
X
′
′


il (−1)l Alm A∗l′ m′ Ylm Yl∗′ m′
(−i)l

 π
l,m
l′ ,m′


Z

X N
X N ′
√ (−i)l Alm Ylm
√ il A∗l′ m′ Yl∗′ m′
=
dω


π
π
l′ ,m′
l,m
Z
2
X
(−i)l Alm Ylm =
dω C =
dω
(2.33)
(2.34)
(2.35)
l,m
where we have taken N =
√
Cπ. Thus
2
X
dP
(−i)l a(l, m)Ylm (θ, φ)
=
dΩ dω
(2.36)
l,m
√
where a(l, m) = Alm / C is some constant amplitude depending on l, m and ω. We will choose
C to get agreement between the classical and the quantum mechanichal distributions.
2.2 Determination of the constants Alm
2.2
40
Determination of the constants Alm
We know that in Fourier (ω) space the scalar field is a solution of
(∇2 + k2 )φ(r) = −ρ(r)
(2.37)
Far outside the source, in the radiation zone where ρ = 0, the solution is (we will call it φ0 ):
φ0 (r → ∞, θ, φ) =
X
l,m
Alm jl (kr ≫ 1)Ylm (θ, φ) = ℜ
(−i)
eikr X
Alm Ylm (θ, φ)
(−i)l
r
k
(2.38)
l,m
where ℜ indicates that we are to take the real part of the expansion. Here we have used the
large-argument limit of the Bessel function. To determine the constants Alm we first look at
a particular solution of eq. (2.37) when ρ 6= 0. We know from theory that the field can be
expressed through an integral of a Green function times the density:
R
φρ (r, θ, φ) = d3 r ′ G(r, r′ )ρ(r′ ),
(∇2 + k2 )G(r, r′ ) = −δ(r − r′ )
(2.39)
Using the result (with r > r ′ )
G(r, r′ ) = ik
X
(1)
(∗)
jl (kr ′ )hl (kr) Ylm (r′ )Ylm (r)
(2.40)
l,m
which we prove in section ??, we obtain
φρ (r, θ, φ) = ik
X
l,m
= ℜ ik
hZ
(1)
hl (kr)
X
(1)
i
(∗)
d3 r ′ ρ(r′ )jl (kr ′ )Ylm (θ ′ , φ′ ) Ylm (θ, φ)
hl (kr)aρ (l, m)Ylm (θ, φ)
(2.41)
(2.42)
l,m
since the field is real. Now, let us push this solution out to a large distance from the source
(1)
such that ρ → 0. Then the Hankel function hl (kr) → (−i)l+1 eikr /(kr) as usual. Thus
φρ (r → ∞, θ, φ) = ℜ
eikr X
(−i)l aρ (l, m)Ylm (θ, φ)
r
(2.43)
l,m
and since it must equal φ0 (r) in this limit it follows upon inspection of eqs. (2.38) and (2.43)
that
Z
(∗)
Alm = ikaρ (l, m) = ik d3 r ′ ρ(r′ )jl (kr ′ )Ylm (θ ′ , φ′ )
(2.44)
We insert this result into the definition of a(l, m) and obtain
Z
1
(∗)
a(l, m) = √ iω d3 r ′ ρ(r′ )jl (ωr ′ )Ylm (θ ′ , φ′ )
C
where we have used k = |k| = ω.
(2.45)
2.3 Quantum angular distribution of radiation from scalar fields
2.3
41
Quantum angular distribution of radiation from scalar fields
Consider a microscopical, quantum mechanical system with density ρ interacting with the
scalar field. The interaction is described by a Hamiltonian density term Hint = ρφ classically.
Let us now study a transition between states through emission of a single scalar particle with
momentum vector k and energy ω = |k|. We label the initial state vector |ii (energy Ei ) and
the final state vector |f i (energy Ef ). The quantum of the scalar field operator φb is a spin-0
boson. The transition probability between the states under the action of a pertubation
Z
b r′ )|ii
Vf i = d3 r ′ hf |b
ρ(r′ ) φ(k,
(2.46)
and where the initial and final states of the emitting system belong to the discrete spectrum
is given by Fermi’s golden rule:
dΓ = 2π|Vf i |2 δ(Ei − Ef − ω)
ω2
d3 k
2
=
2π|V
|
δ(E
−
E
−
ω)
dω dΩ
i
fi
f
(2π)3
8π 3
(2.47)
1
dΓ
= 2 ω 2 |Vf i |2 δ(Ei − Ef − ω)
(2.48)
dω dΩ
4π
where dΓ is the transition rate per unit time intervall and ω is the energy carried away by the
radiated scalar particle. Here we have chosen dω dΩ to be the quantity that describes the state
of the emitted scalar particle. This means that we can accurately determine the angles (θ, φ)
(position in space), but not the quantum numbers l, m (momentum). This is a consequence
of Heiseinberg’s ’old’ uncertainty relation
∆p∆x ≥ h̄/2
(2.49)
We will label each quantum state with the aid of it’s energy, viz:
|ii = |0i|Ei i, |f i = |Ef , ki = |ki|Ef i = b
a†k |0i|Ef i.
(2.50)
Here, |Ef , ki contains not only the system in its final state, but also a scalar particle with
momentum vector k which is created from the vacuum state |0i. We now manipulate the
matrix element into a more managable form:
b i i = hEf |h0|b
b
b
hEf |b
ρ φ|E
ak ρb φ|0i|E
ρ|Ei i h0|b
ak φ|0i
i i = hEf |b
b r′ )|0i
= hf |b
ρ(r′ )|ii h0|b
ak φ(r,
b r′ )|0i
= ρbf i h0|b
ak φ(r,
(2.51)
where ρbf i (r′ ) = hf |b
ρ|ii is the density operator for the source. Expanding the scalar field into
plane waves
Z
o
d3 k 1 n
† −ik·r′
ik·r′
′
b
√
,
(2.52)
+
b
a
e
b
a
e
φ(k, r ) =
k
k
(2π)3 2ω
where k is the momentum vector and r′ is the source position vector, we obtain
Z
Z
d3 k′ 1
d3 k′ 1
′ ′
†
−ik′ ·r′
b
√
√
h0|b
ak φ|0i
=
=
h0|b
a
b
a
|0i
e
e−ik ·r × (2π)3 δ(k′ − k)
′
k
k
3
(2π) 2ω ′
(2π)3 2ω ′
1
′
(2.53)
= √ e−ik·r
2ω
2.3 Quantum angular distribution of radiation from scalar fields
and thus
Therefore
b r′ )|ii = √1 ρbf i (r′ ) e−ik·r′
hf |b
ρ(r′ ) φ(k,
2ω
Vf i
1
=√
2ω
Z
d3 r ′ ρbf i (r′ ) e−ik·r
′
Using the expansion (which we will prove soon)
X
′
′
b (∗) (b
(−i)l jl (kr ′ )Ylm (k)Y
e−ik·r = 4π
lm r )
42
(2.54)
(2.55)
(2.56)
l,m
′
b refers to the scalar particle coordinates (θ, φ) and Y (∗) (b
where Ylm (k)
lm r ) refers to the source
′
′
angles (θ , φ ), we obtain
Z
4π X
(∗) ′ ′
l
3 ′
′
′
Vf i = √
(−i)
(2.57)
d r ρbf i (r ) jl (kr )Ylm (θ , φ ) Ylm (θ, φ)
2ω l,m
where the measure d3 r ′ refers to an integration over the source. Defining the quantum mechanical multipole operator coefficient
Z
(∗)
b
A(l, m, ω) = d3 r ′ ρbf i (r′ ) jl (ωr ′ )Ylm (θ ′ , φ′ )
(2.58)
we can write the transition amplitude compactly as
4π X
b m, ω)Ylm (θ, φ)
(−i)l A(l,
Vf i = √
2ω l,m
(2.59)
where we now have ω = Ef − Ei because of energy conservation. Inserting this result into eq.
(2.48) we obtain
X
2
dΓ
b m, ω)Ylm (θ, φ)
= 2ω (2.60)
(−i)l A(l,
dω dΩ
l,m
To obtain the radiated energy into a solid-angle element dΩ per unit frequency ω, which is
equal to the radiation distribution we multiply the above expression with the particle energy
to obtain
X
2
dP
b m, ω)Ylm (θ, φ)
= 2ω 2 (−i)l A(l,
(2.61)
dω dΩ
l,m
X
2
= (−i)l b
a(l, m)Ylm (θ, φ)
(2.62)
l,m
where
b
a(l, m) =
√
2 iω
Z
(∗)
d3 r ′ ρbf i (r′ ) jl (kr ′ )Ylm (θ ′ , φ′ )
(2.63)
is the multipole moment given in terms
density operator. Comparing with the
√ of the source
√
classical case we see that if we choose C = 1/ 2 in eq. (2.45) we have complete agreement
when we let ρbf i → ρ.
2.4 The Maxwell equations
2.4
43
The Maxwell equations
In this section we follow to a large extent the treatment given in [9], which is a book I highly
recommend for anyone interested in this subject. Maxwell’s equations consitute the foundation
of electromagnetic theory and can be put in a simple form. In vacuum space they have the
following form (using h̄ = c = 1 units):
∇ · E(r, t) = ρ(r, t)
∇ · B(r, t) = 0
∂B(r, t)
=0
∇ × E(r, t) +
∂t
∂E(r, t)
∇ × B(r, t) = J(r, t) +
∂t
(2.64)
where the symbols have their usual meaning. From these equations one can deduce the
continuity equation which ensures charge conservation:
∂ρ(r, t)
+ ∇ · J(r, t) = 0,
∂t
(2.65)
Thus if the charge density ρ = ρ(r) then the divergence of the current density is zero; ∇ ·
J(r, t) = 0. We begin by solving the equations in the static case where the electric and
magnetic fields are homogeneous in time.
2.4.1
Static fields
Let us start by considering the second and third expressions in eqs. (2.64) first. ∇ · B = 0
implies that the magnetic field can be written as the curl of some vector potential; B = ∇×A.
The third equation is ∇ × E = 0, so the electric field should be expressed as the gradient of
some scalar potential φ, thus E = −∇φ. Assuming that the charge density is time independent
we obtain the well-known static Poisson equation
∇2 φ(r) = −ρ(r)
(2.66)
∇ × (∇ × A) = µ0 J(r)
{z
}
|
(2.67)
The last Maxwell equation gives
∇(∇·A)−∇2 A
We impose the Coulomb gauge ∇·A = 0 which gives the final version of the Maxwell equations
in the static field case:
∇2 φ = −ρ, ∇2 A = −J
(2.68)
Introducing the 4-vector potential Aµ = (φ, A) and the 4-current vector j µ = (ρ, J) these
equations can be written compactly as ∂ 2 Aµ = −j µ which is the covariant Maxwell equations.
We will now solve eqs. (2.68) by employing the method of Green functions. The generic form
of the equations is
∇2 ψ(r) = −f (r)
2.4 The Maxwell equations
44
which is an inhomogeneous partial differential equation of second order. We introduce the
representation
Z
ψ(r) = G(r, r′ )f (r′ ) d3 r ′
(2.69)
where G is a Green function satisfying
∇2 G(r, r′ ) = −δ(r − r′ )
(2.70)
which gives
Z
2
∇ ψ(r) =
2
′
′
3 ′
∇ G(r, r ) f (r ) d r = −
| {z }
=−δ(r−r′ )
Z
f (r′ ) δ(r − r′ ) d3 r ′ = −f (r)
(2.71)
Using the integral representation of the delta function:
Z
d3 k ik·(r−r′ )
δ(r − r′ ) =
e
(2π)3
′
′
We see that ∇eik·(r−r ) = ik eik·(r−r ) which means that effectively ∇2 → −k2 . Thus the
equation can be written
Z
′
d3 k eik·(r−r )
2
′
2
∇ G(r, r ) = ∇
(2π)3
k2
Z
′
d3 k eik·(r−r )
′
(2.72)
G(r, r ) =
(2π)3
k2
Switching to spherical coordinates the last integral can be written as
Z ∞ Z 1
Z ∞
′
′
1
1
eik|r−r | − e−ik|r−r |
′
ik|r−r′ | cos θ
G(r, r ) =
d(cos θ) e
=
dk
dk
(2π)2 0
(2π)2 0
ik|r − r′ |
−1
Z ∞
1
sin x
1
=
(2.73)
dx
=
2π 2 |r − r′ | 0
x
4π|r − r′ |
and translating over to our static field problem we obtain
Z
ρ(r′ )
1
d3 r ′
φ(r) =
4π
|r − r′ |
Z
1
J(r′ )
A(r) =
d3 r ′
4π
|r − r′ |
(2.74)
(2.75)
The potentials are determined by integrals over the source charge- and current- densities
respectively. Next we consider fields which vary in time.
2.4.2
Time dependent fields
Keeping the time dependence in Maxwell’s equations we obtain slightly different expressions
for the electromagnetic fields:
E = −∇φ −
B = ∇×A
∂A
∂t
(2.76)
(2.77)
2.4 The Maxwell equations
45
The first equation can then be written
∂
(∇ · A) = ρ
∂t
(2.78)
1 ∂2A
1 ∂φ
−
∇
∇
·
A
+
= −J
c2 ∂t2
c2 ∂t
(2.79)
∇2 φ +
and the last one is then
∇2 A −
Taking advantage of the freedom of choice of the potentials we impose the Lorenz gauge
µ
condition ∇ · A + c12 ∂φ
∂t = 0 which is equivalent to ∂ Aµ = 0 in covariant 4-vector form. One
obtains
1 ∂2φ
c2 ∂t2
1 ∂2A
∇2 A − 2 2
c ∂t
∇2 φ −
= −ρ
(2.80)
= −J
(2.81)
which is a set of equations equivalent to the Maxwell equations. We can solve them analytically
using Fourier transformation. Consider the generic form of eqs (2.80) and (2.81):
1 ∂2
2
(2.82)
∇ − 2 2 ψ(r, t) = −f (r, t)
c ∂t
We Fourier transform both sides:
ψ(r, ω)
f (r, ω)
≡
F{ψ(r, t)} =
Z
∞
ψ(r, t) eiωt dt
Z −∞
∞
f (r, t) eiωt dt
F{f (r, t)} =
−∞
2 ∂ ψ
⇒ F
= −ω 2 F(ψ)
∂t2
≡
(2.83)
We then end up with the inhomogeneous Helmholtz partial differential equation for the Fourier
transformed fields:
∇2 + k2 ψ(r, ω) = −f (r, ω)
(2.84)
As in the previous section we write the solution in terms of an integral over a Green function
times the source function. One then finds that the Green function satisfies
∇2 + k2 G(r, r′ ) = −δ(r − r′ )
(2.85)
If there aren’t any boundary surfaces the solution can only depend on R = |r − r′ | and is
therefore spherically symmetric. We choose to work in spherical coordinates, so the effective
d2
′
Laplacian operator is ∇2 = R1 dR
2 R which gives for r 6= r :
d2 [RG(r, r′ )]
+ k2 [RG(r, r′ )] = 0
dR2
with solution
G(r, r′ ) =
AeikR + Be−ikR
, R>0
R
(2.86)
(2.87)
2.4 The Maxwell equations
46
A sensible condition must be that G(r, r′ ) → 0 when |r − r′ | → ∞. What other restrictions
can we impose on our solution? Comparing with the solution in the static case we see that the
solution of Helmholtz’s equation in the limit k → 0 must approach the solution of Poisson’s
1
. Further specification of the
equation, which translates into demanding that A + B = 4π
constants depends on the actual problem considered. In the solution, the first term represents
a diverging spherical wave propagating from the origin, while the second term represents a
converging wave.
If we have a source of electromagnetic radiation which is switched on at t = 0 then it will
emitt radiation rather than receive. It is then clear that the appropriate solution is the first
term. We embrace this interpretation and, setting B = 0 we have
′
eik|r−r |
G(r, r ) =
4π|r − r′ |
′
(2.88)
This is the Fourier transformed version of the Green’s function and since the Green function
in t-space satisfies
1 ∂2
2
∇ − 2 2 G(r, r′ ; t, t′ ) = −δ(r − r′ )δ(t − t′ )
(2.89)
c ∂t
the source term for the Fourier transformed equation is
Z ∞
Z ∞
′
iωt
δ(r − r′ )δ(t − t′ )eiωt dt = −δ(r − r′ )eiωt
f (r, t)e dt = −
f (r, ω) =
−∞
−∞
and we obtain
′
eik|r−r | iωt′
e
G(r, r ) =
4π|r − r′ |
′
We now perform an inverse Fourier transformation to find the t-space Green function:
Z ∞
Z ∞ ik|r−r′ |
e
1
1
′
eiω(t −t) dt
G(r, r′ ) e−iωt dt =
G(r, r′ ; t, t′ ) =
2π −∞
2π −∞ 4π|r − r′ |
i
h
′
′ − t − |r−r |
h
“
”i
Z ∞
2π
δ
t
|r−r′ |
′
c
1
iω t − t− c
=
dt e
=
8π 2 |r − r′ | −∞
8π 2 |r − r′ |
h
i
′|
δ t′ − t − |r−r
c
(2.90)
=
4π|r − r′ |
where we have used the integral representation of the delta function. We see the similarity
between the time dependent and the static cases, but now we have a delta function which as
we shall see, introduces some notable effects.
This Green function is called the retarded Green function because it displays the same
causal behavior as one would expect from a wave disturbance. The delta function shows that
an effect observed at position r at the time t is due to a signal from an event a distance |r − r′ |
away which was sent at an earlier time t′ .
The formal solution of our problem is
Z
|r − r′ |
d3 r ′
1
′ ′
f
r
,
t
=
t
−
(2.91)
ψ(r, t) =
4π
|r − r′ |
c
2.5 Poynting’s theorem
47
Going back to eqs. (2.80) and (2.81) we see that
Z
ρ(r′ , tr )
1
φ(r, t) =
d3 r ′
4π
|r − r′ |
Z
µ0
J(r′ , tr )
A(r, t) =
d3 r ′
4π
|r − r′ |
(2.92)
(2.93)
where we’ve introduced the retarded time tr ≡ t − |r − r′ |. To find the electromagnetic fields
generated by the two potentials we could use eqs. (2.76) and (2.77) directly; however this
introduces difficult evaluations of partial derivatives which we would rather avoid. Instead,
we use Maxwell’s equations to find wave equations in E and B. Let us take the curl of the
third equation:
∇ × (∇ × E) + ∇ ×
∂
∂B
= 0 ⇐⇒ ∇(∇ · E) − ∇2 E = − (∇ × B)
∂t
∂t
where we have used standard vector identities. We use the Maxwell equations in (2.64) to find
expressions for ∇ × B and ∇ · E. Insertion gives
∇ρ − ∇2 E = −
∂J ∂ 2 E
− 2
∂t
∂t
Rearranging one obtains
∂2
∂J
2
∇ − 2 E = ∇ρ +
∂t
∂t
Similarly one finds for the magnetic field
∂2
2
∇ − 2 B = −∇ × J
∂t
(2.94)
(2.95)
We can now read off the solutions in terms of the electromagnetic fields directly instead of
having to use the potentials. The results are
Z
1
∂J(r′ , t′ )
1
3 ′
′
′ ′
(2.96)
d r
−∇ ρ(r , t ) −
E(r, t) =
4π
|r − r′ |
∂t′
t′ =tr
Z
i
1
1 h ′
′ ′
B(r, t) =
∇
×
J(r
,
t
)
(2.97)
d3 r ′
4π
|r − r′ |
t′ =tr
Here the primed Laplace operator ∇′ operates on the primed coordinates r′ and the quantities in the brackets are evaluated at the retarded time tr . We will consider both static and
time-varying fields later in this thesis, but first, let us find the relation between the electromagnetic fields and the energy content of the system. In electromagnetism the concept of
energy conservation is contained in Poynting’s theorem which we consider in this next section.
2.5
Poynting’s theorem
We continue our developement of the main theorems in electromagnetism. In this section we
consider one of the most basic relations regarding energy conservation. For this purpuose we
use the generalized version of the continuity equation:
1 X dqi
∂ρ
+∇·J=
∂t
δV
dt
i
2.5 Poynting’s theorem
48
where the sum is taken over all particles in a volume element δV , and the qi ’s can represent
an optional particle variable. Here we have assumed that all particles move with the same
velocity and the volume element δV is a co-moving volume element. We introduce the chargeand current- densities:
1 X
qi
(2.98)
ρ =
δV
i
1 X
J =
qi vi
(2.99)
δV
i
where now qi represents the charge of particle i. Consider the fact that the change in a
particle’s kinetic energy T generates power through F · v = dT /dt. We sum over all particles
and use the continuity equation with qi = Ti
1 X dTi
∂um
1 X
Fi · vi =
=
+ ∇ · Sm
δV
δV
dt
∂t
i
(2.100)
i
where we have defined the quantities
1 X
Ti (kinetic energy density)
δV
i
1 X
Ti vi (kinetic energy current)
δV
um =
Sm =
i
The left hand side can be rewritten using the expression for the the Lorentz force of the i′ th
particle Fi = qi (E + vi × B) as
!
1 X
1 X
1 X
Fi · vi =
qi (E + vi × B) · vi =
qi vi · E = J · E
δV
δV
δV
i
i
i
giving
∂um
+ ∇ · Sm = J · E
(2.101)
∂t
which is Poynting’s theorem. It is a mathematical expression of energy conservation. We
eliminate J from the left hand side. This is done by using the fourth equation in (2.64). We
obtain
∂E
∂u
∂B
·B−
·E−E×B=−
−∇·S
(2.102)
J·E=−
∂t
∂t
∂t
where we have defined the energy density
u≡
B2 + E2
2
(2.103)
and the Poynting vector
S ≡ E × B (Real fields)
(2.104)
which represents energy flow. When one considers complex electromagnetic fields one obtains
a complex Poynting vector which is then defined as
S ≡ 12 E × B∗ (Complex fields)
(2.105)
2.6 Multipole radiation
49
where the star denotes complex conjugation. The equation
−J · E =
∂u
+∇·S
∂t
(2.106)
is another way of stating energy conservation. The interpretation is that the energy rate of
change per time unit plus the energy flow through the surface surrounding δV equals the
negative work of the electromagnetic field done on the sources inside δV .
2.6
Multipole radiation
Now that we have developed the mathematical tools needed we can start to consider radiation
from simple charge distributions. One of the most simple systems one can analyze is the linear
antenna, mainly because of it’s simple geometrical shape. The electromagnetic fields far away
from any antenna can be described as a sum of oscillating point multipoles, where the leading
term in the multipole expansion - the dipole term - is of practical interest in many cases. The
fields are independent of the antenna geometry, but, the magnitude of the multipole moments
will be. Close to the antenna we obtain contributions from so-called quasistatic components
as well as radiation terms.
To get started let us consider a Cartesian coordinate system fixed in space. We place a
localized charge distribution with dimension d at the origin O. We supply it with an electric
current making the charge distribution emitt radiation with frequency ω giving it a wavelength
λ = 2π/ω = 2π/k where k is the wave number. We place ourselves at a distance r = |r| from
the the charge distribution with a measuring apparatus. Depending on our distance from it
there are three regions of interest:
• Static zone:
d≪r≪λ
• Radiation zone: d ≪ λ ≪ r
• Induction zone: d ≪ r ∼ λ
We will in almost every case considered in this thesis restrict ourselves to problems in the
radiation zone. Let us first consider the case of fields varying harmonicly in time. We separate
the spatial dependence from the time dependence by Fourier transforming the charge and
current densities. Without loss of generality we write
Z ∞
ρ(r, t) eiωt dt
ρω (r) ≡ F{ρ(r, t)} =
−∞
Z ∞
1
ρω (r) e−iωt dω
(2.107)
ρ(r, t) =
2π −∞
and
Z
∞
J(r, t) eiωt dt
Jω (r) ≡ F{J(r, t)} =
−∞
Z ∞
1
Jω (r) e−iωt dω
J(r, t) =
2π −∞
(2.108)
which means that we are considering one single frequency in ω space. To obtain the time
dependent fields we will have to integrate over all frequencies available and then take the real
2.6 Multipole radiation
50
part of the resulting expressions to represent the physical quantities. These transformations
also implicate that the potentials can be written in the same way. One finds that
Z ∞
1
Aω (r) e−iωt dω
(2.109)
A(r, t) =
2π −∞
Z ∞
1
φω (r) e−iωt dω
(2.110)
φ(r, t) =
2π −∞
where
Aω (r) =
φω (r) =
Z
′
eik|r−r |
d r Jω (r )
|r − r′ |
Z
ik|r−r′ |
1
3 ′
′ e
d r ρω (r )
4πǫ0
|r − r′ |
µ0
4π
3 ′
′
(2.111)
(2.112)
The Fourier transformed version of the continuity equation is then given by
iωρω = ∇ · Jω
(2.113)
and the electromagnetic fields in ω space are given by
Bω (r) = ∇ × Aω (r),
Eω (r) = −∇φω (r) + iωAω (r)
The Fourier transformation of the currents, potentials and fields imply the following relationship between the physical fields in t space and the fields in ω space:
F(r, t) = Re{ Fω (r) e−iωt }
which follow from the definition. Wee see that in going from t-space to ω-space we only need
to make the substitution ∂/∂t → −iω.
From now on we work in ω space and when there is no risk of confusion we will drop the
subscript ω.In the radiation zone where r ≫ λ we have kr ≫ 1. Let us find an expression for
the magnetic vector potential A since E is related to B through a simple relation far away
from the source. There are several ways (all equivalent) of performing this calculation of A.
One way is to start with the exact expression for the retarded vector potential in t-space:
Z
J(r′ , t − |r − r′ |)
1
(2.114)
d3 r ′
A(r, t) =
4π
|r − r′ |
and note that in the radiation zones we have |r − r′ | ≈ r − n · r′ , where n is a unit vector in
the r direction. One can then Taylor expand the current density in powers of r − n · r′ :
Z
2
′
′
1
1
′ 2 ∂ J(r , t)
3 ′
′
′ ∂J(r , t)
+ (r − n · r )
∓ ···
d r J(r , t) − (r − n · r )
A(r, t) ≈
4πr
∂t
2!
∂t2
If we then let J(r′ , t) = J(r′ ) e−iωt then the partial derivatives (∂J(r′ , t)/∂t)n → (−iω)n J(r′ , t)
which gives us
Z
1
1
′ 2
3 ′
′
′
A(r, t) =
d r J(r , t) 1 + iω(r − n · r ) + (iω(r − n · r )) + · · ·
4πr
2!
2.6 Multipole radiation
51
′
where the sum in the brackets can be recognized as eik(r−n·r ) for ω = |k| = k. Then we obtain
Z
eik(r−t
′
(2.115)
A(r, t) =
d3 r ′ J(r′ ) e−ikn·r
4πr
′
We then perform an expansion of e−ikn·r and obtain our final expression for the magnetic
vector potential:
Z
∞
e−iω(t−r) X (−ik)n
d3 r ′ J(r′ )(n · r′ )n
(2.116)
A(r, t) =
4πr n=0 n!
R 3 ′
1
Term number n has an order of magnitude ∼ n!
d r J(r′ )(kd)n and since kd ≪ 1 it suffices
to keep the first few terms in the expansion.
2.6.1
Electric dipole terms
Let us examine the spatial part of expansion (2.116). The lowest order term is usually the
term that dominates, so, choosing n = 0 we obtain for the spatial part:
Z
eikr
iω eikr
A(r) =
(2.117)
d3 r ′ J(r′ ) = − p
4πr
4π r
which is valid everywhere outside the source. Here we have integrated partially and used the
continuity equation. p(r′ ) is the electric dipole moment given by
Z
p(r) = d3 rrρ(r)
(2.118)
It is then rather straight forward to obtain the magnetic field. We take the curl and obtain
eikr
1
k2
(n × p)
1−
(2.119)
B(r) =
4π
r
ikr
where we have used the fact that the ∇ operator acts solely on the unprimed coordinates,
therefore ∇ × p(r′ ) = 0. From the expression above we see that the magnetic field is always
orthogonal to r. The electric field can in general be obtained from Maxwells equations. We
use ∇ × B = −iωE and obtain
1
ik
eikr
1
+ (3n(n · p) − p) 3 − 2 eikr
k2 (n × p) × n
(2.120)
E(r) =
4π
r
r
r
which has components that are both orthogonal and parallell to r. In the radiation zone a
great deal of simplification is possible. We demand that the radiation fields be proportional
to 1/r and obtain
eikr
k2
(n × p)
4π
r
E → B×n
E →
(2.121)
(2.122)
in the limit kr ≫ 1. In this limit the magnetic field B is orthogonal to the position vector
r. Let us find the angular distribution of radiated power with the help of (2.105). We then
obtain
2 2
k
|n × p|2
(2.123)
|B|2 =
4πr
2.6 Multipole radiation
52
which gives
dP
k4
=
|p|2 sin2 θ
(2.124)
dΩ dω
32π 2
The sin2 θ dependence is typical for dipole terms far away from the charge distribution. A
linear dipole antenna placed along the z-axis is an example, and we will later show that the
same contribution to the angular distribution arises when one considers the l = 1, m = 0 term
in the multipole expansion of the electromagnetic fields. More on this later.
2.6.2
Magnetic dipole terms
The term n = 1 in the expansion can be put in the form
Z
eikr 1
− ik
d3 r ′ J(r′ )(n · r′ )
A(r) =
4πr r
(2.125)
where we have added an extra term to make the expression valid everywhere, see [9]. The
electromagnetic fields arising from the n = 1 terms can be put in a similar form as the n = 0
terms if we expand n · r′ in symmetric and anti-symmetric parts:
(n · r′ )J =
1
1
[(n · r′ )J + (n · J)r′ ] + (r′ × J) × n
2
2
(2.126)
where the first term is symmetric. We obtain a division A(r) = AS (r) + AA (r) where
ik
eikr
1
AA =
(n × m)
1−
(2.127)
4π
r
ikr
is the contribution from the anti-symmetric part. We have also defined the magnetization due
to the current:
Z
Z
1
3 ′
′
d3 r ′ r′ × J(r′ )
(2.128)
m = d r M(r ) =
2
The contribution from AA (r) to the electromagnetic fields is
1
1
ik
eikr
ikr
2
BA (r) =
+ (3n(n · m) − m)
−
e
k (n × m) × n
4π
r
r3 r2
eikr
k2
1
EA (r) = − (n × m)
1−
4π
r
ikr
From the expressions above we see that now, the magnetic field has components orthogonal
and parallell to r while the electric field only has a transverse component.
2.6.3
Electric quadrupole terms
Let us now consider the symmetric part of A(r). We integrate the symmetric factor:
Z
Z
iω
1
3 ′
′
′
d r [(n · r )J + (n · J)r ] = −
d3 r ′ r′ (n · r′ )ρ(r′ )
(2.129)
2
2
with the aid of the continuity equation. The expression for the symmetric part of the vector
potential becomes
Z
k2 eikr
1
AS (r) = −
d3 r ′ r′ (n · r′ )ρ(r′ )
(2.130)
1−
8π r
ikr
2.6 Multipole radiation
53
One can then show that the contribution to the magnetic field from the symmetric part of A
in the radiation zone (kr ≫ 1) is
Z
ik3 eikr
d3 r ′ (n × r′ )(n · r′ )ρ(r′ )
(2.131)
BS (r) = −
8π r
At this point we wish to make contact with the quadrupole moment tensor Q, which has
spatial components
Z
Qij = d3 r(3xi xj − r 2 δij )ρ(r′ )
(2.132)
P
and [Q]i = j Qij nj . Let us consider the integral part of BS (r):
Z
n × d3 r r(n · r)ρ(r) ≡ n × T
(2.133)
where we for ease of notation drop the primes. The vector T can be written as
Z
Z
3
2
3
2
d rρ[xynx + y ny + yznz ] ey
d rρ[x nx + xyny + xznz ] ex +
T =
Z
d3 rρ[xznx + yzny + z 2 nz ] ez = Tx ex + Ty ey + Tz ez
+
Let us adopt the notation {x, y, z} → {x1 , x2 , x3 } = {xi }. Then we see that the components
of T can be written as
X
Ti =
Tij nj
(2.134)
j
where the components of the tensor
Tij =
Z
d3 r ρ xi xj
(2.135)
as a symmetric, trace free, tensor of rank two. Let us examine the components of the cross
product n × Q:
X
X
(n × Q)k = ǫijk ni Qj = ǫijk
Qjp np ni = ǫ12k
(Q2p np n1 − Q1p np n2 )
p
p
Let us consider the diagonal elements p = j explicitly. Also choose k = 3 arbitrarily. Then
we obtain
(n × Q)3 = ǫ123 (Q22 n2 n1 − Q11 n1 n2 ) = n1 n2 (Q22 − Q11 )
Using the definition of Qij we obtain
Z
Z
Q22 − Q11 = d3 r ρ[(3y 2 − r 2 ) − (3x2 − r 2 )] = 3 d3 r ρ(y 2 − x2 ) ≡ 3(T22 − T11 )
i.e. all terms containing r 2 vanishes! The occurrence of the kroenecker delta function in Qij
means that we don’t obtain any other contribution from r 2 in the non-diagonal terms of the
cross product. Thus we can replace T by Q/3 and write
ik3 eikr
n×Q
24π r
ik3 eikr
ES (r) = −
(n × Q) × n
24π r
BS (r) = −
2.6 Multipole radiation
54
We notice that the complexity of the expressions increase with n, so that it becomes more and
more cumbersome to evaluate the terms. We therefore content ourselves with the two first
terms in the expansion. The contribution from the electric quadrupole term to the angular
distribution of radiation is then in general
k6
dP
=
|(n × Q) × Q|2
dΩ dω
1152π
(2.136)
which is fairly complicated to evaluate unless the system posesses some kind of symmetry. One
example is an oscillating spheroidal distribution [9]. The contribution from the quadrupole
term then becomes
k6
dP
=
Q2 sin 2 θ cos 2 θ
(2.137)
dΩ dω
512π 0
where Q0 = Q33 . The analogous contribution from the multipole expansion which we perform
later, resides in the l = 2, m = 0 term.
2.6.4
Linear antenna
Let us give a short treatment of the linear center-fed antenna using eq. (2.115). Let us consider
an antenna of length d placed along the z-axis with it’s center at the origin in a Cartesian
coordinate system. We assume that the current is harmonic and consider only the spatial part
of the potential. The current density is then given in ω-space by
kd
J(r) = I sin
− k|z| δ(x)δ(y) ez
(2.138)
2
where we drop the prime notation. With this current density it is possible to solve for the
exact electromagnetic fields. The vector potential becomes
Z
kd
Ieikr d/2
− k|z| e−ikz cos θ
(2.139)
A(r) =
dz sin
4πr −d/2
2
where θ is the angle between n and the source vector r′ . The result of the integration is given
in many textbooks so we will just write down the contribution to the angular distribution,
which is
kd 2
cos
θ
−
cos
I 2 cos kd
dP
2
2 = 2
(2.140)
dΩ dω
8π sin θ
This expression is exact and valid everywhere. If we take the limit kd ≪ 1 and perform Taylor
expansions on the terms in the brackets we obtain
I2
dP
≈ 2
dΩ dω
8π
1 −
1
2
kd
2
2
cos θ − 1 +
sin θ
1
2
2
kd 2 2
k4
|p|2 sin2 θ
=
32π 2
(2.141)
where |p| is the dipole moment of the antenna. This expression is identical to eq. (2.124).
Thus far away from the linear antenna we obtain a dipole pattern. When we later treat the
linear antenna using the multipole formalism developed later on we would thus expect that
to lowest order we obtain only contribution from the l = 1, m = 0 multipole to the angular
distribution.
2.6 Multipole radiation
2.6.5
55
Circular loop antenna
Let us now consider a different example. Let us calculate the radiation fields from a circular
loop antenna, shown in figure (2.1). We feed the antenna across a small gap in the x-axis
r
z = z’
r
φ
θ
θ
z’
n
r’ φ
φ’
ρ’
φ’
Figure 2.1: A two-dimensional current distribution in the form of a loop antenna. The spherical
coordinates (r, θ, φ) describes the field point r and the cylindrical coordinate system (ρ′ , φ′ , z ′ )
describes the source point r′ .
which gives rise to radiation with frequency ω. We choose the circumference of the loop to be
exactly λ = 2π/ω, meaning that the radius is a = 1/ω. The antenna current will then be a
standing wave which can be described by the Fourier transformed current, see Thidé [15];
J(r′ ) = I0 cos φ′ δ(ρ′ − a)δ(z ′ )e′φ
This particular choice of current density implies that although the circular loop is geometrically
symmetric in the φ-plane, the gap in the x-axis and the subsequent current density introduces
a φ-dependence. Jackson [9] considers a circular loop lying in the xy-plane with arbitrary
radius. He obtains a similar current density, but with the angle φ′ = 0 and thus obtains
rotational invariance in this plane. With these remarks in mind, let us proceed.
We are interested in calculating the fields in the radiation zone i.e. far away from the
antenna. By taking the curl of the spatial part A(r) in eq. (2.115) and only keeping the
lowest order terms in r on can show that the magnetic field is
Z
ieikr
′
(2.142)
d3 r ′ e−ik·r J(r′ ) × k
B(r) ≈ −
4πr
′
r−r
where we have defined k = k |r−r
′ | . Using the figure we see that in the radiation zone k ≈ ker
′
′
and r = aeρ The projection of k in the xy-plane is k sin θ and the angle between the projection
of k in the xy- plane and e′ρ is φ − φ′ giving
k · r′ = ka sin θ cos(φ − φ′ )
(2.143)
2.6 Multipole radiation
56
We now concentrate on the cross product e′φ ×er . We first express e′φ in terms of the Cartesian
unit vectors. This is easy since the r′ vector lies in the plane. Then the unit vectors expressed
in cylindrical coordinates correspond to the polar coordinate unit vectors. Since r′ is rotated
an angle φ′ relative the x-axis the unit vectors are related through the rotation matrix:
′ eρ
ex
cos φ′ sin φ′
(2.144)
=
e′φ
− sin φ′ cos φ′
ey
It is trivial to express n = er = r/r through Cartesian unit vectors. We have er =
sin θ cos φ ex + sin θ sin φ ey + cos θ ez . Then it is very easy to perform the cross product.
The result is
e′φ × ker = k[cos(φ − φ′ )eθ − cos θ sin(φ − φ′ )eφ ]
(2.145)
With these relations in place and the volume element d3 r ′ = ρ′ dρ′ dφ′ dz ′ the integral in eq.
(2.142) becomes
Z
Z
Z
′
′
′
′ ′
′
I0 k dz δ(z ) dρ ρ δ(ρ − a) dφ′ cos φ′ e−ika sin θ cos(φ−φ ) e′φ × er
which can be written as
Z 2π
′
dφ′ cos φ′ e−ika sin θ cos(φ−φ ) [cos(φ − φ′ )eθ − cos θ sin(φ − φ′ )eφ ]
I0 ka
(2.146)
0
We see that we obtain two integrals which are in the eθ and eφ directions. This means that
the magnetic field is orthogonal to r. Continuing:
Z 2π
′
= eθ I0 ka
e−ika sin θ cos(φ−φ ) cos(φ − φ′ ) cos φ′ dφ′
{z
}
|0
− eφ I0 ka cos θ
=I1
Z
2π
|0
′
e−ika sin θ cos(φ−φ ) sin(φ − φ′ ) cos φ′ dφ′
{z
}
=I2
We will not perform the integrals explicitly here. A thorough treatment of them can be found
in [15]. The result for the first integral is
and
cos φ
I1 =
2
Z
sin φ
I2 =
2
Z
2π
e−ika sin θ cos x 1 + cos 2x dx
2π
e−ika sin θ cos x 1 − cos 2x dx
0
0
From the theory of Bessel functions we need the general results
Jn (−ξ) = (−1)n Jn (ξ)
Z
Z
(i)−n 2π
(i)−n π
−iξ cos x
dx e
cos(nx) =
dx e−iξ cos x cos(nx)
Jn (−ξ) =
π
2π
0
0
2.7 Multipole expansion and the vector spherical harmonics
57
Comparing with I1 and I2 we obtain
I1
I2
h
i
= π cos φ J0 (ka sin θ) − J2 (ka sin θ)
h
i
= π sin φ J0 (ka sin θ) + J2 (ka sin θ)
(2.147)
(2.148)
The magnetic radiation field in ω-space is then the difference of I1 and I2 times some constants.
Going back to eq. (2.142) we obtain
B(r) = −
i
ika I0 eikr h
I1 eθ − cos θI2 eφ
4π r
(2.149)
From the expression for the magnetic field above it is difficult to extract information about
the angular ditribution, so let us make the approximation ka ≪ 1, and since sin θ ∼ 1 we can
approximate the Bessel function with:
1
ka sin θ ν
lim Jν (ka sin θ) =
(2.150)
ka→0
Γ(ν + 1)
2
then a very crude approximation is J0 (ka sin θ) → 1 and J2 (ka sin θ) → 0, thus I1 → π cos φ
and I2 → π sin φ. Inserting these approximations into the absolute square of eq. (2.149) we
obtain
i
(ka)2 I02 h
2
2
2
|B(r)|2 →
cos
φ
+
sin
φ
cos
θ
(2.151)
16r 2
In the radiation zone the angular distribution is given by eq. (2.192). This gives us
I 2 (ka)2
dP
= 0
(cos2 φ + sin2 φ cos2 θ)
dΩ dω
32
(2.152)
in ω-space.
2.7
Multipole expansion and the vector spherical harmonics
So far we have considered expansions of the electromagnetic potentials in monopoles. These
expansions were based on the solution of the wave equation in t-space. When we worked in
ω-space we were always led to consider the Helmholtz differential equation which had the
generic form
(∇2 + k2 )ψω (r) = 0
(2.153)
in a source-free region. The function ψ(r) was always a scalar function and in the limit k2 → 0
it was spherically symmetric. We wish to generalize this equation and consider solutions of
the Helmholtz equation in ω-space that are vectors. These vector solutions will be easier to
generalize to radiation with higher spin i.e. gravitational radiation.
We saw that it became increasingly difficult to evaluate the different monopoles when we
considered higher order terms in expansion (2.116). Our way was purely brute-force, but when
we consider vector waves and their relations to the sources of radiation we will find that they
provide a more systematic and elegant method to do this. To get started on this we again
consider the Maxwell equations.
2.7 Multipole expansion and the vector spherical harmonics
2.7.1
58
Vector wave equation
We begin by expressing the Maxwell equations in ω-space in a source free region i.e. J(r) = 0
and ρ(r) = 0. They then reduce to
∇ · E = 0,
∇ × E = ikB
∇ · B = 0,
∇ × B = −ikE
From these equations we can form two sets of equivalent equations. They are
(∇2 + k2 )E = 0
(∇2 + k2 )B = 0
(2.154)
and
B = −ik−1 ∇ × E
E = ik−1 ∇ × B
where for both sets the divergence of the fields is zero; ∇ · E = 0 and ∇ · B = 0. Thus E
and B have similiar exansions as the scalar solution in eq. (2.13), but, these expansions must
satisfy the zero divergence requirements imposed by Maxwell’s equations and be vectors. Let
us consider the left set of equations. From vector theory we know that writing a vector as the
curl of some other vector ensures zero divergence. Hence we write
E = ∇ × rψ(r)
(2.155)
where r is a constant position vector and ψ is some scalar function. Then ∇ · E = 0 is
guaranteed. Let us work out some other relations too:
∇2 E = ∇2 [∇ × (rψ)] = ∇ × (r∇2 ψ)
k2 E = k2 ∇ × (rψ) = ∇ × (rk2 ψ)
⇓
h
i
(∇2 + k2 )E = ∇ × r(∇2 + k2 )ψ
which is zero if ψ satisfies the scalar Helmholtz differential equation. One further thing to
notice is that E is transverse to the position vector i.e. r · E = 0. We will take advantage of
this fact soon. Let us examine the consequences for B. With B = −ik−1 ∇ × E we obtain
ik∇2 B = ∇2 [∇ × E] = ∇ × [∇2 E] = ∇ × [−k2 E] = −ik3 B
Hence we can rearrange and obtain
(∇2 + k2 )B = 0
(2.156)
So if E is a solution then so is also B. When a vector is a solution of the Helmholtz equation
we call it the vector wave equation. From the definition we see that the magnetic field is not
transverse to r. We find
i
1
i
r · B = − r · (∇ × E) = − (r × ∇) · E = L · E
k
ω
ω
(2.157)
where L is the angular momentum operator and ω = |k| = k. An important observation is
that if E and B fulfills the Helmholtz differential equation then so does r · E and r · B. This is
2.7 Multipole expansion and the vector spherical harmonics
59
a consequence of the vanishing divergences of E and B. Of course, since r · E = 0 the solution
is trivial, but when we consider the other set of Maxwell equations the situation for the fields
is reversed. Since the electromagnetic fields are proportional to ψ which we now take as a
solution to the scalar Helmholtz differential equation they will have similiar expansions. We
therfore define the magnetic mulitpole of order (l, m) as
1
l(l + 1) gl (kr)Ylm (θ, φ)
ω
(M )
r · Elm (r) = 0
(M )
r · Blm (r) =
(2.158)
(2.159)
(2)
(1)
where gl (kr) = Ahl (kr) + Bhl (kr) is a radial function. Compare with expression (2.13).
Comparing with the expression for r · B eq. (2.157) we see that
(M )
L · Elm (r, θ, φ) = l(l + 1) gl (kr)Yml (θ, φ)
(2.160)
Since the angular momentum operator acts solely on the angular variables, the r dependence
comes from the radial gl (kr) function. Can we find an expression for the electromagnetic field
through this equation? The key is to notice that the angular momentum operator changes the
m values while keeping l constant as it acts on the spherical harmonic. Since we know that
L2 Yml = l(l + 1)Yml we can write E ∝ Cl (r)LYml . Then
(M )
L · Elm
= Cl (r)L2 Ylm = Cl (r)l(l + 1)Ylm
⇓
Cl (r) = gl (kr)
(M )
Elm (r)
⇓
= gl (kr)LYlm (θ, φ)
(2.161)
(M )
(2.162)
(M )
Blm (r) = −iω −1 ∇ × Elm (r)
Sometimes these multipole fields are called transverse electric (TE) fields since the electric
field is orthogonal to the radius vector.
Running through the same procedure again, only this time we use the second set of Maxwell
equations, we obtain similar results. In this situation one defines the electric multipole of order
(l, m) as
(E)
r · Elm (r) = −
1
l(l + 1) fl (kr)Ylm (θ, φ)
ω
(E)
r · Blm (r) = 0
where the fields are called transverse magnetic multipoles (TM). The function fl (kr) is the
same as gl (kr) but with different constants. We then obtain for the fields:
(E)
Blm (r) = fl (kr)LYlm (θ, φ)
(E)
(E)
Em
(r) = ik−1 ∇ × Blm (r)
(2.163)
(2.164)
2.8 Determination of the multipole coefficients aE (l, m) and aM (l, m)
60
The four fields (2.161) to (2.164) constitutes a complete set of solutions to Maxwell’s equations
in a source-free region of space. At this time we can introduce the generaltisation of the
spherical harmonic, namely the vector
Xlm (θ, φ) = p
1
l(l + 1)
(2.165)
LYlm (θ, φ)
which we call the vector sphercial harmonic of order (l, m). They are defined to be exactly
zero for l = 0 i.e. X00 = 0 and obey the following orthonormality conditions;
Z
dΩ X∗l′ m′ Xlm = δl′ l δm′ m
Z
∗
= 0
dΩ Xl′ m′ · r × Xlm
for all l′ , l, m′ and m. We collect all multipole fields and write the total solution of Maxwell’s
equations as
i
Xh
(E)
M (M )
B(r) =
(2.166)
B
+
A
B
AE
lm lm
lm lm
l,m
X
i
=
aE (l, m)fl (kr)Xlm − aM (l, m)∇ × gl (kr)Xlm
k
l,m
i
h
X
(E)
M (M )
E(r) =
AE
lm Elm + Alm Elm
l,m
=
Xi
l,m
k
aE (l, m)∇ × fl (kr)Xlm + aM (l, m)gl (kr)Xlm
(2.167)
(2.168)
(2.169)
where we have defined the reduced coefficents, called the electric (E) and magnetic (M) multipole coefficients:
p
p
(2.170)
aE (l, m) = l(l + 1)AE
l(l + 1)AM
lm and aM (l, m) =
lm
The constants aE (l, m) and aM (l, m) specify the amount of electric and magnetic multipole
fields in the expressions. One can use the scalar equations for r · E and r · B to determine
them. One finds (see [9]) that
Z
k
∗
aM (l, m)gl (kr) = p
dΩ Ylm
(θ, φ) r · B
(2.171)
l(l + 1)
Z
k
∗
(θ, φ) r · E
(2.172)
dΩ Ylm
aE (l, m)fl (kr) = − p
l(l + 1)
The expressions above are valid everywhere outside the source. Let us try to relate the
multipole coefficients to the source’s attributes.
2.8
Determination of the multipole coefficients aE (l, m) and aM (l, m)
When we derived our expressions for the multipoles we assumed that we we’re in a source free
region of space. This lead us to consider homogeneous Helmholtz differential equations for the
2.8 Determination of the multipole coefficients aE (l, m) and aM (l, m)
61
fields. Let us this time keep the source terms and see what happens. The Maxwell equations
in ω space are then
∇·E=ρ
∇ × E = ikB
∇ · B = 0 ∇ × B + ikE = J
It will be advantageous to work with divergence free fields so let us make the substitution
E → E′ = E + ki J where J is the source current density. Outside the source we regain the
physical electric field, E′ = E. We now obtain a new set of equations:
∇ · E′ = 0 ∇ × E′ − ikB = ki ∇ × J
∇·B=0
∇ × B + ikE′ = 0
We can proceed to solve for the fields as we have done in the previous sections in terms of the
source current- and charge- densities. Also, since we want to find expressions for the multipole
coefficients we solve the equations for r · B and r · E. This is not too complicated since we are
working with divergence free fields which means that r · (∇2 + k2 )E′ = (∇2 + k2 )r · E′ and
analogous for B. The results are
1
L · (∇ × J)
k
(∇2 + k2 )r · B = −iL · J
(∇2 + k2 )r · E′ =
where we have introduced the angular momentum vector L = −ir × ∇. We know from earlier
considerations how to solve these equations in terms of an integral over a Green function
times the source terms. The Green function which satisfy the correct boundary conditions i.e.
spherically diverging waves at infinity is given by eq. (2.88). Thus we have
Z
ik|r−r′ |
1
3 ′ e
L′ · (∇′ × J′ )
d r
r·E = −
4πk
|r − r′ |
Z
′
eik|r−r | ′ ′
i
L ·J
d3 r ′
r·B =
4π
|r − r′ |
′
(2.173)
(2.174)
These expressions can be put in a more useful form if we introduce an expansion of the Green
function in terms of spherical harmonics. One can show that [9]
′
X
eik|r−r |
(1)
∗
(θ ′ , φ′ )Ylm (θ, φ)
= ik
jl (kr< )hl (kr> )Ylm
′
4π|r − r |
(2.175)
l,m
where r< , r> means the smaller, larger of r and r ′ . The large argument behavior of gl (kr) and
fl (kr) is determined in eq. (2.193). Let us assume that we are far away from the source i.e.
r< = r ′ and r> = r and kr ≫ 1. Inserting the expansion of the Green’s function we obtain
Z
h
i
X (1)
∗
(θ ′ , φ′ ) (L′ · (∇′ × J′ )
(2.176)
r · E = −i
hl (kr)Ylm (θ, φ) d3 r ′ jl (kr ′ )Ylm
l,m
r · B = −k
X
l,m
(1)
hl (kr)Ylm (θ, φ)
Z
i
h
∗
d3 r ′ jl (kr ′ )Ylm
(θ ′ , φ′ ) L′ · J′
(2.177)
2.8 Determination of the multipole coefficients aE (l, m) and aM (l, m)
62
Outside the source expressions (2.171) and (2.172) are valid. Let us multiply the above
∗ (θ, φ) and integrate over all angles. We obtain
expressions with Ylm
Z
Z
h
i
1
∗
(θ, φ)r · E = −
dΩ Ylm
d3 r ′ (L′ · (∇′ × J′ ) K(l, m, r′ )
k
Z
Z
i
h
∗
(θ, φ)r · B = i d3 r ′ L′ · J′ K(l, m, r′ )
dΩ Ylm
where the quantity
′
K(l, m, r ) = ik
X
jl′ (kr
′
(1)
)hl′ (kr)Yl∗′ m′ (θ ′ , φ′ )
l′ ,m′
= ik
X
Z
′
∗
(θ, φ)Yml ′ (θ, φ)
dΩ Ylm
(1)
jl′ (kr ′ )hl′ (kr)Yl∗′ m′ (θ ′ , φ′ ) δl′ ,l δm′ ,m
l′ ,m′
(1)
∗
(θ ′ , φ′ )
= ikjl (kr ′ )hl (kr)Ylm
Inserting these results we can write down the multipole coefficients:
Z
h
i
ik
∗
(θ ′ , φ′ ) L′ · (∇′ × J′ )
d3 r ′ jl (kr ′ )Ylm
aE (l, m) = p
l(l + 1)
Z
i
h
k2
∗
aM (l, m) = − p
(θ ′ , φ′ ) L′ · J′
d3 r ′ jl (kr ′ )Ylm
l(l + 1)
(2.178)
(2.179)
which are the results we are after. These expressions so far are without approximations and
are valid outside the source. We see that the coefficients are given in terms of integration of
the current density and the angular momentum over the source. There are ways of manipulating the above expressions into slightly more useful ones using vector identities and partial
integration on the quantities inside the brackets. We won’t show the procedure explicitly, we
just state the results of the procedure:
Z
i
∂ h
k2
3
∗
r jl (kr) + ik(r · J)jl (kr)
(2.180)
d r Ylm (θ, φ) ρ
aE (l, m) = p
∂r
i l(l + 1)
Z
k2
∗
aM (l, m) = p
d3 r Ylm
(θ, φ) [∇ · (r × J) jl (kr)]
(2.181)
i l(l + 1)
where ρ is the charge density given through ikρ = ∇ · J. Also, we drop the primes for ease of
notation. Integration is carried out over the current distribution. The first thing to notice is
that if the current density is parallell to r then all magnetic multipole coefficients aM vanish!
2.8.1
The radiation zone limit
We would also like to know if these expressions can be simplified in some particular limit?
The answer is yes. Let us first assume that the source dimensions are of order rmax = a
which is always true in particle or nuclear physics. If the emitted radiation has a wavelength
λ satisfying λ ≫ a then ka ≪ 1. In this case we can expand the Bessel function in powers of
kr and then keep only the lowest order term. A wellknown expansion is
(kr)l
(kr)2
lim jl (kr) =
+ ···
(2.182)
1−
kr→0
(2l + 1)!!
2(2l + 3)
2.8 Determination of the multipole coefficients aE (l, m) and aM (l, m)
63
where (2l + 1)!! = (2l + 1)(2l − 1)(2l − 3) · · · 5 · 3 · 1. Keeping only the first term we can rewrite
the electric multipole coefficient as
Z
i
h
k2
∗
p
ρ(l + 1)(kr)l + ik(r · J)(kr)l (2.183)
d3 r Ylm
aE (l, m) →
{z
} |
{z
}
|
i(2l + 1)!! l(l + 1)
≈
=
k2
i(2l + 1)!!
∼(kr)l
1/2 Z
l+1
∗
d3 r r l Ylm
ρ
l
r
(−1)m−j k2 (2l + 1)(l + 1)
Ql,−m
i(2l + 1)!!
4πl
∼(kr)l+1
(2.184)
(2.185)
The quantity Qlm is defined as
Qlm =
r
4π
2l + 1
Z
d3 r ρ r l Ylm
(2.186)
and is called the electric multipole moment. In the long wavelength approximation we see
that the current density ρ is connected to these multipole moments. By analogy the magnetic
multipole moments which we will introduce in a few lines will be connected to the source
current density J. To see this we perform the same approximation on the Bessel function in
the magnetic multipole coefficient aM (l, m):
Z
i
h
k2
∗
p
(2.187)
∇ · (r × J)(kr)l
d3 r Ylm
aM (l, m) →
i(2l + 1)!! l(l + 1)
r
i(−1)m kl+2 (2l + 1)(l + 1)
=
Ml,−m
(2.188)
(2l + 1)!!
4πl
where Mlm is the magnetic multipole moment mentioned above. And as we promised it is
determined by J. We have defined it as
r
Z
4π
−1
d3 r Ylm r l ∇ · (r × J)
(2.189)
Mlm =
l+1
2l + 1
2.8.2
Angular distribution in the radiation zone
We are now interested in determining how the intensity of radiation, i.e. the power, varies
with the angles. We can derive this relation by making contact with Poyntings theorem, or
more explicitly, the complex Poynting vector
S=
1
E × B(∗)
2
(2.190)
(c = 1). This vector represents the flow of energy from a charge distribution through a surface
surrounding it. It follows that the radiated power per differential suraface area
1
dP
= |S| = |E × B∗ |
dA
2
(2.191)
This suraface area element dA is in general connected to the differential solid angle through
dΩ = dA/r 2 . Let us now consider radiation far away from the source (d ≪ λ ≪ r). Then
2.9 Antennas revisited
64
E → B × n. Combining all relations one obtains the power per solid angle (assuming here
that we work with the ω-space fields):
r2
dP
= |B|2
dΩ dω
2
(2.192)
Let us evaluate expression (2.167) in the limit of kr ≫ 1. All terms are inversely proportional
to some power of r i.e. all terms ∝ r1n and we discard all terms for which n > 1. Let us
consider the behavior of gl (kr) and fl (kr) at kr ≫ 1. Both functions are linear combinations
of spherical Hankel functions, see eqs. (2.14) and (2.15). For real arguments the large argument
(1)
behavior of hl (kr) is
eikr
(1)
hl (kr) → (−i)l+1
(2.193)
kr
(2)
which represents an outgoing wave. Since hl (kr) is the complex conjugate it represents an
ingoing wave which we cannot have at infinity. Therefore we set B = 0 and gl (kr) = fl (kr) =
(1)
hl (kr). Then
∇ × gl (kr)Xlm (θ, φ) ≈ ∇gl × Xlm ≈ (−i)l+1
eikr
n × Xlm
r
(2.194)
where n = r/r. This gives the magnetic radiation field
B→
eikr X
(−i)l+1 [aE (l, m)Xlm + aM (l, m)n × Xlm ]
kr
(2.195)
l,m
Inserting this expression into eq. (2.192) we obtain
2
dP
1 X
= 2
(−i)l aE (l, m)Xlm + aM (l, m)n × Xlm dΩ dω
2k
(2.196)
l,m
We can formally integrate this expression to obtain the total radiated power from any charge
distribution in the radiation zone. Also, if either all magnetic or electric multipole coefficients
vanish then we are left with particularly simple expressions since we won’t have to evaluate
the cross terms.
2.9
Antennas revisited
Let us again calculate the ditributions arising from the different antenna geometries, only this
time we will employ the multipole formalism.
2.9.1
Linear antenna
From earlier considerations we expect the that the contribution to the angular distribution in
the radiation zone to lowest order comes from the l = 1, m = 0 multipole coefficient. Let us
rewrite the current density (2.138) in spherical coordinates (dropping the primes):
J(r) =
i
I(r) h
δ(cos
θ
−
1)
−
δ(cos
θ
+
1)
er
2πr 2
(2.197)
2.9 Antennas revisited
65
where I(r) is a radial current, and the delta functions ensure that it only flows up and down
the z-axis. From the continuity equation one finds
1 dI(r) δ(cos θ − 1) − δ(cos θ + 1)
(2.198)
ρ(r) =
iω dr
2πr 2
Since the current flows radially r×J(r) = 0 and all magnetic multipole coefficients aM (l, m) =
0, so we are left with a pure multipole. The expression for aE (l, m) becomes:
Z d/2
i
k2
1 dI(r) d h
p
aE (l, m) =
r jl (kr)
dr kr jl (kr)I(r) −
k dr dr
2π l(l + 1) 0
Z
h
i
∗
δ(cos θ − 1) − δ(cos θ + 1)
×
dΩ Ylm
|
{z
}
2πδm,0 [Ylm (0,φ)−Ylm (π,φ)]
Due to the delta function we only obtain contributions from m = 0 multipoles! Going back
to the expression for the Legendre polynomial in chapter one we see by inspection that all
multipoles vanish when l = 2, 4, 6, . . . Thus we only obtain contributions from odd multipoles.
Also, let us manipulate the second term in the radial integral:
Z
i Z
i Z
dh
d2 I(r)
dI(r) d h
r jl (kr) = dr
rjl (kr) − dr rjl (kr)
dr
dr dr
dr
dr 2
which is easy verified by differentiating the product rjl (kr)dI(r)/dr and then integrating.
Inserting these results into the expression for aE (l, m) we get
s
Z
i
h d2 I(r)
dI(r) i
dh
4π(2l + 1) d/2
k
2
rjl (kr)
+ k I(r) −
dr rjl (kr)
aE (l = odd, 0) =
2π
l(l + 1)
dr 2
dr
dr
0
Let us specify the current
kd
I(r) = I sin
− k|r|
(2.199)
2
which is the same current as we used when we treated the linear antenna using eq. (2.115),
except that it is written in spherical coordinates. When the current varies sinusoidally we see
that the first term vanishes. Integrating the second term is trivial and we obtain
s
4π(2l + 1) kd 2
I
jl (kd/2)
(2.200)
aE (l = odd, 0) =
πd
l(l + 1)
2
which is an exact expression. Let y = kd/2. The first coefficients are then
p
sin y cos y
−
aE (1, 0) = I(kd)2 6/16πd
y2
y
p
15 1
6
15
2
aE (3, 0) = I(kd) 7/3πd
− 2 sin y −
−
cos y
y4
y
y3
y
The vector harmonics corresponding to the lowest order coefficients are again calculated with
the aid of Maple:
|X10 |2 =
|X30 |2 =
3
sin2 θ
8π
2
63
sin2 θ 5 cos 2 θ − 1
192π
2.9 Antennas revisited
66
As predicted in section 2.6.4 we obtain, to lowest order, a contribution from the l = 1, m = 0
multipole which is associated with the typical sin 2 θ dipole pattern. Let us find the angular
distribution to lowest order in the radiation zone so that we may compare the results from
the different formalisms. Let us first take the long wavelength limit kd ≪ 1. The spherical
Bessel function for l = 1 is given by
j1 (x) =
sin x cos x
−
x2
x
We Taylor expand the trigonometric functions to third order in x to obtain a non-zero result.
We find
x
lim j1 (x) =
x→0
3
and thus jl (kd/2) → kd/6. If we insert this approximation into the first multipole coefficient
we obtain
I 2 k6 d4
|aE (1, 0)|2 →
96π
In order to make a comparison with section 2.6.4 we need to calculate the dipole moment
which is defined in eq. (2.118). The charge density is given in eq. (2.198), and in spherical
coordinates we obtain
Z
p =
d3 r ′ r′ ρ(r′ )
Z
Z d/2
Z
h
i
dI(r) 2π
ez
dr r
dφ dθ cos θ δ(cos θ − 1) − δ(cos θ + 1)
=
2πiω −d/2
dr
0
Z d/2
2 ez
dI(r)
=
dr r
iω 0
dr
When we specify the current in eq. (2.199) the integral is performed without effort and one
obtains
2I 1 − cos(kd/2)
ez
(2.201)
p=
iω
k
In the limit kd ≪ 1 we obtain p → ez Id2 /4i where we used the fact that k = ω in c = 1
units. Thus we have |p|2 = I 2 d4 /42 . Now we insert our approximations into the expression
for the radiated power per solid angle:
dP
dΩ dω
=
≈
1
|aE (1, 0)|2 |X10 |2
2k2
1 4 I 2 d4
k4
1 I 2 k6 d4 3
2
2
sin
θ
=
k
sin
θ
=
|p|2 sin2 θ
2k2 96π 8π
32π 2
42
32π 2
which is exactly the same result as we obtained in 2.6.4 to lowest order.
2.9.2
Circular loop antenna
Let us again return to the situation shown in figure (2.1). This time we write the current
density in spherical coordinates which simplifies the calculations. Then (we as usual drop the
primes the calculation)
J(r) = I0 cos φ δ(r − a)δ(cos θ)eφ = Re I0 eiφ δ(r − a)δ(cos θ)eφ
(2.202)
2.9 Antennas revisited
67
Since r = aer then r · J = 0. Let us calculate some other quantities needed:
r × J = −aI0 eiφ δ(r − a)δ(cos θ)eθ
δ(r − a) δ(cos θ)
∇ · J = iI0 eiφ
r
sin θ
δ(r
− a) δ(cos θ)
=0
∇ · (r × J) = −aI0 eiφ cos θ
r
sin θ
iφ
I0 e
δ(cos θ)
ρ = −Im
δ(r − a)
kr
sin θ
Thus we see that all magnetic multipole coefficients vanish for all l, m because of the presence
of a cos θ term in the integral. Also, the second term in eq. (2.180) vanishes. We are then left
with a pure electric multipole
Z
∂
k2
3
∗
d r Ylm (θ, φ) ρ r jl (kr)
aE (l, m) = p
∂r
i l(l + 1)
Z 2π
Z
Z
∂(r jl (kr))
I0 k
iφ
∗
p
dφ e
dθ δ(cos θ)Ylm (θ, φ) dr δ(r − a)r
= −Im
∂r
i l(l + 1) 0
Z 2π
∂
iI0 ka
∗ π
= Im p
dφ eiφ Ylm
( , φ)
r jl (kr)
(2.203)
2
l(l + 1) ∂r
r=a 0
Let us try to evaluate the integral. Since
s
2l + 1 (l − m)!
∗ π
Plm (0) e−imφ
Ylm ( , φ) =
2
4π (l + m)!
(2.204)
we will only obtain a nonzero value in the integral for m = 1. Explicitly:
s
Z 2π
Z 2π
π
2l + 1 (l − m)!
iφ ∗
dφ e−iφ(m−1)
dφ e Ylm ( , φ) =
Plm (0)
2
4π
(l
+
m)!
0
|0
{z
}
= 2π
s
(2.205)
2πδm,1
2l + 1
Pl,1 (0)
4πl(l + 1)
(2.206)
Let us quote another useful result regarding Pl,1 (0). The Legendre polynomial can be expressed
as a product of gamma functions. They are defined in integral form as
Z ∞
tz−1 e−t dt
(2.207)
Γ(z) =
0
With the aid of the gamma functions one can show that

(−1)n+1 Γ(n+ 3 )

 Γ(n+1)Γ( 3 )2 , l = 2n + 1
2
P1l (0) =


0,
l = 2n
(2.208)
2.9 Antennas revisited
68
where n = 0, 1, 2, . . . Now, let us deal with the derivative of the spherical bessel function. The
quantity in the brackets in eq. (2.203) is in general
h
i
∂ r jl (kr) = kr jl−1 (kr) − ljl (kr)
(2.209)
∂r
where we have used a mathematical identity. We are interested in the far fields i.e. the
radiation fields far away from the source. Let us consider the limit kr ≫ 1. The large
argument behavior of the spherical Bessel function is
lπ
1
sin kr −
lim jl (kr) =
(2.210)
kr→∞
kr
2
which simplifies the derivative considerably. We obtain
∂
lπ
r jl (kr)
= cos ka −
∂r
2
r=a
(2.211)
which means that all multipole fields vanish at ka = π(l+1)
2 . Let us now gather all results a
find a closed form expression for the electric multipole coefficient. It is completely determined
once we specify I0 , ka and l. The result is
r
2πI0 ka 2l + 1 (−1)n+1 Γ(n + 32 )
lπ
aE (l, m) =
cos ka −
(2.212)
l(l + 1)
4π
2
Γ(n + 1)Γ( 32 )
in the radiation zone. This expression can be written more compactly as
aE (l, m) = −2I0 W (l)ka sin(ka)
(2.213)
where W (l) is a function which attaches weight to the different multipoles. It takes discrete
values and we define it as
√
2l + 1 Γ l+2
2
(2.214)
W (l) =
l(l + 1) Γ l+1
2
Let us calculate the first non-vanishing coefficients (l = m = 1) and (l = 3, m = 1). The
results are:
r
3π
aE (1, 1) = −I0
ka sin(ka)
4
r
I0 7π
aE (3, 1) = −
ka sin(ka)
4
4
As a further exercise let us also find the angular distribution to lowest order in aE (l, m). We
know from earlier considerations that for a pure multipole (aE = 0 or aM = 0 ) the radiated
power per solid angle is given by the expression
2
1 X
dP
(2.215)
= 2
(−i)l a(l, m)Xlm dΩ dω
2k
l,m
Since we are only including the lowest order term we obtain
dP
1
= 2 |aE (1, 1)|2 |X11 |2
dΩ dω
2k
(2.216)
2.9 Antennas revisited
69
For this purpuose we calculate |X11 |2 = 12 |LY11 |2 . Since l = m = 1 we obtain by definition
that L+ Y11 = 0. We then use the explicit differential expression of the angular momentum
lowering operator eq. (1.90) to obtain after a simple calculation
|X11 |2 =
3
(1 + cos 2 θ)
16π
(2.217)
Adding these results together we obtain the following expression for the time averaged radiated
power per solid angle:
9 2 2
dP
=
I sin ka (1 + cos2 θ)
(2.218)
dΩ dω
256 0
see figure (2.2) for further details. We see that far away from the source we obtain rotational
invariance in the φ-plane, even though the current density in eq. (2.202) seems to imply
otherwise. The inclusion of higher order terms in the angular distribution leads to a severe
1,0
0,8
0,6
0,4
0,2
0,0
−2
−1
−0,2
0
1
2
−0,4
−0,6
−0,8
−1,0
Figure 2.2: Polarplot of the angular factor (1 + cos2 θ) in the expression for the radiated
power per solid angle. The contribution from the lowest order term produces a fairly uniform
distribution with maxima at θ = 0, π and minima at θ = π/2, 3π/2.
complication of the radiation pattern. With the aid of Maple I found that the next vector
harmonic in the series is
|X31 |2 =
21 1 + 111 cos 2 θ − 305 cos4 θ + 225 cos6 θ
768π
(2.219)
which gives rise to a complicated pattern.
A thing to notice when comparing the results of our calculations. Depending on the
geometry of the antennas we obtain contributions only from a specific value of m and we
2.10 Quantum angular distribution of radiated photons
70
always require that l be an odd non-negative integer. We don’t know if this is a general result,
but the calculations seem to indicate so.
Let us abandon classical electromagnetism in favour of a study of quantum electrodynamics. Our goal is to show the correspondance between the two formulations.
2.10
Quantum angular distribution of radiated photons
We will now consider a quantum system with a current interacting with the electromagnetic
field. The radiated quanta are then the massless spin-1 photons. If the system is initially
in state |ii = |0i|Ei i it may through emission of a single photon with momentum k and
polarization λ undergo a transition to a state |f i = |Ef , k, λi = b
a†kλ |0i|Ef i. The interaction
between the system and the field is governed by the matrix element
Z
b ′ ) · A(k,
b
Vf i = d3 r ′ hf | J(r
r′ ) |ii
(2.220)
b is the current vector operator and A
b is the quantized electromagnetic potential. The
where J
b
b
matrix element hf |J · A|ii is to be integrated over the source distribution. We will repeat the
manipulations of section 2.3 to obtain a more managable form:
b · A|ii
b = . . . = hEf |J|E
b i ih0|b
b
hf |J
akλ A|0i
(2.221)
through elementary operations. We expand the vector potential into plane wave solutions in
ω-space:
o
X Z d3 k 1 n
†
ik· x
∗ −ik· x
b
√
(2.222)
e
b
a
e
e
+
b
a
e
A=
kλ λ
kλ λ
(2π)3 2ω
λ
We are expanding in Cartesian coordinates, hence we use x instead of r. Due to the commutation relations for the creation- and annihilation operators we only obtain a contribution
from the second term, giving
Therefore
1
b
h0|b
akλ A|0i
= √ e∗λ e−ik· x
2ω
(2.223)
b · A|ii
b = √1 J
bf i · e∗ e−ik· x
hf |J
(2.224)
λ
2ω
The expression for Vf i includes a sum over two orthogonal photon polarization directions, so
we obtain no interference terms. It then follows that the probability for photon emission into
a solid-angle element dΩ per unit frequency ω is
Z
2
1 b
ω 2 X dΓ
3
∗ −ik· x √
e
d
x
=
J
(x)
·
e
(2.225)
fi
λ
dΩ dω
4π 2
2ω
λ
Z
2
ω X 3 b∗
ik· x =
d
x
J
(x)
·
e
e
(2.226)
λ
f
i
8π 2
λ
Next on the list is to evaluate this expression. The dot product between the current operator
and the plane wave eλ eik· x with polarization/helicity λ is invariant so we are free to use
spherical coordinates if we want to. Moreover, the factor plane wave eik· x propagating in an
2.10 Quantum angular distribution of radiated photons
71
arbitrary direction in a system K, representing the state |k, λ; θ, φi can be viewed as a rotation
of the state |k, λ; zi representing a plane wave travelling along the z-axis with momentum k
in some coordinate system K ′ . Such a state for the photon can be written in Cartesian
coordinates as
|k, λ = ±1; zi = e±1 eikz = (ex ± iey )eikz
(2.227)
for helicity ±1 respectively, since the massless photon only has two helicity states. This means
that the two states in K and K ′ are connected through the relation
(2.228)
|k, λ = ±1; θ, φi = R(θ, φ) |k, λ = ±1; zi
2.10.1
Plane wave expansion
In the previous section we found an expression for the state describing a photon radiated into
a solid angle Ω in terms of a rotation of the state
|λ = ±1; zi = A± (x) = (ex ± iey )eikz
(2.229)
where the photon is propagating along the z-axis. Now, in section 3.4.1 we found a multipole
expansion of A in spherical coordinates. We write A in eq. (3.174) as (now using j = l):
X
i
(2.230)
a± (l, m)jl (kr)Xlm (θ, φ) − c± (l, m) ∇ × [jl (kr)Xlm (θ, φ)]
A± (x) =
k
l,m
for helicity ± respectively. We have set b = d = 0 since we want the plane wave to be finite
everywhere. To determine the constants we first integrate the dot product X∗lm · A± over all
angles:
Z
Z
Xn
∗
′
′
(2.231)
dΩ Xlm · A± =
a± (l , m )jl′ (kr) dΩ X∗lm · Xl′ m′
l′ ,m′
i
− c± (l′ , m′ )
k
Z
dΩ X∗lm · ∇ × jl′ (kr)Xl′ m′
o
The integral in the last line gives no contribution which is seen from the relations
p
i l(l + 1)
1 ∂
∇ × jl (kr)Xlm = r
jl (kr)Ylm + 2 [rjl (kr)] r × Xlm
2
r
r ∂r
and [9]
Z
dΩ X∗l′ m′ · (r × Xlm ) = 0
Thus we find from the orthogonality of the vector spherical harmonics:
Z
a± (l, m)jl (kr) = dΩ X∗lm · A±
To find c± (l, m) we form the integral
Z
Z
Xn
dΩ X∗lm · (∇ × A± ) =
a± (l′ , m′ ) dΩ X∗lm · ∇ × jl′ (kr)Xl′ m′
l′ ,m′
(2.232)
(2.233)
(2.234)
(2.235)
(2.236)
2.10 Quantum angular distribution of radiated photons
72
Z
o
i
′
′
− c± (l , m )
(2.237)
dΩ X∗lm · ∇ × ∇ × jl′ (kr)Xl′ m′
k
The first integral vanishes by the same arguments as we made above, and the integrand in the
second line can be transformed into
∇ × ∇ × jl (kr)Xlm = −∇2 jl (kr)Xlm = k2 jl (kr)Xlm
(2.238)
since the mode-functions jl (kr)Xlm is divergence free by construction and satisfies the Helmholtz differential equation. We thus find
Z
i
c± (l, m)jl (kr) =
dΩ X∗lm · (∇ × A± )
(2.239)
k
Using expression (2.229) to evaluate the curl of A± in Cartesian coordinates we find ∇×A± =
±kA, giving
Z
c± (l, m)jl (kr) = ±i
dΩ X∗lm · A±
(2.240)
and therefore we have c± (l, m) = ±ia± (l, m). We now proceed to evaluate the integral. Using
the expression for Xlm in terms of the angular momentum operator acting on the spherical
harmonics, and eq. (2.229) for the vector potential we obtain
Z
Z
1
∗
dΩ (ex ± iey ) · (LYlm )∗ eikz
dΩ Xlm · A± = p
l(l + 1)
Z
1
= p
dΩ (L∓ Ylm )∗ eikz
l(l + 1)
p
Z
l(l + 1) − m(m ∓ 1)
∗
p
eikz
(2.241)
dΩ Yl,m∓1
=
l(l + 1)
Using the well-known expansion of eikz in terms of a spherical harmonic,
eikz =
∞
X
il (2l + 1)jl (kr)Pl (cos θ) =
l=0
∞
X
p
il 4π(2l + 1) jl (kr)Yl0 (θ, φ)
(2.242)
l=0
and inserting into (2.241) one obtains
p
Z
∞
l(l + 1) − m(m ∓ 1) X l′ p
∗
′
p
Yl′ ,0
a± (l, m)jl (kr) =
4π(2l + 1) jl′ (kr) dΩ Yl,m∓1
i
l(l + 1)
′
l =0
p
(2.243)
= il 4π(2l + 1) jl (kr)δm,±1
And thus
a± (l, m) = il
p
4π(2l + 1) δm,±1 = ∓ic± (l, m)
When we insert these expressions for the constants into (2.230) we obtain
∞
X
p
1
l
A± =
i 4π(2l + 1) jl (kr)Xl,±1 ± ∇ × jl (kr)Xl,±1
k
(2.244)
(2.245)
l=0
For such a circularly polarized wave the m values of m = ±1 are interpreted as ±1 units of
angular momentum per photon parallel to the propagation direction. This is the expression
2.10 Quantum angular distribution of radiated photons
73
for a photon travelling along the z-axis in the K ′ system. In quantum mechanical terms we
say that when the system undergoes the transition from |ii to |f i it creates a photon with
helicity ±1 travelling along the z-axis, described by the photon wave-function A± as seen
from the K ′ system.
To describe what happens in K we insert this expression into (2.228) to obtain:
b λi
eλ eik·x = |k,
X p
λ
l
=
i 4π(2l + 1) R(θ, φ) jl (kr)Xlλ + ∇ × jl (kr)Xlλ
k
(2.246)
l
The rotation operator acts on Xlλ giving
R Xlλ =
X
(l)
Dmλ (R)Xlm
=
X
m
m
r
4π
Y l∗ Xlm
2l + 1 m,−λ
(2.247)
Thus the plane wave expansion for a photon with helicity λ travelling in an arbitrary direction
is
X λ
l∗
(θ, φ)
(2.248)
eλ eik·x = 4π
il jl (kr ′ )Xlm + ∇ × jl (kr ′ )Xlm Ym,−λ
k
l,m
2.10.2
Angular distribution in terms of spin-±1 harmonics
Now that we have an expression for the state in K we can find the angular distribution of
photons with helicity λ:
Z
2
dΓ
ω b∗ (x) · eλ eik· x = 2 d3 x J
fi
dΩ dω
8π
2
Z
X l
λ
′
′ ′
l∗
′
′
′ ′
3 ′ b∗
= 2ω i
d r Jf i (r ) · jl (kr )Xlm (θ , φ ) + ∇ × jl (kr )Xlm (θ , φ ) Ym,−λ (θ, φ)
k
l,m
X
X
2
2
l b
l
b∗ (l, m, ω)Y l∗ (θ, φ) = 2ω = 2ω il A
(−i)
A
(l,
m,
ω)Y
(θ,
φ)
λ
λ
m,−λ
mλ
l,m
(2.249)
l,m
where we have defined a set of multipole operators which are functions of ω and l, m:
Z
λ
′ ′
′
∗
m
′ ′
3 ′b
′
′
∗
b
Aλ (l, m, ω) = (−1)
d r Jf i (r ) · jl (ωr )Xlm (θ , φ ) + ∇ × jl (ωr )Xlm (θ , φ ) (2.250)
k
where the integration runs over the source, and we have used k = |k| = ω. To obtain the
radiated power into an infinitesimal solid-angle element dΩ per unit frequency we simply
multiply by the energy. The total contribution becomes (when we sum over λ):
X
2 X
2
dP
l
l
l
l
=
(−i) b
a+ (l, m)Ym,1 (θ, φ) + (−i) b
a− (l, m)Ym,−1 (θ, φ)
dΩ dω
l,m
(2.251)
l,m
where we have defined
b
a± (l, m) =
√
b± (l, m, ω)
2 ωA
(2.252)
2.11 Polarization and the Stoke’s parameters
74
(we are suppressing the ω-dependence). This expression for the quantum mechanical angular
distribution is manifestively equivalent (one can show that the operator coefficients agrees
with the classical multipole coefficients in the classical limit, but this will not be done here)
to the classical angular distribution! We also found this agreement between classical electromagnetism and quantum electrodynamics for scalar field radiation, i.e. the quantum formulas
are obtained from the classical formulas by replacing the Fourier components by the matrix
elements of the corresponding transition.
This rule is an expression of Bohr’s correspondance principle and is a particular case of
a general relation between the Fourier components of classical quantities and the quantum
mechanichal matrix elements in the quasi-classical case.
The radiation is quasi-classical for transitions between states that have large quantum
numbers so that the radiated particle energy ω = Ei − Ef is negligable compared to the
energy of the radiating system in it’s initial and final states. This fact has no effect on
the form of the formulas which are valid for all transitions. This explains the fact that the
correspondance principle for radiation intensity is valid in the general, and not only for quasiclassical radiation.
2.11
Polarization and the Stoke’s parameters
Since light i.e. electromagnetic radiation is vectorial by nature we say that it posesses polarization. An electromagnetic wave propagating in the z-direction say, will then have components
Ex (z, t)ex + Ey (z, t)ey = E0x cos(ωt − k · r + δx )ex + E0y cos(ωt − k · r + δy )ey
that are solutions of the wave equation in free space. Here E0x , E0y are the amplitudes and
δx , δy are arbitrary phases. Ex and Ey are then referred to as the polarization components
of the electric field. We will later develop a powerful formalism based on these components
that can account for all types of polarized radiation, but let us first review some background
material which will increase our understanding later.
2.11.1
The polarization ellipse
The polarization ellipse is a locus of points traced out in a plane transverse to a wave’s direction
of propagation and conveniently describes the polarization state. Let us give a brief derivation
of this statement mathematicly. First, we define the propagator τ = ωt − k · r and write the
components as
Ex = E0x cos(τ + δx ), Ey = E0y cos(τ + δy )
(2.253)
With the aid of trigonometic angle identities one easily obtains
Ex
= cos τ cos δx − sin τ sin δx
E0x
Ey
= cos τ cos δy − sin τ sin δy .
E0y
Let us eliminate the propagator τ from these equations by writing
Ey
Ex
sin δy −
sin δx = cos τ sin(δy − δx )
E0x
E0y
Ex
Ey
cos δy −
cos δx = sin τ sin(δy − δx ).
E0x
E0y
(2.254)
(2.255)
(2.256)
(2.257)
2.11 Polarization and the Stoke’s parameters
75
We square both expressions and add them together to obtain
Ey2
Ex Ey
Ex2
cos δ = sin2 δ
+
−2
2
2
E0x E0y
E0x E0y
(2.258)
where δ = δy − δx . Inspection shows that this is the general equation for an ellipse in the
Ex , Ey -plane, as stated in the beginning of this section. Collett [2] deduces many other
properties regarding the polarization ellipse, but, even though it for some choices of E0x , E0y
and δ can describe linearly and circularly polarized light it is inadequate for our purpuoses.
This is were the Stoke’s parameters enter the stage.
2.11.2
The Stoke’s parameters
When the wave propagates through space we now know that the ’light vector’ in a plane
transverse to the direction of propagation traces out som kind of ellipse. The vector rotates
with a very high angular velocity and for light the periodicity is of order 10−15 s. This means
that we are in practice never able to observe the polarization ellipse. Another drawback is that
it only describes fully polarized light. Light is by nature often partially or fully unpolarized
and the polarization ellipse is therefore just an idealization of light, being only correct for
a short period of time. The Stoke’s paramters emerge when we consider the time average
values h. . .i of the polarization ellipse. This in return demands that we represent polarized
light through observables; and it was these observables that were found by Stokes. They are
four by numbers, the first one describes the intensity of light and the other three describe the
polarization state. Let us now follow [2] and derive these. Consider the plane waves in eq.
(2.253) at z = 0. We introduce time dependent quantities through
Ex (t) = E0x (t) cos(ωt + δx (t)), E(t)y = E0y (t) cos(ωt + δy (t)).
(2.259)
We assume that the phases vary slowly in time compared to the rapid oscillations in the
cosines. From the above expressions we once again obtain the equation for the polarization
ellipse eq. (2.258). Let us consider monochromatic waves, this means that the amplitudes and
phases remain constant in time. Thus we have
Ex (t)Ey (t)
Ex2 (t) Ey2 (t)
cos δ = sin2 δ
+
−2
2
2
E
E
E0x
E0y
0x 0y
(2.260)
Let us take the time average of eq. (2.260) during the time of observation. In this way we will
obtain observables. Since the periodicity of the waves is small we can take the observation
time T equal to infinity. We label the time-averaging process of a variable as
Z
1 T
hEi (t)Ej (t)i = lim
dt Ei (t) Ej (t)
(2.261)
T →∞ T 0
and thus obtain
hEx (t)Ey (t)i
hEx2 (t)i hEy2 (t)i
+
−2
cos δ = sin2 δ
2
2
E0x E0y
E0x
E0y
(2.262)
We can rearrange the above equation as
2
2
4E0y
hEx2 (t)i + 4E0x
hEy2 (t)i − 8E0x E0y hEx (t)Ey (t)i cos δ = (2E0x E0y sin δ)2
(2.263)
2.11 Polarization and the Stoke’s parameters
76
Now we calculate the time averages using eq. (2.261). We insert expressions (2.259) with
constant amplitues and phases and obtain
2 ; hE 2 (t)i = 1 E 2 ; hE (t)E (t)i = 1 E E cos δ.
hEx2 (t)i = 12 E0x
x
y
y
2 0y
8 0x 0y
(2.264)
Insertion of these results into eq. (2.263) we find after some algebra that
2
2 2
2
2 2
(E0x
+ E0y
) − (E0x
− E0y
) − (2E0x E0y cos δ)2 = (2E0x E0y sin δ)2
(2.265)
It wasn’t from the above equation that Stoke’s first defined his nowadays famous parameters.
The above equation describes fully polarized light from monochromatic waves, so let us define
2 + E2
I ≡ E0x
0y
2 − E2
Q ≡ E0x
0y
U ≡ 2E0x E0y cos δ V ≡ 2E0x E0y sin δ
where by definition I equals the total intensity of radiation, Q describes the amount of linear
horisontal or vertical polarization, U describes the amount of +45o or −45o polarization, and
finally V describes the amount of left or right circular polarization. As we have defined these
quantities they are all real.
Another way to define the Stoke’s parameters arise when one wishes to bypass the timeaveraging process. If we write electromagnetic fields in terms of complex exponentials i.e.
Ei (t) = E0i ei(ωt+δi ) = E0i eiωt where E0i = E0i eiδi
and i = {x, y}, we can define the Stoke’s parameters in Cartesian and Spherical coordinates
(we will use the Spherical version) as follows:
CARTESIAN COORDINATES SPHERICAL COORDINATES
I:
Ex Ex∗ + Ey Ey∗
Eφ Eφ∗ + Eθ Eθ∗
∗
∗
Eφ Eφ∗ − Eθ Eθ∗
Q:
Ex Ex − Ey Ey
U:
Ex Ey∗ + Ey Ex∗
Eφ Eθ∗ + Eθ Eφ∗
i(Eφ Eθ∗ − Eθ Eφ∗ )
V :
i(Ex Ey∗ − Ey Ex∗ )
These are the Stoke’s parameters for complex monochromatic waves with constant amplitudes
and phases. In the Cartesian case one obtains exactly the same definition (as one should) of
the Stoke’s paramters as in eq. (2.266). It is also common to assemble all parameters into one
column matrix, also often referred to as the Stoke’s vector - though mathematically it is not.
Then

 

Eφ Eφ∗ + Eθ Eθ∗
I
 Q  
Eφ Eφ∗ − Eθ Eθ∗ 

=
(2.266)
S=

 U  
 Eφ Eθ∗ + Eθ Eφ∗ 
V
i(Eφ Eθ∗ − Eθ Eφ∗ )
in sphercial coordinates. To further show the relation between the Stoke’s parameters and the
radiated field we relate the Stoke’s parameter I to the complex Poynting vector in eq. (2.191).
Since far away from the source n × E∗ = B∗ the Poynting vector becomes
1
S = (E · E∗ )n
2
2.12 Stoke’s parameters for a linear antenna
77
As we have defined I as the total time-averaged intensity I = E · E∗ = Eφ Eφ∗ + Eθ Eθ∗ we
see that the magnitude of the Poynting vector is proportional to the first Stoke’s parameter.
This means that upon specifying the first Stoke’s parameter one at the same time obtains the
angular distribution of radiation.
Let us try to specify the polarization state of the two systems we have considered so far.
2.12
Stoke’s parameters for a linear antenna
Let us find the Stoke’s vector i.e. the polarization state in the radiation zone for a linear
antenna. If one evaluates the integral in section 2.6.4 one finds that the magnetic vector
potential becomes
1 2Ieikr cos(kd cos θ/2) − cos(kd/2)
A(r) = ez
4π r
sin2 θ
No approximations have yet been made. Let us simplify our calculations somewhat by taking
the small antenna limit kd ≪ 1. Taylor expansion of the above terms leads to
A(r) ≈
eikr
(kd)2 ez
16πr
By expressing ez through spherical unit vectors and then evaluating the curl of A(r) one finds
B(r) =
eikr
ik3 d2
sin θ
eφ
16π
r
where we have imposed the radiation zone limit kr ≫ 1. Now, in this limit we have E = B × n
where n = r/|r| is a unit vector in the direction of observation. This gives us simpliy
E(r) =
iω
eikr
(kd)2 sin θ
eθ
16π
r
The Stoke’s vector in expression (2.266) follows trivially since we only have the Eθ component:


1
 −1 

S = I0 sin2 θ 
(2.267)
 0 
0
2
ω
(kd)4 . This configuration of the Stokes vector corresponds to vertically
where I0 = 16πr
polarized radiation i.e. the field oscillates in the eθ direction; and as expected we regain the
typical sin2 θ dipole pattern.
2.13
Stoke’s parameters from a circular loop antenna
Let us go through the same process as in the previous section, only this time we use the results
obtained in section 2.9.2 using the multipole formalism. Here we found that
aE (l, m) = −2I0 W (l)ka sin(ka),
aM (l, m) = 0
(2.268)
2.13 Stoke’s parameters from a circular loop antenna
78
where m = 1 and l = 1, 3, 5 . . . We have showed that in the radiation zone the electromagnetic
fields are given by
eikr X
(−i)l+1 aE (l, m)Xlm
kr
B =
(2.269)
l,m
(2.270)
E = B×n
in ω-space. Let us again only keep the lowest order term l = m = 1, this is permissible since
we assume that ka is small. Let us find the explicit form of the electric field only since the
Stokes vector is given by the components of this field. Then
eikr
aE (1, 1) X11 (θ, φ) × n
kr
r
3π a
= I0
sin(ka) eikr X11 × n
4 r
(2.271)
E(r) = −
(2.272)
Here X11 = LY11 . Let us use the angular momentum operator in sphercial coordinates. Then
we have
1 ∂
∂
− eθ
(2.273)
L = −i eφ
∂θ
sin θ ∂φ
r
3
sin θ eiφ
(2.274)
Y11 (θ, φ) = −
8π
It is then trivial to calculate the cross product. One finds
r
9 a
sin(ka) eiα(r) (cos θ eθ + eφ )
E(r) = I0
32 r
where α(r) = kr + φ + π/2. We see already now that the circular loop antenna gives rise to a
more complicated polarization pattern in the radiation zone as compared to the linear antenna
treated in the previous section. This is of course a consequence of it’s more complicated
geometrical shape.
We see that the electric field propagatees in the r direction with components perpendicular
to r. The components are
Eθ = E0 (r) eiα cos θ
Eφ = E0 (r) eiα
r
9 aI0
E0 (r) =
sin(ka)
32 r
We again calculate the Stokes vector

1 + cos2 θ
 1 − cos2 θ
S =I
 2 cos θ
0
in spherical coordinates. One finds




1
1

 0 

2  −1
 ≡ 2I 


 cos θ  − I sin θ  0
0
0




(2.275)
where I = E02 . The first thing we notice is that the circular loop antenna has I ∝ 1 + cos2 θ
which is consistent with our previous calculations. The second thing to notice is that due
2.13 Stoke’s parameters from a circular loop antenna
79
to the additivity of the Stokes vector we can split the total contribution to the polarization
from the electric field in two smaller, less complicated contributions. The first contribution
corresponds to θ dependent linear polarized field and the second is from a linear antenna
(vertically polarized field). The total polarization state is then a superposition of the first
contribution, which has twice the strenght and the second contribution (which is subtracted
from the first).
Chapter 3
Spin weighted harmonics and their
application to electromagnetism
So far we have been performing the multipole expansion of the electromagnetic field using the
vector harmonics Xlm (θ, φ). Although this procedure was successful it demanded a great deal
of mathematical rigour and effort. Another set of basis functions, the so-called spin-s spherical
harmonics which aren’t that well-known, except maybe among some relativists, are spherical
analogues of the vector harmonics which are more natural to implement in radiation problems
from finite sources since the boundary conditions at infinity are spherical in nature. We will
j
use the notation Ym,s
for these functions. The left subscript m is linked to properties in a
fixed inertial reference frame while the s index - as we will show - is connected to a description
of the same function from a rotated coordinate system - the body fixed system.
The purpuose of this chapter is to introduce the reader to this class of functions and
to show how they are used in the multipole expansion. We will show how they naturally
appear when one considers the photon wave function and the requirements imposed on it
from symmetry considerations. The spin-s harmonics are eigenfunctions of a generalized set
of angular momentum operators, which reduce to the ordinary case for s = 0.
In the application of the spin-s harmonics to classical electromagnetism we will find that
only functions with spin weight s = −1, 0, 1 appear in the solutions to Maxwell’s equations.
This is because light consists of massless vector bosons i.e. photons. They are spin-1 particles
which satisfy the transverse gauge condition which means that the wavefunction only has two
degrees of freedom which in turn affects which functions that are allowed.
The term ’spin weight’ is related to the way an object defined on a two-dimensional surface
transforms under rotations through an axis that is orthogonal to the surface (the gradient). If
the object in question is multiplied by a phase factor of the form e−isα (where α is the angle of
rotation) under this rotation, we say that this object has spin-weight s. Some authors define
spin-weight through the phase factor eisα but this is just a matter of convention. The reason
why we choose the minus sign in the phase factor is because we are working with passive
rotations, i.e. rotation of the coordinate axes.
Let us now deal with the derivation of the spin-weighted harmonics. The road to the
desired result is quite long and complicated, but we shall quite soon see the light at the other
end of the tunnel. Let us begin by considering the photon and it’s wave function.
80
3.1 The photon wave function and parity
3.1
81
The photon wave function and parity
The reason why we consider the photon wave function is, in addition to deriving the spinharmonics, because they are the fundamental quanta of the free electromagnetic field. Upon
second quantization the Fourier expansion of the vector potential A in the Heisenberg picture
becomes
o
X Z d3 k 1 n
†
b
√
b
a
A
(x,
t)
+
b
a
A
(x,
t)
(3.1)
A(x,
t) =
kα kα
kα kα
3
(2π)
2ω
α
where Akα is the wave function for a photon with momentum k, polarization eα and energy
ω = |k| (we will from now on drop the subscripts in A). The ’Schrodinger-equation’ for
photons are the Maxwell equations considered in chapter two, which in the transversal gauge
∇ · A = 0 reduce to the wave-equation for the A. How can we determine the allowed states of
the photon? To start with we need to determine the wave functions in the transversal gauge
where A is the vector part of the wavefunction Aµ = (Φ, A). It is custom to set Φ = 0 in this
gauge, and that’s exactly what we will do for now.
We will be conducting our analysis in the momentum representation which we will refer to
as k-space from now on. The reason for doing this is that in relativistic quantum mechanics
the photon wave function in the coordinate representation cannot be regarded as a probability
amplitude since the concept of photon coordinates is phsyically meaningless. The momentum
k of a free photon is however measurable and the wave function in k-space is thus of more
profound physical significance.
In this space the coordinate dependence is replaced by a dependence on the photon momentum k. It then follows naturally that the transversal gauge condition becomes k·A(k) = 0.
We will describe the photon with respect to it’s total angular momentum quantum number
j = l + s. The wave function of a spin-s particle is a symmetrical spinor of rank 2s with 2s + 1
components which transform into linear combinations of one another under rotations of the
coordinate axes. A vector is equivalent to a spinor of rank 2, so the photon has eigenspin
s = 1, [10]. The orbital angular momentum quantum number l describes states with wave
functions containing spherical harmonics of order l. Strictly speaking, only the total angular
momentum quantum number j has any physical meaning, but we can use l and s to classify
the photon transformation properties. s = 1 represents the photons vectorial nature and l
specifies the order of the spherical harmonics Ylm in the wave function as already mentioned.
Any state of the photon can be described with respect to it’s parity P , which is determined
by the order of the spherical harmonics involved. The dependence is
P = (−1)l+1
(3.2)
for spherical harmonics of order l.
To describe these states we consider the action of the inversion operator P (not to be
confused with the partity) on vector- A(k) and scalar- φ(k) functions. We define the parity
of a state through the action of the inversion operator on polar vectors. The action of P on
A and φ is
P A(k) = −A(−k),
P φ(k) = φ(−k)
(3.3)
The minus sign in the first column arises because when we reverse the coordinate axes, the
components of A changes sign. For the vector operator J = L + S we can count the number
of possible states available to the photon when j 6= 0. We find the possible states for spin-1
3.1 The photon wave function and parity
82
particles to be
l = j:
P = (−1)l+1 = (−1)j+1 ,
l = j ± 1: P = (−1)l+1 = (−1)j
(3.4)
If j = 0 the only possible state is the one with l = s = 1 and P = +1.
So far we haven’t taken into account the gauge condition k · A = 0, but assumed that all
components of A are independent. This is obviously wrong. To fix this we must remove all
states which correspond to a longitudinal vector. This vector can be written as kφ(k) where φ
is a scalar. We see that all of it’s three components have transformation properties equivalent
to a single scalar φ under rotations. Scalars have s = 0, and are therefore represented by
spinors of rank 0. Therefore, the state which does not fullfill the transversal gauge condition
has a scalar wave function and the angular momentum j is equal to the order of the spherical
harmonics occuring in φ.
The action of the parity operator on the vector kφ is:
P [kφ(k)] = −(−k)φ(−k) = kφ(−k) = (−1)j kφ(k)
(3.5)
where we in the last step used that φ contains spherical harmonics of order j = l. When we
remove these states we conclude that when j = 0 no states exist, therefore j = 1, 2, . . . When
j 6= 0 we have two states; one even and one odd. A photon with angular momentum j and parity (−1)j is usually called an Ej-photon and if the parity is (−1)j+1 it is called an M j-photon.
We now progress to a more mathematical analysis of the photon wave function. We must
find vector functions which correspond to the eigenvalues j. Since the total angular momentum operator J2 commutes with Jz and the Hamiltonian of the system H we begin by yet
again solving the angular momentum eigenvalue problem for vectors:
J2 Yjm (θ, φ) = j(j + 1)Yjm (θ, φ),
(3.6)
Jz Yjm (θ, φ) = mYjm (θ, φ)
(3.7)
We did this for scalars in chapter one and found that they were the ordinary spherical harmonics Yjm. We thus try Yjm ≡ aYjm where a ∝ k. The orbital angular momentum operator
L in k-space is
L = r × p = −ik × ∂k
(3.8)
which is the same as in the coordinate representation if we substitute k → r. The results we
obtain in the momentum representation carry directly over into the coordinate representation
through a simple substitution! We have in the coordinate representation
(3.9)
[Li , xl ] = ǫijk [xj pk , xl ] = ǫijk xj [pk , xl ] + [xj , xl ] pk = −iǫijk δk,l xj = iǫilj xj
| {z } | {z }
=−iδk,l
=0
and, using the fact that Si xl = −ǫilj xj where S is the eigenspin operator for spin one we
obtain
[Li , xl ] = −Si xl which means that [Li , al ] = −Si al
(3.10)
in momentum space. Now
Ji ak = (Li + Si )ak = Li ak − [Li , ak ] = ak Li .
(3.11)
3.1 The photon wave function and parity
83
This last result means that we have solved the angular momentum eigenvalue problem and
subsequently found the relations:
JYjm = aLYjm
(3.12)
Jz Yjm = aLz Yjm
(3.13)
Thus relations (3.6) and (3.7) are fullfilled. There are three distrinct choices we can make for
a in the momentum representation. They are:
n,
√ ∇n
j(j+1)
,
−in×∇n
√
(3.14)
j(j+1)
where the operator ∇k ≡ |k|∇k is an operator containing only angular derivatives. The
subscript n = k/|k| indicates that we shall differentiate with respect to the photon coordinates.
From now on we will drop it. The numerical factors have been included for future convenience.
We have thus obtained three possible candidates for the photon wave functions. They are then:
(e)
Yjm (n) = √
(m)
Yjm (n) = √
−i
n×
j(j+1)
1
∇Yjm(n),
j(j+1)
1
LYjm (n)
j(j+1)
∇Yj m (n) ≡ √
P = (−1)j
= Xjm , P = (−1)j+1
(l)
(3.15)
P = (−1)j
Yjm (n) = nYjm(n),
The parities of the respective functions are displayed on the right. The corresponding solutions
(m)
in the coordinate representation are simply found by replacing n with er . We see that Yjm
is exactly the same as in eq. (2.165) in chapter one. This means we must be doing something
(e,m)
right here. Both Ylm , where (e)/(m) means that they represent Ej/Mj-photons, are both
(m)
(l)
orthogonal to Ylm . The function Yjm is a linear combination of spherical harmonics of order
(e)
(l)
l = j while Yjm and Yjm contain spherical harmonics of order l = j ± 1. The orthogonality
condition satisfied by all three vectors is
Z
(e,m,l)
(e,m,l)
(3.16)
dΩ Yjm
· Yj ′ m′ ∗ = δj,j ′ δm,m′
Let us now make a choice for the photon wave function. An Ej-photon has parity (−1)j and
(e)
satisfies k · A = 0. Of the above, only Yjm fulfills these demands. An M j-photon with parity
(m)
(l)
(−1)j+1 then requires a wave function proportional to Yjm . The solution Yjm does not
satisfy the transversal gauge condition and cannot represent a physical state of the photon.
Therefore the photon wave functions for electric- and magnetic- photons with definite parities
are:
(e,m)
Aωjm
(e,m)
Φωjm
4π 2
(e,m)
δ(|k| − ω)Yjm (θ, φ)
ω 3/2
= 0
=
(3.17)
(3.18)
where the delta function shows that the energy is ω. The superscripts e, m indicate that the
functions represent electric (e) and magnetic (m) photons. We adopt the same normalization
condition as [10]:
Z
d3 k
(e,m)
(e,m)
(3.19)
ωω ′ Aωjm · Aω′ j ′ m′ = 2πωδ(ω ′ − ω)δj,j ′ δm,m′
(2π)3
3.2 Spherical helicity states
84
hence the numerical factors in front of the vector harmonic. Due to gauge invariance we can
impose the transformations
A → A + nf (n), Φ → Φ + f (n)
(3.20)
without affecting the equations of motion in momentum space. We choose f (n) = CYjm
where C is a constant for electric photons, giving
(e)
Aωjm =
(e)
Φωjm =
i
h
4π 2
(e)
δ(|k|
−
ω)
Y
+
CnY
jm
jm
ω 3/2
4π 2
δ(|k| − ω)CYjm
ω 3/2
(3.21)
(3.22)
For a magnetic photon, this addition to A(m) would render a state without a definite parity
and therefore eq. (3.17) is the only possible choice for a magnetic photon.
3.2
Spherical helicity states
In the previous section we discussed parity and it’s implications on the photon wave function.
We saw that depending on if the state had even or odd parity P it’s corresponding wave
function had to include spherical harmonics of a certain order in j. In this section we will
introduce the concept of helicity. This is because we want to use a set of ’good’ quantum
numbers to enumerate the states of the photon, and as we shall see the helicity eigenvalue λ
is one of those.
In the beginning of section 3.1 we briefly mentioned the polarization vectors eα which are
implicitly included in the expressions for the vector part of the Ej- and the Mj-photons in the
last section. They act as the ’spin-part’ of the wave function and can be resolved into two
circular polarizations which have opposite directions of rotation. The polarization state of the
photon is actually closely related to the helicity. A massless particle does not have any restframe so there is always a preferred direction in space which is along the photon momentum
k. In this case we do not have any rotational symmetry; instead the symmetry is axial. When
there is axial symmetry only the helicity of the particle is conserved. Vectors with only the
components of e1 ± ie2 non-zero (in a reference frame where they are orthogonal to k = e3 )
correspond to λ = ∓1 respectively. Thus the values λ = +1 and −1 correspond to right-hand
and left-hand circulation of the photon.
This means that the component of the angular momentum in the direction of propagation
can only take the values of ±1, the value 0 being impossible! The helicity operator h is
defined as the projection of the particle spin S on the direction of it’s momentum p. Since
the component of total angular momentum J in the direction of propagation is conserved and
the orbital angular momentum operator is L = r × p we have
J · p = (L + S) · p = 0 + S · p = S · p
Therefore
h=
S·p
=S·n
|p|
(3.23)
(3.24)
is conserved. The eigenvalue of h will be denoted by λ and the eigenstates will be called
helicity states. We know that helicity is invariant under Lorentz transformations which don’t
3.2 Spherical helicity states
85
affect the direction of p along which the component of J is taken, so λ is a good quantum
number which we will use to specify the states.
Let us work in a reference frame K ′ with spatial coordinates {x′i } attatched to the particle.
The relativistic solutions ψλ (p) of the eigenvalue equation
(3.25)
(S · n)ψλ (p) = λ ψλ (p)
are up to some constant B given by the spinors
ψλ (p) = Bu(λ) (p)
(3.26)
where u(λ) is the amplitude. In the particle system the x′3 -axis lies in the same direction as
n and is taken as the axis of quantization. In the spinor representation u(λ) is a symmetrical
spinor of rank 2s.
For the massless spin-1 photon there is no rest-frame and the only eigenvalues possible
are λ = ±1. As we will show later on the solutions to these equations are the spherical unit
vectors defined in eq. (1.132)
′
′
e(0) = e′z
e(±) = ± √12 e′x ± ie′y , ,
(3.27)
in the body fixed system K ′ . Let us now find the solutions to eq. (3.25) in k-space. Since S · n
is a scalar it is invariant under rotations. The total angular momentum operators J2 , Jz are
related to infinitesimal changes in the coordinates i.e rotations, so they must commute with
h. Therefore there exist states ψjmλ where j, m, λ all have definite values simultaneously. We
shall name these states spherical helicity states. In k-space the wave function for the photon
becomes
ψλ (p) → ψλ (k) = Bu(λ) (p) δ(p − k)
(3.28)
where the Dirac delta function indicate that we substitute a dependence on the photon momentum p in the coordinate representation for a dependence on the momentum k (which is
taken as an independent variable) in k-space. The spinor u(λ) (k) is a solution of eq. (3.25).
Our goal here is to find the wave function in the space coordinate system K with spatial
coordinates {xi }. The wave function ψjm (k) = hk|j, mi defined with respect to this system
with eigenvalues m and j of the angular momentum operators defined in this system is related
to the body fixed state ψjk (k′ ) = hk′ |j, ki in the following manner:
We will here take on the view that K ′ is the rotated system. To get to K we have to
perform an inverse rotation of the coordinate system:
ψjm (k) = hk|j, mi = hk′ |R−1 |j, mi
=
j
X
k=−j
=
X
hk′ |j, kihj, k|R−1 |j, mi
(j)∗
Dmk ψjk (k′ )
(3.29)
k
since
(j)†
(j)∗
hj, k|R−1 |j, mi = hj, k|R† |j, mi = hj, m|R|j, ki† = Dmk = Dmk
(3.30)
Since K ′ is fixed to the particle the wave function ψjk defined with respect to K ′ will have
a fixed value of the quantum number k independent of orientation and position in space.
3.2 Spherical helicity states
86
We take this value to be k = λ in anticipation of what is to come. This means that we let
ψjk → δkλ ψjk in the sum. We therefore only receive one contribution in eq. (3.30). We define
(j)∗
(3.31)
ψjmλ (k) = ψjλ (k)Dmλ
where we have added the quantum number λ to the wavefunction in K. In the body system the
axis of quantization coincides with the x′3 -axis so ψjλ must be proportional to the amplitude.
We will therefore choose
it to be the same function as in eq. (3.28). The constant B turns
p
out to be equal to (2j + 1)/4π 1 . Then
r
2j + 1 (λ)
(j)∗
u (k) Dmλ (α, β, γ)
(3.32)
ψjmλ (k) =
4π
Since we only need two angles to determine the direction of a point particle we will from now
on take the un-physical Euler angle γ = 0, which makes the other two Euler angles coincide
with the spherical coordinate angles; α = φ and β = θ.
We know that application of the inversion operator P to the spherical helicity states ψjmλ
changes the helicity to the opposite i.e P ψjmλ ∼ ψjm,−λ . Let us determine the prefactor.
With our newly adopted convention γ = 0 we have
(j)∗
(j)∗
(j)∗
P Dmλ (φ, θ, 0) = Dmλ (φ + π, π − θ, π) = eim(φ+π) dmλ (π − θ)eiλπ
(3.33)
Let us compute the action on the reduced matrix elements d(j) . From the definition it follows
that
(j)
(j)
(j)∗
(j)∗
(j)∗
(3.34)
dmλ (π − θ) = dmλ (π) dmλ (−θ) = dmλ (π) dλm (−θ)
A rotation by an angle π about the y-axis gives (−1)j−m δλ,−λ . The Wigner functions are
(j)∗
(j)
(j)
unitary which means that dλm (−θ) = dmλ (θ) = dmλ (θ). We then obtain
(j)∗
(j)∗
dmλ (π − θ) = (−1)j−m dm,−λ (θ)
(3.35)
The following relation also holds, (see [10])
(j)
(j)∗
Dmλ (α, β, γ) = (−1)m−λ D−m,−λ (α, β, γ).
(3.36)
When one combines all these results one finds
(j)∗
(j)∗
P Dmλ (φ, θ, 0) = (−1)j+λ Dm,−λ (φ, θ, 0)
(λ)
To find the action of the inversion oerator on the spinor us
helicity operator h in the body system we use the result
(3.37)
which is an eigenspinor of the
(s)
us(λ) ∼ Dmλ
(3.38)
This result is derived in [10]. Therefore
(s)
(s)
P us(λ) = Dmλ (φ + π, π − θ, π) = e−im(φ+π) dmλ (π − θ)e−iλπ
=
=
(s)
(e−iπ )λ+m (−1)s−m e−imφ dm,−λ (θ)
(−1)λ+s us(−λ)
=
(s)
(−1)λ+s Dm,−λ
(3.39)
(3.40)
(3.41)
Here we adopt the normalization condition dΩ |ψjmλ |2 = 1. The spinors u are assumed normalized
according to u(∗) u = 1. Also, the phase is chosen to be zero.
1
R
3.2 Spherical helicity states
87
What we so far haven’t taken into account is the internal parity of the particle. We represent
this phase factor by ξ. The full expression is
P us(λ) (k) = ξ(−1)λ+s us(−λ) (k)
(3.42)
For the spherical helicity states we then find with the aid of eqs. (3.32), (3.37) and (3.42):
P ψjmλ (k) = ξ(−1)j+s ψjm,−λ (k)
(3.43)
We can extract some valuable insight from above expression. When λ = 0 the helicity states
transform into themselves under P which means they have definite parity. When λ 6= 0
however, the states have no definite parity. To obtain states with definite parity we have to
take linear combinations of states with opposite helicities. Let us define the linear combinations
1
±
= √ (ψjmλ ± ψjm,−λ )
ψjm|λ|
2
(3.44)
where the prefactor is a normalization factor. From this definition we see that
±
±
= ±ξ(−1)j+s ψjm|λ|
P ψjm|λ|
(3.45)
Spin one particles
Let us now consider particles with spin quantum number s = 1. In this case (as we will show
later on) the vector part of the spinors u(λ) are the spherical unit vectors corresponding to the
values λ ∈ {−1, 0, +1}. We will only consider the vector part of the spinors so we will take
the ψ’s to be vectors from now on. We will make the substitutions ψ → Ψ to indicate this.
These vectors are polar so ξ = −1. We then have from eq. (3.45)
j +
P Ψ+
jm|λ| = (−1) Ψjm|λ| ,
P = (−1)j
j+1 Ψ−
P Ψ−
jm|λ| ,
jm|λ| = (−1)
P = (−1)j+1
P Ψjm0 = (−1)j Ψjm0 ,
P = (−1)j
Since the spherical harmonics Yj m have parity (−1)j+1 we see that Ψ+
jm|λ| and Ψjm0 must
−
contain spherical harmonics of order j ± 1 and Ψjm|λ| must contain spherical harmonics of
order j. Since h commutes with J and Jz we know that the solutions to hΨjm|λ| = λΨjm|λ|
must be the same functions that solves J2 Ψjm|λ| = j(j + 1)Ψjm|λ| and Jz Ψjm|λ| = mΨjm|λ| .
Therefore comparing with eqs. (3.15) we have up to some phase factors
(e)
,
Yjm = C1 Ψ+
jm|λ|
(m)
,
Yjm = C2 Ψ−
jm|λ|
(l)
Yjm = C3 Ψjm0
(3.46)
We have already stated earlier that for spin one particles the eigenfunctions of h are the
spherical unit vectors in eq. (3.27) defined with respect to the body fixed system. Let us show
this. In component form we have (denoting the eigenvectors with e):
(S · n)ek = (Si ni e)k = ni (Si ek ) = ni (−iǫikl el ) = −iǫikl ni el = iǫkil ni el = (in × e)k
(3.47)
Therefore
he(λ) = λe(λ) ⇐⇒ in × e(λ) = λe(λ)
(3.48)
3.2 Spherical helicity states
88
In the K ′ system n = e′z so we take e = (e1 , e2 , e3 ) where the unit vectors are defined with
respect to K ′ . For simplicity we drop the primes from now on.
We have three cases. For λ = 0 we obtain e ≡ e(0) = ez . When λ 6= 0 we obtain
ie1 ey + (−ie2 )ex = λ(e1 ex + e2 ey + e3 ez )
(3.49)
We must have e3 =
−ie2 and e2 = ie1 which are equivalent equations.
√ 0. λ = 1 gives e1 = √
We choose e1 = 1/ 2 which gives e2 = i/ 2.
λ =√−1 gives ie1 = −e2 and
√ e1 = ie2 which also are equivalent equations. We choose
e1 = 1/ 2 which gives e2 = −i/ 2. Therefore as promised:
e(±) = ± √12 (ex ± iey , ) ,
e(0) = ez
(3.50)
For us the only interesting solutions here are e(±) since massless particles (=photons) can’t
have spin projection k = λ = 0 which is represented by e(0) . Let us consider a rotation
through ez by an angle α. Since ex , ey span a plane to which ez is orthogonal they transform
under S0(3):
′ ex
cos α sin α
ex
=
(3.51)
e′y
− sin α cos α
ey
Therefore
ex ± iey → ex (cos α ∓ i sin α) ± iey (cos α ∓ i sin α) = e∓iα (ex ± iey )
(3.52)
so
e(±) → e∓iα e(±) ,
e(0) → e(0)
(3.53)
1
√ (Ψjmλ ± Ψjm,−λ )
Ψ±
jm|λ| =
2
(3.54)
The spin-weight s of an object η was defined from the relation η ′ = e−isα η under the rotation
described above. Thus the vectors e(±1) have spin-weight ±1. Let us recapitulate.
When λ = ±1 the helicity eigenfunctions are states with spin-weight ±1. It leads us therefore to suspect that the helicity quantum number λ is connected to transformation properties
of states under rotation about an axis paralell to the particle direction. This should mean
that states with λ = 0 have spin-weight s = 0.
Let us now make us of eq. (3.46). For this purpuose we have to find
From eq. (3.32) we have
Ψjmλ (n) = e
(λ)
r
2j + 1 (j)∗
Dmλ (θ, φ, 0) = e(λ)
4π
r
2j + 1 imφ (j)
e
dmλ (θ)
4π
(3.55)
Insertion gives
(e)
Yjm
(m)
Yjm
(l)
Yjm
r
o
2j + 1 imφ n (j) (+)
(j)
dm,1 e + dm,−1 e(−)
e
8π
r
o
2j + 1 imφ n (j)
(j)
= C2
e
dm,1 n × e(+) + dm,−1 n × e(−)
8π
r
2j + 1 imφ (j) (0)
e dm0 e
= C3
4π
= C1
(3.56)
(3.57)
(3.58)
3.2 Spherical helicity states
89
The cross-products in (3.57) are easily evaluated using eq. (3.48) giving n × e(±) = ∓ie(±) .
Since any vector remains unchanged during a rotation of the coordinate system (in the sense
that xi ei = x′i e′i ) and the e(±1) transform according to eq. (3.53) we realize that under the
rotation described above the components must transform according to
(j)∗
(j)
(j)∗
Dm,∓1 = eimφ dm,∓1 → e∓iα Dm,∓1
(3.59)
to preserve the structure of the vector. Taking the complex conjugate we find that
(j)
(j)
Dm,±1 → e∓iα Dm,±1
(3.60)
Determining the phase factors is just a matter of comparing the left hand sides to the right
hand sides for different values of the quantum numbers and the angles. This is just mathematics so we will only show one of the calculations explicitly. In eq. (3.56) on the left hand
side, form the dot product
(e)
Yjm · eφ = . . . = p
i
h ∂Y
im
1
1
im
jm
+ eφ
Ylm · eφ = p
Yjm
eθ
∂θ
sin θ
j(j + 1)
j(j + 1) sin θ
(3.61)
Expressing the spherical vectors in Cartesian coordinates on the right hand side through
spherical coordinates and taking the dot-product one obtains
r
r
o
i
2j + 1 imφ n (j) (+)
2l + 1 imφ h (j)
i
(j)
(j)
(−)
C1
e
dm,1 e + dm,−1 e
e
dm,1 (θ) + dm,−1 (θ)
· eφ = . . . = √
8π
8π
2
(3.62)
Evaluating both (3.61) and (3.62) at θ = π/2, l = m = 1 one obtains
r
3 iφ
i
(e)
e
(3.63)
Y1,1 (π/2, φ) · eφ = √
2 8π
and
C1
since
r
o
3 iφ n (1)
i
(1)
e d1,1 (π/2)e(+) + d1,−1 (π/2)e(−) · eφ = C1 √
8π
2
(1)
(1)
d1,−1 (π/2) = d1,1 (π/2) = 1/2
r
3 iφ
e
8π
(3.64)
(3.65)
Therefore we immediately obtain C1 = 1. Similarly one obtains C2 = C3 = 1. We can use
eqs. (3.56) and (3.57) to form scalar functions with spin-weight because of the transformation
properties of the spherical unit vectors. We have
r
2j + 1 imφ (j)
(e)
(±)(∗)
e
dm,±1 (θ)
(3.66)
e
· Yjm (θ, φ) =
8π
1
e(±)(∗) · ∇Yjm (θ, φ)
(3.67)
= p
j(j + 1)
The spherical vectors are defined with respect to the body system, so let us express the gradient
in body fixed cartesian coordinates. We have effectively (no z-component in the dot product):
∂ ′
∂
∂ ′
∂
1
√
e(±)(∗) · ∇Yjm = e(±)(∗) ·
e
+
e
∓
i
Yjm
(3.68)
Y
=
±
jm
∂x′ x ∂y ′ y
∂y ′
2 ∂x′
3.2 Spherical helicity states
90
We can switch back to spherical coordinates in the fixed reference frame through the invariance
of the line element ds2 :
ds2 = (dx′ )2 + (dy ′ )2 = dθ 2 + sin2 θdφ2
(3.69)
which defines the two dimensional surface on which the spherical harmonics live. Therfore
∂
∂x′
→
∂
∂θ ,
and
e
(±)(∗)
(e)
· Yjm (θ, φ)
∂
∂y ′
1 ∂
sin θ ∂φ
→
1
= ±p
2j(j + 1)
i ∂
∂
∓
∂θ sin θ ∂φ
(3.70)
Yjm (θ, φ)
(3.71)
Comparing with eqs.(1.148) and (1.149) we see that the quantity in the brackets are just the
raising and lowering operators in the body system expressed in spherical coordinates. If we
choose the negative (positive) sign on the left in (3.71) we obtain a quantity with spin weight
s = 1 (s = −1) since e(±)(∗) has spin-weight s = ∓1. We will therefore define functions which
we will call ’spin-weighted spherical harmonics’ or ’spin-1 harmonics’ through
j
i ∂
∂
−1
√
√ 1 J+ Yjm ,
spin-weight s = 1
Ym,1 =
∂θ + sin θ ∂φ Yjm (θ, φ) =
j(j+1)
j
=√
Ym,−1
1
j(j+1)
j(j+1)
∂
∂θ
−
i ∂
sin θ ∂φ
1
J− Yjm ,
j(j+1)
Yjm(θ, φ) = − √
spin-weight s = −1
Our definition of the spin harmonics agrees with the ones used in [6, 14]. Using eq. (3.66) we
obtain:
q
j
2j+1 (j)∗
√1
(3.72)
8π Dm,±1 = 2 Ym,∓1
or
r
2j + 1 (j)∗
Dm,∓1 (θ, φ, 0)
(3.73)
4π
We see that this is consistent with the transformation properties of the D-functions, see eq.
(3.59). Referring back to eq. (3.60) we deduce from the transformation properties of the
D-functions that
(j)
j
Ym,±1
(θ, φ) = A(j, m) Dm,±1 (θ, φ, 0)
(3.74)
j
Ym,±1
(θ, φ)
=
where the complex constant A could in principle depend on j, m. We demand that the spin-1
harmonics be normalized with respect to integration over all angles. Then
Z
Z
(j)
(j ′ )∗
j
j′∗
∗ ′
′
dΩ Ym,±1 Ym′ ,±1 = A(j, m)A (j , m ) dΩ Dm,±1 Dm′ ,±1
(3.75)
= A(j, m)A∗ (j ′ , m′ )
4π
δj,j ′ δm,m′
2j + 1
4π
=1
2j + 1
= |A(j, m)|2
Therefore
r
(3.76)
(3.77)
2j + 1
(3.78)
4π
where we’ve included a phase-factor. This phase-factor is needed since the D-functions and
their complex-conjugates are related to each other in a non-trivial way. At last we have
−iχ
A(j, m) = A(j) = e
3.3 Spin-weighted spherical harmonics
91
reached our destination, the full expression for the spin-weighted harmonics in terms of the
D-functions. Here they are in all their glory:
r
2j + 1 (j)∗
j
Dm,∓1 (φ, θ, 0)
(3.79)
Ym,±1 (θ, φ) =
4π
r
2j + 1 (j)
=
Dm,±1 (θ, φ, 0) e−iχ
(3.80)
4π
where we remind the reader of our definition of the Wigner D-functions:
(j)
(j)
Dmk (α, β, γ) = e−imα dmk (β) e−ikγ
(3.81)
We will always choose to use (3.79) when we work with these functions. Since there exist
operators which raise and lower the projection of the spin in the body fixed system (represented
by the quantum number k = λ in our calculations) it follows that one can define functions
which can have (almost) arbitrary spin-weight. These can be found by repeated application
of a set of generalized body fixed operators J−s and J+s which reduce to J− and J+ when
s = 0. In our derivation of the spin-weighted harmonics we discovered that there is an intimate
link between the projection of the angular momentum k and our newly introduced quantum
number s which represents spin-weight. In fact k = s. These quantum numbers describe the
same aspect of a function but in different languages or codes.
Now that we have established this connection the generalization of eq. (3.73) to arbitrary
spin-weight is straightforward. We will do this in the following section
3.3
Spin-weighted spherical harmonics
As one might assume from the name the spin-weighted harmonics are a generalization of the
ordinary spherical harmonics discussed earlier. The term ’spin-weight’ is best explained in the
context of rotations. Consider a three dimensional sphere with normal vector n and define a
coordinate system tangential to the sphere surface. Let η(θ, φ) be a quantity defined on the
sphere. Now rotate the coordinate axes an angle α about the n axis. If the relation between
η in the rotated system and the old system can be written in the form
η ′ (θ ′ , φ′ ) = e−isα η(θ, φ)
(3.82)
we say that η has spin-weight s. The derivatives of η may not have well defined spin-weight but
by analogue to quantum mechanics and it’s angular momentum raising and lowering operators
one constructs two operators referred to in the litterature as ð and ð which raises and lowers
the spin-weight respectively, when applied to a quantity of definite spin-weight. They are:
i ∂
∂
s
−s
s
+
ðη ≡ sin θ J+ (sin θη) = − sin θ
sin−s θ η
(3.83)
∂θ sin θ ∂φ
∂
i ∂
ðη ≡ sin−s θ J− (sins θη) = − sin−s θ
−
sins θ η
(3.84)
∂θ sin θ ∂φ
when η has spin-weight s. Notice that in this thesis we will introduce a new notation for the
spin-weight raising and lowering operators. Let us define for a general s:
J+s = sins θ J+ sin−s θ,
J−s = sin−s θ J− sins θ
(3.85)
3.3 Spin-weighted spherical harmonics
92
After the operators (3.85) act on η it will transform according to
(J+s η)′ = e−i(s+1)α (J+s η)
(3.86)
(J−s η)′ = e−i(s−1)α (J−s η)
(3.87)
These operators arise naturally (Torres del Castillo, [4]) when one considers expressions involving derivatives of vector or spinor fields are written in terms of spin-weighted combinations
of the field components. This is exactly what we saw when we formed the spin-weighted components of the spherical harmonic vectors in eq. (3.71). We can therefore assume that these
operators can be introduced when one writes gradients, curls and divergences in terms of
spherical unit vectors. This is exactly what we will do in a little while when we consider the
Maxwell equations.
Let us now define the spin-weighted harmonics of arbitrary order in s and then deduce/write down the properties. The spin-weighted spherical harmonics are defined by acting
(s)
(s)
on the ordinary spherical harmonics s times with J+ and J− . We will denote these functions
j
by Yms
:

q
(1) (0)
(s−1)
(j−s)!

s ∈ [0, j]
· · · J+ J+ Yjm(θ, φ),

+

(j+s)! J

{z
}
|




s times
j
Yms (θ, φ) ≡
q

(−1) (0)
(1−|s|)
(j−|s|)!

s

(−1)
· · · J− J− Yjm(θ, φ), s ∈ [−j, 0]

−
(j+|s|)! J


{z
}
|


|s| times
where J+0 ≡ J+ and J−0 ≡ J− . Maybe a little clarification is needed. When s > 0 we let the
spin weight raising operator act s times on the spherical harmonics. The action of the first
operator yields an object with spin-weight 1. The second operator therefore has s = 1 since it
acts on a spin-1 object. The process is then repeated until the object has spin-weight s. For
s < 0 the process is completely analoguous except that s is negative and we thus obtain an
object with spin weight −|s|. As we see from the above expression the spin spherical harmonics
are not defined for |s| > j and when s = 0 we recover the ordinary spherical harmonics, i.e.
j
Ym0
= Yjm . From the definition above it follows that the complex conjugate of the spin-s
spherical harmonics is
j
j∗
(θ, φ)
(3.88)
Yms
(θ, φ) = (−1)m+s Y−m,−s
∗ = (−1)m Y
which follows from the relation Yjm
j,−m for the ordinary harmonics. You can also
check that
(
m 6= s,
q0,
j
(3.89)
Yms (0, φ) =
2j+1
−isφ , m = −s
(−1)−s
4π e
For each value of s the spin-weighted harmonics form a complete and orthonormal set so any
l :
function f (θ, φ) defined on the sphere can be expanded as a series in Yms
f (θ, φ) =
l
∞
X
X
m=−l l=|s|
j
cjm Yms
(θ, φ)
(3.90)
3.3 Spin-weighted spherical harmonics
93
Table 3.1: The first few normalized spin-s spherical Harmonics
j
m
s
0
0
0
1
±1
0
0
±1
1
1
1
±1
1
−1
±1
j
Yms
(θ, φ)
∓
q
q
±
√1
4π
3
8π
e±iφ sin θ
q
3
8π
sin θ
3
± 16π
(cos θ ∓ 1)eiφ
q
3
(cos θ ± 1)e−iφ
∓ 16π
The coeffients can be determined with the aid of the completeness relations for spin spherical
harmonics. These orthogonality properties are inherited from the properties of the D-functions
and the spherical harmonics. We have
Z
j′∗
j
(3.91)
Yms
(θ, φ) Ym
′ s (θ, φ) dΩ = δj,j ′ δm,m′
and
j
∞
X
X
m=−j j=|s|
j
j∗
Yms
(θ, φ) Yms
(θ ′ , φ′ ) = δ(φ − φ′ )δ(cos θ − cos θ ′ )
which gives
cjm =
Z
j∗
dΩf (θ, φ) Yms
(θ, φ)
(3.92)
(3.93)
As we have shown earlier the spin spherical harmonics are related to the elements of the Wigner
(j)
rotation matrix Dms (R) where R is the rotation operator. This in turn is a representation
of the R3 group. The connection between them when the Euler angle γ 6= 0 is proven in the
article of Goldberg et. al, and we will state the result using the notation introduced in this
thesis. The result is
r
2j + 1 j
−isγ
j
Dm,−s (φ, θ, γ)
(3.94)
=
Yms (θ, φ) e
4π
where the Wigner matrix elements are defined by Goldberg, [6] as
(j)
(j)
Dmk (φ, θ, γ) = eimφ dmk (θ) eikγ
(3.95)
Comparing with our notation for the D-functions, we see that our result in eq. (3.79) are in
complete agreement. Here γ is an Euler angle defined within the context of the y-convention.
We will always take the quantum number in the body fixed system to be the leftmost of
the subscripts in the D-functions. For s = 0 and γ = 0 one relates the ordinary spherical
harmonics to the Wigner matrix elements.
3.3 Spin-weighted spherical harmonics
94
(s)
(s)
We have so far neglected to write down the action of J− and J+ on the spin spherical
harmonics. They satisfy the following angular momentum eigenfunction like relations:
p
j
j
j(j + 1) − s(s + 1) Ym,s+1
J+s Yms
=
,
(3.96)
p
j
j
,
(3.97)
J−s Yms
= − j(j + 1) − s(s − 1) Ym,s−1
j
j
= [s(s + 1) − j(j + 1)] Yms
,
J−s+1 J+s Yms
j
J+s−1 J−s Yms
= [s(s − 1) − j(j +
j
1)] Yms
.
(3.98)
(3.99)
From the last relations we see that spin-s spherical harmonics are eigenfunctions of the J−s J+s
and J+s J−s operators. All these relations follow from eqs. (3.85), (1.143) and (1.144). We can
rewrite the operators in eq. (3.85) as
J+s =
J−s =
i ∂
∂
+
− s cot θ
∂θ sin θ ∂φ
i ∂
∂
−
+ s cot θ
∂θ sin θ ∂φ
(3.100)
(3.101)
It then follows that (3.98) and (3.99) can be written as
1 ∂
∂
1 ∂2
i cos θ ∂
s2
s+1 s
J− J+ η =
sin θ
+
+ 2s 2
−
+ s(s + 1) η
sin θ ∂θ
∂θ sin2 θ ∂φ2
sin θ ∂φ sin2 θ
∂
1 ∂2
s2
i cos θ ∂
1 ∂
s−1 s
sin θ
+
−
+ s(s − 1) η
+ 2s 2
J+ J− η =
sin θ ∂θ
∂θ sin2 θ ∂φ2
sin θ ∂φ sin2 θ
from which the commutator of the spin raising- and lowering- operators becomes
J−s+1 J+s − J+s−1 J−s η = 2sη
(3.102)
Thus if η has spin-weight zero the two operators commute!
In anticipation of our treatment of Maxwell’s equations which consist of expressions involving curls and divergences of vector fields we express the differential operators in spherical
coordinates in terms of spin weighted components of an arbitrary vector field A. Let us now
do this. By quoting for example [13] one has:
i
1 ∂ 2
1 h∂
∂
∇·A =
(r
A
)
+
(A
sin
θ)
+
A
(3.103)
r
θ
φ
r 2 ∂r
r sin θ ∂θ
∂φ
i
i
∂
1h 1 ∂
∂
1 h∂
(Aθ sin θ) −
Aφ er +
Ar − (rAφ ) eθ (3.104)
∇×A =
r sin θ ∂θ
∂φ
r sin θ ∂φ
∂r
h
i
∂
1 ∂
(rAθ ) −
Ar eφ
+
r ∂r
∂θ
To find the spin-weighted components of the vector field we rotate the spherical coordinate
system an angle α about er . It is then straightforward to verify that this operation induces
the transformations
er → er , (eθ ± ieφ ) → e∓iα (eθ ± ieφ )
(3.105)
in the rotated system. Based upon these results we define a new coordiate system related to
the spherical coordinate system by defining the vectors
m=
√1 (eθ
2
+ ieφ ), m =
√1 (eθ
2
− ieφ ), n = er
(3.106)
3.3 Spin-weighted spherical harmonics
95
Based on the definitions the orthogonality properties become
m·m =m·m =m·n=m·n=0
(3.107)
m·m =n·n=1
(3.108)
A = Ar er + Aθ eθ + Aφ eφ = A0 n + A+ m + A− m
(3.109)
and
The vector A can then be written
where A+ and A− are spin weighted components of the vector A with spin weights +1 and
−1 respectively. A0 has spin weight 0. By inserting the expressions for m and m and then
matching the coefficients one can express the spherical components of A through the spin
weighted ones. The results
Ar = A0 , Aθ =
Aφ =
√1 (A+
2
√i (A−
2
+ A− )
− A+ )
(3.110)
Inserting eq. (3.110) into the differential operators and then rearranging the expressions
one finds that the spin raising- and lowering- operators acting on functions with spin-weight
s ∈ [−1, 0, 1] appears. The explicit calculations are done in appendix A, and we write our
results as
1 ∂ 2
1
(r A0 ) + √ (J+ A− + J− A+ )
2
r ∂r
2r
√ ∂
i
i
(∗)
∇ × A = n √ (J+ A− − J− A+ ) + m √
J0 A0 − 2 (rA− )
∂r
2r
2r
√ ∂
i
2 (rA+ ) − J0 A0
+ m√
∂r
2r
∇·A =
(3.111)
(3.112)
where we have introduced the operators
(s=−1)
−J+
≡ J+ =
(s=1)
≡ J− =
−J−
(s=0)
−J+
≡ −J+ ≡ J0 =
i ∂
∂
1
+
sin θ
sin θ ∂θ sin θ ∂φ
1
i ∂
∂
−
sin θ
sin θ ∂θ sin θ ∂φ
i ∂
∂
+
∂θ sin θ ∂φ
(3.113)
(3.114)
(3.115)
This notation has been introduced to agree with [14]. When dealing with Maxwell’s equations
in the next section we will find that certain products of the spin-weight raising and lowering
operators appear. We give their values here:
(∗)
J− J0 = J+ J0
=
1
1
∂θ (sin θ∂θ ) +
∂ 2 ≡ r 2 ∇2Ω
sin θ
sin2 θ φ
where ∇2Ω is the angular part of the Laplace operator in spherical coordinates.
(3.116)
3.4 Solution of the vector Helmholtz equation
3.4
96
Solution of the vector Helmholtz equation
Now that we are more familiar with the concept of spin-s spherical harmonics and the spinraising and lowering operators J+s and J−s we employ our newly developed formalism to the
problem of solving the vector Helmholtz equation. This differential equation appears regularly
when considering radiation problems and also when one quantizes the electromagnetic field,
since one requires the mode-functions in the expression for the quantized magnetic vector
b to be divergenceless solutions of it.
potential A
In this section we solve for the spin-weighted components defined in section 3.3 (eq.
(3.109)) in the general case, and in the next one we restrict ourselves to divergence-free solutions.
The homogeneous Helmholtz equation for a vector field A reads:
∇2 A + k 2 A = 0
(3.117)
We expand A into spin-weighted components:
(3.118)
A(r) = A0 n + A− m + A+ m
From appendix A we have the action of the Laplace operator on the vector field in terms of
the same components:
" #
√
√
∂ 1 ∂
1 1 0
2 1
2 −1
2
2
∇ A =
r A0 + 2 J− A+ + 2 J+ A− + 2 J− J+ A0 n
∂r r 2 ∂r
r
r
r
"
#
√
1 ∂2
1 0 −1
2
+
(rA− ) + 2 J− J+ A− − 2 J−0 A0 m
2
r ∂r
r
r
"
#
√
1 ∂2
1 0 1
2 0
(3.119)
+
(rA+ ) + 2 J+ J− A+ − 2 J+ A0 m
r ∂r 2
r
r
We know that the solution can be found by the method of separation of variables. We write
X
X
X
j
j
j
A(r) = n
g0 (r)Ym,0
(θ, φ) + m
g− (r)Ym,−1
(θ, φ) + m
g+ (r)Ym,1
(θ, φ)) (3.120)
j,m
j,m
or
As =
X
j,m
j
gs (r)Yms
(θ, φ),
j,m
(s = 0, ±1)
(3.121)
where As are the spin-weighted components of A. We suppress the (l m) labels on the gfunctions for clearer notation. Inserting (3.119) into (3.117) and then making use of (3.120)
we obtain the equations:
)
(
√
√
X
1
1
∂
2
2
∂
j
j
j
j
j
+ 2 g0 J−1 J+0 Ym0
+ 2 g− J+−1 Ym,−1
+ k2 g0 Ym0
r 2 g0 + 2 g+ J−1 Ym,1
=0
Ym0
∂r r 2 ∂r
r
r
r
j,m
)
(
p
p
X d 1 d
2j(j
+
1)
2j(j
+
1)
j(j
+
1)
g+ +
g− −
g0 + k2 g0 Yjm = 0
r 2 g0 −
⇐⇒
dr r 2 dr
r2
r2
r2
j,m
3.4 Solution of the vector Helmholtz equation
97
where we have used relations (3.96) to (3.99). We give two example here:
j
J−1 J+0 Ym0
j
J+−1 Ym,−1
(3.98)
=
(3.96)
=
j
−j(j + 1)Ym0
≡ −j(j + 1)Yjm
p
p
j
j(j + 1) Ym0
≡ j(j + 1) Yjm
Performing all angular derivatives we find that the radial functions must satisfy
p
2j(j + 1)
d 1 d 2 j(j + 1)
(g+ − g− ) −
g0 + k2 g0 = 0
r g0 −
2
2
dr r dr
r
r2
Similar manipulations for the other components lead to the differential equations
p
2j(j + 1)
1 d2
j(j + 1)
(rg± ) −
g± ∓
g0 + k2 g± = 0
2
2
r dr
r
r2
We introduce the linear combinations
p
p
G = A j(j + 1)(g+ − g− ), H = A j(j + 1)(g+ + g− )
(3.122)
(3.123)
(3.124)
(3.125)
(3.126)
where A is a constant. They can be inverted to give
g− =
2A
H−G
√
j(j+1)
, g+ =
2A
H+G
√
j(j+1)
Inserting these definitions into eqs. (3.124) and (3.125) we obtain
√
j(j + 1)
2
d 1 d 2 g0 −
G + k2 g0 = 0
r g0 −
2
2
dr r dr
r
Ar 2
√
1 d2
j(j + 1)
k2
2Aj(j + 1)
[r(H
−
G)]
−
(H
−
G)
+
g
+
(H − G) = 0
0
2r dr 2
2r 2
r2
2
√
j(j + 1)
k2
1 d2
2Aj(j + 1)
[r(H
+
G)]
−
(H
+
G)
−
g
+
(H + G) = 0
0
2r dr 2
2r 2
r2
2
Adding eq. (3.129) to eq. (3.130) one obtains
j(j + 1)
1 d2
rH −
H + k2 H = 0
2
r dr
r2
(3.127)
(3.128)
(3.129)
(3.130)
(3.131)
Subtracting (3.129) from (3.130) we obtain a more complicated differential equation which
involves not only G, but also g0 :
√
1 d2
j(j + 1)
2 2Aj(j + 1)
rG −
G−
g0 + k2 G = 0
(3.132)
r dr 2
r2
r2
√
To make the equations agree with those of [4] we set the constant A = −1/ 2. Therefore,
expanding the partial derivatives in eqs. (3.128), (3.131) and (3.132) we have
2
d
j(j + 1)
2 d
2
−
+
+
k
H = 0,
(3.133)
dr 2 r dr
r2
2g0 j(j + 1)
2G
d2 g0 2 dg0
− 2 −
+
g0 + 2 + k2 g0 = 0,
(3.134)
2
2
dr
r dr
r
r
r
d2 G 2 dG j(j + 1)
2j(j + 1)
+
−
G+
g0 + k2 G = 0.
(3.135)
2
2
dr
r dr
r
r2
3.4 Solution of the vector Helmholtz equation
98
Eq. (3.133) for k 6= 0 has as solution a linear combination of spherical Bessel functions, or
equivalently, a linear combination of spherical Hankel functions:
(1)
(2)
H(r) = Ajj (kr) + Bnj (kr) = Ahj (kr) + Bhj (kr)
(3.136)
A, B are arbitrary constants. Following [4] we find that eqs. (3.134) and (3.135) are equivalent
to
2
j(j − 1)
2 d
d
2
+k −
+
(G + jg0 ) = 0,
(3.137)
dr 2 r dr
r2
2
d
2 d
(j + 1)(j + 2)
2
+
+k −
(G − (j + 1)g0 ) = 0,
(3.138)
dr 2
r dr
r2
We see that if we substitute H → G + jg0 and j → j − 1 in the differential operator in
eq. (3.133) we obtain (3.137), and if we substitute H → G − (j + 1)g0 and let j → j + 1
in the differential operator in (3.133) we obtain (3.138). Therefore, performing the same
substitutions in (3.136) we obtain
G(r) + jg0 (r) = Cjj−1 (kr) + Dnj−1 (kr)
(3.139)
G(r) − (j + 1)g0 (r) = Ejj+1 (kr) + F nj+1 (kr)
(3.140)
where C, D, E, F are constants. We now have three equations involving the three unknown
functions g0 , g+ and g− . To ease the notation a bit when we solve these equations, we define
α(j + 1) ≡ Ejj+1 (kr) + F nj+1 (kr)
β(j − 1) ≡ Cjj−1 (kr) + Dnj−1 (kr)
γ(j) ≡ Ajj (kr) + Bnj (kr)
Now, using (3.136) and the definition of H in (3.126) we obtain
s
√
1
2 γ(j)
g+ + g− = −
j(j + 1)
(3.141)
(3.142)
(3.143)
(3.144)
Using (3.139) we find
g− − g+ =
s
√
1
2(β(j − 1) − jg0 )
j(j + 1)
(3.145)
These two equations can be used to solve for g± in terms of g0 . One then obtains after a bit
of manipulation:
i
h
1
(3.146)
− γ(j) − β(j − 1) + jg0
g+ = p
2j(j + 1)
i
h
1
g− = p
(3.147)
− γ(j) + β(j − 1) − jg0
2j(j + 1)
So we see that both g± are determined by g0 . The last equation, eq. (3.140) then gives
r
j(j + 1)
(g− − g+ )
(3.148)
(j + 1)g0 = −α(j + 1) + G = −α(j + 1) +
2
= −α(j + 1) + β(j − 1) − jg0
(3.149)
3.4 Solution of the vector Helmholtz equation
99
where we in the last line used eq. (3.145). Solving for g0 we obtain
i
1 h
g0 =
β(j − 1) − α(j + 1)
2j + 1
i
1 h
Cjj−1 (kr) + Dnj−1 (kr) − Ejj+1 (kr) − F nj+1 (kr)
=
2j + 1
(3.150)
The solution for g0 contains spherical Bessel functions of the first and second kinds, and of
orders j ± 1, but not j. Inserting this result for g0 into eqs. (3.146) and (3.147) we find
i
h
1
j
j+1
g± = ∓ p
β(j − 1) +
α(j + 1)
± γ(j) +
2j + 1
2j + 1
2j(j + 1)
h
j+1 1
± Ajj (kr) ± Bnj (kr) +
Cjj−1 (kr) + Dnj−1 (kr)
= ∓p
2j + 1
2j(j + 1)
i
j
+
Ejj+1 (kr) + F nj+1 (kr)
(3.151)
2j + 1
We see here that we have separation of the spin-weighted components A± . This is not the
case for the spherical coordinate components Aθ and Aφ . So far we haven’t said anything
about which assumptions we have made for j. We have implicitly assumed that j > 1 since
the spin-s harmonics aren’t defined for |s| < j.
When j = 0, A± = 0 and the Helmholtz vector equation reduces to a single differential
equation for A0 with solution
X
A0 (r) =
g0 (r)Ylm (θ, φ)
(3.152)
l,m
where g0 (r) is a solution of the ordinary Bessel equation.
3.4.1
Divergenceless solutions of the vector Helmholtz equation
When we derived the expression for ∇2 F in appendix A we never assumed that the divergence
of F given by the expression
∇·F=
1 1
1 ∂
r 2 F0 − √
J− F+ + J+−1 F−
2
r ∂r
2r
(3.153)
should vanish. In many applications of electromagnetism one imposes gauge conditions which
require the divergence to be zero. We impose this condition on the vector A now, and see
if we can determine some of the constants which appear in the solutions for g0 , g± . We first
calculate the first term:
X 1 ∂
X dg0 (x) 2g0 (x) 1 ∂
2
2
x g0 (x) Yjm = k
+
r A0 = k
Yjm
(3.154)
r 2 ∂r
x2 ∂x
dx
x
j,m
j,m
where x = kr is a dimensionless parameter. Using (3.150) and the relations
dzl (x)
=
dx
1
zl (x) =
x
1
[lzl−1 (x) − (l + 1)zl+1 ]
2l + 1
1
[zl−1 (x) + zl+1 ]
2l + 1
(3.155)
(3.156)
3.4 Solution of the vector Helmholtz equation
(1)
where zl could be any of the functions jl , nl , hl
dg0 (x)
dx
100
(2)
and hl
we obtain
i
1 n 1 h
(j − 1)(Cjj−2 (x) + Dnj−2 (x)) − j(Cjj (x) + Dnj (x))
2j + 1 2j − 1
io
1 h
(j + 1)(Ejj (x) + F nj (x)) − (j + 2)(Ejj+2 (x) + F nj+2 (x)) (3.157)
2j + 3
=
−
1
g0 (x) =
x
−
i
1 n 1 h
C(jj−2 (x) + jj (x)) + D(nj−2 (x) + nj (x))
2j + 1 2j − 1
io
1 h
E(jj (x) + jj+2 (x)) + F (nj (x) + nj+2 (x))
2j + 3
(3.158)
This gives after a lot of algebra:
dg0 2g0
+
dx
x
2−j
1 h j+1
3+j
=
[Cjj−2 (x) + Dnj−2 (x)] +
C−
E jj (x)
2j + 1 2j − 1
2j − 1
2j + 3
i
2−j
3+j
j
+
D−
F nj (x) +
[Ejj+2 (x) + F nj+2 (x)]
(3.159)
2j − 1
2j + 3
2j + 3
For The second and third terms involving the spin- raising and lowering operators we obtain:
1
√ J∓±1 A± =
2r
+
j+1 k X 1h
± Ajj (x) ± Bnj (x) +
Cjj−1 (x) + Dnj−1 (x)
2
x
2j + 1
j,m
i
j Ejj+1 (x) + F nj+1 (x) Yjm
(3.160)
2j + 1
Therefore
X 1h j + 1 1 1
√
Cjj−1 (x) + Dnj−1 (x)
J− A+ + J+−1 A− = k
x 2j + 1
2r
l,m
i
j +
Ejj+1 (x) + F nj+1 (x) Yjm
2j + 1
(3.161)
Calculating the radial part with the help of (3.156) we obtain
X 1 h j+1 1
√ [J−1 A+ + J+−1 A− ] = k
Cjj−2 (x) + Dnj−2 (x)
2j + 1 2j − 1
2r
j,m
j+1
j+1
j
j
+
C+
E jj (x) +
D+
F nj (x)
2j − 1
2j + 3
2j − 1
2j + 3
i
j +
Ejj+2 (x) + F nj+2 (x) Yjm
(3.162)
2j + 3
When we subtract (3.162) from (3.154) we obtain
∇·A
=
−k
X
j,m
i
1 h
!
(C + E)jj (kr) + (D + F )nj (kr) Yjm = 0
2j + 1
⇐⇒ E = −C
& F = −D
(3.163)
3.4 Solution of the vector Helmholtz equation
101
Demanding that (3.163) holds we are able to simplify our solutions a great deal using relations
(3.155) and (3.156). One finds that
g0 =
1
Cjj (kr) + Dnj (kr)
kr
(3.164)
When we demand that A should have zero divergence the Bessel functions of order j ± 1 in
g0 combine to form functions of order j only! We can show that this also happens in g± with
the help of the relation
We then find
i
1 d
1 h
(l + 1)zl−1 − lzl+1
(xzl ) =
x dx
2l + 1
(3.165)
h
1
j+1 g± = ∓ p
± (Ajj (kr) + Bnj (kr)) +
Cjj−1 (kr) + Dnj−1 (kr)
2j + 1
2j(j + 1)
i
j
Cjj+1 (kr) + Dnj+1 (kr)
−
2j + 1
h
1
1 = ∓p
± (Ajj + Bnj ) +
C(j + 1)jj−1 − Cjjj+1
2j + 1
2j(j + 1)
i
1
D(j + 1)nj−1 − Djnj+1
+
2j + 1
h i
1
1 d = ∓p
± Ajj (kr) + Bnj (kr) +
r Cjj (kr) + Dnj (kr) (3.166)
kr dr
2j(j + 1)
We know that the function ψ = [Ajj + Bnj ]Yjm is a solution of the scalar wave equation. We
want to write our vector solution for A in terms of functions of this form. Using the fact that
the spin-s harmonics are eigenfunctions of J−s J+s and J+s J−s we can write
A0 =
1 1 0
J J ψ2
kr − +
1 ∂
i
rJ+0 ψ2
A+ = − √ J+0 ψ1 + √
∂r
2
2kr
i
1 ∂
A− = √ J−0 ψ1 + √
rJ−0 ψ2
2
2kr ∂r
(3.167)
(3.168)
(3.169)
where the scalar functions (which can be viewed upon as potentials or generating functions)
1
Ajj (kr) + Bnj (kr) Yjm
j(j + 1)
j,m
X
1
Cjj (kr) + Dnj (kr) Yjm
ψ2 = −
j(j + 1)
ψ1 = −i
X
(3.170)
(3.171)
j,m
are solutions of the scalar Helmholtz equation. Thus
1 1 0
1 ∂
1 ∂
i 0
i 0
0
0
A=n
J J ψ2 +m √ J− ψ1 + √
rJ− ψ2 +m − √ J+ ψ1 + √
rJ+ ψ2
kr − +
2
2kr ∂r
2
2kr ∂r
(3.172)
l
to the multipole expansion
3.5 Application of Ym,±1
102
but this expression is equivalent to
1
A = r × ∇ψ1 + ∇ × (r × ∇ψ2 )
k
(3.173)
which is easily verified. Thus we can express the vector potential through p
the angular momentum operator L = −ir×∇, and the vector spherical harmonics Xjm = (1/ j(j + 1))LYjm .
Abandoning the ψ notation we thus find
X
i
(3.174)
A(r) =
f (r)Xjm(θ, φ) − ∇ × [g(r)Xjm (θ, φ)]
k
j,m
with
f (r) = ajm jj (kr) + bjm nj (kr), g(r) = cjm jj (kr) + djm nj (kr)
(3.175)
The constants a, b, c, d are some other constants then A, B, C, D.
3.5
l
Application of Ym,±1
to the multipole expansion
(In this section we use j = l). Now that we are more familiar with the spin weighted spherical
harmonics, let us consider it’s application to electromagnetism. We still employ natural units
h̄ = c = 1, and work in ω space where ∂t → −iω = −ik. In this space we write the Maxwell
equations without sources yet again as
∇ · E = 0,
∇ · B = 0,
(3.176)
∇ × E − ikB = 0, ∇ × B + ikE = 0.
This is a set of coupled differential equations for the electromagnetic fields. Our goal here is
l
which transforms according to eq. (3.82)
to make a series expansion of the fields in Ym,±1
with s = ±1. In vacuum we know that the solutions to the Maxwell equations are free plane
waves which travel with the speed of light. We have seen in chapter two that in the radiation
zone B ⊥ E which means that under a rotation about an axis perpendicular to the plane in
which the fields live by an angle α the quantities E ± iB transforms as
(E ± iB) → e∓iα (E ± iB)
(3.177)
i.e. like spin weighted quantities. Thus we would expect that E + iB has a component with
spin-weight +1 and E − iB has a component with spin-weight −1 in the radiation zone. Hence
we expand these fields in terms of spin weighted harmonics. Eqs. (3.176) can be reduced to
two uncoupled equations in terms of the two fiel combinations E ± iB:
∇ · (E ± iB) = 0,
∇ × (E ± iB) = ±k(E ± iB).
(3.178)
(3.179)
We can then recover E and B by taking sums and differences. We see that the equations for
E ± iB only differ in the sign of k. Let us follow the notation used by Scanio [14] and write
the electromagnetic fields as
E + iB = nG0 + mG− + mG+
E − iB = nH0 + mH− + mH+
(3.180)
(3.181)
l
to the multipole expansion
3.5 Application of Ym,±1
103
We know from considerations in chapter two that close to the source the fields contain quasistatic components which implies that E and B are not in general orthogonal to each other.
But we expect that these components have a vanishing contribution when move away from
the source.
From now on we will only work with E + iB and then let k → −k (here one must be
careful in those cases when k = |k|) and G → H to obtain the solution for E − iB. Now we
use eq. (3.111) to write (3.178) as (we employ the operators defined in eqs. (3.113), (3.114)
and (3.115)):
1
1
(3.182)
∂r (r 2 G0 ) + √ (J+ G− + J− G+ ) = 0
2
r
2r
and from (3.179) we get three equations (using (3.112)):
1
√ (J+ G− − J− G+ ) = kG0
2r
1
1 (∗)
√ J0 G0 =
∂r r − ik G−
r
2r
1
1
√ J0 G0 =
∂r r + ik G+
r
2r
(3.183)
(3.184)
(3.185)
We now have a set of four coupled equations and we see that if we know how G0 depends on
the angles θ, φ then we can determine the angular distributions of G+ and G− so we start by
solving for G0 . Let’s multiply eq. (3.182) by i and add it to eq. (3.183). We then obtain
√
1
2
2
− 2 ∂r (r G0 ) −
J+ G− − ikG0 = 0
(3.186)
r
r
Subtracting (3.183) from i times (3.182) produces
√
1
2
2
− 2 ∂r (r G0 ) −
J− G+ + ikG0 = 0
r
r
(3.187)
Now we let J+ act on eq. (3.184) (remember that the J operators only contain angular
derivatives) to obtain:
1
1
(∗)
√ J+ J0 G0 =
∂r r − ik J+ G−
(3.188)
r
2r
Letting J− act on eq. (3.185) we obtain:
1
√ J− J0 G0 =
2r
1
∂r r + ik J− G+
r
(3.189)
Now we use eq. (3.116) in (3.189) and then we use eq. (3.187) to eliminate J− G+ . We are
then left with
∂r2 (r 2 G0 ) + ∇2Ω (r 2 G0 ) + k2 (r 2 G0 ) = 0
(3.190)
Dividing by r and then noting that
1 2
2
∂ r + ∇Ω (rG0 ) = ∇2 (rG0 )
r r
l
to the multipole expansion
3.5 Application of Ym,±1
104
we find that rG0 satisfy the scalar Helmholtz equation
∇2 + k2 (rG0 ) = 0.
(3.191)
We have already solved this equation several times in chapter two in this thesis so it should
come as no surprise to the reader that we can write the solution as
X
(3.192)
G0 =
g0 (r) Ylm (θ, φ)
l,m
(2)
(1)
g0 (r) =
Alm hl (kr) + Blm hl (kr)
r
(3.193)
(1,2)
where hl (kr) are spherical Hankel functions. The sums run over all allowed values of l, m.
Now that we know G0 we can use eqs. (3.186) and (3.187) to find G± . Since they are spin
weighted quantitites we write
X
l
G± =
g± (r) Ym,±1
(θ, φ)
(3.194)
l,m
l
are not defined for l = 0. Let us
where g± are undetermined functions. Notice that Ym,±1
insert G0 and G− into (3.186):
!
√
X
X 1
2
l
(3.195)
g− J+ Ym,−1
− 2 ∂r (r 2 g0 ) − ikg0 Ylm =
r
r
l,m
l,m
l
l
=
we use eq. (3.96) with s = −1. This gives J+ Ym,−1
To find J+ Ym,−1
RHS becomes
!
p
X
2l(l + 1)
g− Ylm .
r
p
l(l + 1)Ylm so the
l,m
Since we have LHS = RHS the terms in the brackets must equal one another. We can then
solve for g− in terms of g0 . The result is
1
1
2
− ∂r (r g0 ) − ikrg0
(3.196)
g− = p
r
2l(l + 1)
The task of finding g+ in terms of g0 is completely analogous. One finds
1
1
∂r (r 2 g0 ) − ikrg0
g+ = p
2l(l + 1) r
(3.197)
We have now performed the complete multipole expansion of the electromagnetic fields in
terms of known radial and angular functions. The constants are determined by the boundary
conditions. Let us gather all results we’ve obtained so far - the expansion of H is found by
letting k → −k. In the Hankel functions however, k is meant to be the positive square root
of k2 so no alteration is made in these functions. The results are:
i
1h
E = n(G0 + H0 ) + m(G− + H− ) + m(G+ + H− )
(3.198)
2
h
i
1
n(G0 − H0 ) + m(G− − H− ) + m(G+ − H+ )
(3.199)
B=
2i
l
to the multipole expansion
3.5 Application of Ym,±1
G0 =
H0 =
g0 (r) =
g± (r) =
P
P
g0 Ylm
G± =
h0 Ylm
h± =
(1)
P
P
h± (r) =
l
g± Ym,±1
l
h± Ym,±1
(2)
Alm hl (kr) + Blm hl (kr)
r
1
1
2
p
± ∂r (r g0 ) − ikrg0
r
2l(l + 1)
(1)
h0 (r) =
105
(3.200)
(3.201)
(2)
Clm hl (kr) + Dlm hl (kr)
r
1
1
p
± ∂r (r 2 g0 ) + ikrg0
r
2l(l + 1)
(3.202)
(3.203)
These results are valid everywhere in space. In this thesis we have been working in the
radiation zone where kr ≫ 1. Let us examine the properties of the fields here and see which
simplifications can be made.
3.5.1
The radiation zone fields
In the radiation zone we know that the waves are outgoing and decay as 1/r. Therefore we
ignore all components inversely proportional to higher powers of r. In chapter two we stated
that
e±ikr
(1,2)
hl (kr) → (∓i)l+1
∝ 1/r
kr
for kr ≫ 1. Outgoing waves are proportional to eikr so
eikr
1
(1)
g0 → Alm hl (kr) → Alm 2 ∝ 1/r 2
r
kr
We thus ignore it’s contribution in the radiation zone. By an identical argument we can show
that also h0 (r) → 0 in the radiation zone. The expressions for g± (r) and h± (r) become:
g± (r) →
h± (r) →
Therefore
iAlm (−i)l+1
p
(±eikr − eikr )
r 2l(l + 1)
iClm (−i)l+1
p
(±eikr + eikr )
r 2l(l + 1)
l+1
√lm (−i)
g− (r) = −h+ (r) = − 2iA
2l(l+1)
eikr
r ,
g+ (r) = h− (r) = 0
When we insert these results into eqs. (3.198) and (3.199) we find
(
)
Alm
ieikr X
Clm
l+1
l
l
p
(−i)
E = −√
Ym,−1 m − p
Ym,1 m
2r l,m
l(l + 1)
l(l + 1)
(
)
A
eikr X
C
lm
lm
l
l
(−i)l+1 p
B = −√
m
Ym,−1
m+ p
Ym,1
2r l,m
l(l + 1)
l(l + 1)
(3.204)
(3.205)
(3.206)
(3.207)
(3.208)
l
to the multipole expansion
3.5 Application of Ym,±1
106
By analogy to what we did in section 2.7.1 with the spherical vector harmonics Xlm we now
define another set of reduced multipole coefficients through
a− (l, m) ≡ √Alm ,
l(l+1)
a+ (l, m) ≡ √Clm
l(l+1)
(3.209)
Let us use the expression for the B-field to see what the angular distribution dP/dΩ of the
radiated power becomes. Using eq.(2.192) and the orthogonality properties in eqs.(3.107) and
(3.108) we find




X
X
2
2
1 dP
l
l (−i)l a− (l, m)Ym,−1
=
(−i)l a+ (l, m)Ym,1
(3.210)
+

dΩ
4
l,m
3.5.2
l,m
Determination of a±
To see the explicit angular dependence one needs to evaluate the distribution coefficients. Let
us now turn to this task. One way of doing this is by repeating the same procedure as in
chapter two section 2.8. This is the method of choice in [14] which is straight forward but
involves a lot of algebra so we will take a different route by exploiting the close relationship
between expression (4.124) and eq.(2.195) in chapter two. This will allow a much quicker
derivation of the expressions for the reduced multipole coefficents a± . We have from (2.195)
that
eikr X
(−i)l+1 {aE (l, m)Xlm + aM (l, m)n × Xlm }
(3.211)
B=
kr
l,m
in the radiation zone. Since the two expressions for the magnetic field in the radiation zone
are identical (they just utilize different sets of basis functions) we should be able to relate the
different multipole coefficients with one another. To see how this is done we first form the
scalar m · B and then m · B. It is not hard to show the following four relations:
l ,
m · Xlm = − √12 Ym,1
m · Xlm =
√1 Y l
2 m,−1
l
l
.
m · (n × Xlm ) = − √i2 Ym,1
m · (n × Xlm ) = − √i2 Ym,−1
Therefore:
m·B
=
(4.124)
=
eikr X
l+1 −(aE + iaM )
l
√
(−i)
Ym,1
kr
2
l,m
ikr
e X
l+1 −ka+
l
√
(−i)
Ym,1
kr
2
l,m
where we in the second line used eq.(4.124). We can then deduce that
1
a+ =
aE + iaM
k
Similarly we obtain
eikr X
l+1 aE − iaM
l
√
m·B
=
(−i)
Ym,−1
kr
2
l,m
ikr
X
(4.124) e
l+1 −ka−
l
√
(−i)
Ym,−1
=
kr
2
l,m
(3.212)
(3.213)
(3.214)
(3.215)
(3.216)
3.6 Angular distribution of photon emission
107
which means that
1
(3.217)
aE − iaM
k
These results are identical to those in [14]. We can now write down the radiation zone fields
in terms of the reduced multipole coefficients (given by expressions (2.180 and (2.181)):
a− = −
E + iB =
√
2m
eikr X
l
(−i)l (aE − iaM )Ym,−1
kr
(3.218)
l,m
E − iB =
√
2m
eikr
kr
X
l
(−i)l (aE + iaM )Ym,1
(3.219)
l,m
Of course, in the radiation zone the multipole coefficients reduced to eqs.(2.185) and (2.189).
These results are in accordance with the predictions we made earlier in the beginning of section
3.5. These predictions are given mathematically in eq.(3.177).
3.6
Angular distribution of photon emission
We consider emission of a single photon with specific values of the quantum numbers j (total
angular momentum) and m (projection of the angular momentum) along the z-axis, from an
electron initially at rest in an electricmagnetic field. The emitted photon can be in one of two
states; one state with positive, and one state of negative parity for any value of j 6= 0. The
parity of the state is determined once the photon wave function is specified, which in turn is
determined by the gauge chosen for the electromagnetic potentials.
In this calculation we will choose to work in k-space. The interaction operator inducing
the transition of the electron from it’s initial state |ii to it’s final state |f i through emission
of a single photon is in relativistic quantum mechanics the matrix element
Vf i (t) = Vf i δ(Ei − Ef − ω)
Z
Vf i = hf |V |ii = d3 r jfµi (r)A(∗)
µ (r)
(3.220)
(3.221)
where jfµi = ψf γ µ ψi is the transition current and Aµ is the photon 4-potential. Ef , Ei and
ω are the energies of the final state, initial state and the photon respectively. We have also
assumed the time-dependence to be harmonic - hence the delta function. The spinors ψf and
ψi specify the final and initial states of the electron respectively. We Fourier transform the
4-potential;
Z
d3 k (∗)
(∗)
Aµ (r) =
A (k) e−ik·r
(3.222)
(2π)3 µ
and insert into the matrix element. Let Aµ = (Φ, A) and the flat Minkowski metric ηµν =
(∗)
diag(−, +, +, +) so that Aµ = (−Φ, A). This gives us j µ Aµ = ρf i Φ(∗) + jf i · A(∗) . We will
consider the two terms and their contributions to the amplitude. For the first term we obtain:
Z
Z
d3 k (∗)
(e)
3
Φ (k) e−ik·r
(3.223)
Vf i = e d rρf i (r)
(2π)3
Now, the scalar potential Φ(k) is determined when we specify the gauge. In k-space we have
the following gauge transformations;
A → A + nf (k), Φ → Φ + f (k)
(3.224)
3.6 Angular distribution of photon emission
108
where n = k/|k| and f (k) is a scalar function to be specified. Landau & Lifshitz [10] choose
f (k) so that the potentials are proportional to the ordinary spherical harmonics Yjm and thus
have definite parity. We then have
s
j + 1 4π 2
δ(|k| − ω)Yjm (n)
(3.225)
Φ=−
j ω 3/2
Here [10] has employed h̄ = c = 1. Here, n specifies the angles θγ , φγ which determine the
photon direction. The normalization condition satisfied by these functions are
Z
d3 k
(3.226)
ωω ′ Φωjm Φω′ j ′ m′ = 2πωδ(ω ′ − ω)δj,j ′ δm,m′
(2π)3
Insertion the expression for the scalar potential into (3.223) gives
s
Z
Z
j + 1 ω 1/2
(e)
(∗)
Vf i = −e
d3 rρf i (r) dΩn e−ik·r Yjm (n)
j
2π
(3.227)
where we have integrated out the delta function δ(|k| − ω). By dΩn mean integration over the
photon coordinate angles. To advance further we use the known expansion
X
(∗)
e−ik·r = 4π
i−j jj (kr)Yjm (n)Yjm (θ, φ)
(3.228)
j,m
which let’s us compute the inner integral. We obtain
Z
Z
X
(∗)
(∗)
−ik·r (∗)
−j ′
′
′
′
dΩn e
Yjm (n) = 4π
dΩn Yj m (n)Yjm (n)
i jj (kr)Yj ′ m′ (θ, φ)
j ′ ,m′
{z
}
|
δj,j ′ δm,m′
=
(∗)
4πi−j jj (kr)Yjm (θ, φ)
(3.229)
Consequently the transition amplitude becomes
s
Z
j + 1 1/2 −j
(∗)
ω i
d3 rρf i (r)jj (kr)Yjm (θ, φ)
Vf i = −2e
j
(3.230)
We are to take the integral over the source distribution, but the integrand is quite complex.
To make contact with classical electromagnetism (and to simplify the integration) we choose
to consider a source whose dimensions are small compared to the wavelength λ of the emitted
photon. Then we have kr ≪ 1 in the integral. Using the approximation
lim jj (kr) =
kr→0
(kr)j
(2j + 1)!!
(3.231)
(∗)
and Yj,m = (−1)m−j Yj,−m. we obtain
Vf i =
s
j + 1 2(−1)m−j+1 i−j ω 1/2
j
(2j + 1)!!
Z
d3 r j ρf i Yj,−m
(3.232)
3.6 Angular distribution of photon emission
109
We can write (−1)m−j+1 i−j = (−1)m+1 ij and identify
r
Z
2j + 1
(Qj,m )f i = d3 r j ρf i Yj,m
4π
(3.233)
as the matrix elements of the electric multipole moment found earlier, see eq.(2.185). Our
final expression for the transition operator is then
s
(2j + 1)(j + 1) ω j+1/2
m+1 j
Vf i = (−1)
i
e(Qj,−m )f i
(3.234)
πj
(2j + 1)!!
This operator depends only on the quantum numbers j and m which is entirely logical. We
can not expect to find an angle dependence in the amplitudes for photon emission along a
specified axis. The quantum number m is fixed through the relation Mi − Mf = m where Mi
and Mf is the projection of the spin for the entire initial and final system respectively. We
can use this expression to calculate the probability for photon emission - Wjm . For a photon
with specified momentum k this is [10]:
Wjm = 2π|Vf i |2 =
2(2j + 1)(j + 1) 2j+1 2
ω
e |(Qj,−m )f i |2
j[(2j + 1)!!]2
(3.235)
This expression represents the probability for a photon to be emitted from a source with small
dimensions along the z-axis, with definite quantum numbers j, m. For our later purpuoses this
is not enough. We want to know the amplitudes Vf′i for photon emission in a given direction,
′ . This dependence
specified by the spherical coordinate angles θ, φ. We will calle them Vjm
can be found by applying the rotation operator R(α, β, γ) defined in eq. (1.106) on Vf i . From
the y-convention it follows that if we take γ = 0 then α, β coinside with the spherical angles
′ ) = R(φ, θ, 0)(V
φ, θ. Thus the relationship rquired is (Vjm
jm ), which is essentially a matrix
equation. In component form we have
′
Vjm
=
j
X
(j)
Dm′ ,m Vjm′
(3.236)
m′ =−j
Remember; the index we sum over - m′ - in this case refers to the fixed system. Using eq.
(1.108) with γ = 0 we obtain
′
Vjm
=
j
X
′
(j)
e−im φ dm′ m (θ)Vjm′
(3.237)
m′ =−j
We are intetersted in seing how the spin weighted spherical harmonics come in to play, so we
use eq.(3.79) to replace the D-functions. We then need to take the square modulus of the
whole expression. We have:
2
r
X
j
4π
j
′ 2
(3.238)
Ym′ m Vjm′ |Vjm | = 2j + 1
′
m =−j
Of special interest is the case when j = 1. The amplitudes along the z-axis are for j = 1:
p
V1,m = (−1)m+1 i 2/3πω 3/2 (Q1,−m )f i
(3.239)
r Z
4π
(Q1,−m )f i =
d3 r ρf i rY1,−m
(3.240)
3
3.6 Angular distribution of photon emission
110
Using table 1.1 and the definition of the electric dipole moment eq. (2.118) in spherical and
cartesian coordinates, we obtain (dropping the f i subscripts):
Z
Z
Q1,0 =
d3 r ρ r cos θ = d3 r ρ z = pz
(3.241)
Z
1
1
(3.242)
d3 rρ (x ± iy) = ∓ √ (px ± ipy )
Q1,±1 = ∓ √
2
2
where we in the second line have expressed Y1,±1 in cartesian coordinates. The probabilities
for radiation along the z-axis for different values of m are then calculated to be
W1,0 =
4ω 3
2
3 |epz | ,
W1,±1 =
2ω 3
2
3 (|epx |
+ |epy |2 )
(3.243)
The total probability is obtained when we sum over m. The result is
W1 (radiation along z-axis) =
4ω 3 e2 2
|p|
3
(3.244)
which is exactly what we should expect to lowest order. Since we don’t have an angle dependence we don’t obtain the sin2 θ factor.
Let us now calculate how the probability changes when we introduce an angle dependence
through rotation. We need to evaluate expression (3.238) for j = 1. Using eq.(3.239) we
obtain (we will drop the j = 1 index in the following to ease notation):
2
2 ω 3 X
8e
′
(3.245)
|Vm′ |2 =
(−1)m +1 Ym′ m Q−m′ 9 ′
m
(∗)
8e2 ω 3 Y−1,m Q1 − Y0,m Q0 + Y1,m Q−1 Y−1,m Q1 − Y0,m Q0 + Y1,m Q−1
=
9
2
8e ω 3 n
(∗)
(∗)
(∗)
(∗)
=
|Y−1,m Q1 |2 − Y−1,m Y0,m Q1 Q0 + Y−1,m Y1,m Q1 Q−1
9
(∗)
(∗)
(∗)
(∗)
− Y0,m Y−1,m Q0 Q1 + |Y0,m Q0 |2 − Y0,m Y1,m Q0 Q−1
o
(∗)
(∗)
(∗)
(∗)
+ Y1,m Y−1,m Q−1 Q1 − Y1,m Y0,m Q−1 Q0 + |Y1,m Q−1 |2
So far we haven’t made any assumptions regarding the dipole moments pi . When we work
in ω-space the dipole moment is purely imaginary pi = iIm pi . Comparing with eqs.(3.241),
(3.242) we see that complex conjugation of the transition moments only introduces a minus
sign:
(∗)
(∗)
(3.246)
Q0 = −Q0 , Q±1 = −Q±1
At this point we also need to have explicit expressions for the spin weighted harmonics. The
ones we need are the following:
q
q
3
3
(3.247)
(cos θ ± 1) eiφ Y0,±1 = ∓ 8π
sin θ
Ym,0 = Y1m , Y1,±1 = ± 16π
(∗)
All other harmonics needed can be found from Ym,s = (−1)m+s Y−m,−s . Let’s evaluate some
special cases. First assume that the dipole moment is along the z-axis so that Q±1 = 0 and
|Q0 |2 = |p|2 . We then have:
8e2 ω 3 2
|p| |Y0,m |2
(3.248)
|Vm′ |2 =
9
3.6 Angular distribution of photon emission
Then
|V0′ |2 =
2e2 ω 3
2
2
3π |p| cos θ,
111
′ |2 =
|V±1
e2 ω 3
2
2
3π |p| sin θ
(3.249)
in agreement with earlier results since the total probability (which one obtains when summing
over all m) is independent
of θ. Now assume that py = pz = 0 and |Q±1 |2 = |p|2 /2 with
√
Q1 = −Q−1 = −px 2. We obtain
|Vm′ |2 =
o
8e2 ω 3 n
(∗)
(∗)
(∗)
(∗)
|Y−1,m Q1 |2 + |Y1,m Q−1 |2 + Y−1,m Y1,m Q1 Q−1 + Y1,m Y−1,m Q−1 Q1
(3.250)
9
which gives
|V0′ |2 =
e2 ω 3
2
2
3π |p| sin θ
′ |2 =
1 + cos 2φ , |V±1
e2 ω 3
2
6π |p| (1
Summing over all possible spin-projections one obtains
W1 (radiation along x-axis) =
+ cos2 θ) 1 + cos 2φ
8e2 ω 3 2
|p| cos2 φ
3
(3.251)
(3.252)
where we used some wellknown trigonometric identities to simplify. Of course, these situations are just special cases of the more general situation where the transition moments Q are
unspecified.
Chapter 4
Gravitational radiation and spin-2
fields
4.1
Linearized field equations
Let us now consider pertubations in the spacetime metric gµν . We can think of the linearized
version of general relativity as describing a theory of a symmetric tensor field hµν propagating
on a flat space-time background given by the Minkowski metric. Einsteins field equations in
general read
1
(4.1)
Gµν = Rµν − gµν R = 8πGTµν
2
where Rµν is the Ricci-tensor and R is the Ricci-scalar. On the right hand side we have the
enery-momentum tensor Tµν which is formally given by, [1]
1 δSM
Tµν = −2 √
−g δgµν
(4.2)
where SM is the action for matter and g = det(gµν ). Let us now also assume that we reside in
a coordinate frame where the geometry is nearly Euclidean (which is true almost everywhere
in our solar system). Euclidean space is of course described by the flat Minkowski metric
ηµν = diag(−1, +1, +1, +1). We will allow small pertubations to occur in the metric, which
we can represent by writing
gµν = ηµν + hµν ,
|hµν | ≪ 1
(4.3)
The field equations can now be expressed in terms of hµν to first order (we neglect all higher
order terms). To do this we first calculate the Christoffel symbols [3]:
Γαβγ =
1 αµ
g (gµβ,γ + gµγ,β − gβγ,µ )
2
(4.4)
where we now introduce the Einstein comma notation
,µ
= ∂µ =
∂
∂xµ
(4.5)
We will switch between the different notations, all depending on which is more convenient to
use in the given situation. Since the Minkowski metric is constant all derivatives of ηµν vanish
112
4.1 Linearized field equations
113
and we are left with
Γαβγ
=
=
1 αµ
η (hµβ,γ + hµγ,β − hβγ,µ )
2
1 α
(h β,γ + hαγ,β − hβγ ,α )
2
(4.6)
(4.7)
From the Christoffel symbols one obtains the Ricci tensor:
Rµν
= Rαµαν = Γαµν,α − Γαµα,ν + O(|hµν |2 )
1 α
1
=
(h µ,να + hαν,µα − hµν,αα ) − (hαν,να + hµα,ν α − h,µν )
2
2
1 α
=
(h ν,µα + hµα,ν α − ∂ 2 hµν − h,µν )
2
(4.8)
(4.9)
(4.10)
where h = hαα is the trace of the matrix hµν . Since hµν is a symmetric we see that Rµν also
is. It can actually be derived by varying
1 µν
1 µν
1
ρσ
µν
ρσ
µ
(4.11)
(∂µ h )(∂ν h) − (∂µ h )(∂ρ h σ ) + η (∂µ h )(∂ν hρσ ) − η (∂µ h)(∂ν h)
L=
2
2
2
with respect to hµν The Ricci scalar is then the trace of the Ricci tensor:
R = η µν Rµν = hαβ ,αβ − h,β β
(4.12)
We can now insert these results into eq. (4.1). One then obtains (after multiplying by 2):
hαν,µα + hµα,ν α − ∂ 2 hµν − h,µν − ηµν (hαβ ,αβ − h,ββ ) = 16πTµν
(4.13)
Now we see that the number of terms have increased considerably in the equations. To reduce
them we define
1
hµν ≡ hµν − ηµν h
(4.14)
2
From this relation one infers that
1
hµν = hµν − ηµν h
2
(4.15)
Once inserted into the field equations one finds that the number of terms is reduced by two,
yielding
αβ
α
α
(4.16)
−∂ 2 hµν − ηµν h ,αβ + hµα,ν + hνα,µ = 16πTµν
On the left hand side of the equations we can make the three rightmost terms disappear if we
impose the ’gauge condition’
,α
hµα = 0
(4.17)
which is an analogue to the Lorenz gauge condition ∂ α Aα = 0 in the theory of electromagnetism. The above condition can actually be imposed without any loss of generalization [3]. The
reason why we can impose such a condition is because the curvature tensor is invariant under
the local gauge transformation
hµν → hµν + ∂µ χν + ∂ν χµ
(4.18)
4.2 Gravitational radiation in the weak-field limit
114
where χµ = χµ (x) is an arbitrary vector function, [12]. Thus, if we demand that eq. (4.17)
should hold then we obtain the standard linearized field equations for the pertubed metric:
−∂ 2 hµν = 16πGTµν
(4.19)
Our next task is to solve it and apply the solutions to physical problems. We might for example
ask ourselves how graviational distortions (waves) travelling through space affect matter (test
particles) in their vincinity. We are also interested in the angular distribution of graviational
radiation from a binary star system. Both problems presuppose that we solve eq. (4.19). In
the first case Tµν = 0. In the latter case Tµν 6= 0, which complicates things. But as we also
know: complex is more interesting. Let us now deal with the first case.
4.2
Gravitational radiation in the weak-field limit
We will find the problem of solving the linearized field equations very similar to the corresponding radiation problem in electromagnetism which we dealt with in chapter two. The
methods we will use are exactly the same, except that instead of vectors we are now working
with tensors. To start off let’s solve eq. (4.19) in vacuum where the energy-momentum tensor
is zero. Expanding the ∂ 2 operator we obtain
∂2
2
(4.20)
− 2 + ∇ hµν = 0
∂t
There are essentially two ways (relevant to our inquiry) of solving this equation. We could
Fourier-transform away the t-dependence by going to ω-space as in ch. two, to obtain the
Helmholtz equation (∇2 + k2 )hµν = 0 and then decompose it into spin-weighted components.
Then we would solve it by using the spin-s harmonics as basis functions. This proceedure is
highly relevant, and we will come back to this idea later, but for now let’s just take the easy
way out and assume that the solutions are complex waves:
hµν = Aµν eik·x = Aµν eikµ x
µ
(4.21)
Here, Aµν is a constant symmetric tensor and kµ is the wave-vector. We can check that this
is a solution:
!
∂ 2 hµν = η ρσ ∂ρ ∂σ hµν = . . . = −kσ kσ hµν = 0
(4.22)
Thus, if the solution is non-trivial then kσ kσ = 0. This is always true if the time-like component (the frequency ω) is related to the spatial components k = (k1 , k2 , k3 ) by ω 2 = k · k,
which means it is a null-vector. This is an indication that the graviational waves travel at the
speed of light.
4.2.1
The TT-gauge
We still have a lot of parameters to specify. A has ten indendent components, and the three
components of kσ remains to be specified. This is a result of gauge- and coordinate-freedom.
To eliminate them we start by considering the gauge condition. Then
∂µ hµν
=
Aµν ∂µ (eik
µν
⇒ kµ A
σx
σ
σ
!
) = ikµ Aµν eikσ x = 0
= 0 (4 equations)
(4.23)
(4.24)
4.2 Gravitational radiation in the weak-field limit
115
This result reduces the number of independent parameters in A from ten to six.
We still have some coordinate freedom left. The Lorenz gauge condition is still satisfied
under a coordinate transformation xµ → xµ + ζ µ provided that ∂ 2 ζ µ = 0 which is a wave
equation. Since the full exploration of this digression would take to long I will not go into
details here. It can be shown (this is done in [1], chapter 6) that this condition reduces the
number of independent coefficients to two. It follows from the condition A0ν = 0. It also
means that Aµν will be traceless.
Having exhausted the gauge- and coordinate-freedom we are left with two numbers which
contain all physical informations about the wave in this gauge. Let us now choose kµ =
(ω, 0, 0, k3 ) where ω = k3 . In this case we have a wave travelling in the z-direction and the
two equations
kµ Aµν = 0, A0ν = 0
(4.25)
imply A3ν = 0. Since it is symmetric and traceless we obtain


0 0
0
0
 0 A11 A12 0 

Aµν = 
 0 A12 −A11 0 
0 0
0
0
(4.26)
The fact that the metric pertubation is traceless and orthogonal to the wave-vector implies that
we are in a sub-gauge to the Lorenz-gauge called the ’transverse traceless gauge’ (TT-gauge).
Since trhµν = 0 in this gauge we have
TT
hµν = hTµνT
(4.27)
When we are in this gauge we will drop the bar over hµν
4.2.2
Matter in the prescence of gravitational waves
Let us now see what happens to test particles in the prescence of gravitational wwaves. To
obtain a coordinate independent description of the waves we will consider the relative motion of
nearby particles following geodesics. This naturally leads us to consider the geodesic deviation
equation. Let the separation 4-vector be given by ξ µ (x) and let the particles be travelling with
4-velocitites uµ (x). Then we have
D2 ξ µ
dτ 2
Dξ µ
dτ
= Rµαβγ uα ξ β uγ
=
∂ξ µ
∂τ
+ ξj
∂ξ µ
∂xj
(4.28)
(4.29)
and τ is the proper time. We will assume that the test particles are non-relativistic in the sense
that their spatial speed v ≪ 1 (c = 1 units). This means τ ≈ t the coordinate time. In the
epxression for the total derivative we then neglect the small contribution from the convective
derivative. The Riemann-tensor is to first order in the metric pertubation, so all corrections
in the 4-velocity uµ = (1 + ≀(|hµν |), 0, 0, 0) produce quantities of second order. So we drop
them. Thus,
D2 ξ µ
= Rµ00β ξ β
(4.30)
2
dτ
4.2 Gravitational radiation in the weak-field limit
116
Here we have used the fact that the Riemann-tensor is symmetric in it’s third and fourth lower
indices. It is calculated to be
Rµ00β =
1
(∂0 ∂0 hµβ + ∂β ∂µ h00 − ∂β ∂0 hµ0 − ∂µ ∂0 hβ0 )
2
But hµ0 = 0 so
(4.31)
1
Rµ00β = ∂02 hµβ
2
(4.32)
So to lowest order we obtain
∂2ξµ
1 β ∂2 µ
=
ξ
h
(4.33)
∂t2
2 ∂t2 β
The wave has two degrees of freedom given by A11 ≡ A+ and A12 ≡ A× . First assume that
A× = 0. We then find
∂2ξ1
∂t2
∂2ξ2
∂t2
To lowest order the solutions are [1]
1 1 ∂2 ikσ xσ
ξ
A
e
+
2 ∂t2
1 ∂2 σ
= − ξ 2 2 A+ eikσ x
2 ∂t
(4.34)
=
ξ 1 (t) = ξ 1 (0) 1 + 12 A+ eikσ x
σ
(4.35)
, ξ 2 (t) = ξ 2 (0) 1 − 12 A+ eikσ x
σ
(4.36)
How do we interpret the solutions? We see that particles initially separated by some distance
in the x-direction will oscillate back and forth relative to each other, and the same thing
happens in the y-direction. The standard picutre is that of particles in a ring in the xy-plane.
When the wave passes they bounce back and forth in the shape of a ′ +′ . See the figure.
If now A+ = 0 and A× 6= 0 we obtain the solutions
σ
σ
ξ 1 (t) = ξ 1 (0) + 12 A× eikσ x ξ 2 (0), ξ 2 (t) = ξ 2 (0) + 21 A× eikσ x ξ 1 (0)
(4.37)
In this case the ring of particles oscillate back and forth in the shape of an ′ ×′ . In the solutions
we interpret the two coefficients A+ and A× as the amplitudes, or the measure of modes of
linear polarization of the gravitational wave. One can then (just as one does in the case of
electromagnetic radiation) define right- and left-handed circular polarization amplitudes, but
we will postpone this until later, the reason being that we want to discuss these in terms of
spin-weight.
We can actually relate the polarization states of the gravitational waves to the kinds of
particles we should expect to find when we quantize the theory. The electromagnetic field
also has two independent modes of polarization (given by ex and ey ) which are described by
vectors in the xy-plane. The modes are invariant under a rotation by 2π, or 360◦ in this plane.
When we quantize we obtain massless spin-1 particles - the photons. The general rule is that
the spin s of the massless particles is related to the angle θ by which the polarization modes
are invariant through
360◦
(4.38)
s=
θ
where θ is given in degrees. From the figures (COMMING) one can infer that this angle
is θ = 180◦ for gravitational waves. Thus we should expect that upon quantization of the
gravitational field one should obtain massless spin-2 particles - the gravitons! As of today
these particles have yet to be discovered, and perhaps they will never be detected directly.
4.3 Constructing spin-weighted tensors
4.3
117
Constructing spin-weighted tensors
We have already seen that for electromagnetism, that certain vector combinations transform
as spin-weighted objects, explicitly
ex ∓ iey → e∓iα (ex ∓ iey )
(4.39)
Let us now do this in general. First consider an orthonormal basis {e1 , e2 , e3 } in threedimensional space where
e1 ∓ ie2 → e∓iα (e1 ∓ ie2 ), e3 → e3
(4.40)
under a rotation through the e3 -axis by an angle α. Using these three combinations we can
construct obbjects with higher rank and with arbitrary integer spin-weight s by taking tensor
products. Let us now first construct a tensor with s = 0. This is easy. Let tij (only the spatial
components are needed) be a symmetric, traceless tensor of rank 2. Then the component with
spin-weight 0 is simply
t0 ≡ e3 ⊗ e3 = t33
(4.41)
The components with spin-weight ±1 are
t±1 ≡ (e1 ± ie2 ) ⊗ e3 = t13 ± it23
(4.42)
t±2 ≡ (e1 ± ie2 ) ⊗ (e1 ± ie2 ) = t11 − t22 ± 2it12
(4.43)
and finally
since t is traceless and symmetric. Also tij = ei ⊗ ej . Notice that t has five independent
components tij which let us construct five spin-weighted ones.
4.4
Solution of the linearized Einstein equation
In the following sections we will utilize two different methods to solve eqs. (4.19) in the
TT-gauge.
4.4.1
The method of Green functions
Let us remind ourselves what the field equations for the pertubed metric looks like:
∂ 2 hµν = −16πGTµν
(4.44)
It beares a striking resemblance to the corresponding equation in electromagnetism which is
∂ 2 Aµ = −J µ
(4.45)
(this is just a compactification of eqs. (2.80) and (2.81) in chapter two). We will now solve
the linearized equations just as we did in section 2.6 in ch. two. The solution is then formally
Z
hµν (xσ ) = 16πG d4 x′ G(xσ − x′σ )Tµν (x′σ )
(4.46)
where the four-dimensional Green function satisfies
∂ 2 G(xσ − x′σ ) = −δ(4) (xσ − x′σ )
(4.47)
4.4 Solution of the linearized Einstein equation
118
This equation was solved in chapter two with the solution given by eq. (2.90) and we will
quote it here (in c = 1 units):
G(xσ − x′σ ) =
δ [t′ − (t − |r − r′ |)]
4π|r − r′ |
(4.48)
Inserting into eq. (4.46) we find
Z
Tµν (t′ , r′ )
d4 x′ δ t′ − t − |r − r′ |
|r − r′ |
Z
′
′
Tµν (r , t − |r − r |)
= 4G d3 r ′
|r − r′ |
hµν (r, t) = 4G
(4.49)
(4.50)
Let us now perform a Taylor expansion of the energy-momentum tensor. For this purpuose
we impose the radiation zone requirement r ≫ r ′ so that
|r − r′ |−1 ≈ 1/r, |r − r′ | ≈ r − n · r′
(4.51)
where n = r/|r|. Hence (tr ≡ t′ = t − r) we expand in powers of n · r′ :
Tµν (r′ , tr + n · r′ ) = Tµν (r′ , t′ ) +
∂Tµν (r′ , t′ )
1 ∂ 2 Tµν (r′ , t′ )
′
(n
·
r
)
+
(n · r′ )2 + · · ·
∂t′
2
∂t′2
(4.52)
Inserting into eq. (4.50) yields
hµν (r, t) =
+
Z
Z
∂Tµν (r′ , t′ )
4G n
d3 r ′ Tµν (r′ , t′ ) + n · d3 r ′ r′
r
∂t′
Z
∂ 2 Tµν (r′ , t′ ) o
1
d3 r ′ (n · r′ )2
2
∂t′2
(4.53)
(4.54)
In the weak-field approximation we have ∂µ T µν = 0. We will in the appendix show that this
formula makes the two first terms disappear in the TT-gauge. The spatial components are
therefore
Z
Z
∂Tij (r′ , t′ )
4G n
TT
3 ′
′ ′
hij (r, t) =
(4.55)
d r Tij (r , t ) + ni · d3 r ′ x′i
r
∂t′
Z
∂ 2 Tij (r′ , t′ ) o
1
nk nk d3 r ′ (n · r′ )2
(4.56)
+
2
∂t′2
2G
1
4G
Q̈ij (t − r)
(4.57)
0 + 0 + Q̈ij (t′ ) =
=
r
6
3r
where we have defined the quadrupole moment
Z
Qij (t) = 3 d3 r ′ xi xj T 00 (r, t)
(4.58)
We know that the left hand side is traceless, but the right hand side isn’t, so we have to
subtract the trace of Qij . When this is done one obtains
2G d2
1
TT
′
k
′
hij (t − r) =
Qij (t ) − δij Q k (t )
(4.59)
3r dt′2
3
t′ =t−r
|
{z
}
T
=QT
ij (t−r)
4.5 Angular distribution of gravitational radiation
4.5
119
Angular distribution of gravitational radiation
Let us now concentrate on obtaining the angular distribution of power from the gravitational
waves in the radiation zone. We start by writing down the expression we will be working on
and then take it from there.
dP
dE
=
= r 2 ei t0i
(4.60)
dΩ dt
dΩ dt
This formula states that the flow of flow of energy per time interval dt into a solid angle dΩ
in direction ei is equal to the distance squared times the dot product of ei and t0i . Here, t0i
is the energy-momentum tensor for gravitation in the weak-field limit. We will work with the
time-average of eq. (4.60). The time-averaging operation will be denoted by h· · · i.
So far we have only being keeping the lowest order terms of the metric pertubation hµν ;
hence in order to keep track of the energy carried away by the gravitational waves we need to
include at least second order terms. Actually we have been sort of cheating all along. When
we discussed the effect gravitational waves had on test particles we assumed that the test
particles move along geodesics. This is derived from the covariant conservation law of energy
momentum:
∇µ T µν = 0
(4.61)
To first order however ∇µ = ∂µ which implies that they (the test particles) move along straight
lines in Minkowski space. Now, the inconsistency of the weak-field limit is revealed.
So what should one do? The proper thing to do is to solve the Einstein field equations to
some given order and then justify the validity of the solution. With these thoughts in mind
we now proceed to solve the vacuum field equations to second order and see how the result
can be interpreted in terms of an energy-momentum tensor for the gravitational field.
If now the metric is written as gµν = ηµν + hµν then to first order
1
(1)
(1)
=0
G(1)
µν [η + h] = Rµν − ηµν R
2
(4.62)
(2)
Let us now expand to second order, i.e. let gµν = ηµν + hµν + hµν where h(2) is a second order
pertubation. We obtain the expansions
Rµν
(0)
(1)
(2)
= Rµν
+ Rµν
+ Rµν
+ ···
R = R(0) + R(1) + R(2) + · · ·
(4.63)
(4.64)
(0)
(of course Rµν = 0 since there is no curvature in flat space) which means that
(2)
(2)
G(1)
µν [η + h ] + Gµν [η + h] = 0
(4.65)
Gµν = 8πGtµν
(4.66)
This can be written as
with
1
1
1
(2)
G(2)
(Rµν
− ηµν R(2) )
(4.67)
µν [η + h] = −
8πG
8πG
2
To obtain it we must first go through a lengthy calculation of the second order Ricci-tensor
and the second order Ricci-scalar. This is done in the next section.
tµν = −
4.5 Angular distribution of gravitational radiation
4.5.1
120
Calculating the energy-momentum tensor
The angular distribution requires the time-averaged energy-momentum operator. We start
with the Ricci-tensor to second order in hµν . From [3] we have
(2)
hRµν
i=
1 Dh 1
∂µ hαβ ∂ν hαβ + hαβ (∂µ ∂ν hαβ + ∂α ∂β hµν − ∂ν ∂β hαµ − ∂µ ∂β hαν )
2 |2
{z
}
{z
} |
(4.68)
(B)
(A)
iE
1
+ ∂ β hν α (∂β hαµ − ∂α hβµ ) − (∂β hαβ − ∂ α h)(∂ν hαµ + ∂µ hαν − ∂α hµν )
2
{z
} |
|
{z
}
(C)
(4.69)
(D)
Let us now recast this akward looking expression into a more manageable form. We impose
the TT-gauge conditions given by
∂α hµα = 0
α
tr h = h
(4.70)
= 0
(4.71)
h0µ = hµ0 = 0
(4.72)
α
We will also take advantage of the fact that we are in the radiation zone where r = |r| ≫ |r′ |.
In this case the metric pertubation hµν is approximately a function of the retarded time
tr = t − r where r 2 = xi xi . Let us define
tr ≡ kσ xσ = t − r
(4.73)
with kσ = (1, −e), ei = xi /r. xµ = (t, r) is the ordinary position 4-vector. We can check that
this definition gives the correct result:
kσ xσ = k0 x0 + ki xi = t − ei xi = t − xi xi /r = t − r
(4.74)
Over a sufficiently small region in space kσ is approximately constant which means that the
metric pertubation is very close to a plane wave. The functional dependence is then dominated
by tr ; we can express this explicitly by converting all partial derivatives into time derivatives:
∂σ hµν =
∂
∂tr dhµν
∂xλ
ḣµν = kσ ḣµν
h
=
=
k
µν
λ
∂xσ
∂xσ dtr
∂xσ
|{z}
(4.75)
λ
δσ
where
ḣµν =
dhµν
dtr
(4.76)
Far outside the source the energy-momentum tensor Tµν vanishes and we have ∂ 2 hµν = 0
(remember that we are in the TT-gauge now) which means that
!
∂α ∂ α hµν = kα kα ḧµν = 0 ⇒ kα kα = 0
(4.77)
We will take advantage of this result when we try to simplify the expression for the second
order Ricci-tensor. The first TT-gauge condition becomes in our new notation
kα ḣµα = 0 ⇔ kα ḣµα = 0 ⇔ kα ḣµα = 0
(4.78)
4.5 Angular distribution of gravitational radiation
121
We will now apply these results to expression (4.69). We see immediately that in term number
(D) the terms in the first bracket becomes
1
1
∂β hαβ − ∂ α h = kβ ḣαβ − ∂ α h = 0
2
2
(4.79)
because of (4.78) and tr h = 0. Thus the last term vanishes. Now look at term (C):
∂ β hν α (∂β hαµ − ∂α hβµ ) = kβ ḣν α (kβ ḣαµ − kα ḣβµ )
β
α
β
(4.80)
α
= k kβ ḣν ḣαµ − k kα ḣν ḣβµ
| {z }
| {z }
(4.81)
=0
=0
= 0
(4.82)
Thus both the two last terms vanish in the TT-gauge! Now let us take a look at the
time-average of term (B):
hhαβ (∂µ ∂ν hαβ + ∂α ∂β hµν − ∂ν ∂β hαµ − ∂µ ∂β hαν )i
These terms involves expressions of the form
Z
d2 hαβ (tr )
1 T
αβ
hh ḧαβ i =
dtr hαβ (tr )
T 0
dt2r
Z
dhαβ (tr ) T
1
1 T
dhαβ dhαβ
= hαβ (tr )
dtr
−
T
dtr
T 0
dtr dtr
0
(4.83)
(4.84)
(4.85)
If the metric pertubation and it’s derivative are well-behaved functions then the first term of
the partial integration can be made arbitrarily small as we let T → ∞. Let us assume that
this is a valid assumption and drop the first term, which means that
hhαβ ḧαβ i = −hḣαβ ḣαβ i
(4.86)
which is exactly the result we need to evaluate term (B). Let us now proceed with that. If we
multiply out the terms we have:
hhαβ ∂µ ∂ν hαβ i + hhαβ ∂α ∂β hµν i − hhαβ ∂ν ∂β hαµ i − hhαβ ∂µ ∂β hαν i
|
{z
}
{z
} |
{z
} |
{z
}
|
=−kµ kν hḣαβ ḣαβ i
∝kβ ḣαβ =0
= −kµ kν hḣαβ ḣαβ i
∝kβ ḣβα =0
(4.87)
∝kβ ḣαβ =0
(4.88)
Finally, the time average of term (A) is
1
kµ kν hḣαβ ḣαβ i
2
(4.89)
Inserting all of these results into the time-average of (4.69) we finally obtain
but since h0µ = 0
1
(2)
hRµν
i = − kµ kν hḣαβ ḣαβ i
4
(4.90)
1
1
(2)
hRµν
i = − kµ kν hḣij ḣij i = − h∂µ hij ∂ν hij i
4
4
(4.91)
4.6 Angular distribution of tensor radiation
122
One nice feature with our notation is that we can explicitly see that the Ricci-scalar gives no
contribution to the energy-momentum pseudo tensor tµν :
(2)
R(2) = η µν hRµν
i = η µν kµ kν [· · · ] = kν kν [· · · ] = 0
(4.92)
The expression for tµν becomes in the TT-gauge
1
1
(2)
hRµν
− ηµν R(2) i
8πG
2
1
kµ kν hḣij ḣij i
32πG
htµν i = −
=
4.6
(4.93)
(4.94)
Angular distribution of tensor radiation
We have found the expression for the time-averaged energy-momentum tensor as a function
of time and coordinates:
ht0i i =
1
1
k0 ki hḣmn ḣmn i =
h∂ 0 hmn ∂ i hmn i
32πG
32πG
(4.95)
where kµ = (1, −e). The dots mean differentiation with respect to the retarded time tr = t−r.
The time-average of a function f (t) is defined as
1
hf (t)i =
T
Z
T
dt f (t)
(4.96)
0
where T is some time-intervall. It will be nessecary to evaluate
k0 ei ki = (−1)ei (−ei ) = 1
(4.97)
so we can immediately write down the time-averaged angular distribution in the radiation
zone:
1
dP
= r 2 hei t0i i =
r 2 hḣij ḣij i
(4.98)
dΩ
32πG
where the metric pertubation is given in terms of the time-derivative of the quadrupole-moment
tensor:
2G
hij =
Q̈ij (t − r)
(4.99)
3r
Since the metric is given in the TT-gauge, the subscripts ij refer to a coordinate system K ′
where the plane wave disturbance travels along the z-axis, . The angular distribution should
be calculated with respect to a fixed space coordinate system K so we need to transform h so
that the indices refer to said coordinate system.
Let us choose K in such a way that the the wave travels in the yz-plane and where the
position vector from the origin to h makes an angle θ with the z-axis in K.
The quadrupole-moment Q̃αβ (we take αβ to be purely spatial indices throughout this
calculation) defined with respect to K is then related to the quadrupole-moment Qij defined
with respect to K ′ through an inverse rotation about the x-axis:
T
T
Qij
Q̃αβ = Riα
Rjβ
(4.100)
4.7 Binary star system
123
where R is the rotation matrix corresponding to a rotation through an angle θ about the x-axis
TR
T
in K. The rotation matrix is orthogonal so that Rij
ik = δjk where R is the transposed
rotation matrix. For said rotation it is given by


1
0
0
RT =  0 cos θ − sin θ 
(4.101)
0 sin θ cos θ
The properties of Q are in the TT-gauge: tr Q = 0 and Qµ0 = Q0µ = 0. So the spatial part
of Q is


Q11 Q12 0
Q =  Q12 Q22 0 
(4.102)
0
0 0
where Q11 = −Q22 . From
expressions:
Q̃11
Q̃22
Q̃33
relations (4.100), (4.101) and (4.102) one obtains the following
T RT Q , Q̃
T
T
= R11
12 = Q̃21 = R11 R22 Q12
11 11
T
T
T RT Q , Q̃
= R22
23 = Q̃32 = R22 R32 Q22
22 22
T
T
T
T
= R32 R32 Q22 , Q̃13 = Q̃31 = R11 R32 Q12
(4.103)
Here we see that we have a mixing of components due to the rotation, and consequently all
matrix elements of Q̃ are non-zero. Inserting the matrix elements from RT we obtain


Q11
cos θ Q12
sin θ Q12
cos2 θ Q22
sin θ cos θ Q12 
Q̃ =  cos θ Q12
(4.104)
2
sin θ Q12 sin θ cos θ Q12
sin θ Q22
We notice directly that tr Q̃ = Q11 + Q22 = tr Q = 0. As expected most of the symmetries
possessed by Q are retained after the rotation, the only difference being that Q̃ doesn’t have
non-zero matrix elements. Since the indices of Q̃ are defined with respect to K we can now
write down the angular distribution of radiation
1
2G 2 X ... ...αβ
dP
=
r2
hQ̃αβ Q̃ i
(4.105)
dΩ
32πG
3r
α,β
G X ... ...αβ
(4.106)
hQ̃αβ Q̃ i
=
72π
α,β
where Q̃αβ is given by (4.104). This expression should be evaluated at the retarded time t − r
according to previous calculations; and the dots mean derivation with respect to time.
4.7
Binary star system
A system of two masses in orbit around each other will emmitt gravitational radiation with a
corresponding loss of energy. In this case eq. (4.106) is applicable.
Let the star with mass m1 and m2 have coordinates (d1 cos ψ, d1 sin ψ, 0) and
(−d2 cos ψ, −d2 sin ψ, 0). The motion is planar as we can see. We place the origin of the
coordinate system at the center of mass, which means that
d1 =
m2
m1 +m2
d, d2 =
m1
m1 +m2
d
(4.107)
4.7 Binary star system
124
where d is the distance between m1 and m2 . The angle ψ is defined as the angle between the
x-axis and the position vector of m1 .
We assume that the motion is Keplerian, i.e. the motion is elliptic, and we can use
Newtonian mechanics to describe the motion. The energy-momentum tensor for point masses
is then
X
T µν (r) =
mn δ(r − rn )(1 − v 2 ) uµ uν
(4.108)
X
≈
mn δ(r − rn ) uµ uν
(4.109)
So The 00-component of the energy-momentum tensor can then be approximated by
X
T 00 (r) =
mn δ(r − rn )
(4.110)
which gives for the quadrupole-moment:
Z
X
Qij = 3 d3 r xi xj T 00 (r) = 3
mn δ(r − rn )xin xjn
(4.111)
n
In our case the non-zero components of this tensor are
Q11 = 3µd2 cos2 ψ
Q22 = 3µd2 sin2 ψ
Q12 = Q21 = 3µd2 sin ψ cos ψ
(4.112)
where µ is the reduced mass m1 m2 /(m1 + m2 ) of the system. The transverse traceless part
of the quadrupole moment tensor is


cos2 ψ − 1/3 sin ψ cos ψ
0
QT T = 3µd2  sin ψ cos ψ sin2 ψ − 1/3
0 
(4.113)
0
0
−1/3
For Kepler motion the orbit equation is
d=
a(1 − e2 )
1 + e cos ψ
where the angular velocity is given by
p
ψ̇ = G(m1 + m2 )a(1 − e2 )/d2
(4.114)
(4.115)
Using (4.114) and (4.115) we can obtain the time-derivatives of the quadrupole-moment quite
straight-forwardly. The results are
...
Q11 = 3β(1 + e cos ψ)2 [2 sin 2ψ + 3e sin ψ cos2 ψ]
...
Q22 = −3β(1 + e cos ψ)2 [2 sin 2ψ + e sin ψ(1 + 3 cos2 ψ)]
...
...
Q12 = Q21 = −3β(1 + e cos ψ)2 [2 cos 2ψ − e cos ψ(1 − 3 cos2 ψ)]
(4.116)
4.7 Binary star system
125
where we now define the constant β through
β2 =
4G3 m21 m22 (m1 + m2 )
a5 (1 − e2 )5
(4.117)
From these one can find the corresponding matrix elements of the quadrupole-moment tensor
defined in the K system. We don’t obtain extra factors since the rotation angle is contant
with respect to time.
...
The angular distribution involves the square of Q̃αβ . Performing the summation in (4.106)
one obtains
X ... ...αβ
...2
...2
...2
...2
...2
...2
= Q11 + cos4 θ Q22 + sin4 θ Q22 + 2[cos2 θ Q12 + sin2 θ Q12 + sin2 θ cos2 θ Q12
Q̃αβ Q̃
α,β
...2
...2
...2
...2
...2
= [Q11 + 2Q12 + Q22 ] + 2 sin2 θ cos2 θ [Q12 − Q22 ]
(4.118)
We have lowered all the indices with the spatial Minkowski metric ηij = δij so we don’t obtain
any extra minus signs. The second line is gotten when one uses the trigonometric identity
1 = sin4 θ + cos4 θ + 2 sin2 θ cos2 θ.
We still need to average this result over one orbit. Since ψ is a function of time we perform
the following trick:
Z 2π
Z
T −1 a2 (1 − e2 )2
1 T
(4.115)
dψ f [t(ψ)](1 + e cos ψ)−2
(4.119)
f (t) dt = p
T 0
G(m1 + m2 )a(1 − e2 ) 0
q 3
a (1−e2 )3
The period T = 2π G(m
for Kepler motion. When we insert this we obtain for the
1 +m2 )
averaging operation:
Z 2π
1
g(ψ)(1 + e cos ψ)−2 dψ
(4.120)
hg(ψ)i =
2π 0
where g(ψ) = f [t(ψ)]. We will use this expression to average the square of the quadrupolemoment. We now obtain
Z 2π
h
i
dP
G
2
2
−2
A
+
2B
sin
θ
cos
θ
(4.121)
=
dψ
(1
+
e
cos
ψ)
dΩ
144π 2 0
where
...2
...2
...2
...2
...2
A = Q11 + 2Q12 + Q22 , B = Q12 − Q22
(4.122)
We calculate A and B from expressions (4.116). This requires a lot of boring algrebra so we
just state the result of the calculations:
h
i
(4.123)
A = 9β 2 (1 + e cos ψ)4 8 + 16e cos ψ + e2 (1 + 7 cos2 ψ)
h
B = 9β 2 (1 + e cos ψ)4 4 − 8 sin2 ψ cos2 ψ − e 4 cos ψ + 36 cos3 ψ − 48 cos5 ψ
i
−e2 1 + 4 cos2 ψ + 9 cos4 ψ − 18 cos6 ψ]
(4.124)
Since the trigonometric functions sin ψ and cos ψ are 2π-periodic, any odd power of of them
will vanish upon averaging. We therefore define an effective integrand based on this fact.
4.8 Classical angular distribution of tensor radiation
126
Inserting eqs. (4.123) and (4.124) into (4.121) and then multiplying out the factors, and then
throwing away all odd factors, one obtains
Z
h
i
G4 m21 m22 (m1 + m2 ) 2ψ
dP
2
4
2
2
2
4
=
dψ
8
+
αe
+
βe
+
2
sin
θ
cos
θ
γ
+
δe
−
κe
(4.125)
dΩ
4π 2 a5 (1 − e2 )5
0
where
α = 1 + 47 cos2 ψ,
δ = 146 cos 6 ψ − 113 cos4 ψ − 8 cos2 ψ − 1,
γ = 4 − 8 sin2 2ψ,
2
4
2
4
6
8
β = cos ψ + 7 cos ψ, κ = cos ψ + 4 cos ψ + 9 cos ψ − 18 cos ψ.
(4.126)
One easily performs the integration over ψ with the result
i
dP
4G4 m21 m22 (m1 + m2 ) h
2
2
=
1
+
f
(e)
+
g(e)
sin
θ
cos
θ
dΩ
πa5 (1 − e2 )5
(4.127)
where the functions f and g are defined as
f (e) = (49/16)e2 + (25/64)e4 , g(e) = −(7/16)e2 + (14/512)e4 .
4.8
(4.128)
Classical angular distribution of tensor radiation
We now want to examine the radiation pattern from a mass distribution described by an
energy-momentum tensor Tµν . The source interaction with the gravitational field is described
in the weak-field limit by the equation
∂ 2 hµν = −16πGTµν
(4.129)
The theory of gravitational radiation from a pertubation in the flat space Minkowksi metric
stems from a Lagrangian density
L=
1
1
(∂ν hαβ )2 − (∂α h)2 + ∂α hαν ∂ν h − ∂α hαν ∂β hβ ν + Lint
2
2
(4.130)
where Lint = −(1/2)Tµν hµν and the barred pertubation is defined as
1
hµν = hµν − ηµν h
2
(4.131)
µν
In the Hilbert gauge ∂µ h = 0. Going to second order in the pertubation one obtains the
Einstein equation, and from it one can define an energy-momentum tensor:
Gµν = 8πGtµν
where tµν is the energy-momentum pseudo tensor, given by
1
1
ij
ij
hij ∂µ ∂ν h − ∂µ hij ∂ν h
tµν = −
16πG
2
(4.132)
(4.133)
in the TT-gauge. The angular distribution is then given in terms of this pseudo-tensor:
Z ∞
dP
dt r 2 ei t0i (t − r)
(4.134)
=
dΩ
−∞
4.8 Classical angular distribution of tensor radiation
127
Fourier-transforming the fields in (4.133) we find
1
dP
=
(kr)2 |hij (ω, r)|2
dΩdω
64π 2 G
(4.135)
where the Fourier-transformed fields satisfy the spin-2 Helmholtz equation ∇2 hij + k2 hij = 0.
By analogy to classical electromagnetism one writes the solution in Cartesian coordinates in
terms of plane waves:
o
X Z d3 k
1 n
∗
−ik· x
ik· x
∗
√
hij (ω, x) =
(4.136)
(k)e
(k,
λ)
e
a
(k)e
(k,
λ)
e
+
a
ij
λ
ij
λ
(2π)3 2ω
λ
where eij is the polarization tensor orthogonal to the direction of momentum: eij ki = 0.
The polarization tensors are also orthogonal to each other for each value of λ meaning
eij (k, λ)eij (k, λ′ ) = δλ,λ′ . This fact ensures that each polarization mode is independent of
the others.
We express the fields in spherical coordinates:
X
hij (ω, r) = ℜ
ajmλ uij (ωλ; j, m)
(4.137)
λ
where the tensor mode-function uij is a solution of the spin-2 Helmholtz equation
∇2 uij + k2 uij = 0. The sum over j and m are included into u.
4.8.1
Divergenceless solutions of the spin-2 Helmholtz equation
(This entire section is written with [5] as reference, so the reader interested in more details
should consult this excellent article and the references cited therein). Since the tensor hij is
symmetric and traceless so is uij . We can take the five independent components of u and form
spin-weighted components:
1
1
(u11 − u22 ± 2iu12 ) = (u33 + 2u11 ± 2iu12 )
2
2
1
= ∓ (u13 ± iu23 ),
2
1
=
u33 ,
2
u±2 =
(4.138)
u±1
(4.139)
u0
(4.140)
where us has spin-weight s. The components are written with respect to the orthonormal
basis {eθ , eφ , er }. If the components uij are real then u∗s = (−1)s u−s .
Writing the Helmholtz equation in terms of the spin-weighted components is indeed a
difficult task, but luckily for us this has been done before in the article by Torres del Castillo,
[5]. Quoting the result from him and the re-writing the equations using our notation we find
1 1 2
4 1
1 ∂ 2∂
2
r
u
+2 + 2 J+ J− u+2 + 2 J+ u+1 + k u+2 = 0
2
r ∂r
∂r
r
r
1 ∂ 2∂
4
1
1
3
r
− 2 u+1 + 2 J+0 J−1 u+1 − 2 J−2 u+2 + 2 J+0 u0 + k2 u+1 = 0
u
+1
2
r ∂r
∂r
r
r
r
r
6
2
1 −1 0
1 ∂ 2∂ r
(4.141)
u0 − 2 u0 + 2 J+ J− u0 + 2 (J+−1 u−1 − J−1 u+1 ) + k2 u0 = 0
r 2 ∂r
∂r
r
r
r
4.8 Classical angular distribution of tensor radiation
128
1 ∂ 2∂
4
1
3
1
r
− 2 u−1 + 2 J−0 J+−1 u−1 + 2 J+−2 u−2 − 2 J−0 u0 + k2 u−1 = 0
u
−1
2
r ∂r
∂r
r
r
r
r
1 −1 −2
4 −1
1 ∂ 2∂
r
u−2 + 2 J− J+ u−2 − 2 J− u−1 + k2 u−2 = 0
r 2 ∂r
∂r
r
r
Here we see that five independent components give five coupled partial differential equations
for the five spin-weighted components of u. As for vector-fields we search for solutions of the
form
s
X
j(j + 1)
j
u±2 =
g±2 (r)Ym,±2
(θ, φ)
(j − 1)(j + 2)
j,m
Xp
j
j(j + 1) g±1 (r)Ym,±1
u±1 =
(θ, φ)
(4.142)
j,m
u0 =
X
j(j + 1)g0 (r)Yjm (θ, φ)
j,m
where j is an integer greater than 1. Inserting eqs. (4.142) into eqs. (4.141) and then using
the action of the spin- raising and lowering operators on the various spin-s harmonics one
obtains a set of five coupled ordinary differential equations for the radial functions gs (r):
2
2 d
(j − 1)(j − 2)
4(j − 1)(j + 2)
d
2
+
−
+
k
g±1 = 0
g±2 +
dr 2 r dr
r2
r2
2
d
2 d
j(j + 1) + 4
3j(j + 1)
1
2
+
−
+ k g±1 + 2 g±2 +
g0 = 0
(4.143)
2
2
dr
r dr
r
r
r2
2
2 d
j(j + 1) + 6
2
d
2
+
−
+ k g0 + 2 (g+1 + g−1 ) = 0
dr 2
r dr
r2
r
We now form linear combinations to decouple the equations. They are
g+2 − g−2 − 2(j + 2)(g+1 − g−1 ),
g+2 − g−2 + 2(j − 1)(g+1 − g−1 ),
g+2 + g−2 − 4(j + 2)(g+1 + g−1 ) + 6(j + 1)(j + 2)g0 ,
(4.144)
g+2 + g−2 − 2(g+1 + g−1 ) − 2j(j + 1)g0 ,
g+2 + g−2 + 4(j − 1)(g+1 + g−1 ) + 6j(j − 1)g0
The solutions for these combinations are spherical Bessel functions of different orders (provided
that k 6= 0). We then obtain from eqs. (4.142):
n
1p
(j − 1)(j + 1)(j + 2) ajj+2 (kr) + bnj+2 (kr) − 2[cjj (kr) + dnj (kr)]
u±2 =
2
o
u±1
u0
j
+ [ejj−2 (kr) + f nj−2 (kr)] ± 2[−Ajj+1 (kr) − Bnj+1 (kr) + Cjj−1 (kr) + Dnj−1 (kr)] Ym,±2
n
1p
j(j + 1) − (j + 2)[ajj+2 (kr) + bnj+2 (kr)] + cjj (kr) + dnj (kr)
=
2
+ (j − 1)[ejj−2 (kr) + f nj−2 (kr)]
(4.145)
o
j
± (j + 2)[Ajj+1 (kr) + Bnj+1 (kr)] ± (j − 1)[Cjj−1 (kr) + Dnj−1 (kr)] Ym,±1
n (j + 1)(j + 2)
j(j + 1)
[ajj+2 (kr) + bnj+2 (kr)] +
[cjj (kr) + dnj (kr)]
=
2
3
o
j(j − 1)
[ejj−2 (kr) + f nj−2(kr)] Yjm
+
2
4.8 Classical angular distribution of tensor radiation
129
where a, b, c, d, e, f, A, B, C and D are arbitrary constants. The cases j = 0 and j = 1 are
j
treated separately since Yms
= 0 for |s| > j. We find, [5], that also in these cases the solutions
of eqs. (4.141) are the separable solutions given by eqs. (4.145).
The divergence of a symmetric traceless tensor field u of rank 2 with respect to the basis
{eθ , eφ , er } is (again, see [5] and the references cited therein):
2 ∂ 3
1 s−1
1
1 s+1
(div u)s = √
− J− us+1 − 3 (r us ) + J+ us−1
(4.146)
r
r ∂r
r
2
Substituting eqs. (4.145) into eqs. (4.146) and then using recurrence relations for the spinweighted harmonics and the spherical Bessel functions, one finds that the divergence of u
vanishes if (and only if)
a=
j(2j−1)c
3(j+2)(2j+1) ,
f=
(j+1)(2j+3)d
3(j−1)(2j+1) ,
b=
j(2j−1)d
3(j+2)(2j+1) ,
e=
(j+1)(2j+3)c
3(j−1)(2j+1) ,
(4.147)
A=
j−1
j+2 C,
B=
j−1
j+2 D.
Hence we only need the four constants c, d, C, D to describe the divergence-free tensor field u.
Here we are suppressing the l, m indices on these constants. As in the case of spin-1 radiation
we can express the spin-weighted solutions of the Helmholtz equation of higher rank through
operators acting on solutions of rank 0 (scalar fields) to the Helmholtz equation. In terms of
the scalar functions ψ1 and ψ2 (definition follows below) we have
1 1 ∂2 2
ik ∂ 2 1 0
2
r J+ J+ ψ1 +
r − k J+1 J+0 ψ2
u+2 = − 2
r ∂r
2 r 2 ∂r 2
ik 2 1 0
1 ∂
u+1 =
J− J+ J+ ψ1 − 2
rJ 2 J 1 J 0 ψ2
2r
2r ∂r − + +
1
J 1 J 2 J 1 J 0 ψ2
(4.148)
u0 =
2r 2 − − + +
ik −2 −1 0
1 ∂
u−1 =
J+ J− J− ψ1 + 2
rJ −2 J −1 J−0 ψ2
2r
2r ∂r + − ik ∂ 2 −1 0
1 1 ∂2 2
u+2 =
J
ψ
+
r
J
r − k2 J−−1 J−0 ψ2
− 1
−
r 2 ∂r
2 r 2 ∂r 2
where the generating functions are
i
X i(2j + 1) h
C
j
(kr)
+
D
n
(kr)
Yjm (θ, φ),
jm
j
jm
j
k2 (j + 2)
j,m
i
X (2j − 1)(2j + 3) h
c
j
(kr)
+
d
n
(kr)
Yjm (θ, φ).
=
jm
j
jm
j
3k2 (j − 1)(j + 2)
ψ1 =
ψ2
(4.149)
(4.150)
j,m
By virtue of the completeness of the spin-weighted spherical harmonics, any divergence-free
solution of the spin-2 Helmholtz equation can be expressed as a superposition of separable
solutions of the form given in (4.148), where ψ1 and ψ2 are solutions of the scalar Helmholtz
equation. If they are real then u is real.
The components (4.148) can be written in terms of tensor operators U and V [5], whose
components with respect to the orthonormal spherical basis mentioned earlier are
Uij (ψ) = Li Xj ψ + Lj Xi ψ,
Vij = ǫimn ∂m Unj (ψ)
(4.151)
4.8 Classical angular distribution of tensor radiation
130
where
L = r × ∇,
X = ∇ × L − ∇.
(4.152)
U and V are symmetric, traceless and divergence-free tensors. By computing the spin-weighted
components of them we find that expressions (4.148) are equivalent to
hij = kUij (ψ1 ) + Vij (ψ2 )
(4.153)
This is a decomposition of the components of h into gradient plus curl components, an expansion which is completely general.
4.8.2
The radiation zone fields
We are interested in the far fields, so we let r → ∞ and only keep the lowest order terms
which are proportional to 1/r. In the radiation zone we know that the fields are outgoing and
thus proportional to eikr , so we set d = D = 0 in (4.149) and (4.150). We also impose the
large-argument limit in the spherical Bessel function jj (kr). We then obtain
ψ1
ψ2
2j + 1
1 eikr X
j
(−i)
Cjm Yjm
=
k2 kr
j+2
j,m
i eikr X
j (2j − 1)(2j + 3)
(−i)
= − 2
cjm Yjm
3k kr
(j − 1)(j + 2)
(4.154)
(4.155)
j,m
where we are to take the real part of the expressions. Thus ψ1 ∼ 1/r and ψ2 ∼ 1/r infinitely
far away from the source. Inserting these expressions for the scalar functions into eqs. (4.148)
we find that
lim u0 = lim u±1 = 0
(4.156)
r→∞
and
lim u±2 =
r→∞
r→∞
eikr X
j
(−i)j B±2 (j, m)Ym,±2
(θ, φ)
kr
(4.157)
j,m
where the multipole coefficients have the structure
B±2 (j, m) = ±M (j) [Cjm ± iN (j)cjm ]
(4.158)
It should be possible to relate them to the attributes of the source, but this will not be done
here, since we already know from previous calculations how it should be done (and for tensors
this would require a lot of computational effort) and we do not need any explicit expressions
for these multiple coefficients to proceed further.
4.8.3
Angular distribution in the radiation zone
We will now use expressions (4.135) and (4.137) to find the distribution of gravitational radiation. We will drop the constants alm (k, λ) in the expression for hij (we are absorbing them
into u):
X
hij (ω, r) =
uij (ωλ; jm)
(4.159)
λ
4.8 Classical angular distribution of tensor radiation
131
The sum over polarization modes is equivalent to a sum over helicity since the symmetry
is axial rather than rotational. We have shown earlier (see chapter three) that helicity λ
corresponds to spin-weight λ in a reference system with axial symmetry.
The basis tensors for circular polarization are the right- and left-handed polarization
matrices which are constructed from the basis tensors for linear polarization:
i
1 h
1
√
√
ex ⊗ ex − ey ⊗ ey + i(ex ⊗ ey + ey ⊗ ex ) (4.160)
[e(+) + ie(×)] =
eR =
2
2
i
1 h
1
eL = √ [e(+) − ie(×)] = √ ex ⊗ ex − ey ⊗ ey − i(ex ⊗ ey + ey ⊗ ex ) (4.161)
2
2
It is straight-forward to show that eR has spin-weight 2 and that eL has spin-weight −2. Hence
the scalar |hij (ω, r)|2 = hij (ω, r)h∗ij (ω, r) is a sum over indepndent modes of spin-weight s:
s 2
X
|hij (ω, r)| =
uλ (ω, j, m)
2
(4.162)
λ=−s
Note that these arguments are only valid for massless particles travelling at the speed of
light, with eigenspin s. For massless gravitons, which are spin-2 particles, we only obtain
contributions from s = ±2 in the radiation zone:
2 2
lim |hij (ω, r)|2 = u+2 (ω, l, m) + u−2 (ω, l, m)
r→∞
(
)
X
2 X
2
1
l
l
=
(−i)l B+2 (l, m)Ym,+2
(θ, φ) + (−i)l B−2 (l, m)Ym,−2
(θ, φ)
(kr)2
l,m
(4.163)
l,m
Inserting (4.163) into (4.135) we finally find
)
(
2
2 X
X
1
dP
l
l
(−i)l B−2 (l, m)Ym,−2
(θ, φ)
=
(−i)l B+2 (l, m)Ym,+2
(θ, φ) + dΩdω
64π 2 G
l,m
l,m
(4.164)
which is the main result of this section. This result should by now not come as a surprise to
the reader. Surely, the classical angular distribution in the radiation zone from spin-0, spin-1
and spin-2 fields have the same canonical form:
X
2 X
2
dP
l l
∼
(−i)l αs (l, m)Yms
(−i)l β−s (l, m)Ym,−s
(4.165)
+
dΩdω spin-s field
l,m
l,m
where αs and β−s are multipole coefficients which can be expressed as integrals over the source
densities. This is due to the fact that all quasi-static components of the fields disappear far
away from the source of the radiation.
When we carry out the quantum mechanical computations, we know from the correspondence principle that we will obtain the same result, but with the Fourier coefficents replaced
with matrix elements.
4.9 Quantum angular distribution of tensor radiation
4.9
132
Quantum angular distribution of tensor radiation
The interaction between the gravitational field and a mass-distribution (from now on referred
to as the source) described by an energy-momentum tensor Tµν is described by the term
Lint = −(1/2)T ij hij
(4.166)
in the classical weak-field theory. Here we assume that only the spatial components contribute. The transition to quantum mechanics follows the usual path of promoting the fields to
operators. The interaction Hamiltonian in the quantum theory is thus
bint = (1/2)Tbij b
H
hij
(4.167)
Consider now, a quantum mechanical system undergoing a transition from a state |ii = |Ei i|0i
where Ei is the initial energy of the system, to a state |f i = |Ef ; k, λi = |Ef ib
a†k,λ |0i through
emission of a graviton with momentum k, polarization λ and energy ω = |k|.
The transition is described by Fermi’s Golden rule:
2
Z
3
bint |ii δ(Ei − Ef − ω) d k
(4.168)
dΓ = 2π d3 r hf |H
(2π)3
We start by evaluating the matrix element. We will expand the field into plane waves using
Cartesian coordinates. By analogy to quantum electrodynamics this expansion is
o
X Z d3 k
1 n
†
ik· x
∗
−ik· x
b
√
(4.169)
b
a
e
(k,
λ)
e
+
b
a
e
(k,
λ)
e
hij (ω, r) =
k,λ ij
k,λ ij
(2π)3 2ω
λ
where eij is the polarization tensor. This canonical form of the expansion ensures the coma†k,λ ] = (2π)3 δ(k′ − k) δλ′ ,λ . Inserting this expansion into the matrix
mutation relations [b
ak′ ,λ′ , b
element in (4.168), we find
b ij b
hf |Q
hij |ii = hEf |Tbij |Ei i h0|b
ak,λ b
hij |0i
1
= Tbfiji √ e∗ij (k, λ) e−ik· x
2ω
(4.170)
b f i is the transition energy-momentun tensor of the source. Inserting the result in
where Q
(4.170) into (4.168) and then using d3 k = ω 2 dω dΩ we obtain
Z
2
ω X dΓ
3 ′ bij ′ ∗
−ik·r′ d
r
T
(r
)
e
(k,
λ)
e
(4.171)
=
ij
fi
dΩ dω
16π 2
λ
where we have inserted a sum over all possible graviton polarization modes. We have also
labeled all source coordinates with primes to better distinguish which coordinates that should
be integrated over. The integral itself runs over the source dimensions. The angular distribution of radiated power is gotten as usual by multiplying with the graviton energy:
Z
2
dP
ω 2 X 3 ′ b∗ij ′
ik·r′ =
d
r
T
(r
)
e
(k,
λ)
e
(4.172)
ij
fi
dΩ dω
16π 2
λ
This is the expression we will work with. As in the case of QED we see that eik· r = R(θ, φ) eikz
where R is the rotation operator. Thus we first find an expansion of a gravitaton travelling
along the z-axis and then rotate it to obtain the state of a graviton travelling in an arbitrary
direction described by the spherical angles θ, φ.
4.9 Quantum angular distribution of tensor radiation
4.9.1
133
Expansion of the graviton field
A circularly polarized plane wave travelling in the z-direction with helicity ±2 has the following
expansion in Cartesian coordinates:


1 ±i 0
(hij ) = (eij ) eikz =  ±i −1 0  eikz
(4.173)
0
0 0
The multipole expansion of the graviton field was earlier found to be
hij = kUij (ψ1 ) + Vij (ψ2 )
and ψ1 and ψ2 are given by eqs. (4.149) and (4.150). We will need the curl of h:
(curl h)ij = ǫimn ∂m kUnj (ψ1 ) + Vnj (ψ2 )
(4.174)
(4.175)
With the expression for V we find
ǫimn ∂m Vnj (ψ) = ǫimn ǫnpk ∂m ∂p Ukj (ψ) = (δip δmk − δik δmp )∂m ∂p Ukj (ψ)
= ∂i ∂k Ukj (ψ) − ∂m ∂m Uij (ψ) = −Uij (∇2 ψ) = k2 Uij (ψ)
(4.176)
since U is divergence-free and linear. The action of the curl-operator thus implies that we
multiply by the wave number k and then substitute ψ1 with ψ2 , and vice versa. Therefore
(curl h)ij = kǫimn ∂m Unj (ψ1 ) + k2 Uij (ψ2 ) = k2 Uij (ψ2 ) + kVij (ψ1 )
(4.177)
From eqs. (4.148) we see that
2r 2 h0 = J−1 J−2 J+1 J+0 ψ2 .
(4.178)
Taking the curl of both sides we obtain
(curl h)0 =
k
J 1 J 2 J 1 J 0 ψ1
2r 2 − − + +
(4.179)
Our goal here is to determine the constants in ψ1 and ψ2 using (4.173). Since the plane wave
should be finite everywhere we have
ψ1 =
X i(2j + 1)
Cjm jj (kr)Yjm
k2 (j + 2)
(4.180)
j,m
ψ2 =
X (2j − 1)(2j + 3)
cjm jj (kr)Yjm
3k2 (j − 1)(j + 2)
(4.181)
j,m
From the definition of h0 in (4.140) and (4.173) we find
2r 2 h0 = xi xj hij = (x ± iy)2 eikz = r 2 sin2 θe±2iφ eikr cos θ
d2
= (−1/k2 ) sin2 θe±2iφ
eikr cos θ
d(cos θ)2
(4.182)
when we switch to spherical coordinates. To determine the constants we expand the exonential,
eikr cos θ =
∞
X
j=0
ij (2j + 1)jj (kr)Pj (cos θ)
(4.183)
4.9 Quantum angular distribution of tensor radiation
134
which is a well-known expansion that we have already used in this thesis. Inserting into (4.182)
we find
d2
1 X j
i (2j + 1)jj (kr)e±2iφ sin2 θ
2r 2 h0 = − 2
Pj (cos θ)
(4.184)
k
d(cos θ)2
j
As it stands now this expression looks quite confusing, but the exponential factors suggest
that the angular factors could sum up to some kind of spherical harmonic. Indeed, when one
considers the definition of the associated Legendre functions Pjm (x) in terms of the Legendre
polynomial (for m > 0)
Pjm (x) = (−1)m (1 − x2 )m/2
dm
Pj (x)
dxm
where x = cos θ we see that when comparing (4.184) with (4.185) that
s
(2j + 1)
e±2iφ Pj,±2 (cos θ)
Yj,±2 =
4π(j − 1)j(j + 1)(j + 2)
s
d2
(2j + 1)
e±2iφ sin2 θ
Pj (cos θ)
=
4π(j − 1)j(j + 1)(j + 2)
d(cos θ)2
(4.185)
(4.186)
and therefore
2r 2 h0 = −
=
1 X jp
i 4π(2j + 1)(j − 1)j(j + 1)(j + 2) jj (kr)Yj,±2
k2
j
1 2 1 0
J− J− J+ J+ ψ2
(4.187)
Evaluating the action of the spin- raising and lowering operators we find that we can choose
s
∞
X
4π(2j + 1)
1
ij
jj (kr)Yj,±2
(4.188)
ψ2 = − 2
k
(j − 1)j(j + 1)(j + 2)
j=2
This would correspond to
j
cjm = −3 i
s
4π(2j + 1)(j − 1)(j + 2)
δm,±2
j(j + 1)[(2j − 1)(2j + 3)]2
(4.189)
for a wave with helicity ±2 respectively, in eq. (4.181). Taking the curl of (4.173) we find that
curl h = ±kh, and if we compare eq. (4.178) to eq. (4.179) we find that ψ1 = ±ψ2 . Inserting
these results into eqs. (4.148) or eq. (4.174) we obtain the desired multipole expansion of a
plane wave travelling in the z-direction:
i
h
eij eikz
= kUij (±ψ2 ) + Vij (ψ2 )
(4.190)
±2
where ψ2 is given by eq. (4.188). To get to the desired result we have to act on (4.190) with
the rotation operator:
eij eik·r = eij R(θ, φ)eikz
= kUij (±Rψ2 ) + Vij (Rψ2 )
(4.191)
4.9 Quantum angular distribution of tensor radiation
135
We find the action of R on ψ2 :
∞
1 X j
i
Rψ2 = − 2
k
j=2
RYj,±2 (θ, φ) =
j
X
s
4π(2j + 1)
jj (kr)RYj,±2
(j − 1)j(j + 1)(j + 2)
(4.192)
s
(4.193)
(j)
Dm,±2 Yjm (θ, φ)
=
m
(−1)
m=−j
m=−j
Thus
j
X
4π
Y j∗ Yjm(θ, φ)
(2j + 1) m,∓2
#
"
4π X j
(−1)m
j
Rψ2 = − 2
jj (kr)Yjm (θ, φ) Ym,±2
i p
k
(j − 1)j(j + 1)(j + 2)
j,m
(4.194)
All coordinates in the brackets refer to the source coordinates and shall be integrated over.
The angles (not explicitly shown) in the spin-weighted harmonics are the angles describing
the photon direction.
We now have all the ingredients to find the angular distribution. Inserting result (4.191)
into (4.172) we get
Z
2
ω 2 X dP
3 ′ b∗ij ′
d
r
T
(r
)[kU
(±Rψ
)
+
V
(Rψ
)]
=
ij
2
ij
2 fi
dΩ dω
16π 2
λ∈±2
( Z
2
2
ω
3 ′ b∗ij ′
=
d
r
T
(r
)[kU
(+Rψ
)
+
V
(Rψ
)]
ij
2
ij
2
f
i
2
16π
λ=+2
)
Z
2
′
(4.195)
+ d3 r ′ Tbf∗ij
(r
)[kU
(−Rψ
)
+
V
(Rψ
)]
ij
2
ij
2
i
λ=−2
Let us obtain a more detailed expression. We can write the integral as
Z
′
d3 r ′ Tbf∗ij
i (r )[kUij (±Rψ2 ) + Vij (Rψ2 )]λ=±2 =
−
where
4π X
b±2 (j, m)Y j
(−i)j G
m,±2 (θ, φ)
k
(4.196)
j,m
b±2 (j, m) = f (j, m)
G
Z
∗
d3 r ′ Tbfiji Fij±2 jj (kr ′ )Yjm
(θ ′ , φ′ )
(4.197)
where the function f and the tensor Fij±2 are given by
f (j, m) = √
(−1)m
,
(j−1)j(j+1)(j+2)
Fij±2 = ∓Uij +
1
k
Vij
(4.198)
Thus
2
2 X
X
dP
j
jb
b+2 (j, m)Y j
(−i)
G
(j,
m)Y
(θ,
φ)
(−i)j G
(θ,
φ)
+
=
−2
m,−2
m,+2
dΩ dω
j,m
j,m
(4.199)
Appendix A
Calculation of ∇2F
Let us first remind ourselves on how we construct spin-weighted components of a vector. In
spherical coordinates any vector F can be written as
F = Fr er + Fθ eθ + Fφ eφ
(A.1)
In terms of the spin-weighted combinations er and eθ ± ieφ which tranform under rotations
around an axis parallell to er according to
er → er , eθ ± ieφ → e∓iα (eθ ± ieφ )
(A.2)
the vector field may be rewritten as
1
1
F = F0 er + √ F− (eθ + ieφ ) + √ F+ (eθ − ieφ )
2
2
(A.3)
where Fs has spin-weight s, and
F0 = F · er = Fr , F±1 = √12 F · (eθ ± ieφ )
√
√
In terms of n = er , m = (1/ 2)(eθ + ieφ ) and m = (1/ 2)(eθ − ieφ ) one has
F = F0 n + mF− + mF+
From expressions (A.1) and (A.4) one finds that
√
√
Fr = F0 , Fθ = (1/ 2)(F+ + F− ), Fφ = (i/ 2)(F− − F+ )
er = n,
√
eθ = (1/ 2)(m + m),
√
eφ = (i/ 2)(m − m)
(A.4)
(A.5)
(A.6)
From [13] we have the expressions
∇f
=
∇·F =
∇×F =
+
∂f
1 ∂f
1 ∂f
er +
eφ +
eθ
∂r
r sin θ ∂φ
r ∂θ
1 ∂Fφ
1 ∂
1 ∂
2
(sin θ Fθ ) +
r
F
+
r
r 2 ∂r
r sin θ ∂θ
r sin θ ∂φ
1
∂Fθ
∂
(sin θ Fφ ) −
er
r sin θ ∂θ
∂φ
1
1 ∂
∂Fr
∂
1 ∂Fr
− sin θ (rFφ ) eθ +
(rFθ ) −
eθ
r sin θ ∂φ
∂r
r ∂r
r ∂θ
where f is some scalar function.
136
(A.7)
(A.8)
(A.9)
A.1 The gradient
A.1
137
The gradient
Let us first find the gradient in terms of the spin-weighted basis vectors. We insert eqs. (A.6)
for the unit vectors into eq. (A.7) and find
∂f
∂f
1
i
∂f
+√
(m − m)
+ (m + m)
∇f = n
(A.10)
∂r
∂φ
∂θ
2r sin θ
∂
∂f
∂
1
i ∂
i ∂
= n
+√
−
+
m
+m
f
(A.11)
∂r
∂θ sin θ ∂φ
∂θ sin θ ∂φ
2r
Remembering the definition of the spin-weight raising and lowering operators J+s and J−s we
find
1
1
∂f
− √ mJ−0 f − √ mJ+0 f
(A.12)
∇f = n
∂r
2r
2r
Since we can write ∇2 f = ∇ · ∇f we can combine (A.12) with (A.17) to find
1 ∂
1 2 ∂f
∇2 f =
r
+ 2 J−1 J+0 + J+−1 J−0 f
2
r ∂r
∂r
2r
∂f
1
1 ∂
r2
+ 2 J−1 J+0 f
=
2
r ∂r
∂r
r
(A.13)
(A.14)
since f has spin-weight 0. This result came with relatively little effort. As we soon shall
find out calculating the effect of ∇2 on a vector field in terms of spin-weight is much more
complicated. Let us carry on.
A.2
The divergence
The calculation of the divergence proceeds along a similar path, the only difference being
that we now also insert the expressions for the vector components in terms of spin-weighted
components. Using (A.6) for the components of F in eq. (A.8) we find that
1 ∂
i ∂
1 ∂ 1
2
∇·F = 2
sin θ (F+ + F− ) +
(F− − F+ )
r F0 + √
r ∂r
sin θ ∂φ
2r sin θ ∂θ
1 ∂
= 2
r 2 F0
(A.15)
r ∂r
1
∂
∂
i ∂
i ∂
1
1
−
+
(sin θF+ ) +
(sin θF− )
+ √
sin θ ∂θ sin θ ∂φ
2r sin θ ∂θ sin θ ∂φ
1 ∂
1
1
= 2
(A.16)
r 2 F0 − √ J−1 F+ − √ J+−1 F−
r ∂r
2r
2r
Therefore
∇·F=
A.3
1 ∂
1 1
r 2 F0 − √
J− F+ + J+−1 F−
2
r ∂r
2r
(A.17)
The curl
The curl of the vector field is for obvious reasons the quantity which involves most calculational
effort. But there are no short cuts in science, so we might as well get started. Using (A.6) for
A.4 The Laplace operator ∇2
138
the components and unit vector of F we obtain for the first term:
1 ∂
i ∂ 1
√
sin θ (F− − F+ ) − √
(F+ + F− ) n
(∇ × F)r =
r sin θ
2 ∂θ
2 ∂φ
i
∂
∂
i ∂
i ∂
1
1
= √
−
+
−
(sin θF+ ) +
(sin θF− ) n
sin θ ∂θ sin θ ∂φ
sin θ ∂θ sin θ ∂φ
2r
i
i h 1
(A.18)
= √
J− F+ − J+−1 F− n
2r
The second and third term together give:
(∇ × F)θ,φ =
+
∂F0 i sin θ ∂ 1
√
(m + m)
r(F− − F+ )
− √
∂φ
2r sin θ
2 ∂r
∂F 1 ∂ i
0
√ (m − m) √
r(F+ + F− ) −
∂θ
2r
2 ∂r
(A.19)
(A.20)
i
i ∂
1 ∂
1 ∂
1 ∂
1 ∂
∂
= m√
−
(rF+ ) − √
(rF− ) − √
(rF− ) + √
(rF+ )
F0 − √
∂θ sin θ ∂φ
2r
2 ∂r
2 ∂r
2 ∂r
2 ∂r
i ∂
1 ∂
1 ∂
1 ∂
∂
1 ∂
i
+
(rF+ ) + √
(rF− ) − √
(rF− ) + √
(rF+ )
−
F0 + √
+ m√
∂θ sin θ ∂φ
2r
2 ∂r
2 ∂r
2 ∂r
2 ∂r
√ ∂
√ ∂
i
i
0
0
= −m √
(A.21)
J− F0 + 2 (rF− ) + m √
J+ F0 + 2 (rF+ )
∂r
∂r
2r
2r
Adding all these terms one obtains the full expression:
i
i h 1
∇ × F = n√
J− F+ − J+−1 F−
2r
√ ∂
√ ∂
i
i
− m√
J−0 F0 + 2 (rF− ) + m √
J+0 F0 + 2 (rF+ ) (A.22)
∂r
∂r
2r
2r
We will need the spin-weighted components of the vector A = ∇ × F for later calculations.
They are inferred from eq. (A.22) to be
i
i
h
h
√ ∂
(rF± )
(A.23)
A0 = √i2r J−1 F+ − J+−1 F− , A± = ± √i2r J±0 F0 + 2 ∂r
In the next section we combine all our previous results to find ∇2 F in terms of spinweighted components.
A.4
The Laplace operator ∇2
The wellknown identity ∇2 F = ∇(∇ · F) − ∇ × (∇ × F) makes it possible to find what we are
looking for. Let us first calculate the first term. We have from eqs. (A.12) and (A.17)
∂
1 1
1 ∂
−1
2
∇(∇ · F) = n
r F0 − √
(A.24)
J− F+ + J+ F−
∂r r 2 ∂r
2r
1 ∂
1 1
1 −1
2
√
mJ−0 + mJ+0
−
F
r
F
J
F
+
J
(A.25)
−√
−
0
+
− +
r 2 ∂r
2r
2r
A.4 The Laplace operator ∇2
The first line (A.24) is
∂ 1 ∂
2
r F0 −
= n
∂r r 2 ∂r
∂ 1 ∂
2
= n
r F0 −
∂r r 2 ∂r
139
1
1
1 ∂
−1 ∂
√ J−
F+ /r − √ J+
F− /r
∂r
∂r
2
2
h
h
1
1
1
1 ∂F+ i
1 ∂F+ i
1
1
1
√ J− − 2 F+ +
− √ J− − 2 F+ +
r
r ∂r
r
r ∂r
2
2
(A.26)
In the second line (A.25) the terms proportional to m and m are
h
i
1
1
1
0 1 ∂
0 1
0 −1
2
r F0 + 2 J− J− F+ + 2 J− J+ F−
= m − √ J− 2
r ∂r
2r
2r
2r
h
i
1
1
1
0 1 ∂
0 1
0 −1
2
+ m − √ J+ 2
r F0 + 2 J+ J− F+ + 2 J+ J+ F−
r ∂r
2r
2r
2r
Adding our results we obtain
h
1
∂ 1 ∂
2
1
√
−
r
F
−
∇(∇ · F) = n
J
0
−
∂r r 2 ∂r
2
h1 ∂
i
1
1
+ m − √ J−0 2
r 2 F0 + 2
r ∂r
2r
2r
h
i
1
1
1 ∂
+ m − √ J+0 2
r 2 F0 + 2
r ∂r
2r
2r
(A.27)
(A.28)
h
1
1 ∂F+ i
1 ∂F+ i
1
1
1
− √ J− − 2 F+ +
F+ +
r2
r ∂r
r
r ∂r
2
1
J−0 J−1 F+ + 2 J−0 J+−1 F−
2r
1
0 1
0 −1
J+ J− F+ + 2 J+ J+ F−
2r
(A.29)
We now proceed to calculate ∇ × (∇ × F) ≡ ∇ × A. From eq. (A.22) we have
i
i h 1
J− A+ − J+−1 A−
∇ × A = n√
2r
√ ∂
√ ∂
i
i
0
0
√
√
− m
J− A0 + 2 (rA− ) + m
J+ A0 + 2 (rA+ ) (A.30)
∂r
∂r
2r
2r
and from eqs. (A.23) we have the components. We first find the term proportional to n.
Inserting the expressions for A± one obtains
h
h
i
i
√ ∂
√ ∂
i 0
i 0
i
−1
1
J+ F0 + 2 (rF+ ) − J+ − √
J− F0 + 2 (rF− )
n√
J− √
∂r
∂r
2r
2r
2r
h
h
i
√
√
∂F− i
∂F+
1
−1
−1 0
1
1 0
+ J+ J− F0 + 2 J+ F− + r
= − 2 n J− J+ F0 + 2 J− F+ + r
2r
∂r
∂r
h
1
∂F− i
1
∂F+
= n − 2 J−1 J+0 F0 − √
+ J+−1 F− + r
J−1 F+ + r
r
∂r
∂r
2r 2
(A.31)
The term proportional to m is
√ ∂ i h
i
√ ∂
i 1
i
−1
0
0
J− F+ − J+ F− + 2
J− √
− √ J− F0 + 2 (rF− )
− m√
∂r
∂r
2r
2r
2
2
∂F0
1 0 1
1
1
1 ∂
J J F+ − 2 J−0 J+−1 F− − √ J−0
−
(rF− )
(A.32)
= m
2r 2 − −
2r
∂r
r ∂r 2
2r
A.4 The Laplace operator ∇2
140
Similarly we obtain for the term proportional to m:
i
√ ∂ i h
√ ∂
i
i 1
−1
0
0
√ J+ F0 + 2 (rF+ )
m√
J− F+ − J+ F− + 2
J+ √
∂r
∂r
2r
2r
2
2
1
1 0 −1
1 ∂
1 0 1
0 ∂F0
−
(rF+ )
(A.33)
= m − 2 J+ J− F+ + 2 J+ J+ F− − √ J+
2r
2r
∂r
r ∂r 2
2r
Collecting all results we find (after a lot of effort)
∂F− i
∂F+ 1 h 1
1 1 0
−1
+ J+ F− + r
J− F+ + r
∇ × (∇ × F) = n − 2 J− J+ F0 − √
r
∂r
∂r
2r 2
2
1 0 1
1
1 0 −1
1 ∂
0 ∂F0
+ m
J J F+ − 2 J− J+ F− − √ J−
−
(rF− )
2r 2 − −
2r
∂r
r ∂r 2
2r
1 0 1
1
1 0 −1
1 ∂2
0 ∂F0
+ m − 2 J+ J− F+ + 2 J+ J+ F− − √ J+
−
(rF+ )
2r
2r
∂r
r ∂r 2
2r
(A.34)
We are to take eq. (A.29) minus eq. (A.34) to find the Laplace operator. We will do this by
comparing the components (with the sign reversed in (A.34)). Doing this for n we find that
the terms involving derivatives of F± with respect to r vanishes and we are left with
√
√
1
2 1
2
∂ 1 ∂
2
2
(A.35)
r F0 + 2 J− F+ + 2 J+−1 F− + 2 J−1 J+0 F0
(∇ F) · n =
2
∂r r ∂r
r
r
r
For the m-component we find that upon expanding the partial derivatives in r that
√
1 0 −1
2
1 ∂2
2
(∇ F) · m =
(rF− ) + 2 J− J+ F− − 2 J−0 F0
r ∂r 2
r
r
(A.36)
The same calculation for the m-component gives
√
1 0 1
1 ∂2
2
(rF+ ) + 2 J+ J− F+ − 2 J+0 F0
(∇ F) · m =
2
r ∂r
r
r
2
Therefore we finally find
" #
√
√
∂
∂
1
1
2
2
∇2 F =
r 2 F0 + 2 J−1 F+ + 2 J+−1 F− + 2 J−1 J+0 F0 n
∂r r 2 ∂r
r
r
r
"
#
√
1 ∂2
1
2
+
(rF− ) + 2 J−0 J+−1 F− − 2 J−0 F0 m
2
r ∂r
r
r
"
#
√
2 0
1 ∂2
1 0 1
+
(rF+ ) + 2 J+ J− F+ − 2 J+ F0 m
r ∂r 2
r
r
This result is in exact agreement with the result found in [4]!
(A.37)
(A.38)
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141