APPENDIX A
Spherical Geometry, Spherical Harmonics and Tensor Calculus
1.
Introduction
As the Earth is almost a spherical body, many problems in geophysics require the use of spherical
coordinates rather than Cartesian coordinates. Furthermore we often need to be able to specify scalar,
vector and tensor fields anywhere within this body and this is most commonly done using expansions in
spherical harmonics. Unfortunately, tensor calculus in anything but Cartesian coordinates is algebraically
tedious and many techniques have been devised to alleviate the labor. You may be familiar with the use
of Christoffel symbols in covariant differentiation of tensors. Another technique, familiar in quantum
mechanics, is expansion in generalized spherical harmonics (GSH). This technique was made popular in
the geophysical literature in a paper by Phinney and Burridge (1973) and some detail can also be found
in the book by Edmonds (1960).
We use standard spherical polar coordinates :
x1 = r sin θ cos φ and
x2 = r sin θ sin φ and
x3 = r cos θ
where θ is colatitude and φ is east longitude.
The spherical harmonics in most common use in the free oscillation literature are those defined by
Edmonds (1960), i.e.,
Ylm (θ, φ) = (−1)m
2l + 1 (l − m)!
4π (l + m)!
1
2
Plm ( cos θ)eimφ
where the Plm are associated Legendre functions which are progressively wigglier functions of θ as l
increases. We often need to compute Ylm ’s and their derivatives with respect to θ. This can easily be
done using the recursion relationships given by Edmonds (1960) for the Plm .
Some properties of the Ylm ’s are:
∗
Yl−m = (−1)m Ylm
where ∗ denotes complex conjugation
1
∇21 Ylm = −l(l + 1)Ylm
where ∇21 =
2
∂2
∂
2 ∂
+
cot
θ
θ
+
cosec
∂θ2
∂θ
∂φ2
This relationship allows us to cast higher order θ derivatives in terms of Ylm and ∂Ylm /∂θ. Note that φ
derivatives are trivial, i.e.,
∂ m
Y = imYlm
∂φ l
Finally the Ylm ’s are fully normalized, i.e.,
∗
Ylm
Ylm dS = δmm δll
etc.
where
dS = sin θdθdφ
S
If we wish to specify an appropriately smooth scalar function of position inside the Earth, ρ(r, θ, φ)
say, then we can make an expansion in spherical harmonics, e.g.,
ρ(r, θ, φ) =
∞ l
m
ρm
l (r)Yl (θ, φ)
(A.1)
l=0 m=−l
and the field is completely specified by the coefficients, ρm
l (r).
The situation is a little more complicated if we have a vector field and it turns out that it is not useful
to expand the r, θ, φ components in a way similar to the scalar expansion. The usual way to represent a
vector field in spherical geometry is to use the form (Morse and Feshback, 1953):
u(r, θ, φ) = r̂U (r, θ, φ) + ∇1 V (r, θ, φ) − r̂ × (∇1 W (r, θ, φ))
(A.2)
where U , V , and W are scalars, ∇1 is the surface gradient operator defined by
∇1 = θ̂θ
∂
∂
φ
+ cosec θφ̂
∂θ
∂φ
φ are unit vectors in the r, θ, φ directions. Note that r̂ × ∇1 (× is vector cross product) is
and r̂, θ̂θ, φ̂
r̂ × ∇1 = −θ̂θ cosec θ
∂
∂
φ
+ φ̂
∂φ
∂θ
The scalars U, V and W can now be expanded as in A1. Equation A2 is the standard form for the
expansion of the displacement field in low-frequency seismology.
Canonical components and generalized spherical harmonics
When we come to tensor fields, the algebra gets a little more awkward and it turns out that things
simplify if we abandon the r, θ, φ coordinates and introduce new ones. We label these new directions
−, 0, +. If u(r, θ, φ) has components ur , uθ , uφ then the new directions are defined as
1
u− = √ (uθ + iuφ )
2
u0 = ur
1
u+ = √ (−uθ + iuφ )
2
2
It is convenient to represent this coordinate transformation as a matrix operation. Suppose we let
u1 = uθ ,
u2 = u φ
and
u3 = ur
†
Cαi
ui
α = −, 0, +
then
uα =
3
(A.3)
i=1
where

√1
2
√i
2


†
=
Cαi
 0

− √12
0
√i
2
0



1


0

√1
2


with Hermitian conjugate Ciα =  − √i2

0
0
0
− √12



− √i2 

1
(A.4)
0
A tensor of any order can be put into the new coordinate system by repetition of the operation A3, i.e.,
†
†
†
mαβγ... = Cαi
Cβj
Cγk
· · · mijk...
(A.5)
where summation over repeated indices is implied. Remember that i, j, k, etc. go from 1 to 3 corresponding to θ, φ, r directions respectively. mαβγ... can be returned to the original coordinate system by
the operation
mijk... = Ciα Cjβ Ckγ · · · mαβγ...
(summation implied)
(A.6)
Tensor calculus becomes much simpler if we expand the contravariant canonical components in
generalized spherical harmonics rather than the Ylm ’s defined earlier, i.e., we set
mαβγ... (r, θ, φ) =
∞ l
Mlαβγ...,m (r)YlN,m (θ, φ)
(A.7)
l=0 m=−l
where N = α + β + γ + .... For the contravariant canonical components of the displacement field we
have
α
u (r, θ, φ) =
∞ l
Ulα,m Ylα,m (θ, φ) where
α = −, 0, +
(A.8)
l=0 m=−l
The generalized spherical harmonics are given by
YlN,m = PlN,m ( cos θ)eimφ
where the PlN,m are defined by Phinney and Burridge who also give recursion relations for their calculation. We note two of their formulae here:

√
ΩlN +1 YlN +1,m + ΩlN YlN −1,m = 2[N cot θ − m cosec θ]YlN,m 

N,m
(A.9)
√ dY


ΩlN YlN −1,m − ΩlN +1 YlN +1,m = 2 l
dθ
3
1
where ΩlN = [ 12 (l + N )(l − N + 1)] 2 . Note that Ωl−N +1 = ΩlN so that Ωl0 = Ωl1 , Ωl−1 = Ωl2 , etc.
It is convenient to write down some relationships between the GSH and the ordinary Ylm ’s. Let
2l+1 γl =
4π . Then


γl Yl0,m = Ylm




1

−1,m
+1,m
l
m

√ γl Ω1 [Yl

+ Yl
] = −m cosec θYl


2



m

∂Y
1

−1,m
+1,m
l
l
√ γl Ω1 [Yl
− Yl
]=
(A.10)
∂θ
2

m 

l(l + 1) m
∂Y 
−2,m
l l
1

]Yl − cot θ l 
+ Yl+2,m ] = [m2 cosec 2 θ −

2 γl Ω1 Ω2 [Yl

2
∂θ



m

∂Yl

−2,m
+2,m
l l
m
1

γ
Ω
Ω
[Y
−
Y
]
=
m
cosec
θ
cot
θY
−

l
1
2
l
l
l
2
∂θ
Relationships for higher N can be found from the recursion formulae A9. These relationships are useful
if one wants to convert from the canonical components back to expressions in r, θ, φ involving ordinary
spherical harmonics.
Equation A5 defines the contravariant canonical component of mijk... and we can also define covarient
components of mijk... (though these are less useful), i.e.,
mαβγ = Ciα Cjβ Ckγ · · · mijk...
(summation implied)
(A.11)
We will need the covarient components of the tensor δij . We define these as
∆αβ = Ciα Cjβ δij = Ciα Ciβ = C1α C1β + C2α C2β + C3α C3β
From A4 we find that
∆00 = 1 and ∆−+ = ∆+− = −1 and all other components are zero
(A.12)
As an example of the use of ∆, we consider the trace of a second-order tensor, i.e.,
Tr (mij ) = mij δij = mii = m11 + m22 + m33
mii is a scalar and is equivalent to
mii = ∆αβ mαβ
00
=m
−m
(summation implied)
+−
− m−+
(as all other components of ∆αβ are zero)
In elasticity we have to worry about the derivatives of vectors and tensors. Differentiation is straightforward in the new coordinate system. For definiteness we consider a second-order tensor mij , differentiation gives a third-order tensor which we write as mij,k .
From equation A7 we have the expansion
mαβ (r, θ, φ) =
∞ l
(α+β),m
Mlαβ,m (r)Yl
(θ, φ)
l=N m=−l
To reduce the number of indices we have to write out, we consider a single l, m component of mαβ and
write this as
4
mαβ = M αβ Y α+β
( l and m understood)
Differentiation results in a higher rank tensor which we write as
mαβ,γ = M αβ|γ Y α+β+γ
The coefficients M αβ|γ can be found in terms of the M αβ by using the following recipe which we give
for a tensor of any order. Consider γ = −, 0, + separately, then we have
M
αβ...|−
1
=
r
ΩlN
M
αβ...
−
terms obtained from M αβ... by changing
+ into 0, 0 into − one at a time
dM αβ

dr



αβ...

1
by changing
terms obtained from M

l
αβ...


=
−
ΩN +1 M
r
− into 0, 0 into + one at a time
M αβ...|0 =
M αβ...|+









(A.13)
Here N = α + β + ... and when calculating N , regard −, 0, + as shorthand for −1, 0 and +1.
The use of A13 is best illustrated by example. For a second rank tensor we consider the +0 component
only so N = 1. Thus
1 l +0
Ω1 M − M 00 − M +−
r
+0
dM
M +0|0 =
dr
1 l +0
+0|+
M
=
Ω2 M − M ++
r
As another example of the use of ∆αβ , we consider the divergence of a second rank tensor which would
be written mij,j . This is equivalent to
M +0|− =
∆βγ mαβ,γ = mα0,0 − mα+,− − mα−,+
Another example of the use of A13 is given by the computation of the Laplacian of a scalar field. A
single l, m component of the expansion of a scalar field φ can be written
φ = Φ Y 0,m
where Φ is an expansion coefficient. Differentiation gives
φ,α = Φ|α Y α,m
which using A13 becomes
1 l
Ω Φ Y −1,m
r 0
d
φ,0 =
Φ Y 0,m
dr
1 l
,+
φ = Ω1 Φ Y +1,m
r
φ,− =
5
The Laplacian is given by
∆αβ φ,αβ = φ,00 − φ,−+ − φ,+−
but using A13 again gives
d2
Φ Y 0,m
dr2
1
1
=
Ωl0 Ωl0 Φ −
r
r
1
1
=
Ωl1 Ωl1 Φ −
r
r
φ,00 =
,−+
φ
φ,+−
d
Φ Y 0,m
dr
d
Φ Y 0,m
dr
so the Laplacian of a single l, m component of φ is
2
l(l + 1)
d
2 d
−
+
Φ Y 0,m
∇2 φ = ∇2 (Φ Y 0,m ) =
dr2
r dr
r2
(A.14)
The canonical components of displacement
These definitions are sufficient for our purposes so now we illustrate how to use these formulae with
the free oscillation equations. Suppose we have represented the displacement field by the expression in
equation A2, i.e.,
u(r, θ, φ) = r̂
∞ l
Ulm (r)Ylm (θ, φ) + θ̂θ
l=0 m=−l
φ
+ i cosec θφ̂
l
l
l
φ
− φ̂
l
l
m=−l
∞ l
∂Ylm
(θ, φ)
∂θ
mVlm (r)Ylm (θ, φ) + iθ̂θ cosec θ
m=−l
Wlm (r)
m=−l
Vlm (r)
l
l
mWlm (r)Ylm (θ, φ)
m=−l
∂Ylm
(θ, φ)
∂θ
For simplicity we consider a single l, m component so that, on collecting terms, we obtain
∂Ylm
∂Y m
φ im cosec θV Ylm − W l
u(r, θ, φ) = r̂U Ylm + θ̂θ V
+ im cosec θW Ylm + φ̂
(A.15)
∂θ
∂θ
where the indices l and m on U , V , and W are understood. Applying equation A3 gives
∂Ylm
1
u− = √ (V − iW )
− m cosec θYlm
∂θ
2
0
m
u = U Yl
∂Y m
1
u+ = √ (V + iW ) − l − m cosec θYlm
∂θ
2
6
(A.16)
Equation A8 for a single l, m component gives
u− = U − Yl−1,m
u0 = U 0 Yl0,m
+
u =U
+
(A.17)
Yl+1,m
Comparison of A16 and A17 and use of the relationships between YlN,m and Ylm gives
U − = (V − iW )γl Ωl0
U 0 = U γl
+
U
= (V +
(A.18)
iW )γl Ωl0
We now have the relationship between the new and old coordinate system and so can turn to giving
examples of tensor differentiation for a single l, m component.
The Strain Tensor
In Cartesian coordinates the strain tensor is given by
∂uj
∂ui
1
ij = 2
+
=
∂xj
∂xi
1
2
(ui,j + uj,i )
In spherical polar coordinates, the answer is not so obvious. In our canonical coordinate system we have
αβ =
1
2
(uα,β + uβ,α )
where
uα,β = U α|β Y α+β
Using equation A13 gives
U −|− = 1r Ωl−1 U −
U −|0 =
dU −
dr
U −|+ = 1r [Ωl0 U − − U 0 ]
U 0|− = 1r [Ωl0 U 0 − U − ]
U 0|0 =
dU 0
dr
U 0|+ = 1r [Ωl1 U 0 − U + ]
U +|− = 1r [Ωl1 U + − U 0 ]
U +|0 =
dU +
dr
U +|+ = 1r Ωl2 U +
where
E αβ =
Thus
αβ = E αβ Y α+β
1
2
(U α|β + U β|α )
and we obtain for the components of αβ
−−
00
++
1
= Ωl−1 U − Yl−2,m
r
dU 0 0,m
=
Y
dr l
1
= Ωl2 Yl2,m
r
−0
+0
+−
dU −
U−
Ωl0 U 0
=
=
−
+
Yl−1,m
dr
r
r
+
dU
U+
Ωl1 U 0
0+
1
=
= 2
−
+
Yl+1,m
dr
r
r
1
= −+ = 12 [Ωl0 U − + Ωl1 U + − 2U 0 ]Yl0,m
r
0−
7
1
2
If we use the fact that Ωl−1 = Ωl2 and Ωl0 = Ωl1 (as pointed out earlier) and we substitute for U − , U 0
and U + from equation A18, we finally obtain







00 = U γl Yl0,m
1
±± = (V ± iW )γl Ωl1 Ωl2 Yl±2,m
r
±0
= 0± = 12 [X ± iZ]γl Ωl1 Yl±1,m
(A.19)






±∓ = − 12 F γl Yl0,m
where prime ( ) denotes differentiation with respect to radius and
X =V+
U −V
,
r
Z = W −
W
r
and F =
1
[2U − l(l + 1)V ]
r
For reference, we note that inspection of A19 gives the expansion coefficients, E αβ :
E 00 = U γl
1
E ±± = (V ± iW )γl Ωl1 Ωl2
r
E ±0 = E 0± = 12 [X ± iZ]γl Ωl1
E ±∓ = − 12 F γl







(A.20)






Often we need go no further as the canonical components can be used throughout the calculations. If
we wish to obtain expressions for in the original r, θ, φ coordinate system we must use equation A6,
i.e.,
ij = Ciα Cjβ αβ
= Ci− Cj− −− + Ci− Cj0 −0 + Ci− Cj+ −+
+ Ci0 Cj− 0− + Ci0 Cj0 00 + Ci0 Cj+ 0+
+ Ci+ Cj− +− + Ci+ Cj0 +0 + Ci+ Cj+ ++
Remember that i and j go from 1 to 3 and 1 ≡ θ, 2 ≡ φ, 3 ≡ r directions respectively. As an example,
consider the 11 = θθ component. We have from A4 that
1
C1− = √ ,
2
C10 = 0,
and
1
C1+ = − √
2
Substitution gives
θθ =
1 −−
2 −
1 −+
2 −
1 +−
2 +
1 ++
2 iW
V
+2,m
−2,m
+2,m
−2,m
l l
l l
+
+
= γl Ω0 Ω2 Yl
+ Yl
− Yl
γl Ω0 Ω2 Yl
2r
2r
where we have made use of equation A19. A complete list of strain components is
8
1
2
F γl Yl0,m

rr = U K0





V + iW − F

θθ = K2 −
K2 + K0 


r
r
2




V + iW − F
K2 + K0 
φφ = − K2 +
r
r
2


2
rθ = XK1− − iZK1+




+
−


2
rφ = −iXK1 − ZK1





i2V − 2W +

2
θφ = −
K2 −
K2
r
r
(A.21)
where
K0 = γl Yl0,m
1
K1± = √ γl Ωl1 (Yl−1,m ± Yl+1,m )
2
1
K2± = γl Ωl1 Ωl2 (Yl−2,m ± Yl+2,m )
2
1
±
K3 = √ γl Ωl1 Ωl2 Ωl3 (Yl−3,m ± Yl+3,m )
2 2
Equation A20 can be cast in terms of ordinary spherical harmonics using the relationships in A10. The
result is:

rr = U Ylm




m

1
∂Yl

m
2
2
m
m

U Yl + V m cosec θYl − l(l + 1)Yl − cot θ
θθ =



r
∂θ



m


∂Yl
W
m


− cot θYl
+ i m cosec θ


r
∂θ




m

1
∂Yl

m
2
2
m

φφ =
U Yl + V cot θ
− m cosec θYl


r
∂θ




m

W
∂Yl
m
− i m cosec θ
− cot θYl
r
∂θ



m

∂Yl

m


2
rθ = X
+ im cosec θZYl


∂θ


m

∂Y

m
l


2
rφ = im cosec θXYl − Z


∂θ



m

2mV
∂Yl

m

2
θφ = i
cosec θ
− cot θYl


r
∂θ




m

W
∂Yl

2
2
m
m


+
2 cot θ
− 2m cosec θYl + l(l + 1)Yl
r
∂θ
(A.22)
where X and Z are defined in equation A19.
It is sometimes convenient to work in “epicentral coordinates,” i.e., with the pole of the coordinate
system at the earthquake epicenter. The strain tensor must then be evaluated at θ, φ → 0. Equation A20
makes this evaluation particularly simple as generalized spherical harmonics have the property
9
YlN,m (0, 0) = δN m
(A.23)
At (θ, φ) = (0, 0), it therefore follows that
K0 = d0l
for m = 0 only
Kk+ = dkl
for m = ±k only
Kk− = ∓dkl
for m = ±k only





(A.24)
where
1
dkl = k
2
2l + 1 (l + k)!
4π (l − k)!
1
2
(A.25)
and the nonzero values of the strain tensor can be easily evaluated (Table A1).
The Divergence of the Displacement Vector ∇ · u
In Cartesian coordinates ∇ · u = ∂ui /∂xi = ii (summation implied). In canonical components we
have ii = ij δij ≡ ∆αβ αβ and from equation A12 this becomes
∇ · u = 00 − +− − −+
= (U + F ) γl Yl0,m = (U + F ) Ylm
(A.26)
where we have used equation A19. This is easily verified by summing rr + θθ + φφ in equation A21.
10
The gradient of the strain tensor
When computing the excitation of modes by extended sources, we need to know the strain tensor
of a mode throughout the source volume. One way of doing this is to expand the strain tensor in a
Taylor series about some fiducial point. This requires (at least) the first derivative of the strain tensor.
Fortunately, the calculation is not too onerous in canonical components. Our starting point is (for a
single l, m component)
αβ,γ = E αβ|γ Y α+β+γ
Using equation A13 gives
E 00|− =
1
r
l 00
Ω0 E − 2E −0
E 00|0 = E 00
E ++|− =
1
r
l ++
− 2E +0
Ω2 E
E ++|0 = E ++
E −−|− = 1r Ωl3 E −−
E +0|− =
1
r
E −−|0 =
d
−−
dr E
l +0
Ω0 E − E 00 − E +−
E +0|0 =
d
+0
dr E
E −0|− =
1
r
l −0
Ω2 E − E −−
E −0|0 =
d
−0
dr E
E +−|− =
1
r
l +−
− E 0−
Ω0 E
E +−|0 =
d
+−
dr E
E 00|+ =
1
r
l 00
Ω0 E − 2E +0
E ++|+ = 1r Ωl3 E ++
E −−|+ =
1
r
l −−
− 2E −0
Ω0 E
E +0|+ =
1
r
l +0
Ω2 E − E ++
E −0|+ =
1
r
l −0
Ω0 E − E 00 − E −+
E +−|+ =
1
r
l +−
− E +0
Ω0 E
so we obtain
00,0 = U γl Y 0,m
1
00,± = [U − X ∓ iZ] γl Ωl0 Y ±1,m
r
1
(V ± iW )
±±,0
=
(V ± iW ) −
γl Ωl0 Ωl2 Y ±2,m
r
r
1
±±,± = 2 (V ± iW )γl Ωl0 Ωl2 Ωl3 Y ±3,m
r
l l
Ω2 Ω 2
1
±±,∓
(V ± iW ) − (X ± iZ) γl Ωl0 Y ±1,m
=
r
r


(X ± iZ )γl Ωl0 Y ±1,m





1 1
1
l l ±2,m 


(V
±
iW
)
γ
=
(X
±
iZ)
−
Ω
Ω
Y
l 0 2

2

r
r



l l


1 Ω0 Ω 0

0,m
1

(X ± iZ) − U + 2 F γl Y
=



r
2



0,m
1


= − 2 F γl Y




1

l ±1,m

= − [F + X ± iZ] γl Ω0 Y
2r
±0,0 =
±0,±
±0,∓
+−,0
+−,±


































1
2
Equation A6 can now be used to retrieve the r, θ, φ components giving
11
(A.27)
rr,r = U K0
1
iZ +
K
rr,θ = (U − X)K1− +
r
r 1
i
Z
rr,φ = − (U − X)K1+ + K1−
r
r
1 V
iZ
θθ,r = (V − )K2+ −
K2− + 12 F K0
r
r
r V − iW +
1 (Ω2 )2
i (Ω2 )2
−
θθ,θ = 2 K3 − 2 K3 −
V − 2X − F K1 +
W − 2Z K1+
r
r
2r
r
2r
r
iV + W −
i (Ω2 )2
1 (Ω2 )2
+
V + F K1 −
W K1−
θθ,φ = − 2 K3 − 2 K3 −
r
r
2r
r
2r
r
1
V
iZ − 1 φφ,r = − (V − )K2+ +
K2 + 2 F K0
r
r
r V − iW +
1 (Ω2 )2
i (Ω2 )2
−
V + F K1 −
W K1+
φφ,θ = − 2 K3 + 2 K3 +
r
r
2r
r
2r
r
iV + W −
i (Ω2 )2
1 (Ω2 )2
+
V − 2X − F K1 +
W − 2Z K1−
φφ,φ = 2 K3 + 2 K3 +
r
r
2r
r
2r
r
2
rθ,r = X K1− − iZ K1+
1
i
1
2V
2W
2
rθ,θ = (X −
)K2+ − (Z −
)K2− −
(Ω0 )2 X − 2U + F K0
r
r
r
r
r
2V
2W
i
1
(Ω0 )2
)K2− − (Z −
)K2+ −
ZK0
2
rθ,φ = − (X −
r
r
r
r
r
2
rφ,r = −iX K1+ − Z K1−
i
1
(Ω0 )2
2V
2W
2
rφ,θ = − (X −
)K2− − (Z −
)K2+ +
ZK0
r
r
r
r
r
2V
2W
1
i
1
)K2+ + (Z −
)K2− −
(Ω0 )2 X − 2U + F K0
2
rφ,φ = − (X −
r
r
r
r
r
2i V
2Z
K2+
2
θφ,r = − (V − )K2− −
r
r
r 2iV + 2W − i (Ω2 )2
1 (Ω2 )2
+
V − X K1 +
W − Z K1−
2
θφ,θ = − 2 K3 − 2 K3 +
r
r
r
r
r
r
2V − 2iW + 1 (Ω2 )2
i (Ω2 )2
−
2
θφ,φ = − 2 K3 + 2 K3 −
V − X K1 +
W − Z K1+
r
r
r
r
r
r
(A.28)
Finally, we can evaluate this at the pole using equation A24. The non-zero values are given in Table A1.
The Divergence of the Elastic Stress Tensor
In the equations of motion we require ∇ · τ . As pointed out before, in canonical components this can
be written
∇ · τ ≡ ∆βγ τ αβ,γ = τ α0,0 − τ α+,− − τ α−,+ ≡ P α
say
(A.29)
Now suppose that the stress tensor has been expanded in generalised spherical harmonics as in A7 with
a single l, m component being
12
13
-
2
rφ
2
θφ
-
2
θφ,θ
2
θφ,φ
+F
|m|
Note that each column should be multiplied by dl .
-
1
2
r (Ω0 ) Z
2
1
− r (Ω0 ) X − 2U 2
θφ,r
2
rφ,φ
2
rφ,θ
-
(Ω2 )2
r V
-
-
−iX ± Z
2
− X ∓ 1r (Ωr2 ) W − Z
2
2
± 1r (Ωr2 ) V − X + ri (Ωr2 ) W − Z
i
r
-
2
rθ,φ
2
rφ,r
-
− 1r (Ω0 )2 Z
2
rθ,θ
∓X − iZ 2
1
− r (Ω0 ) X − 2U + F
2
rθ,r
2
-
φφ,φ
i
2r
i
− 2r
2
(Ω2 )
i
− 2X − F + 2r
W
−
2Z
2
r 2 (Ω2 )
(Ω2 )
1
V
+
F
±
r
2r
r W
(Ω2 )2
r V
iZ
r
iZ
r
2 (Ω2 )
(Ω2 )
1
i
∓ 2r
W
r V + F − 2r
2
r 2
(Ω2 )
(Ω2 )
1
r V − 2X − F ∓ 2r
r W − 2Z
1
± 2r
− X) ∓
-
F
− ri (U ∓ 1r (U − X) +
-
-
−
-
-
)∓
iZ
r
iZ
r
2V
r
1
− r (X − 2V
r
± 2i
(V
r
± ri (X −
)−
)∓
V
r
-
-
−
−
2Z
r
i
r (Z
) − 1r (Z −
2W
r
2W
r
)
)
1
2V
i
2W
r (X − r ) ± r (Z − r )
1
2W
± ri (X − 2V
r ) − r (Z − r )
-
-
-
2W
r
)±
V
r
V
r
-
− 1r (V −
1
r (V
-
-
−
-
−iX ± Z
± 2iV
r
-
V
iW
r ± r
V
− r ∓ iW
r
-
m = ±2
∓X − iZ
-
-
-
m = ±1
φφ,θ
1
2
-
θθ,φ
φφ,r
-
θθ,θ
F
-
rr,φ
θθ,r
1
2
U
rr,r
rr,θ
-
2
rθ
F
F
φφ
θθ
U
1
2
1
2
rr
m=0
TABLE A1. The strain tensor and its gradient at the origin
−
-
W
r2
∓
2W
r2
2iW
r2
± 2V
r2 +
-
-
-
-
-
-
-
iW
r2
+
-
W
r2
iW
r2
±
− 2iV
r2
iV
r2
± rV2
− iV
r2 ±
∓ rV2
-
-
-
-
-
-
-
-
-
m = ±3
τ αβ = T αβ Ylα+β
(A.30)
The covariant derivative is therefore
τ αβ,γ = T αβ|γ Ylα+β+γ
where the coefficients T αβ|γ can be determined from the T αβ using the recipe of equation A13. Substituting this last equation into A29 gives
P − = (T −0|0 − T −+|− − T −−|+ )Yl−1,m
P 0 = (T 00|0 − T 0+|− − T 0−|+ )Yl0,m
P + = (T +0|0 − T ++|− − T +−|+ )Yl+1,m
and using equation A13 gives
T −+|− = 1r [Ωl0 T −+ − T −0 ]
T −0|0 =
d −0
dr T
T −−|+ = 1r [Ωl−1 T −− − T −0 − T 0− ]
T 0+|− = 1r [Ωl1 T 0+ − T 00 − T −+ ]
T 00|0 =
d 00
dr T
T 0−|+ = 1r [Ωl0 T 0− − T 00 − T +− ]
T ++|− = 1r [Ωl2 T ++ − T 0+ − T +0 ]
T +0|0 =
d +0
dr T
T +−|+ = 1r [Ωl1 T +− − T +0 ]
Combining these and remembering that the stress tensor is symmetric (so T αβ = T βα ) gives

d −0 1 l −+

−
−0
l −−

− 3T + Ω2 T ) Yl−1,m
P =
T − (Ω0 T


dr
r




d
1
0,m
0
00
l 0+
00
−+
l 0−
P =
+ Ω0 T ) Yl
T − (Ω0 T − 2T − 2T

dr
r




d
1

+1,m
+
+0
l −+
+0
l ++


T − (Ω0 T
P =
− 3T + Ω2 T ) Yl
dr
r
(A.31)
Finally, the r, θ, φ components can be recovered from the −, 0, + components by applying equation A6.
Thus
Pi = Ciα P α = Ci− P − + Ci0 P 0 + Ci+ P +
so
1
Pθ = P1 = √ (P − − P + )
2
i
Pφ = P2 = − √ (P − + P + )
2
Pr = P3 = P 0
To proceed further we must specify the form of the stress-strain relation. The case of an isotropic
elastic solid is particularly simple as the canonical components of stress are simply proportional to the
canonical components of strain except for the terms involving the dilatation which require a little care.
We have
τ = 2µ
+ λ(∇ · u) I
14
where I is the identity tensor. In canonical components the identity tensor is given by
†
†
†
†
Cβj
δij = Cαi
Cβi
∆αβ = Cαi
so that using equation A3 gives
∆00 = 1,
∆−+ = ∆+− = −1
and all other elements are zero
(Note that ∆αβ = ∆αβ .) We can now write out the canonical components of τ , i.e.,
τ αβ = 2µ
αβ + λ(∇ · u) ∆αβ
Thus

τ 00 = [2µU + λ (U + F )] γl Yl0,m 



2µ
±±
l l ±2,m 

τ
=
(V ± iW )γl Ω0 Ω2 Yl
r


τ ±0 = µ[X ± iZ]γl Ωl0 Yl±1,m



0,m 
±∓
τ
= [−µF − λ (U + F )] γl Yl
(A.32)
Comparison with equation A30 gives the expansion coefficients, T αβ i.e.,
T 00 = [2µU + λ (U + F )] γl
2µ
(V ± iW )γl Ωl0 Ωl2
T ±± =
r
T ±0 = T 0± = µ[X ± iZ]γl Ωl0
T ±∓ = − [µF + λ (U + F )] γl













(A.33)
Substituting these into equation A31 gives

d
d


P =
(µX) − i (µZ)


dr
dr





2µ
1
l l
l −1,m 

(V − iW )Ω2 Ω2 γl Ω0 Yl
+ µF + λ(U + F ) + 3µ(X − iZ) −



r
r




d

0


P =
(2µU + λ(U + F ))

dr

1


− 2µXΩl0 Ωl0 − 4µU + 2µF γl Yl0,m


r





d
d

+

(µX) + i (µZ)
P =



dr
dr




2µ
1
l l
l +1,m 


+ µF + λ(U + F ) + 3µ(X + iZ) −
(V + iW )Ω2 Ω2 γl Ω0 Yl
r
r
−
so that the (r, θ, φ) components are
15
(A.34)

d
µ
m


(2µU + λ(U + F )) + [4U − 2F − l(l + 1)X] Yl
Pr =


dr
r




m

d
∂Yl
µV
1


Pθ =
(µX) +
(l + 2)(l − 1)
µF + λ(U + F ) + 3µX −


dr
r
r
∂θ




µ
W
d
m
(A.35)
(µZ) +
3Z −
(l + 2)(l − 1) im cosec θYl
+

dr
r
r




1
d
µV
m

(µX) +
µF + λ(U + F ) + 3µX −
(l + 2)(l − 1) im cosec θYl 
Pφ =


dr
r
r




m

∂Yl
d
µ
W


(µZ) +
3Z −
(l + 2)(l − 1)
−
dr
r
r
∂θ
The case of a more general elastic tensor is much more complicated. The most general relationship
is
τij = Cijkl kl
where the fourth-order tensor C has the symmetries
Cijkl = Cjikl = Cijlk = Cklij
and so has only 21 independent components. Applying A5 gives the tensor in canonical components
with the same symmetries i.e.,
C αβγδ = C βαγδ = C αβδγ = C γδαβ
The 21 independent components are





1
1

C
−
C
+
C

θφθφ
4 φφφφ
2 θθφφ



1
1
1

= 4 Cθθθθ + 4 Cφφφφ + 2 Cθθφφ




1
1


= − 2 Cθθrr − 2 Cφφrr




1
1

= − 2 Cθrθr − 2 Cφrφr





1
i


= ∓ √ Cθrrr + √ Cφrrr


2
2




i
1


= ± √ (Cθθθr + 2Cθφφr − Cφφθr ) + √ (Cθθφr − 2Cθφθr − Cφφφr ) 
2 2
2 2

1
i



= ± √ (Cθθθr + Cφφθr ) − √ (Cθθφr + Cφφφr )


2 2
2 2




1
1

= 2 Cθθrr − 2 Cφφrr ∓ iCθφrr




1
1

= 2 Cθrθr − 2 Cφrφr ∓ iCθrφr





i

1
1

= − 4 Cθθθθ + 4 Cφφφφ ± (Cθθθφ + Cθφφφ )



2



1
i

= ∓ √ (Cθθθr − 2Cθφφr − Cφφθr ) + √ (Cθθφr + 2Cθφθr − Cφφφr ) 



2 2
2 2



1
1
1
= 4 Cθθθθ + 4 Cφφφφ − 2 Cθθφφ − Cθφθφ ∓ i (Cθθθφ − Cθφφφ )
C 0000 = Crrrr
C ++−− = 14 Cθθθθ +
C +−+−
C +−00
C +0−0
C ±000
C ±±∓0
C
+−±0
C ±±00
C ±0±0
C ±±+−
C ±±±0
C ±±±±
16
(A.36)
We are particlularly interested in the case of transverse isotropy when only the five components with
N = α + β + γ + δ = 0 are non-zero. In the usual notation, we have

C 0000 = C




++−−

= 2N
C


+−+−
C
=A−N


C +−00 = −F 




+0−0
C
= −L
(A.37)
The double contraction of the elastic tensor with the strain tensor to give the stress tensor is written in
canonical components as
τ αβ = C αβγδ ηξ ∆γη ∆δξ
Performing this contraction gives
τ αβ = C αβ−− ++ + C αβ00 00 + C αβ++ −−
− 2C αβ−0 +0 − 2C αβ+0 −0 + 2C αβ−+ +−
(A.38)
and specialising to the case of a transversely isotropic solid gives
τ 00 = C 0000 00 + 2C 00−+ +−
τ ±± = C ++−− ±±
τ ±0 = τ 0± = −2C +0−0 ±0
(A.39)
τ ±∓ = C +−00 00 + 2C +−+− +−
It is interesting to compare this with the isotropic case and, using the notation of equation A37, we get
τ 00 = C
00 − 2F
+−
τ ±± = 2N
±±
versus
τ ±0 = τ 0± = 2L
±0
τ ±∓ = −F
00 + 2(A − N)
+−
τ 00 = (λ + 2µ)
00 − 2λ
+−
τ ±± = 2µ
±±
τ ±0 = τ 0± = 2µ
±0
(A.40)
τ ±∓ = −λ
00 + 2(λ + µ)
+−
from which we see that the isotropic case can be recovered by setting A = C = λ + 2µ,
µ and F = λ. The expansion coefficients are (by comparison with A30)

T 00 = [CU + FF ] γl





2N
±±
l l
T
=
(V ± iW )γl Ω0 Ω2 
r

±0

T = T 0± = L[X ± iZ]γl Ωl0 



±∓
T
= − [(A − N)F + FU ] γl
Substituting these into equation A31 gives
17
L=N=
(A.41)

d
d


P =
(LX) − i (LZ)


dr
dr





2N
1
l l
l −1,m 

(V − iW )Ω2 Ω2 γl Ω0 Yl
+ (A − N)F + FU + 3L(X − iZ) −



r
r




d

0


P =
(CU + FF )

dr

1


− 2LXΩl0 Ωl0 + 2(F − C)U + 2(A − N − F)F γl Yl0,m


r





d
d

+

P =
(LX) + i (LZ)



dr
dr




2N
1
l l
l +1,m 


+ (A − N)F + FU + 3L(X + iZ) −
(V + iW )Ω2 Ω2 γl Ω0 Yl
r
r
−
(A.42)
so that the (r, θ, φ) components are

1
d


Pr =
(CU + FF ) − [2(F − C)U + 2(A − N − F)F + l(l + 1)LX] Ylm


dr
r




m

d
NV
1
∂Y

l

Pθ =
(LX) +
(A − N)F + FU + 3LX −
(l + 2)(l − 1)


dr
r
r
∂θ




1
NW
d
m
(A.43)
(LZ) +
3LZ −
(l + 2)(l − 1) im cosec θYl
+

dr
r
r




1
d
NV


(LX) +
(A − N)F + FU + 3LX −
(l + 2)(l − 1) im cosec θYlm 
Pφ =


dr
r
r




m

d
1
NW
∂Yl


−
(LZ) +
3LZ −
(l + 2)(l − 1)
dr
r
r
∂θ
The Traction Vector
φτrφ can also be expanded in vector spherical harmonics (as
The traction vector t = r̂τrr + θ̂θτrθ + φ̂
we have implicity showed in the last section). We can recover the usual expression for a single l, m
component of the elements of t from equation A30 along with equations A3 and A5, i.e.,
τij = Ciα Cjβ τ αβ
Thus
τrr = τ33 = C3α C3β τ αβ = τ 00
1
τrθ = τ31 = C3α C1β τ αβ = √ (τ 0− − τ 0+ )
2
i
τrφ = τ32 = C3α C2β τ αβ = − √ (τ 0− + τ 0+ )
2
For an isotropic solid we get
18
τrr = [(λ + 2µ)U + λF ] K0
τrθ = µXK1− − iµZK1+
τrφ = −iµXK1+ − µZK1−
so

τrr = [(λ + 2µ)U + λF ] Ylm




m

∂Yl
m
τrθ = µX
+ im cosec θµZYl
∂θ


∂Ylm 

m

τrφ = im cosec θµXYl − µZ
∂θ
and for a transversely isotropic solid we have

dU

m

τrr = C
+ FF Yl


dr



m
∂Yl
m
τrθ = LX
+ im cosec θLZYl 
∂θ


m 

∂Y
m
l 

τrφ = im cosec θLXYl − LZ
∂θ
Inspection of A.44 and A.45 shows that defining
R = CU + FF
and
S = LX
and
T = LZ
(A.44)
(A.45)
(A.46)
gives the usual vector spherical harmonic expansion of t :
t = r̂R(r)Ylm + S(r)∇1 Ylm − T (r)r̂ × ∇1 Ylm
(A.47)
Dot products and contraction
We often have to evaluate integrals of functions which involve dot products and contractions, e.g.
ρ0 u∗ · u dV
Transformation to the canonical basis can make such integrals easy but we have to be careful about the
complex conjugation since the new basis is complex and not real. For a real basis (ui )∗ = (u∗ )i and
since
ui = Ciα uα
we have
∗
(ui )∗ = Ciα
(uα )∗
and
(u∗ )i = Ciα (u∗ )α
Thus
†
†
∗
Cβi
Ciα (u∗ )α = Cβi
Ciα
(uα )∗
which evaluates to
19
δαβ (u∗ )α = ∆αβ (uα )∗
or
(u∗ )β = ∆αβ (uα )∗
Now consider the dot product (u · v∗ ) which in canonical components is
(u · v∗ ) = uα (v ∗ )β ∆αβ = uα (v γ )∗ ∆γβ ∆αβ = uα (v γ )∗ δαγ
(A.48)
which is the conventional form for the inner product. We shall also need the contraction of τ with the
complex conjugate of the strain tensor which can be written:
τ : ∗ = τ αβ (
γδ )∗ δαγ δβδ
(A.49)
Integrals of the products of spherical harmonics
The major result for the integral of the products of two generalized spherical harmonics is
2
m ∗
YlN m YlN
dΩ = δll δmm
γl
(A.50)
Ω
where γl = (2l + 1)/4π and dΩ = sin θdθdφ
Now consider
ρ0 u∗ · u dV
V
In canonical components, we have using A48
∗
∗
u∗ · u = u− u− + u0 u0 + u+ u+
∗
∗
∗
= (V − iW )(V + iW )γl2 12 l(l + 1) Yl−1,m Yl−1,m + Yl+1,m Yl+1,m
(A.51)
+ U 2 γl2 Yl0,m Yl0,m∗
Substituting A51 into the integral and using A50 gives
∗
ρ0 u · u dV = ρ0 U 2 + l(l + 1)V 2 r2 dr
V
(spheroidals)
(A.52)
=
ρ0 l(l + 1)W 2 r2 dr
(toroidals)
As a second example, consider the elastic energy density integral:
∗ · ·C : dV
V
20
(A.53)
Now
∗
∗ · ·C : = ∗ · ·ττ = ∗ : τ = τ αβ γδ δαγ δβδ
∗
∗
∗
= τ −− −− + 2τ −0 −0 + 2τ −+ −+
∗
∗
∗
+ τ 00 00 + 2τ 0+ 0+ + τ ++ ++
Substituting the form for τ αβ for an isotropic solid gives
∗
∗
∗
∗ ∗ : τ = 2µ
−− −− + 4µ
−0 −0 + 2 −λ
00 −+ + 2(λ + µ)
+− +−
∗
∗
∗
∗
+ (λ + 2µ)
00 00 − 2λ
+− 00 + 4µ
0+ 0+ + 2µ
++ ++
∗
2
2
2µ
= 2 (V − iW )(V − iW )∗ γl2 Ωl1 Ωl2 Yl−2,m Yl−2,m
r
∗
2
2
2µ
+ 2 (V + iW )(V + iW )∗ γl2 Ωl1 Ωl2 Yl+2,m Yl+2,m
r
∗
2
+ µ(X − iZ)(X − iZ)∗ γl2 Ωl1 Yl−1,m Yl−1,m
∗
+ µ(X + iZ)(X + iZ)∗ γl2 Ωl1 Yl+1,m Yl+1,m
∗
+ 2λU F + (λ + µ)F 2 + (λ + 2µ)U 2 γl2 Yl0,m Yl0,m
2
Substituting into A53 and using A50 gives
µ
l(l + 1)(l − 1)(l + 2)(V 2 + W 2 ) + µl(l + 1)(X 2 + Z 2 )
r2
+2λU F + (λ + µ)F 2 + (λ + 2µ)U 2 r2 dr
This can be rearranged in terms of the compressional and shear energy densities by substituting the bulk
modulus K instead of λ using the relationship K = λ + 2/3µ yielding
∗ · ·C : dV =
µ
2
2
r2 dr
l(l
+
1)(l
−
1)(l
+
2)W
+
µl(l
+
1)Z
r2
(toroidals)
(A.54)
µ
µ
l(l + 1)(l − 1)(l + 2)V 2 + µl(l + 1)X 2 + (2U − F )2 + K(U + F )2 r2 dr
=
r2
3
V
for spheroidals. We can also evaluate the integral A53 for a transversely isotropic solid by using A40.
The result is
N 2
2
· ·C : dV =
l(l + 1)(l − 1)(l + 2) 2 W + l(l + 1)LZ r2 dr (toroidals)
r
V
(A.55)
N 2
=
l(l + 1)(l − 1)(l + 2) 2 V + l(l + 1)LX 2 + 2FU F + (A − N)F 2 + CU 2 r2 dr
r
∗
for spheroidals.
21