or
i=
V
R
with R =
l
.
σA
R is the resistance, measured in V /A or Ohms, Ω.
We can generalise Ohm’s Law by writing
ji = σij Ej ,
where σ is the conductivity tensor (by the quotient theorem).
• Layered material of conductor and insulator where the current can only flow
in the layers.
1111111111111
0000000000000
0000000000000
1111111111111
0000000000000
1111111111111
0000000000000
1111111111111
0000000000000
1111111111111
0000000000000
1111111111111
0000000000000
1111111111111
z
y
x
1111111111111
0000000000000
0000000000000
1111111111111
0000000000000
1111111111111


σ0 0 0
σ =  0 σ0 0 
0 0 0

 

j1
σ0 E1
 j2  =  σ0 E2  .
j3
0
giving
• Uniform material, no prefered direction, isototropic, so σij = σ0 δij and j = σ0 E
(the original Ohm’s Law when j and E are always parallel to each other.
5.3
5.3.1
The Inertia Tensor
Angular momentum and kinetic energy
Suppose a rigid body of arbitrary shape rotates about a fixed axis with angular
velocity ω = ωn. Consider a blob of matter dm at point P , with position vector
ω
P
O
~r
41
dm
r relative to O. The velocity of a little element dm = ρdV where ρ ≡ ρ(r) is the
(mass) density is
v = ω ×r.
Repeating previous results:
• Simple proof δr = δθr sin φ = |δθn × r| and so v = δr/δt = ω × r where
~ω
δθ
~r
φ
~ω
ω = δθ/δtn.
• More complicated proof:
Use previous result for rotation matrix R(θ, n),
δri = Rij (δθ, n)rj − ri
= (−ǫijk nk δθ + O((δθ)2 ))rj ,
δθ
δri
= (n × r)i ,
δt
δt
or v = ω × r again.
Angular momentum
The angular momentum dh (or dL) of an element of mass dm = ρdV at r is dh =
ρ(r)dV r × v giving
Z
ρr × (ω × r)dV ,
h =
body
giving
hi =
Z
ρǫijk rj ǫklm ωl rm dV
body
=
Z
body
ρ(δil δjm − δim δjl )rj ωl rm dV .
Thus
hi = Iij ωj ,
Iij =
Z
body
42
ρ(r 2 δij − ri rj )dV .
The geometric factor I(O) (where O is the origin) is the moment of Inertia tensor.
It is a tensor because h is pseudovector, ω is a pseudovector and hence from the
quotient theorem I is a tensor.
[Furthermore note that Iij is symmetric and independent of the n axis chosen.]
Kinetic energy
For the kinetic energy, T , we have dT = 12 (ρdV )(ω × r)2 or
Z
1
ρǫijk ωj rk ǫilm ωl rm dV
T =
2 body
Z
1
=
ρ(δjl δkm − δjm δkl )ωj rk ωl rm dV
2 body
Z
1
ρ(ω 2 r 2 − (r · ω)2 )dV
=
2 body
Z
1
=
ρ(r 2 δij − ri rj )dV ωi ωj ,
2 body
or
1
1
T = Iij ωi ωj ≡ ωIω .
2
2
Alternative (more familiar) forms
Often write
L =
T =
I (n) ω
1 (n) 2
ω
2I
with L = h · n ,
I (n) = Iij ni nj .
where I (n) is now the moment of inertia with respect to n,
Z
Z
(n)
2
2
2
I = Iij ni nj =
ρ(r − (r · n )) dV ≡
ρr⊥
dV ,
body
body
with r⊥ being the perpendicular distance from the n-axis.
An example
For example, a cube of side a of constant density M = ρa3 ,
Z
Iij (O) = ρ(r 2 δij − ri rj )dV .
43
z
y
a
x
O
This gives
So by symmetry
a
dxdydz (x2 + y 2 + z 2 ) − x2
a
10 3
= ρ 3 y xz + 13 z 3 xy 0 = 23 ρa5 = 23 Ma2 ,
Z a
= ρ
dxdydz(−xy)
0
a
= −ρ 21 x2 21 y 2 z 0 = − 41 ρa5 = − 14 Ma2 .
I11 = ρ
I12
Z

I(O) = Ma2 
5.3.2
2
3
− 14
− 14

− 41 − 14
2
1 
.
3 −4
1
2
−4
3
Parallel Axes Theorem
Often simpler to find the moment of inertia tensor about the centre of gravity, G,
rather than an arbitrary point, O. There is, however, a simple relationship between
them. Taking O to be the origin, and OG = R we have r = R + r ′ ,
ω
P
dm
~r′
O
~r
~
R
G
giving
Iij (O) =
=
Z
Z
ρ(r) r 2 δij − ri rj dV
ρ′ (r ′ ) (R + r ′ )2 δij − (Ri + ri′ )(Rj + rj′ ) dV ′
44
′
where we have set ρ′ (r′ ) ≡ ρ(r) and changed integration variables
R ′ to′ r′ . ′Upon
expanding the integrand and using the definition of G, namely ρ (r )r dV = 0
gives
Z
Iij (O) = ρ′ (r′ ) (R2 + r ′2 )δij − Ri Rj − ri′ rj′ dV ′ ,
or
Iij (O) = Iij (G) + M(R2 δij − Ri Rj ) ,
R
where M = ρ′ (r ′ )dV ′ is the total mass of the body. This is a general result, given
I(G) then we can easily find the moment of inertia tensor about another point.
The Parallel Axes Theorem technically refers to the result about the same axis n as
the original axis,
2
I (n) (O) = I (n) (G) + MR⊥
,
2
where R⊥ (R⊥
= R2 − (R · n)2 ) is the perpendicular distance from the n axis.
Previous example
Using the previous example of the cube, then G is the centre of the cube and
R = ( 12 a, 12 a, 21 a). Then
z
y
G
a
x
O
I11 (G) = ρ
h
a/2
−a/2
dxdydz (x2 + y 2 + z 2 ) − x2
a/2
a/2
1 3 a/2
3 [y ]−a/2 [x]−a/2 [z]−a/2
a/2
a/2
a/2
+ 31 [z 3 ]−a/2 [x]−a/2 [y]−a/2
= ρ 31 · 2(a/2)3 2(a/2)2(a/2) · 2 = 16 ρa5 = 16 Ma2 ,
= ρ
and
Z
I12 (G) = ρ
Z
a/2
dxdydz (−xy)
−a/2
= 0,
as
45
1
2 a/2
2 x −a/2
= 0.
i
Thus

0 0
Iij (G) = Ma2  0 61 0  ,
0 0 61 ij

1
6
and

M(R2 δij − Ri Rj ) = Ma2 
As
1
2
+
5.3.3
1
6
=
2
3
1
2
− 14
− 14

− 41 − 41
1
1 
.
2 −4
1
1
−4
2
ij
then this gives Iij (O) as before.
Diagonalisation of rank two tensors
Question: Are there any directions for ω
b such that h is parallel to ω?
Now h = λω means that
(Iij − λδij ) ωj = 0 .
Three equations. For a non-trivial solution, we must have det(Iij −λδij ) = 0. Simply
expanding, or first writing this as
det(Iij − λδij ) = 61 ǫijk ǫlmn (Iil − λδil ) (Ijm − λδjm ) (Ikn − λδkn ) = 0 ,
and then expanding gives
P − Qλ + Rλ2 − λ3 = 0 ,
where
P = 61 ǫijk ǫlmn Iil IjmIkn
= det I
Q = 61 ǫijk ǫlmn (δil Ijm Ikn + Iil δjm Ikn + Iil Ijm δkn )
= 61 (δjm δkn − δjn δkm )Ijm Ikn × 3
= 12 (TrI)2 − TrI 2
R = 16 ǫijk ǫlmn (δil δjm Ikn + δil Ijm δkn + Iil δjm δkn )
= 61 (δjm δkn − δjn δkm )δjm Ikn × 3
= TrI .
As det A, TrA are invariant then P , Q, R are thus invariants of the tensor I (ie
values do not depend on choice of frame).
46
The three values of λ (ie solutions of the cubic equation) are called eigenvalues and
ω are called eigenvectors. (The usual notation for eigenvector is e.)
As Iij ωj = λωi then
T
(LI L
L)ij ωj = λ(Lω)i ,
|{z}
giving Iij′ (Lω)j = λ(Lω)i ,
1
so comparing with Iij′ ω ′ = λ′ ωi′ we see that eigenvectors ω are vectors (ie transform
as vectors as ωi′ = lij ωj ) and eigenvalues are scalars (ie transform as scalars as
λ′ = λ).
Note that only the (±) direction and not the magnitude is determined by the eigenvalue equation.
So the answer to the question posed above is that we must find the eigenvalues λ(i) ,
i = 1, 2, 3 and the corresponding eigenvectors ω (i) whence h(i) = λ(i) ω (i) (no sum).
Eigenvalues/eigenvectors of a real symmetric tensor
Theorem
• The eigenvalues of a real symmetric matrix are real
• The corresponding eigenvectors are orthogonal.
Or can be chosen to be so, if the eigenvalues are degenerate as
– the eigenvector subspace corresponding to the degenerate eigenvalues is
orthogonal to the other eigenvectors
– within this subspace, the eigenvectors can be chosen to be orthogonal by
the Gram-Schmidt procedure
The proof will not be given here.
Diagonalisation of a real symmetric tensor
Let T be a real second rank symmetric tensor with real eigenvalues λ(1) , λ(2) , λ(3)
and normalised eigenvectors l(1) , l(2) , l(3) so that T l(i) = λ(i) l (i) (no summation) and
l(i) · l (j) = δij . Let


(1)
(1)
(1)
l 1 l2 l 3


(i)
lij = lj ≡  l1(2) l2(2) l3(2)  = l (i) · ej .
(3)
(3)
(3)
l 1 l2 l 3
This is a rotation matrix as
(i) (j)
(LLT )ij = lim ljm = lm
lm = δij .
47
and can always be arranged to be RH,
det L = ǫijk l1i l2j l3k = l(3) · (l(1) × l (2) ) = +1 ,
Thus L can always be taken as a rotation matrix for S to S ′ .
Hence we have (only summing over α, β),
Tij′ = liα ljβ Tαβ
(j)
= lα(i) Tαβ lβ
= lα(i) λ(j) lα(j) ,
or

λ(1) 0
0
Tij′ = λ(j) δij =  0 λ(2) 0  .
0
0 λ(3) ij

Thus we have found a frame of reference, S ′ , in which the tensor T takes diagonal
form, the diagonal elements being the eigenvalues of T .
Moment of Inertia Tensor
Thus for the inertia tensor, I, we can diagonalise the tensor. Taking these axes as
the principal axes of I then


A 0 0
I′ =  0 B 0  ,
0 0 C
where A, B, C are the principal moments of inertia
h = Aω1′ e′ω1 + Bω2′ e′ω2 + Cω3′ e′ω3
T = 12 Aω1′2 + Bω2′2 + Cω3′2 .
A ‘geometric’ interpretation is given by the Inertia Ellipsoid, which is defined by
Iij ωi ωj = 1 ,
(ie absorb
√
2T into ω). If we rotate to the principal axes (ωi′ = liα ωα ) then
Aω1′2 + Bω2′2 + Cω3′2 = 1 ,
the normal form. This
is an ellipsoid as A, B, C are all > 0. (We see this from the
R
definition, eg A = ρ(y 2 + z 2 )dV .)
48