15Invariant and isotropic tensors
IA Vector Calculus
15.2 Application to invariant integrals
We have the following very useful theorem. It might seem a bit odd and arbitrary
at first sight — if so, read the example below first (after reading the statement
of the theorem), and things will make sense!
Theorem. Let
T
ij···k
=
Z
V
f(x)x
i
x
j
···x
k
dV.
where f(x) is a scalar function and V is some volume.
Given a rotation
R
ij
, consider an active transformation:
x
=
x
i
e
i
is mapped
to
x
0
=
x
0
i
e
i
with
x
0
i
=
R
ij
x
i
, i.e. we map the components but not the basis, and
V is mapped to V
0
.
Suppose that under this active transformation,
(i) f(x) = f(x
0
),
(ii) V
0
= V (e.g. if V is all of space or a sphere).
Then T
ij···k
is invariant under the rotation.
Proof.
First note that the Jacobian of the transformation
R
is 1, since it is
simply the determinant of
R
(
x
0
i
=
R
ip
x
p
⇒
∂x
0
i
∂x
p
=
R
ip
), which is by definition
1. So dV = dV
0
.
Then we have
R
ip
R
jq
···R
kr
T
pq···r
=
Z
V
f(x)x
0
i
x
0
j
···x
0
k
dV
=
Z
V
f(x
0
)x
0
i
x
0
j
···x
0
k
dV using (i)
=
Z
V
0
f(x
0
)x
0
i
x
0
j
···x
0
k
dV
0
using (ii)
=
Z
V
f(x)x
i
x
j
···x
k
dV since x
i
and x
0
i
are dummy
= T
ij···k
.
The result is particularly useful if (i) and (ii) hold for any rotation
R
, in
which case T
ij···k
is isotropic.
Example. Let
T
ij
=
Z
V
x
i
x
j
dV,
with
V
being a solid sphere of
|r| < a
. Our result applies with
f
= 1, which,
being a constant, is clearly invariant under rotations. Also the solid sphere is
invariant under any rotation. So
T
must be isotropic. But the only rank 2
isotropic tensor is αδ
ij
. Hence we must have
T
ij
= αδ
ij
,
and all we have to do is to determine the scalar α.
Taking the trace, we have
T
ii
= 3α =
Z
V
x
i
x
i
dV = 4π
Z
a
0
r
2
· r
2
dr =
4
5
πa
5
.
So
T
ij
=
4
15
πa
5
δ
ij
.
Normally if we are only interested in the
i 6
=
j
case, we just claim that
T
ij
= 0
by saying “by symmetry, it is 0”. But now we can do it (more) rigorously!
There is a closely related result for the inertia tensor of a solid sphere of
constant density ρ
0
, or of mass M =
4
3
πa
3
ρ
0
.
Recall that
I
ij
=
Z
V
ρ
0
(x
k
x
k
δ
ij
− x
i
x
j
) dV.
We see that
I
ij
is isotropic (since we have just shown that
R
x
i
x
j
d
V
is isotropic,
and x
k
x
k
δ
ij
is also isotropic). Let I
ij
= βδ
ij
. Then
I
ij
=
Z
V
ρ
0
(x
k
x
k
δ
ij
− x
i
x
j
) dV
= ρ
0
δ
ij
Z
V
x
k
x
k
dV −
Z
V
x
i
x
j
dV
= ρ
0
(δ
ij
T
kk
− T
ij
)
= ρ
0
4
5
πa
5
δ
ij
−
4
15
πa
5
δ
ij
=
8
15
ρ
0
πa
5
δ
ij
=
2
5
Ma
2
δ
ij
.