15Invariant and isotropic tensors
IA Vector Calculus
15.1 Definitions and classification results
Definition
(Invariant and isotropic tensor)
.
A tensor
T
is invariant under a
particular rotation R if
T
0
ij···k
= R
ip
R
jq
···R
kr
T
pq···r
= T
ij···k
,
i.e. every component is unchanged under the rotation.
A tensor
T
which is invariant under every rotation is isotropic, i.e. the same
in every direction.
Example. The inertia tensor of a sphere is isotropic by symmetry.
δ
ij
and
ε
ijk
are also isotropic tensors. This ensures that the component
definitions of the scalar and vector products
a·b
=
a
i
b
j
δ
ij
and (
a×b
)
i
=
ε
ijk
a
j
b
k
are independent of the Cartesian coordinate system.
Isotropic tensors in R
3
can be classified:
Theorem.
(i) There are no isotropic tensors of rank 1, except the zero tensor.
(ii) The most general rank 2 isotropic tensor is T
ij
= αδ
ij
for some scalar α.
(iii)
The most general rank 3 isotropic tensor is
T
ijk
=
βε
ijk
for some scalar
β
.
(iv)
All isotropic tensors of higher rank are obtained by combining
δ
ij
and
ε
ijk
using tensor products, contractions, and linear combinations.
We will provide a sketch of the proof:
Proof.
We analyze conditions for invariance under specific rotations through
π
or π/2 about coordinate axes.
(i) Suppose T
i
is rank-1 isotropic. Consider a rotation about x
3
through π:
(R
ij
) =
−1 0 0
0 −1 0
0 0 1
.
We want
T
1
=
R
ip
T
p
=
R
11
T
1
=
−T
1
. So
T
1
= 0. Similarly,
T
2
= 0. By
consider a rotation about, say x
1
, we have T
3
= 0.
(ii) Suppose T
ij
is rank-2 isotropic. Consider
(R
ij
) =
0 1 0
−1 0 0
0 0 1
,
which is a rotation through π/2 about the x
3
axis. Then
T
13
= R
1p
R
3q
T
pq
= R
12
R
33
T
23
= T
23
and
T
23
= R
2p
R
3q
T
pq
= R
21
R
33
T
13
= −T
13
So T
13
= T
23
= 0. Similarly, we have T
31
= T
32
= 0.
We also have
T
11
= R
1p
R
1q
T
pq
= R
12
R
12
T
22
= T
22
.
So T
11
= T
22
.
By picking a rotation about a different axis, we have
T
21
=
T
12
and
T
22
= T
33
.
Hence T
ij
= αδ
ij
.
(iii)
Suppose that
T
ijk
is rank-3 isotropic. Using the rotation by
π
about the
x
3
axis, we have
T
133
= R
1p
R
3q
R
3r
T
pqr
= −T
133
.
So T
133
= 0. We also have
T
111
= R
1p
R
1q
R
1r
T
pqr
= −T
111
.
So
T
111
= 0. We have similar results for
π
rotations about other axes and
other choices of indices.
Then we can show that T
ijk
= 0 unless all i, j, k are distinct.
Now consider
(R
ij
) =
0 1 0
−1 0 0
0 0 1
,
a rotation about x
3
through π/2. Then
T
123
= R
1p
R
2q
R
3r
T
pqr
= R
12
R
21
R
33
T
213
= −T
213
.
So
T
123
=
−T
213
. Along with similar results for other indices and axes of
rotation, we find that
T
ijk
is totally antisymmetric, and
T
ijk
=
βε
ijk
for
some β.
Example. The most general isotropic tensor of rank 4 is
T
ijk`
= αδ
ij
δ
k`
+ βδ
ik
δ
j`
+ γδ
i`
δ
jk
for some scalars
α, β, γ
. There are no other independent combinations. (we
might think we can write a rank-4 isotropic tensor in terms of
ε
ijk
, like
ε
ijp
ε
k`p
,
but this is just
δ
ik
δ
j`
− δ
i`
δ
jk
. It turns out that anything you write with
ε
ijk
can be written in terms of δ
ij
instead)