13Tensors and tensor fields
IA Vector Calculus
13.5 Tensor calculus
Tensor fields and derivatives
Just as with scalars or vectors, we can define tensor fields:
Definition
(Tensor field)
.
A tensor field is a tensor at each point in space
T
ij···k
(x), which can also be written as T
ij···k
(x
`
).
We assume that the fields are smooth so they can be differentiated any
number of times
∂
∂x
p
···
∂
∂x
q
T
ij···k
,
except for where things obviously fail, e.g. for where
T
is not defined. We now
claim:
Proposition.
∂
∂x
p
···
∂
∂x
q
| {z }
m
T
ij ···k
| {z }
n
, (∗)
is a tensor of rank n + m.
Proof.
To show this, it suffices to show that
∂
∂x
p
satisfies the tensor transfor-
mation rules for rank 1 tensors (i.e. it is something like a rank 1 tensor). Then
by the exact same argument we used to show that tensor products preserve
tensorness, we can show that the (
∗
) is a tensor. (we cannot use the result of
tensor products directly, since this is not exactly a product. But the exact same
proof works!)
Since x
0
i
= R
iq
x
q
, we have
∂x
0
i
∂x
p
= R
ip
.
(noting that
∂x
p
∂x
q
= δ
pq
). Similarly,
∂x
q
∂x
0
i
= R
iq
.
Note that R
ip
, R
iq
are constant matrices.
Hence by the chain rule,
∂
∂x
0
i
=
∂x
q
∂x
0
i
∂
∂x
q
= R
iq
∂
∂x
q
.
So
∂
∂x
p
obeys the vector transformation rule. So done.
Integrals and the tensor divergence theorem
It is also straightforward to do integrals. Since we can sum tensors and take
limits, the definition of a tensor-valued integral is straightforward.
For example,
R
V
T
ij···k
(
x
) d
V
is a tensor of the same rank as
T
ij···k
(think of
the integral as the limit of a sum).
For a physical example, recall our discussion of the flux of quantities for a
fluid with velocity
u
(
x
) through a surface element — assume a uniform density
ρ
. The flux of volume is
u · nδs
=
u
j
n
j
δS
. So the flux of mass is
ρu
j
n
j
δS
.
Then the flux of the
i
th component of momentum is
ρu
i
u
j
n
j
δS
=
T
ij
n
j
kδS
(mass times velocity), where
T
ij
=
ρu
i
u
j
. Then the flux through the surface
S
is
R
S
T
ij
n
j
dS.
It is easy to generalize the divergence theorem from vectors to tensors. We
can then use it to discuss conservation laws for tensor quantities.
Let
V
be a volume bounded by a surface
S
=
∂V
and
T
ij···k`
be a smooth
tensor field. Then
Theorem (Divergence theorem for tensors).
Z
S
T
ij···k`
n
`
dS =
Z
V
∂
∂x
`
(T
ij···k`
) dV,
with n being an outward pointing normal.
The regular divergence theorem is the case where
T
has one index and is a
vector field.
Proof.
Apply the usual divergence theorem to the vector field
v
defined by
v
`
= a
i
b
j
···c
k
T
ij···k`
, where a, b, ··· , c are fixed constant vectors.
Then
∇ · v =
∂v
`
∂x
`
= a
i
b
j
···c
k
∂
∂x
`
T
ij···k`
,
and
n · v = n
`
v
`
= a
i
b
j
···c
k
T
ij···k`
n
`
.
Since
a, b, ··· , c
are arbitrary, therefore they can be eliminated, and the tensor
divergence theorem follows.