14Tensors of rank 2

IA Vector Calculus



14.1 Decomposition of a second-rank tensor
This decomposition might look arbitrary at first sight, but as time goes on, you
will find that it is actually very useful in your future career (at least, the lecturer
claims so).
Any second rank tensor can be written as a sum of its symmetric and
anti-symmetric parts
T
ij
= S
ij
+ A
ij
,
where
S
ij
=
1
2
(T
ij
+ T
ji
), A
ij
=
1
2
(T
ij
T
ji
).
Here
T
ij
has 9 independent components, whereas
S
ij
and
A
ij
have 6 and 3
independent components, since they must be of the form
(S
ij
) =
a d e
d b f
e f c
, (A
ij
) =
0 a b
a 0 c
b c 0
.
The symmetric part can be be further reduced to a traceless part plus an isotropic
(i.e. multiple of δ
ij
) part:
S
ij
= P
ij
+
1
3
δ
ij
Q,
where
Q
=
S
ii
is the trace of
S
ij
and
P
ij
=
P
ji
=
S
ij
1
3
δ
ij
Q
is traceless. Then
P
ij
has 5 independent components while Q has 1.
Since the antisymmetric part has 3 independent components, just like a usual
vector, we should be able to write
A
i
in terms of a single vector. In fact, we can
write the antisymmetric part as
A
ij
= ε
ijk
B
k
for some vector
B
. To figure out what this
B
is, we multiply by
ε
ij`
on both
sides and use some magic algebra to obtain
B
k
=
1
2
ε
ijk
A
ij
=
1
2
ε
ijk
T
ij
,
where the last equality is from the fact that only antisymmetric parts contribute
to the sum.
Then
(A
ij
) =
0 B
3
B
2
B
3
0 B
1
B
2
B
1
0
To summarize,
T
ij
= P
ij
+ ε
ijk
B
k
+
1
3
δ
ij
Q,
where B
k
=
1
2
ε
pqj
T
pq
, Q = T
kk
and P
ij
= P
ji
=
T
ij
+T
ji
2
1
3
δ
ij
Q.
Example.
The derivative of a vector field
F
i
(
r
) is a tensor
T
ij
=
F
i
x
j
, a tensor
field. Our decomposition given above has the symmetric traceless piece
P
ij
=
1
2
F
i
x
j
+
F
j
x
i
1
3
δ
ij
F
k
x
k
=
1
2
F
i
x
j
+
F
j
x
i
1
3
δ
ij
· F,
an antisymmetric piece A
ij
= ε
ijk
B
k
, where
B
k
=
1
2
ε
ijk
F
i
x
j
=
1
2
( × F)
k
.
and trace
Q =
F
k
x
k
= · F.
Hence a complete description involves a scalar
· F
, a vector
× F
, and a
symmetric traceless tensor P
ij
.