Integral theorems for tensors

TENSORS
Further, Poisson’s ratio is defined as σ = −e22 /e11 (or −e33 /e11 ) and is thus
Ee11
λθ
λ
λ
1
1
=
=
.
σ=
e11 2µ
e11
2µ 3λ + 2µ
2(λ + µ)
(26.49)
Solving (26.48) and (26.49) for λ and µ gives finally
σE
E
pij =
ekk δij +
eij . (1 + σ)(1 − 2σ)
(1 + σ)
26.13 Integral theorems for tensors
In chapter 11, we discussed various integral theorems involving vector and scalar
fields. Most notably, we considered the divergence theorem, which states that, for
any vector field a,
0
∇ · a dV =
V
a · n̂ dS,
(26.50)
S
where S is the surface enclosing the volume V and n̂ is the outward-pointing unit
normal to S at each point.
Writing (26.50) in subscript notation, we have
0
∂ak
dV = ak n̂k dS.
(26.51)
V ∂xk
S
Although we shall not prove it rigorously, (26.51) can be extended in an obvious
manner to relate integrals of tensor fields, rather than just vector fields, over
volumes and surfaces, with the result
0
∂Tij···k···m
dV = Tij···k···m n̂k dS.
∂xk
V
S
This form of the divergence theorem for general tensors can be very useful in
vector calculus manipulations.
A vector field a satisfies ∇ · a = 0 inside some volume V and a · n̂ = 0 on the boundary
surface S. By considering the divergence theorem applied to Tij = xi aj , show that
a dV = 0.
V
Applying the divergence theorem to Tij = xi aj we find
0
∂Tij
∂(xi aj )
dV =
dV = xi aj n̂j dS = 0,
∂xj
V ∂xj
V
S
since aj n̂j = 0. By expanding the volume integral we obtain
∂(xi aj )
∂xi
∂aj
dV =
aj dV +
xi
dV
∂xj
∂xj
V
V ∂xj
V
=
δij aj dV
V
=
ai dV = 0,
V
where in going from the first to the second line we used ∂xi /∂xj = δij and ∂aj /∂xj = 0. 954