Physical applications of tensors

TENSORS
Using (26.40), show that Aij = ijk pk .
By contracting both sides of (26.40) with ijk , we find
ijk pk = 12 ijk klm Alm .
Using the identity (26.30) then gives
ijk pk = 12 (δil δjm − δim δjl )Alm
= 12 (Aij − Aji ) = 12 (Aij + Aij ) = Aij ,
where in the last line we use the fact that Aij = −Aji . By a simple extension, we may associate a dual pseudoscalar s with every
totally antisymmetric third-rank tensor Aijk , i.e. one that is antisymmetric with
respect to the interchange of every possible pair of subscripts; s is given by
1
(26.41)
s = ijk Aijk .
3!
Since Aijk is a totally antisymmetric three-subscript quantity, we expect it to
equal some multiple of ijk (since this is the only such quantity). In fact Aijk = sijk ,
as can be proved by substituting this expression into (26.41) and using (26.36).
26.12 Physical applications of tensors
In this section some physical applications of tensors will be given. First-order
tensors are familiar as vectors and so we will concentrate on second-order tensors,
starting with an example taken from mechanics.
Consider a collection of rigidly connected point particles of which the αth,
which has mass m(α) and is positioned at r(α) with respect to an origin O, is
typical. Suppose that the rigid assembly is rotating about an axis through O with
angular velocity ω.
The angular momentum J about O of the assembly is given by
r(α) × p(α) .
J=
α
But p(α) = m(α) ṙ(α) and ṙ(α) = ω × r(α) , for any α, and so in subscript form the
components of J are given by
m(α) ijk xj(α) ẋk(α)
Ji =
α
=
(α)
m(α) ijk xj(α) klm ωl xm
α
=
(α)
m(α) (δil δjm − δim δjl )xj(α) xm
ωl
α
=
m(α)
2
r (α) δil − xi(α) xl(α) ωl ≡ Iil ωl ,
(26.42)
α
where Iil is a symmetric second-order Cartesian tensor (by the quotient rule, see
950
26.12 PHYSICAL APPLICATIONS OF TENSORS
section 26.7, since J and ω are vectors). The tensor is called the inertia tensor at O
of the assembly and depends only on the distribution of masses in the assembly
and not upon the direction or magnitude of ω.
A more realistic situation obtains if a continuous rigid body is considered. In
this case, m(α) must be replaced everywhere by ρ(r) dx dy dz and all summations
by integrations over the volume of the body. Written out in full in Cartesians,
the inertia tensor for a continuous body would have the form

 2
(y + z 2 )ρ dV − xyρ dV
− xzρ dV
I = [Iij ] =  − xyρ dV
(z 2+ x2 )ρ dV − yzρ dV  ,
− xzρ dV
− yzρ dV
(x2 + y 2 )ρ dV
where ρ = ρ(x, y, z) is the mass distribution and dV stands for dx dy dz; the
integrals are to be taken over the whole body. The diagonal elements of this
tensor are called the moments of inertia and the off-diagonal elements without the
minus signs are known as the products of inertia.
Show that the kinetic energy of the rotating system is given by T = 12 Ijl ωj ωl .
By an argument parallel to that already made for J, the kinetic energy is given by
T = 12
m(α) ṙ(α) · ṙ(α)
α
=
1
2
=
1
2
(α)
m(α) ijk ωj x(α)
k ilm ωl xm
α
α
=
1
2
(α)
m(α) (δjl δkm − δjm δkl )x(α)
k xm ωj ωl
2
(α)
ωj ωl
m(α) δjl r(α) − x(α)
j xl
α
= 12 Ijl ωj ωl .
Alternatively, since Jj = Ijl ωl we may write the kinetic energy of the rotating system as
T = 12 Jj ωj . The above example shows that the kinetic energy of the rotating body can be
expressed as a scalar obtained by twice contracting ω with the inertia tensor. It
also shows that the moment of inertia of the body about a line given by the unit
vector n̂ is Ijl n̂j n̂l (or n̂T In̂ in matrix form).
Since I (≡ Ijl ) is a real symmetric second-order tensor, it has associated with it
three mutually perpendicular directions that are its principal axes and have the
following properties (proved in chapter 8):
(i) with each axis is associated a principal moment of inertia λµ , µ = 1, 2, 3;
(ii) when the rotation of the body is about one of these axes, the angular
velocity and the angular momentum are parallel and given by
J = Iω = λµ ω,
i.e. ω is an eigenvector of I with eigenvalue λµ ;
951
TENSORS
(iii) referred to these axes as coordinate axes, the inertia tensor is diagonal
with diagonal entries λ1 , λ2 , λ3 .
Two further examples of physical quantities represented by second-order tensors
are magnetic susceptibility and electrical conductivity. In the first case we have
(in standard notation)
(26.43)
Mi = χij Hj ,
and in the second case
ji = σij Ej .
(26.44)
Here M is the magnetic moment per unit volume and j the current density
(current per unit perpendicular area). In both cases we have on the left-hand side
a vector and on the right-hand side the contraction of a set of quantities with
another vector. Each set of quantities must therefore form the components of a
second-order tensor.
For isotropic media M ∝ H and j ∝ E, but for anisotropic materials such as
crystals the susceptibility and conductivity may be different along different crystal
axes, making χij and σij general second-order tensors, although they are usually
symmetric.
The electrical conductivity σ in a crystal is measured by an observer to have components
as shown:


√
2 0
√1
[σij ] =  2
(26.45)
3
1 .
0
1
1
Show that there is one direction in the crystal along which no current can flow. Does the
current flow equally easily in the two perpendicular directions?
The current density in the crystal is given by ji = σij Ej , where σij , relative to the observer’s
coordinate system, is given by (26.45). Since [σij ] is a symmetric matrix, it possesses
three mutually perpendicular eigenvectors (or principal axes) with respect to which the
conductivity tensor is diagonal, with diagonal entries λ1 , λ2 , λ3 , the eigenvalues of [σij ].
As discussed in chapter 8, the eigenvalues of [σij ] are given by |σ − λI| = 0. Thus we
require
√
1−λ
2
0 √
2
3−λ
1 = 0,
0
1
1−λ from which we find
(1 − λ)[(3 − λ)(1 − λ) − 1] − 2(1 − λ) = 0.
This simplifies to give λ = 0, 1, 4 so that, with respect to its principal axes, the conductivity
tensor has components σij given by


4 0 0
[σij ] =  0 1 0  .
0 0 0
Since ji = σij Ej , we see immediately that along one of the principal axes there is no current
flow and along the two perpendicular directions the current flows are not equal. 952
26.12 PHYSICAL APPLICATIONS OF TENSORS
We can extend the idea of a second-order tensor that relates two vectors to a
situation where two physical second-order tensors are related by a fourth-order
tensor. The most common occurrence of such relationships is in the theory of
elasticity. This is not the place to give a detailed account of elasticity theory,
but suffice it to say that the local deformation of an elastic body at any interior
point P can be described by a second-order symmetric tensor eij called the strain
tensor. It is given by
1 ∂ui
∂uj
+
,
eij =
2 ∂xj
∂xi
where u is the displacement vector describing the strain of a small volume element
whose unstrained position relative to the origin is x. Similarly we can describe
the stress in the body at P by the second-order symmetric stress tensor pij ; the
quantity pij is the xj -component of the stress vector acting across a plane through
P whose normal lies in the xi -direction. A generalisation of Hooke’s law then
relates the stress and strain tensors by
pij = cijkl ekl
(26.46)
where cijkl is a fourth-order Cartesian tensor.
Assuming that the most general fourth-order isotropic tensor is
cijkl = λδij δkl + ηδik δjl + νδil δjk ,
(26.47)
find the form of (26.46) for an isotropic medium having Young’s modulus E and Poisson’s
ratio σ.
For an isotropic medium we must have an isotropic tensor for cijkl , and so we assume the
form (26.47). Substituting this into (26.46) yields
pij = λδij ekk + ηeij + νeji .
But eij is symmetric, and if we write η + ν = 2µ, then this takes the form
pij = λekk δij + 2µeij ,
in which λ and µ are known as Lamé constants. It will be noted that if eij = 0 for i = j
then the same is true of pij , i.e. the principal axes of the stress and strain tensors coincide.
Now consider a simple tension in the x1 -direction, i.e. p11 = S but all other pij = 0.
Then denoting ekk (summed over k) by θ we have, in addition to eij = 0 for i = j, the three
equations
S = λθ + 2µe11 ,
0 = λθ + 2µe22 ,
0 = λθ + 2µe33 .
Adding them gives
S = θ(3λ + 2µ).
Substituting for θ from this into the first of the three, and recalling that Young’s modulus
is defined by S = Ee11 , gives E as
E=
µ(3λ + 2µ)
.
λ+µ
953
(26.48)