Exercises

26.23 EXERCISES
Writing out the covariant derivative, we obtain
i
dt
duk
+ Γijk t j
ei = 0.
ds
ds
But, since t j = du j /ds, it follows that the equation satisfied by a geodesic is
j
k
d2 ui
i du du
= 0.
+
Γ
jk
ds2
ds ds
(26.101)
Find the equations satisfied by a geodesic (straight line) in cylindrical polar coordinates.
From (26.83), the only non-zero Christoffel symbols are Γ122 = −ρ and Γ212 = Γ221 = 1/ρ.
Thus the required geodesic equations are
2
2
2
d2 u1
d2 ρ
dφ
1 du du
+
Γ
−
ρ
= 0,
=
0
⇒
22
ds2
ds ds
ds2
ds
d2 u2
du1 du2
+ 2Γ212
=0
ds2
ds ds
2 3
du
=0
ds2
⇒
⇒
2 dρ dφ
d2 φ
+
= 0,
ds2
ρ ds ds
2
dz
= 0. ds2
26.23 Exercises
26.1
Use the basic definition of a Cartesian tensor to show the following.
(a) That for any general, but fixed, φ,
(u1 , u2 ) = (x1 cos φ − x2 sin φ, x1 sin φ + x2 cos φ)
are the components of a first-order tensor in two dimensions.
(b) That
x22
x1 x2
2
x1 x2
x1
is not a tensor of order 2. To establish that a single element does not
transform correctly is sufficient.
26.2
The components of two vectors, A and B, and a second-order tensor, T, are given
in one coordinate system by






√
1
3 0
0
√2





0
1
A=
, B=
, T=
3
4
0 .
0
0
0
0
2
In a second coordinate system, obtained from the first by rotation, the components
of A and B are

 √ 

−1
1 3 
1
.
, B =
A =
0
√0
2
2
1
3
Find the components of T in this new coordinate system and hence evaluate,
with a minimum of calculation,
Tij Tji ,
Tki Tjk Tij ,
977
Tik Tmn Tni Tkm .
TENSORS
26.3
26.4
In section 26.3 the transformation matrix for a rotation of the coordinate axes
was derived, and this approach is used in the rest of the chapter. An alternative
view is that of taking the coordinate axes as fixed and rotating the components
of the system; this is equivalent to reversing the signs of all rotation angles.
Using this alternative view, determine the matrices representing (a) a positive
rotation of π/4 about the x-axis and (b) a rotation of −π/4 about the y-axis.
Determine the initial vector r which, when subjected to (a) followed by (b),
finishes at (3, 2, 1).
Show how to decompose the Cartesian tensor Tij into three tensors,
Tij = Uij + Vij + Sij ,
26.5
26.6
where Uij is symmetric and has zero trace, Vij is isotropic and Sij has only three
independent components.
Use the quotient law discussed in section 26.7 to show that the array

 2
y + z 2 − x2
−2xy
−2xz
2
2
2


−2yz
−2yx
x +z −y
−2zx
−2zy
x2 + y 2 − z 2
forms a second-order tensor.
Use tensor methods to establish the following vector identities:
(a)
(b)
(c)
(d)
(e)
26.7
(u × v) × w = (u · w)v − (v · w)u;
curl (φu) = φ curl u + (grad φ) × u;
div (u × v) = v · curl u − u · curl v;
curl (u × v) = (v · grad)u − (u · grad)v + u div v − v div u;
grad 12 (u · u) = u × curl u + (u · grad)u.
Use result (e) of the previous question and the general divergence theorem for
tensors to show that, for a vector field A,
[A divA − A × curl A] dV ,
A(A · dS) − 12 A2 dS =
S
26.8
V
where S is the surface enclosing volume V .
A column matrix a has components ax , ay , az and A is the matrix with elements
Aij = −ijk ak .
(a) What is the relationship between column matrices b and c if Ab = c?
(b) Find the eigenvalues of A and show that a is one of its eigenvectors. Explain
why this must be so.
26.9
Equation (26.29),
|A|lmn = Ali Amj Ank ijk ,
is a more general form of the expression (8.47) for the determinant of a 3 × 3
matrix A. The latter could have been written as
|A| = ijk Ai1 Aj2 Ak3 ,
whilst the former removes the explicit mention of 1, 2, 3 at the expense of an
additional Levi–Civita symbol. As stated in the footnote on p. 942, (26.29) can
be readily extended to cover a general N × N matrix.
Use the form given in (26.29) to prove properties (i), (iii), (v), (vi) and (vii)
of determinants stated in subsection 8.9.1. Property (iv) is obvious by inspection.
For definiteness take N = 3, but convince yourself that your methods of proof
would be valid for any positive integer N.
978
26.23 EXERCISES
26.10
A symmetric second-order Cartesian tensor is defined by
Tij = δij − 3xi xj .
Evaluate the following surface integrals, each taken over the surface of the unit
sphere:
(a)
Tij dS; (b)
Tik Tkj dS; (c)
xi Tjk dS.
26.11
Given a non-zero vector v, find the value that should be assigned to α to make
Pij = αvi vj
26.12
and
Qij = δij − αvi vj
into parallel and orthogonal projection tensors, respectively, i.e. tensors that
satisfy, respectively, Pij vj = vi , Pij uj = 0 and Qij vj = 0, Qij uj = ui , for any vector
u that is orthogonal to v.
Show, in particular, that Qij is unique, i.e. that if another tensor Tij has the
same properties as Qij then (Qij − Tij )wj = 0 for any vector w.
In four dimensions, define second-order antisymmetric tensors, Fij and Qij , and
a first-order tensor, Si , as follows:
(a) F23 = H1 , Q23 = B1 and their cyclic permutations;
(b) Fi4 = −Di , Qi4 = Ei for i = 1, 2, 3;
(c) S4 = ρ, Si = Ji for i = 1, 2, 3.
26.13
26.14
26.15
Then, taking x4 as t and the other symbols to
have their usual meanings in
electromagnetic theory, show that the equations j ∂Fij /∂xj = Si and ∂Qjk /∂xi +
∂Qki /∂xj + ∂Qij /∂xk = 0 reproduce Maxwell’s equations. In the latter i, j, k is any
set of three subscripts selected from 1, 2, 3, 4, but chosen in such a way that they
are all different.
In a certain crystal the unit cell can be taken as six identical atoms lying at the
corners of a regular octahedron. Convince yourself that these atoms can also be
considered as lying at the centres of the faces of a cube and hence that the crystal
has cubic symmetry. Use this result to prove that the conductivity tensor for the
crystal, σij , must be isotropic.
Assuming that the current density j and the electric field E appearing in equation
(26.44) are first-order Cartesian tensors, show explicitly that the electrical conductivity tensor σij transforms according to the law appropriate to a second-order
tensor.
The rate W at which energy is dissipated per unit volume, as a result of the
current flow, is given by E · j. Determine the limits between which W must lie for
a given value of |E| as the direction of E is varied.
In a certain system of units, the electromagnetic stress tensor Mij is given by
Mij = Ei Ej + Bi Bj − 12 δij (Ek Ek + Bk Bk ),
26.16
where the electric and magnetic fields, E and B, are first-order tensors. Show that
Mij is a second-order tensor.
Consider a situation in which |E| = |B|, but the directions of E and B are
not parallel. Show that E ± B are principal axes of the stress tensor and find
the corresponding principal values. Determine the third principal axis and its
corresponding principal value.
A rigid body consists of four particles of masses m, 2m, 3m, 4m, respectively
situated at the points (a, a, a), (a, −a, −a), (−a, a, −a), (−a, −a, a) and connected
together by a light framework.
(a) Find the inertia tensor at the √
origin and show that the principal moments of
inertia are 20ma2 and (20 ± 2 5)ma2 .
979
TENSORS
(b) Find the principal axes and verify that they are orthogonal.
26.17
A rigid body consists of eight particles, each of mass m, held together by light
rods. In a certain coordinate frame the particles are at positions
±a(3, 1, −1),
±a(1, −1, 3),
±a(1, 3, −1),
±a(−1, 1, 3).
Show that, when the body rotates about an axis through the origin, if the angular
velocity and angular momentum vectors are parallel then their ratio must be
40ma2 , 64ma2 or 72ma2 .
26.18
The paramagnetic tensor χij of a body placed in a magnetic field, in which its
energy density is − 21 µ0 M · H with Mi = j χij Hj , is

2k
 0
0
0
3k
k

0
k .
3k
Assuming depolarizing effects are negligible, find how the body will orientate
itself if the field is horizontal, in the following circumstances:
(a) the body can rotate freely;
(b) the body is suspended with the (1, 0, 0) axis vertical;
(c) the body is suspended with the (0, 1, 0) axis vertical.
26.19
A block of wood contains a number of thin soft-iron nails (of constant permeability). A unit magnetic field directed eastwards induces a magnetic moment in
the block having components (3, 1, −2), and similar fields directed northwards
and vertically upwards induce moments (1, 3, −2) and (−2, −2, 2) respectively.
Show that all the nails lie in parallel planes.
26.20
For tin, the conductivity tensor is diagonal, with entries a, a, and b when referred
to its crystal axes. A single crystal is grown in the shape of a long wire of length L
and radius r, the axis of the wire making polar angle θ with respect to the crystal’s
3-axis. Show that the resistance of the wire is L(πr2 ab)−1 a cos2 θ + b sin2 θ .
26.21
By considering an isotropic body subjected to a uniform hydrostatic pressure
(no shearing stress), show that the bulk modulus k, defined by the ratio of the
pressure to the fractional decrease in volume, is given by k = E/[3(1 − 2σ)] where
E is Young’s modulus and σ is Poisson’s ratio.
26.22
For an isotropic elastic medium under dynamic stress, at time t the displacement
ui and the stress tensor pij satisfy
pij = cijkl
∂ul
∂uk
+
∂xl
∂xk
and
∂pij
∂ 2 ui
=ρ 2 ,
∂xj
∂t
where cijkl is the isotropic tensor given in equation (26.47) and ρ is a constant.
Show that both ∇ · u and ∇ × u satisfy wave equations and find the corresponding
wave speeds.
980
26.23 EXERCISES
26.23
A fourth-order tensor Tijkl has the properties
Tjikl = −Tijkl ,
Tijlk = −Tijkl .
Prove that for any such tensor there exists a second-order tensor Kmn such that
Tijkl = ijm kln Kmn
and give an explicit expression for Kmn . Consider two (separate) special cases, as
follows.
(a) Given that Tijkl is isotropic and Tijji = 1, show that Tijkl is uniquely determined and express it in terms of Kronecker deltas.
(b) If now Tijkl has the additional property
Tklij = −Tijkl ,
show that Tijkl has only three linearly independent components and find an
expression for Tijkl in terms of the vector
Vi = − 14 jkl Tijkl .
26.24
26.25
Working in cylindrical polar coordinates ρ, φ, z, parameterise the straight line
(geodesic) joining (1, 0, 0) to (1, π/2, 1) in terms of s, the distance along the line.
Show by substitution that the geodesic equations, derived at the end of section
26.22, are satisfied.
In a general coordinate system ui , i = 1, 2, 3, in three-dimensional Euclidean
space, a volume element is given by
dV = |e1 du1 · (e2 du2 × e3 du3 )|.
Show that an alternative form for this expression, written in terms of the determinant g of the metric tensor, is given by
√
dV = g du1 du2 du3 .
26.26
26.27
Show that, under a general coordinate transformation to a new coordinate
system u i , the volume element dV remains unchanged, i.e. show that it is a scalar
quantity.
By writing down the expression for the square of the infinitesimal arc length (ds)2
in spherical polar coordinates, find the components gij of the metric tensor in this
coordinate system. Hence, using (26.97), find the expression for the divergence
of a vector field v in spherical polars. Calculate the Christoffel symbols (of the
second kind) Γijk in this coordinate system.
Find an expression for the second covariant derivative vi; jk ≡ (vi; j ); k of a vector
vi (see (26.88)). By interchanging the order of differentiation and then subtracting
the two expressions, we define the components R l ijk of the Riemann tensor as
vi; jk − vi; kj ≡ R l ijk vl .
Show that in a general coordinate system ui these components are given by
R l ijk =
∂Γl ij
∂Γlik
−
+ Γmik Γl mj − Γmij Γlmk .
j
∂u
∂uk
By first considering Cartesian coordinates, show that all the components R l ijk ≡ 0
for any coordinate system in three-dimensional Euclidean space.
In such a space, therefore, we may change the order of the covariant derivatives
without changing the resulting expression.
981