Hints and answers

TENSORS
26.28
A curve r(t) is parameterised by a scalar variable t. Show that the length of the
curve between two points, A and B, is given by
B
dui du j
L=
gij
dt.
dt dt
A
Using the calculus of variations (see chapter 22), show that the curve r(t) that
minimises L satisfies the equation
du j duk
d2 ui
s̈ dui
+ Γijk
=
,
dt2
dt dt
ṡ dt
26.29
where s is the arc length along the curve, ṡ = ds/dt and s̈ = d2 s/dt2 . Hence, show
that if the parameter t is of the form t = as + b, where a and b are constants,
then we recover the equation for a geodesic (26.101).
[ A parameter which, like t, is the sum of a linear transformation of s and a
translation is called an affine parameter. ]
We may define Christoffel symbols of the first kind by
Γijk = gil Γljk .
Show that these are given by
Γkij =
1
2
∂gjk
∂gij
∂gik
+
− k
∂u j
∂ui
∂u
.
By permuting indices, verify that
∂gij
= Γijk + Γjik .
∂uk
Using the fact that Γl jk = Γlkj , show that
gij; k ≡ 0,
i.e. that the covariant derivative of the metric tensor is identically zero in all
coordinate systems.
26.24 Hints and answers
26.1
26.3
26.5
26.7
26.9
(a) u1 = x1 cos(φ − θ) − x2 sin(φ − θ), etc.;
(b) u11 = s2 x21 − 2scx1 x2 + c2 x22 = c2 x22 + csx1 x2 + scx1 x2 + s2 x21 .
√
√
√ √
1, 1). (b) (1/ 2)(1, 0, −1; 0, 2, 0; 1, 0, 1).
(a) (1/√ 2)( 2, 0, 0;
√ 0, 1, −1; 0,√
r = (2 2, −1 + 2, −1 − 2)T .
Twice contract the array with the outer product of (x, y, z) with itself, thus
obtaining the expression −(x2 + y 2 + z 2 )2 , which is an invariant and therefore a
scalar.
Write Aj (∂Ai /∂xj ) as ∂(Ai Aj )/∂xj − Ai (∂Aj /∂xj ).
(i) Write out the expression for |AT |, contract both sides of the equation with lmn
and pick out the expression for |A| on the RHS. Note that lmn lmn is a numerical
scalar.
(iii) Each non-zero term on the RHS contains any particular row index once and
only once. The same can be said for the Levi–Civita symbol on the LHS. Thus
interchanging two rows is equivalent to interchanging two of the subscripts of
lmn , and thereby reversing its sign. Consequently, the magnitude of |A| remains
the same but its sign is changed.
(v) If, say, Api = λApj , for some particular pair of values i and j and all p then,
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26.24 HINTS AND ANSWERS
in the (multiple) summation on the RHS, each Ank appears multiplied by (with
no summation over i and j)
ijk Ali Amj + jik Alj Ami = ijk λAlj Amj + jik Alj λAmj = 0,
since ijk = −jik . Consequently, grouped in this way all terms are zero and
|A| = 0.
(vi) Replace Amj by Amj + λAlj and note that λAli Alj Ank ijk = 0 by virtue of
result (v).
(vii) If C = AB,
|C|lmn = Alx Bxi Amy Byj Anz Bzk ijk .
26.11
26.13
26.15
26.17
26.19
26.21
26.23
26.25
26.27
26.29
Contract this with lmn and show that the RHS is equal to xyz |AT |xyz |B|. It then
follows from result (i) that |C| = |A||B|.
(2)
α = |v|−2 . Note that the most general vector has components wi = λvi +µu(1)
i +νui ,
where both u(1) and u(2) are orthogonal to v.
Construct the orthogonal transformation matrix S for the symmetry operation
of (say) a rotation of 2π/3 about a body diagonal and, setting L = S−1 = ST ,
construct σ = LσLT and require σ = σ. Repeat the procedure for (say) a rotation
of π/2 about the x3 -axis. These together show that σ11 = σ22 = σ33 and that
all other σij = 0. Further symmetry requirements do not provide any additional
constraints.
The transformation of δij has to be included; the principal values are ±E · B.
The third axis is in the direction ±B × E with principal value −|E|2 .
The principal moments give the required ratios.
The principal permeability, in direction (1, 1, 2), has value 0. Thus all the nails lie
in planes to which this is the normal.
Take p11 = p22 = p33 = −p, and pij = eij = 0 for i = j, leading to −p =
(λ + 2µ/3)eii . The fractional volume change is eii ; λ and µ are as defined in (26.46)
and the worked example that follows it.
Consider Qpq = pij qkl Tijkl and show that Kmn = Qmn /4 has the required property.
(a) Argue from the isotropy of Tijkl and ijk for that of Kmn and hence that it
must be a multiple of δmn . Show that the multiplier is uniquely determined and
that Tijkl = (δil δjk − δik δjl )/6.
(b) By relabelling dummy subscripts and using the stated antisymmetry property,
show that Knm = −Kmn . Show that −2Vi = min Kmn and hence that Kmn = imn Vi .
Tijkl = kli Vj − klj Vi . √
Use |e1 · (e2 × e3 )| = g.
√
√
Recall that g = |∂u/∂u | g and du 1 du 2 du 3 = |∂u /∂u| du1 du2 du3 .
l
(vi; j ); k = (vi; j ), k − Γ ik vl; j − Γl jk vi; l and vi; j = vi, j − Γmij vm . If all components of a
tensor equal zero in one coordinate system, then they are zero in all coordinate
systems.
Use gil g ln = δin and gij = gji . Show that
∂gij
gij;k =
− Γjik − Γijk ei ⊗ ej
k
∂u
and then use the earlier result.
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