26.24 HINTS AND ANSWERS
in the (multiple) summation on the RHS, eachAnkappears multiplied by (with
no summation overiandj)
ijkAliAmj+jikAljAmi=ijkλAljAmj+jikAljλAmj=0,
sinceijk=−jik. Consequently, grouped in this way all terms are zero and
|A|=0.
(vi) ReplaceAmjbyAmj+λAljand note thatλAliAljAnkijk= 0 by virtue of
result (v).
(vii) IfC=AB,
|C|lmn=AlxBxiAmyByjAnzBzkijk.Contract this withlmnand show that the RHS is equal toxy z|AT|xy z|B|.Itthen
follows from result (i) that|C|=|A||B|.26.11 α=|v|−^2. Note that the most general vector has componentswi=λvi+μu(1)i+νu(2)i ,
where bothu(1)andu(2)are orthogonal tov.
26.13 Construct the orthogonal transformation matrixSfor the symmetry operation
of (say) a rotation of 2π/3 about a body diagonal and, settingL=S−^1 =ST,
constructσ′=LσLTand requireσ′=σ. Repeat the procedure for (say) a rotation
ofπ/2 about thex 3 -axis. These together show thatσ 11 =σ 22 =σ 33 and that
all otherσij= 0. Further symmetry requirements do not provide any additional
constraints.
26.15 The transformation ofδijhas to be included; the principal values are±E·B.
The third axis is in the direction±B×Ewith principal value−|E|^2.
26.17 The principal moments give the required ratios.
26.19 The principal permeability, in direction (1, 1 ,2), has value 0. Thus all the nails lie
in planes to which this is the normal.
26.21 Takep 11 =p 22 =p 33 =−p,andpij=eij=0fori=j, leading to−p=
(λ+2μ/3)eii. The fractional volume change iseii;λandμare as defined in (26.46)
and the worked example that follows it.
26.23 ConsiderQpq=pijqklTijkland show thatKmn=Qmn/4 has the required property.
(a) Argue from the isotropy ofTijklandijkfor that ofKmnand hence that it
must be a multiple ofδmn. Show that the multiplier is uniquely determined and
thatTijkl=(δilδjk−δikδjl)/6.
(b) By relabelling dummy subscripts and using the stated antisymmetry property,
show thatKnm=−Kmn. Show that− 2 Vi=minKmnand hence thatKmn=imnVi.
Tijkl=kliVj−kljVi.
26.25 Use|e 1 ·(e 2 ×e 3 )|=
√
g.
Recall that√
g′=|∂u/∂u′|√
ganddu′^1 du′^2 du′^3 =|∂u′/∂u|du^1 du^2 du^3.
26.27 (vi;j);k=(vi;j),k−Γlikvl;j−Γljkvi;landvi;j=vi, j−Γmijvm. If all components of a
tensor equal zero in one coordinate system, then they are zero in all coordinate
systems.
26.29 Usegilgln=δniandgij=gji. Show that
gij;k=(
∂gij
∂uk−Γjik−Γijk)
ei⊗ejandthenusetheearlierresult.