Appendix: Exercises... 391
yields the same matrix as in (A.4). Thus one has also
AasyμλAasyμλ=1
2
.
Due to
AμνAμν= 1 , AμνAνμ= 0 ,this is in accord with (3.10), viz.
AμνBνμ=1
3
AλλBκκ+AasyμνBνμasy+ Aμν Bνμ,withBνμ=Aμν.
3.2 Symmetric Traceless Dyadics in Matrix Notation(p.39)
(i)Write the symmetric traceless parts of the dyadic tensor Cμν=Cμν(α)= 2 aμbν
in matrix form for the vectorsa=a(α):{c,−s, 0 }andb=b(α):{s,c, 0 },where
c and s are the abbreviations c=cosαand s=sinα.Discuss the special cases
α= 0 andα=π/4.
The desired tensor is
Cμν=⎛
⎝
2 cs c^20
−s^2 − 2 cs 0
000⎞
⎠, (A.5)
and consequently, due to 2cs=sin 2α,c^2 −s^2 =cos 2α, one obtains
Cμν(α)=⎛
⎝
sin 2α cos 2α 0
cos 2α−sin 2α 0
000⎞
⎠. (A.6)
Forα=0 andα=π/4, this tensor reduces to
⎛
⎝010
100
000
⎞
⎠,
⎛
⎝
100
0 − 10
000
⎞
⎠, (A.7)
respectively. The diagonal expression follows from the first of these tensors when
the Cartesian components of the vectors and tensors are with respect to a coordinate
system rotated by 45◦.
(ii)Compute the product Bμν(α)=Cμλ( 0 )Cλν(α),determine the trace and the
symmetric traceless part of this product. Determine the angleα,for wich one has
Bμμ=0.