3.2 Dyadics 39
The further contractionμ=νleads to
ab:cd=aμbλcλdμ =1
2
(a·cb·d+a·db·c)−1
3
a·bc·d. (3.25)For the special casea=bandc=d, relation (3.25) reduces to
aa :cc=aμaλ cλcμ =(a·c)^2 −1
3
a^2 c^2=a^2 c^2(
(̂a·̂c)^2 −1
3
)
=a^2 c^2(
cos^2 φ−1
3
)
, (3.26)
wherêaand̂care unit vectors, andφis the angle between the vectorsaandc.
Clearly, fora=c, corresponding toφ=0, one finds
aa :aa=aμaλaλaμ =2
3
a^4. (3.27)Notice, the double dot product of two symmetric traceless dyadic tensors constructed
from orthogonal vectors is not zero. Forcperpendicular toa, corresponding to
φ= 90 ◦, relation (3.26) implies
a⊥c =⇒ aa:cc =aμaλ cλcμ =−1
3
a^2 c^2. (3.28)On the other hand, the two dyadic tensors aa and cc are “orthogonal”, in the
sense that their double dot product vanishes, when the angle between the two vectors
is given by the “magic angle”φ=arccos( 1 /
√
3 )≈ 54. 7 ◦. Applications of these
relations for the double dot product of dyadics are discussed later.
3.2 Exercise: Symmetric Traceless Dyadics in Matrix Notation
Write the symmetric traceless parts of the dyadic tensorCμν=Cμν(α)= 2 aμbνin
matrix form for the vectorsa=a(α):{c,−s, 0 }andb=b(α):{s,c, 0 }, where
candsare the abbreviationsc=cosαands=sinα. Discuss the special cases
α=0 andα=π/4. Compute the productBμν(α)= Cμλ( 0 )Cλν(α), determine
the trace and the symmetric traceless part of this product. Determine the angleα,for
which one hasBμμ=0.