9.1 Definition and Examples 157
In many cases, the explicit form of the symmetric traceless higher rank tensors are
not needed. It is one of the advantages of tensor calculus that general properties of
tensors often suffice in applications.
Aremarkonnotationisappropriate.AsmentionedbeforeinSect.3.1.2,alternative
ways to write and display irreducible tensors are found in the literature. In additions
to the double arrow, the symbols[...] 0 or[[...]] 0 ,aswellas{...} 0 were used to
indicate the symmetric traceless part of a tensor.
9.2 Products of Irreducible Tensors.
LetAμνandBλκbe irreducible tensors. Multiplication of the irreducible fourth rank
tensor constructed from the product of the second rank tensorAμν,cf.(9.7), byBλκ
yields
AμνAλκ Bλκ=1
3
Aμν(AλκBλκ)+4
105
Bμν(AλκAλκ)+2
3
AμλBλκAκν−8
21
BμλAλκAκν. (9.8)For the special caseB=A,(9.8), reduces to
AμνAλκAλκ=13
35
Aμν(AλκAλκ)+2
7
AμλAλκAκν=18
35
Aμν(AλκAλκ).
(9.9)The last equality follows from AμλAλκAκν=^12 AμνAλκAλκ,cf.(5.51).
9.1 Exercise: Verify the Required Properties of the Third and Fourth Rank
Irreducible Tensors(9.5)and(9.6)
First, verify by inspection, that the tensors explicitly given by (9.5) and (9.6)are
symmetric against the interchange of indices, within any pair of components. Then
putλ=ν,in(9.5)toshowaμaνaν =0. Likewise, useκ=λin (9.6)toverify
aμaνaλaλ =0.9.3 Contractions, Legendre Polynomials
The multiplication of two tensors of rankconstructed from the components of
two vectorsaandband their subsequent total contraction yields a scalar which is
proportional to the Legendre polynomialP, depending on the cosine of the angle
between these two vectors. More specifically:
P(a,b)≡aμ 1 aμ 2 ...aμbμ 1 bμ 2 ...bμ =abNP(̂a·̂b), (9.10)