Tensors for Physics

194 11 Isotropic Tensors

Now, letAbe the traceless tensor constructed from the components of the vector

b,viz.Aν 1 ν 2 ···ν=bμ 1 bμ 2 ···bμ. Then, the second term on the right hand side of
(11.49), involving the cross product, vanishes. Due to

bμbμbμ 1 bμ 2 ···bμ− 1 =



( 2 − 1 )

b^2 bμ 1 bμ 2 ···bμ− 1 ,

cf. (10.19), the relation (11.49), with (11.50) and (11.52), reduces to

bμbμ 1 bμ 2 ···bμ =bμbμ 1 bμ 2 ···bμ (11.53)

+



( 2 + 1 )

b^2 Δ()μ 1 μ 2 ···μ,μ ν 1 ν 2 ···ν− 1 bν 1 bν 2 ···bν− 1.

For=1, this relation reduces tobμbν=bμbν+^13 δμν. The case=2 yields

bμbνbλ =bμbνbλ+

2

5

b^2 Δνλ,μκbκ. (11.54)

Multiplication of (11.53) by the componentsaμaμ 1 ···aμof the vectora,useof

aμaμaμ 1 aμ 2 ···aμ− 1 =/( 2 − 1 )a^2 aμ 1 aμ 2 ···aμ− 1 ,cf.(10.19), and of the

abbreviation, cf. (9.10),P(a,b)=aμ 1 aμ 2 ...aμ bμ 1 bμ 2 ...bμ yields

a·bP(a,b)=P+ 1 (a,b)+



( 2 + 1 )



( 2 − 1 )

a^2 b^2 P− 1 (a,b). (11.55)

Due toP(a,b)=abNP(x), withx=̂a·̂bandN=!/( 2 − 1 )!!,cf.(9.11),
the (11.55) is equivalent to the recursion relation for the Legendre polynomials

xP(x)=

+ 1

2 + 1

P+ 1 (x)+



2 + 1

P− 1 (x). (11.56)

11.6 Coupling of Second Rank Tensors with Irreducible Tensors

Te n s o r s

The product of second rank tensor with an irreducible tensor of rankcan be decom-
posed by analogy to (11.49), where now irreducible tensors of ranks+2,+1,,
−1,−2 occur. Here, the special case is considered where both the second rank
tensor and the-th rank tensor are constructed from the components of the vectorb.
Then the tensors of ranks±1 vanish and the desired result can be obtained by a
twofold application of (11.53). Thus one finds