Tensors for Physics

200 12 Integral Formulae and Distribution Functions

12.1.1 Integrals of Products of Two Irreducible Tensors

Integrals

∫

...d^2 ̂rof products of irreducible tensorŝrμ 1 ̂rμ 2 ···̂rμand̂rν 1 ̂rν 2 ···̂rν′
over the surface of the unit sphere must be proportional to an isotropic tensor, since
a sphere possesses no preferential direction. Furthermore, the particular isotropic
tensor must have the same subscripts with the same symmetry properties as the
integrands. There is no isotropic tensor which is symmetric traceless in both sets
of subscripts, unless′=holds true. So the integral must be proportional to the
Kronecker deltaδ′and to theΔ()-tensor. Thus the ansatz

∫

̂rμ 1 ̂rμ 2 ···̂rμ̂rν 1 ̂rν 2 ···̂rν′d^2 ̂r=Cδ′Δ()μ 1 μ 2 ···μ,ν 1 ν 2 ···ν.

The coefficientCis determined via the total contraction{μ 1 μ 2 ···μ}={ν 1 ν 2 ···ν}.

Witĥrμ 1 ̂rμ 2 ···̂rμ ̂rμ 1 ̂rμ 2 ···̂rμ′ =!/(( 2 − 1 )!!, cf. Sect.9.3,

∫

d^2 ̂r= 4 πand

Δ()μ 1 μ 2 ···μ,μ 1 μ 2 ···μ= 2 +1, one obtainsC= 4 π!/(( 2 + 1 )!!, and consequently

1

4 π

∫

̂rμ 1 ̂rμ 2 ···̂rμ ̂rν 1 ̂rν 2 ···̂rν′d^2 ̂r=

!

( 2 + 1 )!!

δ′Δ()μ 1 μ 2 ···μ,ν 1 ν 2 ···ν.

(12.1)

Notice, the integral 41 π

∫

...d^2 ̂ris an orientational average. The orientational average
of corresponding tensors constructed from the vectorrrather than the unit vector̂r
is given by the right hand side of (12.1), multiplied byr^2 .
Some special cases of (12.1)are
∫
̂rμd^2 ̂r=

∫

̂rμ̂rνd^2 ̂r=

∫

̂rμ̂rν̂rλd^2 ̂r= 0 , (12.2)

and

1

4 π

∫

̂rμ̂rνd^2 ̂r=

1

3

δμν, (12.3)

1

4 π

∫

̂rμ̂rν ̂rλ̂rκ =

2

15

Δμν,λκ. (12.4)

The numerical factor occurring here can be checked immediately by observing that

δμμ=3,̂rμ̂rν̂rμ̂rν= 2 /3, andΔμν,μν=5.
In terms of theY···-tensors defined in Sect.10.1.4, equation (12.1) is equivalent to

1

4 π

∫

Yμ 1 μ 2 ···μYν 1 ν 2 ···ν′d^2 ̂r=!( 2 − 1 )!!δ′

1

2 + 1

Δ()μ 1 μ 2 ···μ,ν 1 ν 2 ···ν.

(12.5)