Tensors for Physics

166 10 Multipole Potentials

Multiplication ofXμ 1 μ 2 ···μby the components ofryields a tensor of rank+1.

The contraction with the position vector, on the other hand, gives a tensor of rank

−1, which is proportional to a multipole potential, viz.

rμXμ 1 μ 2 ···μ− 1 μ=Xμ 1 μ 2 ···μ− 1. (10.12)

The proof is as follows. On account of (10.3) andrμ∂r∂μ

=r∂∂r, the left hand
side of (10.12) is equal to−r∂∂rXμ 1 μ 2 ···μ− 1. The multipole potential of rank−1is
proportional tor−,cf.(10.10). Thus the differentiation with respect to the magnitude
ryields the result (10.12).
Letg=g(r)be a function which depends onronly via the magnituder=|r|.
Then the Laplace operator applied on the product ofg(r)and the multipole potential
of rankis equal to

Δ

(

g(r)Xμ 1 μ 2 ···μ

)

=

(

g′′− 2 r−^1 g′

)

Xμ 1 μ 2 ···μ(r)=r^2 

(

r−^2 g′

)′

Xμ 1 μ 2 ···μ(r)

≥ 1 , (10.13)

where the prime′indicates the derivative with respect tor. When the case=0is

associated with the monopole functionr−^1 , then (10.13) also applies for=0. The

proof of (10.13) is transferred to the following exercise.

10.1 Exercise: Prove the Product Rule (10.13) for the Laplace Operator

Hint:useΔ(fg)=fΔg+ 2 (∇κf)(∇κg)+gΔf, for any two functionsfandg.

10.2 Exercise: Multipole Potentials in D Dimensional Space

In D dimensions,r(^2 −D)is the radially symmetric solution of the Laplace equation, cf.

Exercise7.6,forD≥3. By analogy with (10.2), D dimensional multipole potential

tensors are defined by

XμD 1 μ 2 ···μ≡(− 1 )

∂

∂rμ 1 ∂rμ 2 ···∂rμ

r(^2 −D)=(− 1 )∇μ 1 ∇μ 2 ···∇μr(^2 −D),

(10.14)

where now∇is the in D dimensional nabla operator. ForD=2,r(^2 −D)is replaced

by−lnr. Compute the first and second multipole potentials, forD≥3 and for

D=2.

10.2 Ascending Multipoles

The factor(g′′− 2 r−^1 g′)=r^2 (r−^2 g′)′in (10.13) is equal to zero not only for

g=1, but also forg=r(^2 +^1 ). This implies: the tensors

X ̃μ 1 μ 2 ···μ≡r(^2 +^1 )Xμ 1 μ 2 ···μ=rYμ 1 μ 2 ···μ (10.15)