186 11 Isotropic Tensors
11.2.3 Δ-Tensors as Derivatives of Multipole Potentials
The-fold spatial derivative of the-th rank irreducible tensor constructed from
the components ofris proportional to the isotropicΔ()-tensor. More specifically,
one has
∇μ 1 ∇μ 2 ···∇μrν 1 rν 2 ···rν =!Δ()μ 1 μ 2 ···μ,ν 1 ν 2 ···ν. (11.14)Notice, the tensor...with theν-indices imposes its property of being traceless onto
theμ-indices. As an exercise, verify (11.14), for=1 and=2. Due to
rμ 1 rμ 2 ···rμ =1
( 2 − 1 )!!
r(^2 +^1 )Xμ 1 μ 2 ···μ,see (10.9), and with the multipole potential given by (10.2), (11.14) leads to
!( 2 − 1 )!!(− 1 )Δ()μ 1 μ 2 ···μ,ν 1 ν 2 ···ν=∇μ 1 ∇μ 2 ···∇μ
(
r(^2 +^1 )∇ν 1 ∇ν 2 ···∇νr−^1)
.
(11.15)
It is remarkable that the differentiation, with unrestricted subscripts, on the right
hand side of (11.15), yields an expression, on the left hand side, which is traceless
with respect to any pair of eitherμ-orν-subscripts.
11.2 Exercise: DetermineΔ(μνλ, μ^3 ) ′ν′λ′
Hint: computeΔ(μνλ,μ^3 ) ′ν′λ′, in terms of triple products ofδ-tensors, from (11.15)for
=3.
11.3 Generalized Cross Product,h-Tensors.
11.3.1 Cross Product via theh-Tensor
The isotropic tensor of rank 2+1, defined by
()μ
1 μ 2 ···μ,λ,μ′ 1 μ′ 2 ···μ′
≡Δ()μ 1 μ 2 ···μ,ν 1 ν 2 ···ν− 1 νενλν′Δ()ν′
ν^1 ν^2 ···ν−^1 ,μ′ 1 μ′ 2 ···μ′,
(11.16)
allows to link a vector with the symmetric traceless part of a tensor of ranksuch
that the result is a symmetric traceless, i.e. irreducible tensor of rank. Here and in
the following, it is understood that≥1.