Tensors for Physics

Appendix: Exercises... 399

which is the relation (7.83), and

LμLν−LνLμ=rν∇μ−rμ∇ν.

The right hand side of the last equation can be written as−εμνλLλ, since one has

−εμνλLλ=−εμνλελαβrα∇β =−(rμ∇ν−rν∇μ). This proves the commutation

relation (7.82).

7.6 Determine the Radial Part of the Laplace Operator in D Dimensions(p.108)

Let f = f(r)a function of the magnituder =|r|. It does not depend on the

direction ofr, it has no angular dependence. Thus only the radial partΔrof the

Laplace operatorΔ≡∇μ∇μgives a contribution, whenΔis applied onf(r).Due

to∇μ∇μf=∇μ(∇μf)and∇μf=ddrf∇μr=ddrfr−^1 rμ, one obtains∇μ∇μf=

rμ∇μ(r−^1 ddrf)+r−^1 ddrf∇μrμ. Notice thatrμ∇μisrtimes the spatial derivative in
radial direction, here equal torddr. On account of∇μrμ=D,forDdimensions, one
finds∇μ∇μf=d

(^2) f
dr^2 +(D−^1 )r
− 1 df
dr. For the case where the Laplace operator is
applied to a function which also depends on the direction ofr, the derivativeddrwith
respect toris replaced by the partial derivative∂∂r, in the previous expression. Thus
the radial part of the Laplace operator is inferred to be
Δr=

∂^2

∂r^2

+(D− 1 )r−^1

∂

∂r

=r−(D−^1 )

∂

∂r

(

r(D−^1 )

∂

∂r

)

.

ProveΔr(^2 −D)= 0

Application of the expression given above yieldsΔrn=n(D+n− 2 )r(n−^2 ), where

the exponentnis a real number. Apart from the trivial solutionn=0, the requirement

Δrn=0 impliesn= 2 −D. Thusr−^1 is a solution of the Laplace equation, for

D=3. One findsr−^2 forD=4.

The caseD=2 has to be considered separately. Here ln(r/rref)is a solution of the

Laplace equation, please check it. The quantityrrefis a reference length introduced

such that the argument of the logarithm ln is dimensionless.

Exercises Chapter 8

8.1 Compute Path Integrals Along a Closed Curve for Three Vector Fields

(p.116)

The differential drneeded for the integration is recalled to bedx{ 1 , 0 , 0 },for the

straight line C 1 ,andρdφ{−sinφ,cosφ, 0 }for the semi-circle C 2 .The start and

end points are x=−ρand x=ρ,for C 1 .Fo r C 2 one hasφ= 0 andφ=π,see

Fig.8.3.