Tensors for Physics

404 Appendix: Exercises...

Due torκ∇κ=r∂/∂randXμ 1 μ 2 ···μ∼r−(+^1 ), the second term in the preceding
equation is equal to

− 2 g′r−^1 ((+ 1 )X....

Use ofΔg=g′′+ 2 r−^1 g′then leads to (10.13). The condition≥1 is obvious for
the validity of the present considerations. When the functionr(−^1 )stands forXin
the case=0, the value=0 is also included in (10.13).

10.2 Multipole Potentials in D Dimensional Space(p.166)
In D dimensions, r(^2 −D)is the radially symmetric solution of the Laplace equation, cf.
exercise7.6,for D≥3.By analogy with(10.2),D dimensional multipole potential
tensors are defined by

Xμ(D 1 )μ 2 ···μ≡(− 1 )

∂

∂rμ 1 ∂rμ 2 ···∂rμ

r(^2 −D)=(− 1 )∇μ 1 ∇μ 2 ···∇μr(^2 −D),
(A.11)
where now∇is the in D dimensional Nabla operator. For D=2,r(^2 −D)is replaced
by−lnr.Compute the first and second multipole potentials, for D≥ 3 and for
D=2.
The first descending multipole is the D dimensional vector

X(μD)=(D− 2 )r(^1 −D)r̂μ=(D− 2 )r−Drμ, D≥ 3 , X(μ^2 )=r−^1 r̂μ=r−^2 rμ.

The resulting second multipole potentialX(μνD)=−∇νX(μD),D≥2, is

X(μνD)=(D− 2 )r−(D+^2 )(Drμrν−δμνr^2 ), D≥ 3 , Xμ(^2 )=r−^4 ( 2 rμrν−δμνr^2 ).

It is understood thatδμνis the D dimensional unit tensor withδμμ=D. Thus the

tensorsX(μνD)are traceless.

10.3 Compute the Torque on a Rotating Sphere(p.181)
A sphere rotating with the angular velocityΩexperiences a friction torque

Tμ=− 8 πηR^3 Ωμ.

To derive this result from the creeping flow equation, considerations similar to those
used for the Stokes force, should be made.
The distortion of the pressure and the flow velocity should be linear inΩand the
respective expressions should have the appropriate parity. The only scalar available
forpis proportional toXνΩν. This term, however has the wrong parity behavior,
thus one hasp=0, in this case. The possible vectors are proportional toΩμ,XμνΩν
andεμλνΩλXν. The first two of these expressions have the wrong parity. The only