Tensors for Physics

92 7 Fields, Spatial Differential Operators

Clearly, due to∇μvμ=εμνλ∇μ∇νAλ=0, the divergence of a vector field given by
(7.39) vanishes, when the symmetry∇μ∇ν=∇ν∇μapplies for the second spatial
derivatives of the vector potentialA.
The vector potentialAis not unique, in the sense that also the vector potential
A′=A+∇φ(r), with a scalar functionφ(r),used(7.39), yields the same vector field
v. The reason is that∇×∇φ(r)=0. The freedom for the choice of a vector potential
is by far greater than that of a scalar potential, which is determined except for an
additive constant. In some applications, a source-free vector potential is required,
i.e. the extra condition∇·A=0 is imposed.

Some Examples for Vector Potentials

(i) Homogeneous Vector Field

A homogeneous vector fieldv=const. is obtained as the rotation of a vector potential
A, which should be linear in the position vectorr. Furthermore,Amust contain the
information on the constantv. A plausible ansatz isA=cv×r, with a coefficient
c, which has to be determined. To this purpose, one computes, with the help of the
properties of the epsilon-tensor, cf. Sect.4.1.2,

vμ=(∇×A)μ=εμνλ∇νcελσ τvσrτ=cεμνλελσ τδντvσ

=cεμνλελσ νvσ=c 2 δμσvσ= 2 cvμ

and consequently,c= 1 /2. Thus the constant vector fieldvis represented as the
rotation of the vector potential

Aμ=

1

2

εμνλvνrλ. (7.40)

Notice that the constant vector field can be derived both from a scalar potential and
from a vector potential. A scalar potential does not exist for the next example.

(ii) Solid-like Rotational Flow

The solid-like rotational flow is described by the vector field

vμ=εμνλwνrλ,

with the constant angular velocityw. The pertaining vector potential must be linear
inwand of second order inr. A guess isAλ=c 1 r^2 wλ+c 2 rλrκwκ, with two
coefficientsc 1 andc 2 , which have to be determined such thatεμνλ∇νAλ=vμ.The
direct computation gives

εμνλ∇νAλ=εμνλ( 2 c 1 rν+c 2 δνλrκwκ+c 2 rλδνκwκ)=(− 2 c 1 +c 2 )εμνλwνrλ.