Fundamentals of Plasma Physics

266 Chapter 9. MHD equilibria

or two directions (theθandzdirections for the cylindrical example here), and exponen-

tially growing or decaying in the remaining direction or directions (therdirection for the

cylindrical example here).

Non-vacuum magnetic fields are more complicated than vacuum fields, and unlike vac-

uum fields, are not uniquely determined by the surface boundary conditions. This is be-

cause non-vacuum fields are determined by both the current distribution within the volume

and the surface boundary conditions. Vacuum fields are distinguished from non-vacuum

fields because vacuum fields are the lowest energy fields satisfying given boundary condi-

tions on the surfaceSof a volumeV. Let us now prove this statement.

Consider a volumeV bounded by a surfaceSover which boundary conditions are

specified. LetBmin(r)be the magnetic field having theloweststored magnetic energy of

all possible magnetic fields satisfying the prescribed boundary conditions.We use methods

of variational calculus to prove that this lowest energy field is the vacuum field.

Bmin

B

B

W

Figure 9.1: Magnetic field energy for different configurations,Bminis the configuration
with minimum magnetic energy for a given boundary condition.

Consider some slightly different field denoted asB(r)which satisfies the same bound-

ary conditions asBmin.This slightly different field can be expressed asB(r) =Bmin(r) +

δB(r)whereδB(r)is a small arbitrary variation aboutBmin.This situation is sketched in a

qualitative fashion in Fig.9.1 where the horizontal axis represents a continuumof different

allowed choices for the vector functionB(r). SinceB(r)satisfies the same boundary con-

ditions asBmin,δB(r)must vanish onS.The magnetic energyWassociated withB(r)

can be evaluated as

2 μ 0 W =

∫

V

(Bmin+δB)^2 d^3 r

=

∫

V

B^2 mind^3 r+ 2

∫

V

δB·Bmind^3 r+

∫

V

(δB)^2 d^3 r. (9.7)

If the middle term does not vanish,δBcould always be chosen to be antiparallel toBmin

insideV so thatδB·Bminwould be negative. SinceδBis assumed small,δB·Bmin

would be larger in magnitude than(δB)^2. Such a choice forδBwould makeWlower than

the energy of the assumed lowest energy field, thus contradicting the assumption thatBmin