11.2 Topological interpretation of magnetic helicity 337BecauseBvanishes outside of the twoflux tubes, the helicity integral is finite only in the
volumesV 1 andV 2 of the two tubes and so the helicity can be expressed as (Moffatt 1978)K=∫
V 1A·Bd^3 r+∫
V 2A·Bd^3 r. (11.2)The contribution to the helicity from integrating over the volume offlux tube #1 isK 1 =∫
V 1A·Bd^3 r. (11.3)In order to evaluate this integral it is recalled that the magneticflux through a surfaceS
with perimeterCcan be expressed asΦ=∫
SB·ds=∮
CA·dl. (11.4)Influx tube #1,d^3 r=dl·Bˆ∆S,where∆Sis the cross-sectional area offlux tube #1 anddl
is an element of length alongC 1 .It is therefore possible to recast the integrand in Eq.(11.3)
as
A·Bd^3 r = A·Bdl·Bˆ∆S
= A·dlΦ 1 (11.5)
sincedlis parallel toBandB∆S=Φ 1 .BecauseΦ 1 is by definition constant along the
length offlux tube #1, it may be factored from theK 1 integral, givingK 1 =Φ 1∫
C 1A·dl. (11.6)However, theflux linked by contourC 1 is precisely theflux in tube #2, i.e.,∫
C 1 A·dl=Φ^2
and so
K 1 =Φ 1 Φ 2. (11.7)
The same analysis applied toflux tube#1leads toK 2 =Φ 1 Φ 2 and so the helicity content
of the linkedflux tubes is therefore
K=K 1 +K 2 =2Φ 1 Φ 2. (11.8)flux 1
contourC 1
volume V 1flux 2
contourC 2
volume V 2B BFigure 11.1: Two linked thin untwistedflux tubes. Magnetic field is zero outside theflux
tubes.