\frac{D B}{D t} = (B \cdot \nabla) u - B \nabla \cdot u,

\frac{D u}{D t} = - \nabla (h + Φ) + T \nabla s + \frac{( \nabla \times B ) \times B}{4 π ρ} .

The magnetic force is perpendicular to

B

. For

h_{c} = u \cdot B

, the two equations therefore give

\frac{D h _{c}}{D t} = - B \cdot \nabla (h + Φ - \frac{u ^{2}}{2}) + T B \cdot \nabla s - h_{c} \nabla \cdot u .

Since

\nabla \cdot B = 0

\frac{\partial h _{c}}{\partial t} + \nabla \cdot F_{c} = S_{c},

with

F_{c} = u h_{c} + B (h + Φ - \frac{u ^{2}}{2}), S_{c} = T B \cdot \nabla s .

\nabla s = 0

, so the source vanishes.

b

Solution

u

is parallel to

B

, write

u = κ B

. Then

u (u \cdot B) = u^{2} B,

and the flux from part a reduces to

F_{c} = B (\frac{u ^{2}}{2} + h + Φ) = C_{B} B,

where

C_{B}

is the gravitational Bernoulli function. If it is constant along field lines,

B \cdot \nabla C_{B} = 0

, then

\nabla \cdot F_{c} = B \cdot \nabla C_{B} + C_{B} \nabla \cdot B = 0.

The source also vanishes because the flow is homentropic. Hence

\partial_{t} h_{c} = 0

, and every fixed volume satisfies

\frac{d}{d t} \int_{V} u \cdot B d V = 0 .

Thus the cross-helicity is conserved even for this time-dependent aligned flow.

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