Appendix D: Coulomb-gauged helicity flux

In the $\phi =0$ gauge we have $\dot{\vec{A}}=-\mbox{\boldmath$E$ } {}$. In that gauge, the evolution of the magnetic helicity density is given by

$${\partial\over\partial t}(\vec{A}\cdot\mbox{\boldmath$B$ } {}... ...ec{A}) =-2\mbox{\boldmath$E$ } {}\cdot\mbox{\boldmath$B$ } {}.$$ (D.1)

Note that the term on the right hand side of this equation is gauge-invariant, but the terms on the left hand side are not. In the Coulomb gauge we have $\mbox{\boldmath$\nabla$ } {}\cdot\vec{A}_{\rm C}=0$, where $\vec{A}_{\rm C}\equiv\vec{A}-\vec{A}_0$. In order to maintain $\mbox{\boldmath$\nabla$ } {}\cdot\vec{A}_{\rm C}=0$ for all times, we have to add the term $-\mbox{\boldmath$\nabla$ } {}\phi\equiv\mbox{\boldmath$E$ } {}_0=\mbox{\boldmat... ...abla$ } {}(\nabla^{-2}\mbox{\boldmath$\nabla$ } {}\cdot\mbox{\boldmath$E$ } {})$ to the right hand side of the uncurled induction equation,

$${\partial\over\partial t}\vec{A}_{\rm C}=-(\mbox{\boldmath$E$ } {}-\mbox{\boldmath$E$ } {}_0)\cdot$$ (D.2)

In this gauge the evolution equation for the helicity density becomes

$${\partial\over\partial t}(\vec{A}_{\rm C}\cdot\mbox{\boldmath... ...$ } {}-\mbox{\boldmath$E$ } {}_0)\cdot\mbox{\boldmath$B$ } {}.$$ (D.3)

Unlike Eq. (D.1), the right hand side of Eq. (D.3) is gauge dependent owing to the extra term $2\mbox{\boldmath$E$ } {}_0\cdot\mbox{\boldmath$B$ } {}$. However, because of $\mbox{\boldmath$\nabla$ } {}\cdot\mbox{\boldmath$B$ } {}=0$, we can write this as the divergence of another contribution to the helicity flux density,

$$2\mbox{\boldmath$E$ } {}_0\cdot\mbox{\boldmath$B$ } {}=-2(\mb... ...box{\boldmath$\nabla$ } {}\cdot(2\phi\mbox{\boldmath$B$ } {}),$$ (D.4)

which should be included in the expression for the Coulomb gauged helicity flux density,

$$\mbox{\boldmath$J$ } {}_{\rm H}^{\rm Cou}=(\mbox{\boldmath$E$... ...} {}_0)\times(\vec{A}-\vec{A}_0)+2\phi\mbox{\boldmath$B$ } {},$$ (D.5)

so Eq. (D.3) becomes

$${\partial\over\partial t}(\vec{A}_{\rm C}\cdot\mbox{\boldmath... ...rm Cou}=-2\mbox{\boldmath$E$ } {}\cdot\mbox{\boldmath$B$ } {}.$$ (D.6)

Now the right hand sides of Eqs. (D.1) and (D.6) agree and are gauge-invariant. Note, however, that

$$\phi\mbox{\boldmath$B$ } {}=\phi\mbox{\boldmath$\nabla$ } {}\... ...ec{A}_{\rm C})+\mbox{\boldmath$E$ } {}_0\times\vec{A}_{\rm C},$$ (D.7)

so Eq. (D.5) can also be written as

$$\mbox{\boldmath$J$ } {}_{\rm H}^{\rm Cou}= (\mbox{\boldmath$E... ...+\mbox{\boldmath$\nabla$ } {}\times[2\phi(\vec{A}-\vec{A}_0)],$$ (D.8)

which is identical to Eq. (11).