Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 64 / 1

Let ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{v}t$, $t^{\prime }=t$ for a constant relative velocity $\mathbf{v}$. The Galilean transformation gives ${\nabla }^{\prime }=\nabla$ and ${\partial }_{t^{\prime }}={\partial }_t+\mathbf{v}\cdot \nabla$. In the magnetic Galilean limit of the pre-Maxwell equations, takeThe Ampère's law without displacement current and the solenoidal constraint are immediately unchanged. For Faraday's law, the constant-velocity vector identity is $\nabla \times (\mathbf{v}\times \mathbf{B})=\mathbf{v}\nabla \cdot \mathbf{B}-(\mathbf{v}\cdot \nabla )\mathbf{B}$. ThusThese are transformations of the magnetic nonrelativistic approximation. The exact charge-current transformation also contains the convective term $-{\rho }_e\mathbf{v}$ in ${\mathbf{J}}^{\prime }$; ignoring it requires the quasineutral magnetic limit. They are not exact transformations of the full Maxwell equations.

Since ${\mathbf{u}}^{\prime }=\mathbf{u}-\mathbf{v}$, the combination ${\mathbf{E}}^{\prime }+{\mathbf{u}}^{\prime }\times {\mathbf{B}}^{\prime }$ equals $\mathbf{E}+\mathbf{u}\times \mathbf{B}$. For an ideally conducting fluid the rest-frame electric field vanishes, so the ideal moving-conductor Ohm law is $\mathbf{E}+\mathbf{u}\times \mathbf{B}=0$. Substitution in Faraday's law givesThe ideal magnetohydrodynamic induction equation requires nonrelativistic fluid motion, negligible displacement current, and negligible nonideal terms in the electric-field relation. In particular, magnetic diffusion must be slow compared with advection: the magnetic Reynolds number $\mathrm{Rm}=UL/{\eta }_m\gg 1$, where ${\eta }_m=({\mu }_0\sigma {)}^{-1}$ is the magnetic diffusivity. The single-fluid moving-conductor Ohm law must apply on the scales considered, with corrections to that relation negligible; viscosity is not itself excluded by this induction equation.

For a smooth ideal evolution, magnetic flux freezing states that the magnetic flux through every material surface is conserved and that magnetic field lines are carried with the fluid. Their connectivity is preserved: ideal smooth advection does not reconnect them. These implications are stated here without proof, as requested.

Put $D_t={\partial }_t+\mathbf{u}\cdot \nabla$. The continuity equation and the expanded ideal magnetohydrodynamic induction equation areso $D_t(B_i/\rho )=(B_j/\rho ){\partial }_j{u}_i$. Because $\mathbf{B}=\nabla \times \mathbf{A}$, the induction equation permits the magnetic vector potential equation ${\partial }_t\mathbf{A}=\mathbf{u}\times \mathbf{B}-\nabla \phi$, with an arbitrary scalar electric potential $\phi$ carrying the gauge choice. Using $(\mathbf{u}\times \nabla \times \mathbf{A}{)}_i=u_j{\partial }_i{A}_j-u_j{\partial }_j{A}_i$ yieldsContracting with $B_i/\rho$ and differentiating that factor givesThe last two terms cancel by exchanging their dummy indices. Thus the requested material derivative isThis is material transport of magnetic helicity density.

For a material volume, mass conservation gives $d{\int }_{V(t)}\rho hdV/dt={\int }_{V(t)}\rho D_{t}hdV$. Apply this with $h=\mathbf{A}\cdot \mathbf{B}/\rho$, and use $\nabla \cdot \mathbf{B}=0$ and the divergence theorem:If the magnetic field lines do not penetrate the boundary, $\mathbf{B}\cdot \mathbf{n}=0$, so $\boxed{d{H}_m/dt=0}$.

For uniqueness under a regular, single-valued gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \Lambda$,Consequently this magnetic helicity is both conserved and gauge independent. As usual, vector potentials are understood to be globally regular potentials for the same field, rather than unrelated harmonic-circulation choices on a multiply connected restricted domain.

For $ab\ne 0$, a magnetic field line stays at fixed $R$ and satisfies $d\varphi /dz=B_{\varphi }/(R{B}_z)=a/b$. It is therefore a helix with axial pitch $2\pi b/a$; the sign of $a/b$ fixes its handedness. The axial magnetic flux is $\pi b{R}_0^2$, and the toroidal magnetic field supplies the twist. If $a=0$ the field is untwisted; if $b=0$ it consists of azimuthal circles rather than helical lines carrying axial flux.

A convenient globally regular magnetic vector potential has $A_R=0$ andIndeed, the cylindrical curl gives $B_z=R^{-1}d(R{A}_{\varphi })/dR$ and $B_{\varphi }=-d{A}_z/dR$, with $B_R=0$. The tangential potential components are continuous at $R_0$, so their derivatives introduce no unwanted delta-function magnetic field at the tube boundary. The $C$ term is a constant axial potential.

The most general regular potential is this one plus $\nabla \Lambda (R,\varphi ,z)$ for an arbitrary single-valued gauge function. A putative extra $C_{\varphi }/R$ term in the inner $A_{\varphi }$ is excluded by regularity at the axis; the exterior $1/R$ circulation is already fixed by the enclosed magnetic flux. Since the potential is globally defined across the axis and exterior, any remaining difference is a curl-free vector field, hence a gradient. The constant $C$ can itself be absorbed into a gauge function proportional to $z$.

For the displayed gauge family, inside the tubeand henceFor example, requiring $A_z=0$ outside sets $C=a{R}_0^2/2$ and gives $\boxed{H_m/L=\pi ab{R}_0^4/2}$. The equally regular choice $C=0$ gives zero. More generally, for a length-$L$ segment the gauge transformation contributesIn particular, $\Lambda =\Delta Cz$ shifts $H_m/L$ by $\pi b{R}_0^2\Delta C$. This gauge dependence of helicity in an open magnetic flux tube is possible because field lines penetrate the end caps when $b\ne 0$. There is no contradiction with part (b), whose no-penetration condition fails at those caps. A gauge or end-cap convention is needed to give a unique value per unit length. For $b=0$ the ordinary helicity is zero and this end-cap ambiguity disappears.