Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 65 / 1

Write horizontal components with a subscript $h$. The steady ideal magnetohydrodynamic induction equation gives $\nabla \times (\mathbf{u}\times \mathbf{B})=0$. Since ${\partial }_z=0$, its horizontal components implyThus $(\mathbf{u}\times \mathbf{B}{)}_z=C$ is a spatial constant. A vanishing axial electric field, $E_z=-C=0$, is needed to infer horizontal alignment. It is satisfied, for example, if the constant electric field vanishes at a boundary or at infinity. With this additional condition and ${\mathbf{B}}_h\ne 0$, define the magnetohydrodynamic mass loading $k$ by $\rho {\mathbf{u}}_h=k{\mathbf{B}}_h$, and put $U=u_z-k{B}_z/\rho$. ThenThe steady continuity equation and the solenoidal magnetic-field constraint giveAlso $\mathbf{u}\times \mathbf{B}=U{\mathbf{e}}_z\times \mathbf{B}$. Using the vector identity for the curl of a cross product, together with ${\partial }_z\mathbf{B}=0$ and $\nabla \cdot \mathbf{B}=0$, its curl is ${\mathbf{e}}_z\mathbf{B}\cdot \nabla U$. HenceThese are steady planar ideal-MHD field-line invariants: both quantities are constant along a magnetic field line.

The zero-electric-field condition does not follow from the printed stationarity assumptions alone. Constant $\rho ,p,s$, $\Phi =0$, $\mathbf{u}=u_0{\mathbf{e}}_x$ and $\mathbf{B}=B_0{\mathbf{e}}_y$, with $u_0{B}_0\ne 0$, satisfy every displayed ideal magnetohydrodynamics equation but have perpendicular horizontal velocity and magnetic field. Their $C=u_0{B}_0$ is nonzero, so no $k,U$ give the stated decomposition. The field-line decomposition is valid on the $E_z=0$ branch; as an unrestricted implication, the first claim has this counterexample. The subsequent calculations use that branch.

Continue on the branch of steady planar ideal-MHD field-line invariants, with $k\ne 0$ and positive $\rho ,T$. Adiabatic entropy transport gives $\mathbf{u}\cdot \nabla s=0$. As all quantities are $z$-independent,The axial component of the ideal magnetohydrodynamic momentum equation has no pressure or gravitational term. Magnetic tension givesSince $k$ is constant along a magnetic field line, this becomesSubtracting the two axial invariants yields the useful compatibility relationThe squared ratio of horizontal flow speed to horizontal Alfvén speed is $M_h^2={\mu }_0{k}^2/\rho$; vector equality with the horizontal Alfvén velocity additionally requires $k>0$. If $V-U\ne 0$, the right side is a nonzero constant along the field line, so a regular finite $B_z$ cannot permit $M_h^2=1$ anywhere. This is the nondegenerate exclusion.

However, $B_z\not\equiv 0$ does not imply $V-U\ne 0$. The Alfvénic degeneracy of steady planar MHD is a genuine exception: take constant $\rho ,p,s$, $\Phi =0$, and a constant magnetic field with nonzero horizontal and axial components, and setEvery equation holds, but the horizontal velocity equals the horizontal Alfvén velocity everywhere despite $B_z\ne 0$. Therefore the printed pointwise exclusion needs a nondegeneracy hypothesis, such as $V\ne U$. The general conclusion is the boxed compatibility relation; when $V=U$, it says $(M_h^2-1)B_z=0$ and allows the Alfvénic branch.

Let $b=\frac{1}{2}{u}^2+\Phi +w$. Because the specific entropy is constant along the horizontal flow, the specific enthalpy identity gives $\mathbf{u}\cdot \nabla w={\rho }^{-1}\mathbf{u}\cdot \nabla p$. Dotting the ideal magnetohydrodynamic momentum equation with $\mathbf{u}$ therefore givesOn the field-aligned horizontal branch, $\mathbf{u}\times \mathbf{B}=U{\mathbf{e}}_z\times \mathbf{B}$. Directly evaluating the horizontal components of the curl givesThe scalar triple-product identity consequently makes the magnetic work $U{\mathbf{B}}_h\cdot \nabla B_z/{\mu }_0$. Meanwhile $\rho \mathbf{u}\cdot \nabla b=k{\mathbf{B}}_h\cdot \nabla b$. The previously established field-line constancy of $U,k$ now gives the planar ideal-MHD Bernoulli invariantThis is conservation of total energy transported along a magnetic field line, including electromagnetic work. To see its flux interpretation, the ideal magnetohydrodynamic energy conservation law has fluxThe second term is the Poynting vector, since $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$. Substitution of $\mathbf{u}=k\mathbf{B}/\rho +U{\mathbf{e}}_z$ givesThe steady divergence of this horizontal energy flux is zero. Magnetic stress can change the ordinary hydrodynamic quantity $b$ in the Bernoulli equation by doing work; the term $-U{B}_z/({\mu }_{0}k)$ accounts for the accompanying Poynting flux. When $U=0$, the velocity is completely field-aligned, $\mathbf{E}=0$, and the usual hydrodynamic Bernoulli quantity is constant. These conclusions require the field-aligned branch and $k\ne 0$; the arbitrary constant-$E_z$ flow from the first part is not covered by this particular invariant.