The quantities $B_x$ and $u_xB_y-u_yB_x=-E_z$ are each constant across the shock. A single tangential Galilean boost therefore changes the common $E_z$ to zero on both sides. This is the <de Hoffmann–Teller frame>. Ideal MHD then gives
$$
\mathbf u_i\times\mathbf B_i=0,
\qquad
\boxed{\mathbf u_i\parallel\mathbf B_i}.
$$

Let $j=\rho u_x$ and retain the common $B_x$. The independent jump conditions simplify to
$$
[j]=0,
\qquad [B_x]=0,
$$
$$
\left[\rho u_x^2+p+\frac{B_y^2}{2\mu_0}\right]=0,
$$
$$
\left[\rho u_xu_y-\frac{B_xB_y}{\mu_0}\right]=0,
$$
$$
\left[\frac12u^2+h\right]=0,
\qquad
\frac{u_y}{u_x}=\frac{B_y}{B_x}.
$$
Terms involving the common $B_x^2$ cancel from normal momentum, and the electromagnetic term vanishes from energy because $E_z=0$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-314/3/b/solution