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 Incoming links: de Hoffmann–Teller frame

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 3 / b / Solution Created 2026-09-24 Updated 2026-09-25

B_{x}

and

u_{x} B_{y} - u_{y} B_{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

u_{i} \times B_{i} = 0, u_{i} ∥ B_{i} .

Let

j = ρ u_{x}

and retain the common

B_{x}

. The independent jump conditions simplify to

[j] = 0, [B_{x}] = 0,

[ρ u_{x}^{2} + p + \frac{B _{y}^{2}}{2 μ _{0}}] = 0,

[ρ u_{x} u_{y} - \frac{B _{x} B _{y}}{μ _{0}}] = 0,

[\frac{1}{2} u^{2} + h] = 0, \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

.

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