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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-54/2/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 54 / 2 / c / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-06

At the initially resting instant

K_{ij} = 0

, and the cold-fluid assumption removes the pressure stress. Sum the

xx

and

yy

components of the magnetized-fluid tensor virial theorem. The magnetic and gravitational traces are

M_{xx} + M_{yy} = - \frac{B _{z}^{2}}{μ _{0}}, W_{xx} + W_{yy} = \frac{g _{z}^{2}}{4 π G} .

Since

I_{⊥} = I_{xx} + I_{yy}

and

g_{z} = - \partial_{z} Φ

, this yields the horizontal virial balance of a cold magnetized fluid:

\frac{1}{2} \ddot{I}_{⊥} = \frac{1}{4 π G} \int [\frac{4 π G B _{z}^{2}}{μ _{0}} - (\partial_{z} Φ)^{2}] d τ .

Being at rest sets the instantaneous velocity to zero; it does not assert equilibrium or zero acceleration. The fields throughout space contribute to this stress integral, including their vacuum exterior.

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