 Solution

ID: past-exam-of-the-advanced-fluid-dynamics-course-of-the-university-of-oxford/2016/1/a/solution

Use the source's Gaussian units and neglect the displacement current, as in nonrelativistic magnetohydrodynamics. Write

Δ p = p_{⊥} - p_{∥}

P = p_{⊥} I - Δ p bb

- \nabla \cdot P = - \nabla p_{⊥} + \nabla \cdot (Δ p bb) .

\frac{j \times B}{c} = \frac{( \nabla \times B ) \times B}{4 π} = - \nabla \frac{B ^{2}}{8 π} + \frac{( B \cdot \nabla ) B}{4 π} = - \nabla \frac{B ^{2}}{8 π} + \nabla \cdot \frac{BB}{4 π} .

Adding these two force densities and using

BB = B^{2} bb

yields

ρ \frac{D u}{D t} = - \nabla (p_{⊥} + \frac{B ^{2}}{8 π}) + \nabla \cdot [(Δ p + \frac{B ^{2}}{4 π}) bb] .

T = B^{2} / (4 π) + Δ p

. At fixed magnetic field, positive pressure anisotropy increases resistance to bending, while negative pressure anisotropy reduces it. When

T

becomes negative, bending experiences an amplifying force rather than a restoring force. Gradients of

Δ p

can also drive motion, so the curvature interpretation refers specifically to the bending contribution.

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