Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 314 / 3

Set $A={\Phi }_0{R}_0^{\beta }<0$, $q={\lambda }^2$ and $s=R^2+q{z}^2$. The flattened power-law gravitational potential is $\Phi =A{s}^{-\beta /2}$. In axisymmetric cylindrical coordinates, the Poisson equation for Newtonian gravity givesThe first derivatives are ${\partial }_R\Phi =-\beta AR{s}^{-(\beta +2)/2}$ and ${\partial }_z\Phi =-\beta Aqz{s}^{-(\beta +2)/2}$. Differentiating again and collecting powers givesTherefore the required mass density isThis is a formal density for arbitrary $\lambda$, but a physical dark matter distribution must be nonnegative. The density positivity for a flattened power-law potential condition isNecessity follows by evaluating on the midplane and symmetry axis; sufficiency follows because both numerator coefficients are then nonnegative. In this range the origin is a locally integrable density cusp: the mass enclosed near radius $r$ scales as $r^{1-\beta }$ and tends to zero, so no point mass needs adding. The scale-free distribution has infinite total mass at large radius and represents an idealized background, not a finite isolated halo.

For the cold, non-self-gravitating disk, radial balance of a circular orbit is $R{\Omega }^2={\partial }_R\Phi (R,0)$. ThusThe displayed positive root chooses the rotation orientation; the opposite orientation has the negative of this angular frequency.

Use the poloidal magnetic flux function convention ${\mathbf{B}}_p=R^{-1}\nabla \Psi \times {\mathbf{e}}_{\varphi }$, so $2\pi \Psi$ is magnetic flux up to a reference constant. Its poloidal magnetic field components areThe prescribed surface flux therefore givesThe constant offset ${\Psi }_1$ does not affect either field component. If instead $\Psi$ denotes the full physical flux rather than flux divided by $2\pi$, both component formulas acquire a common $1/(2\pi )$ factor; the subsequent logarithmic derivative relation is unchanged.

The Gauss's law for magnetism constraint is $R^{-1}{\partial }_R(R{B}_R)+{\partial }_z{B}_z=0$. Since $\alpha$ is constant with radius on the surface, it gives the power-law poloidal field near a disk surface relationFor a smooth one-sided field above the surface, $B_z(R,z)=B_z(R,0)[1-(\delta -1)\tan \alpha z/R]+O(z^2)$. Thus the field initially decreases with height if $\delta >1$, increases if $0<\delta <1$, and has zero first height derivative if $\delta =1$.

For local cold magnetocentrifugal acceleration, the field must guide the gas and enforce approximate corotation with its disk footpoint in the sub-Alfvénic launching region. Let that footpoint be $R=R_f$, and keep its field-line angular velocity ${\Omega }_f=\Omega (R_f)$ constant along the field. In this rotating frame the relevant effective potential isThe footpoint is a critical point because ${\partial }_R\Phi (R_f,0)=R_f{\Omega }_f^2$ and ${\partial }_z\Phi (R_f,0)=0$. The meridional Hessian matrix there isTake distance $\ell$ along a smooth outward magnetic field line whose initial tangent is $(\sin \alpha ,\cos \alpha )$ in the $(R,z)$ plane. Since the gradient of $V$ vanishes at the footpoint, field-line curvature does not contribute to its second derivative. ConsequentlyA negative quadratic coefficient gives a downhill displacement and acceleration without thermal assistance. The magnetocentrifugal launching criterion in a flattened power-law potential is thereforeAngles are measured from the vertical, with $0\le \alpha <\pi /2$. Below the critical angle the footpoint is a local minimum along the field, so a cold particle cannot cross the initial potential barrier. At equality the quadratic test is marginal; higher derivatives and the unspecified field curvature can decide the outcome. The threshold is a local condition for launch, not proof that the wind escapes along an arbitrary global field geometry. The spherical point-mass limit $\lambda =1$, $\beta \to 1$ recovers ${\alpha }_{\mathrm{crit}}=30^{\circ }$, consistent with Ogilvie's discussion of cold disk winds, section 9.9.