Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 68 / 2

Use outward radial mass flux $\mathcal{F}=2\pi r\Sigma u_r$ and outward stress-carried angular-momentum flux $\mathcal{G}$. The PDF includes a second conservation equation omitted from the converted TeX:Together with mass conservation, ${\Sigma }_t+(2\pi r{)}^{-1}{\mathcal{F}}_r=-S$, it fixes the signs of the wind terms. Subtract h times the mass equation, noting that the prescribed $h(r)$ is independent of time. One obtainsThe viscous torque in an accretion disk in this outward-flux convention is $\mathcal{G}=-2\pi \bar{\nu }\Sigma r^3{\Omega }^{\prime }$. Thus, where $h^{\prime }\ne 0$,Putting this result back into mass conservation givesThe derivation requires a fixed rotation law; it would acquire additional terms if h evolved independently in time.

For Keplerian rotation, $h=\sqrt{GMr}$, $h^{\prime }=h/(2r)$, and $r^3{\Omega }^{\prime }=-(3/2)h$. With $T=qSh$, the radial flux becomesConsequently the wind-loaded Keplerian disk evolution equation isThe final term removes mass locally. The extra flux term describes the disc's radial response when the outflow removes more or less than the local specific angular momentum. In particular, $q=1$ cancels this additional transport term without canceling the local mass sink.

Set $S=T=0$ and choose $\boxed{\bar{\nu }(r)={\nu }_{0}r}$, with ${\nu }_0>0$ constant. Define $y=\sqrt{r}$ and $g=r^{3/2}\Sigma$. Then $\sqrt{r}\bar{\nu }\Sigma ={\nu }_{0}g$, and ${\partial }_r=(2y{)}^{-1}{\partial }_y$. Directly transforming the Keplerian viscous diffusion equation givesThe heat-equation transform for linear-radius disk viscosity is completed by $x=2y/\sqrt{3{\nu }_0}$:Equivalently one can use $g=\sqrt{r}\bar{\nu }\Sigma$, which differs only by a constant. This is the viscosity law making the transformed diffusivity constant under this square-root-radius reduction. Initial and boundary conditions transform too; the change of variables does not supply unspecified disc-edge data.

The physical diffusion arises from angular momentum transport. Differential rotation creates a viscous shear stress, which transfers angular momentum between neighboring annuli. Gas that loses angular momentum moves inward, while gas that gains it can move outward. Combining conservation of angular momentum with mass conservation turns the stress gradient into a radial mass flux and its divergence into a second radial derivative. Positive kinematic viscosity smooths the appropriately weighted profile; matter need not all move in one direction, and an accreting disc can simultaneously spread outward. Viscosity dissipates orbital energy but redistributes angular momentum internally except for boundary losses.

Put $\beta =2(q-1)>0$. With zero viscosity and $S=f(r)\Sigma$, the wind-loaded Keplerian disk evolution equation becomesOn an interval with $f>0$, introduce the inward travel-time coordinate and weighted densityPrimes on $x^{\prime }$ and $g^{\prime }$ denote the new variables, not differentiation. To verify the transformation, let $W=f{r}^{2-1/\beta }$ and $w=\beta rf$. Its logarithmic derivative satisfiesSince ${\partial }_{x^{\prime }}=w{\partial }_r$, multiplication of the density equation by W gives exactlyThis establishes inward transport by an angular-momentum-enhanced disk wind, including the amplitude weighting rather than only a characteristic speed.

The general local advection equation solution is $g^{\prime }=F(x^{\prime }+t)$; an initial profile fixes F on the interval swept out by its characteristics. A constant wave phase therefore obeys $d{x}^{\prime }/dt=-1$, orThus the transformed wave propagates inward. The physical surface density is not a rigid profile in r: its weighting accounts for radial compression and loss of mass into the outflow. The same velocity follows independently from the mass flux in part (a), $u_r=\mathcal{F}/(2\pi r\Sigma )=-\beta rf$. If f vanishes, the coordinate change applies only on the neighboring positive-loss intervals; at a loss-free point this wind-driven velocity vanishes.

A magnetocentrifugal acceleration mechanism can produce $q>1$. A poloidal field anchored to the disc transmits a torque to the outflow and enforces approximate corotation below its Alfvén surface. The wind's total specific angular momentum is ${\Omega }_0{r}_A^2$, where $r_A$ is its cylindrical Alfvén radius, compared with the footpoint value $h_0={\Omega }_0{r}_0^2$. Hence $q=(r_A/r_0{)}^2>1$ for a lever arm beyond the footpoint. Magnetic stresses remove the surplus angular momentum and drive accretion without an effective internal viscosity.