Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 57 / 1

The circular Keplerian orbit has speed $v_K=(GM/r{)}^{1/2}$. Equating it with the heated gas sound speed givesThis is the photoevaporative gravitational radius, where thermal and orbital binding energies have the same order of magnitude. A circular orbit has specific mechanical energy $-GM/(2r)$. Heating adds thermal energy and, in a fluid outflow, available specific enthalpy of order $c^2$. For example, if $c$ is the adiabatic sound speed, an ordinary ideal gas has enthalpy $c^2/(\gamma -1)$; for $\gamma =5/3$ this is $3c^2/2$. At $r\ge r_g$, this more than compensates the circular-orbit binding energy. Equivalently the hot hydrostatic scale height satisfies $H_{\mathrm{hot}}/r\sim c/v_K\gtrsim 1$, so a thin bound surface layer cannot be maintained. With continued irradiation, the gas can expand into a thermal wind: photoevaporation removes disk material.

The condition is a thermal binding scale, rather than an assertion that the sound speed equals the ballistic escape speed, which is $\sqrt{2}{v}_K$. Detailed wind launching can change the numerical critical radius by factors of order unity. Using exactly the supplied numerical estimates, $c=10^6\mathrm{cm}{\mathrm{s}}^{-1}$, and therefore

The wind removes $2\pi rW(r)dr$ of mass per unit time from an annulus; $W$ is already the surface-density loss term in the supplied mass conservation equation, so there is no additional two-face factor. Integrating the photoevaporation profile givesThe convergence at infinity is important: the loss is concentrated near the photoevaporative gravitational radius. Using $r_g=1.4\times 10^{14}\mathrm{cm}$ yields ${\dot{M}}_w\simeq 2.46\times 10^{17}\mathrm{g}{\mathrm{s}}^{-1}$, or about $3.9\times 10^{-9}{M}_{\odot }\mathrm{y}{\mathrm{r}}^{-1}$. The initial disk mass is $2\times 10^{30}\mathrm{g}$, so the wind-only depletion time isThis estimate treats the heated area and wind normalization as fixed and neglects additional removal through stellar accretion. Once the disk shrinks, its wind rate and geometry need not remain constant.

Positive $F_{\mathrm{acc}}$ denotes inward flow. In steady state, mass conservation gives $d{F}_{\mathrm{acc}}/dr=2\pi rW$. In the remote-feeding approximation, take the incoming flux at large radius inside the feeding region to approach ${\dot{M}}_d$. The cumulative loss exterior to $r\ge r_g$ is ${\dot{M}}_w\sqrt{r_g/r}$, so the steady photoevaporating-disk mass flux isIt is constant inside $r_g$, continuous at $r_g$, and increases outwards towards ${\dot{M}}_d$. Its right derivative is positive and decreases as $r^{-3/2}$. The curve has a change of slope at $r_g$ because the wind turns on there. Physically less mass survives to cross successively smaller radii, while the inner disk has no wind sink. The star accretes at ${\dot{M}}_d-{\dot{M}}_w$.

For constant kinematic viscosity $\nu$, set $Y=\nu \Sigma \sqrt{r}$. The supplied viscous evolution of an accretion disk relation becomes $Y^{\prime }=F_{\mathrm{acc}}/(6\pi \sqrt{r})$. The viscous torque in an accretion disk is proportional to $Y$, so the zero-stress condition at the idealized origin sets $Y(0)=0$. ThusFor $r<r_g$ the integral is $2({\dot{M}}_d-{\dot{M}}_w)\sqrt{r}$. For $r\ge r_g$, separating the two intervals givesConsequently the surface density of a steady photoevaporating disk isBoth the density and its first derivative match at $r_g$. Let $x=r/r_g$ and take the special case ${\dot{M}}_d={\dot{M}}_w$. Then $\Sigma =0$ throughout the inner disk, while outsideThe profile leaves zero with zero slope at $x=1$, rising as $(x-1{)}^2/8$ locally, and approaches ${\dot{M}}_w/(3\pi \nu )$ from below at large radius. The requested sketches are:

The analytic profiles describe the region interior to a distant feeding boundary. A finite outer radius and the disk outside the feeding point need additional boundary conditions. For example, if the wind is truncated at $R\gg r_g$ and $F_{\mathrm{acc}}(R)={\dot{M}}_d$, the exact interior formulas above replace ${\dot{M}}_d$ by ${\dot{M}}_d+{\dot{M}}_w\sqrt{r_g/R}$; the total wind inside $R$ is ${\dot{M}}_w(1-\sqrt{r_g/R})$. The displayed profiles are their remote-boundary limit. Extending the prescribed wind to infinity is useful for estimating mass loss but is not a complete global angular-momentum boundary condition for a finite feeding radius.

As the viscous supply decreases towards the wind rate, the inner accretion flux approaches zero. Wind removal near $r_g$ can cut off replenishment of the inner disk, which then drains on its own viscous timescale, leaving a gap or inner hole. This is photoevaporative gap opening. If supply falls below wind loss, the formal steady profile would require negative inner density and is unphysical; time-dependent depletion must replace it. The exposed outer edge may then be removed rapidly. Equality of the two rates indicates the onset of rapid clearing, rather than a negative steady surface density.