Blowoff Threshold

Assumptions for below: photon momentum is directed purely radially. Full extent of torus is optically thick (all photons are absorbed). Torus moves uniformly.

The initial gravitational binding energy of the gas torus is given by

E_0 = \frac{G M_{gas} M_\star}{a_0}, with a_0 the initial average orbital radius (major radius of the torus) and M_{gas} = \pi R^2 \theta a_0 \rho the mass of gas contained in a torus that extends for an angle \theta around the star with minor radius R.

The area presented to Lyman-alpha radiation is essentially the rectangle of height 2R and width \theta a_0, so that

A_0 = 2 \theta a_0 R.

For some major radius a, the flux incident on the torus is

F = F_0 \left(\frac{a_0}{a}\right)^2

(and area is A = 2 \theta a R).

At this radius, the power provided by the photons is

P_{\gamma} = F A e_{\gamma}. (This part is potentially bothersome - the photons are absorbed as momentum, not energy.)

We can calculate the blowoff timescale by determining the amount of time required for the photons to deposit the binding energy of the torus:

\tau_{blowoff} = \frac{E_0}{P_{\gamma}}

We can also calculate the time required for the torus to be fully replenished:

\tau_{rep} = \frac{M}{\dot{M}}

Taking the ratio of these timescales:

\frac{\tau_{blowoff}}{\tau_{rep}} = \frac{G \dot{M} M_\star (a_0 + \Delta a)}{2 a_0^3 e_\gamma F_0 R \theta}, where \Delta a is the change in major radius we consider sufficient for the material to be "blown off." If \frac{\tau_{blowoff}}{\tau_{rep}} \gg 1, the torus will remain essentially unaffected. If \frac{\tau_{blowoff}}{\tau_{rep}} \ll 1, the torus will be completely blown away. The timescales are equal for a mass loss rate of 3x10¹⁰ g/s at a flux of ~8.5x10¹² phot/cm²/s.

Including Ablation

As a first attempt, I've taken the energy required to unbind a thick parabola (seen in the steady state at the end of ​this movie) compared to the time required to refill it. This needs to be modified still to account for the fact that the material being pushed around the torus pulls new material in behind it, increasing the fill rate compared to the mass loss rate of the planet alone.

The edge of the torus is approximately 4 R_P from the planet, so the area exposed to Lyman-\alpha radiation is

A_P = 2 \pi (a_0 - 4 R_P) [\sqrt{a_1^2 + 4h_1^2} + \frac{a_1}{2 h_1} \mbox{arcsinh}{\frac{2 h_1}{a_1}}],

where the term in square brackets is the arc length of the outside edge of the thick parabola (with a₁ the half-width (approximately from planet to edge in -z direction) and h₁ the height (approximately from planet to edge in -x direction). Additionally, we can calculate the volume of the thick portion:

V_P = 2 \pi (a_0 - 4 R_P) [\frac43 (a_1 h_1 - a_2 h_2)],

where the term in square brackets here is the difference in area between the outer and inner bounding parabolas.

We can calculate the power delivered by the Lyman-\alpha photons as before.

P_{gamma} = F_0 (\frac{a_0}{a_0 - 4 R_P})^2 A_P e_{\gamma}

And the ablation timescale is then

\tau_{abl} = \frac{G M_{\star} \rho V_P}{2(a_0 - 4R_P)}/P_{\gamma} = 6800 s

And the timescale for replenishing (taking into account only the mass loss rate of the planet) is the same:

\tau_{rep,P} = \frac{\rho V_P}{\dot{M}} = 320000 s

Both of these timescales are linearly related to \rho, so an increase in density would not explain the factor of 50 difference (though we don't see an increase in density, in any case). However, material appears to accelerate to the outer edge as it reaches steady state. It seems as though the ablated material is pulling new material in to fill the vacuum left by it, which could set up a steady state (although it seems like this would require extra material from somewhere?).

Lyman-\alpha Line Transfer

This is basically identical to the line transfer for ionizing radiation, except:

\sigma_H = 5.9\times 10^{-14} \sqrt{\frac{10^4 \text{K}}{T}}

and the energy per photon is instead the momentum per photon,

p_\gamma = \frac{h \nu}{c}.

To test, I've put neutral hydrogen of uniform density and temperature 10⁴ K in a uniform gravitational field opposed by radiation pressure. To balance, equate volumetric force:

F_g = n_H m_H g$ $F_\gamma = J \sigma_H n_H p_\gamma$ ,

with g = 980 \text{ cm/s}^2. Clearly the value of n_H doesn't affect the result, and I've chosen a small enough length scale that absorption doesn't matter and we can take \tau = 0 across the grid.

With a flux J = 5.09\times10^{13}, we get a result of (tweaking some decimals - flux in physics.data for this is actually 5.0999972\times10^{13}):

v \approx 0.1 \text{ cm/s} after ~0.1 seconds, when acceleration under only gravity would give v = g t = 980(0.1) = 98 \text{ cm/s}. Radiation pressure and gravity are clearly not perfectly matched, but it also seems clear that radiation pressure is working as it should.

Dynamic tests

Lyman-\alpha flux	5.1d13 phot/cm2/s
lScale	1

Slab

Slab density	1 CD
Ambient density	1d-4 CD
Slab extent	(0,0.3)
Domain	{(0,1),(0,1)}

Clump

Clump density	1 CD
Ambient density	1d-4 CD
Clump extent	Circle around (0,0) with R = 0.5
Domain	{(-1,1),(-1,1)}

Slab with ionization

Slab density	1d4\cdot10^{-20(0.5-y)^2} CD
Ionizing flux	2d8	←- Need to test value on smaller length scale