Blog Post 24, Worksheet 7.2, Problem 1

1. Recall that on Worksheet 6.1, you found that many galaxies host a central supermassive black hole and derived the mass of the super-massive black hole (SMBH) in Andromeda (M31). In galaxies that contain large amounts of gas rotating along with the stars in their disks, some of the gas will end up surrounding the black hole. As the galls falls inward, centrifugal forces will cause it to settle into a much smaller, denser disk which rotates around the central black hole, known as an "accretion disk."

The orbital period of the gas in the accretion disk changes with radius (just like the orbital periods of the planets). This means that adjacent packets of gas in the disk will rub against each other, generating frictional heat which gets released as radiation. The energy loss results in the material moving closer to the black hole, eventually falling onto it (hence, accretion).

Let's figure out how hot and bright this material ends up being around a black hole of mass \(M_{BH}\).

(a) First, imagine a gas packet of mass dM in the accretion disk. The gas packet falls from a radius r + dr to a radius r during the accretion process, losing potential energy. With the help of the Virial Theorem, write down the thermal energy the packet gains as a result (dE). Then, simply divide this dE by dt to express the luminosity of the packet, dL, in terms of of \(M_{BH}\), r, and the mass accretion rate dM/dt = \(\dot{M}\).

We know that potential energy of this system comes from gravitational potential energy. Thus, the initial potential energy between the mass dM and the total mass M is given by \[U_i = \frac{GM_{\text{BH}}dM}{r+dr}\] and the final potential energy is given by \[U_f = \frac{GM_{\text{BH}}dM}{r}\] The change in this potential energy is thus just given by: \[\frac{d}{dr}(U) = -\frac{GM_{\text{BH}}dM}{r^2}\] From the Virial Theorem, we know that \(K = -\frac{1}{2}U\), so: \[K = --\frac{1}{2}U = \frac{GM_{\text{BH}}dM}{2r^2} = dE\] Now, we divide each side by dt to get: \[\frac{dE}{dt} = dL = \frac{GM_{\text{BH}}dM}{2r^2dt} = \frac{\dot{M}GM_{\text{BH}}}{2r^2}\]
(b) Now let's assume the disk gas radiates like a blackbody at the same radius where the potential energy is released. Using the Stefan-Boltzmann law, write down an expression for the luminosity from a given annulus in the disk (between r and r + dr) in terms of the temperature in that annulus, T(r). 

The Stefan-Boltzmann law says that the luminosity of a blackbody is given by \(L = A\sigma T^4\), where A is the area of the object. Because we are looking at an annulus ring, we know that the area of the annulus is given by \(A = π((r+dr)^2-r^2)\). However, because we will be integrating over the annulus, each small ring of the annulus at some radius r will have an area of 2πr. Thus, the luminosity from a given annulus in the disk is given by: \[L = 2π\sigma T(r)^4 r\]
(c) Now, set dL from (a) equal to the expression you just derived in (b). Solve for T(r), and express in terms of the mass accretion rate, \(\dot{M}\). 
\[dL = 2π\sigma T(r)^4 r\] Now solve for T(r): \[T(r) = (\frac{dL}{2π\sigma r})^{1/4}\] From part a, we know that \[dL = \frac{\dot{M}GM_{\text{BH}}}{2r^2},\] which we can plug in to get: \[T(r) = (\frac{ \frac{\dot{M}GM_{\text{BH}}}{2r^2}}{2π\sigma r})^{1/4}\] \[T(r) = (\frac{GM_{\text{BH}}\dot{M}}{4π\sigma r^3})^{1/4}\]
(d) Finally, integrate your expression for dL over dr to find the total disk luminosity, \(L_{\text{disk}}\). Let's say here that the disk extends from inner radius \(r_{\text{in}}\) to outer radius \(r_{\text{out}}\). If \(r_{\text{out}}\) >> \(r_{\text{in}}\), what is \(L_{\text{disk}}\)?

From before, we know that \(dL = \frac{\dot{M}GM_{\text{BH}}}{2r^2}\). Now we integrate:  \[\int dL = \int_{r_{\text{out}}}^{r_{\text{in}}} \frac{\dot{M}GM_{\text{BH}}}{2r^2}dr\] \[L = \frac{GM_{\text{BH}}\dot{M}}{2} \int_{r_{\text{out}}}^{r_{\text{in}}} r^{-2}dr\] \[L = \frac{GM_{\text{BH}}\dot{M}}{2} [-r^{-1}]|_{r_{\text{out}}}^{r_{\text{in}}}\] \[L = \frac{GM_{\text{BH}}\dot{M}}{2r_{\text{in}}}\]
(e) Astronomers find it useful to express the \(L_{\text{disk}}\) you found in 1d) as a fraction, \(\eta\), of the total luminosity that would be released if the entire rest mass of the disk were converted to energy, \(\dot{M}c^2\): \[L_{\text{disk}} = \eta \dot{M} c^2\] Show that \(\eta\) is given by \[\eta = \frac{1}{2} \frac{GM_{\text{BH}}}{c^2r_{\text{in}}}\] We think of this as the radiative efficiency of the disk. And this efficiency apparently has a strong dependence on the inner disk radius (\(r_{\text{in}}\)). 

If \(L_{\text{disk}} = \eta\dot{M}c^2\), then \[\eta = \frac{L_{\text{disk}}}{\dot{M}c^2} = \frac{\frac{GM_{\text{BH}}\dot{M}}{2r_{\text{in}}}}{\dot{M}c^2} = \frac{1}{2}\frac{GM_{\text{BH}}}{c^2r_{\text{in}}}\]
(f) Accretion disks around massive black holes tend to have low efficiencies of \(\eta \approx 0.1\), and can accrete up to \(\dot{M} = 1 \:M_\odot \text{ yr}^{-1}\). What is the resulting disk luminosity? How hot would an accretion disk around the SMBH in Andromeda be?

We are given that \(\eta\)  = 0.1, and \(\dot{M} = 1 M_\odot \text{ yr}^{-1}\) . We plug this into the formula \(L_{\text{disk}} = \eta\dot{M}c^2\) to get:
\[L_{\text{disk}} = (0.1)(\frac{1 M_\odot}{1 yr})(\frac{1 yr}{3.2*10^7 \text{ s}})(\frac{2*10^{33} \text{ g}}{1 M_\odot})(3*10^10 \text{ cm/s})^2\] \[L_{\text{disk}} = 5.7 * 10^{45} \text{ erg/s}\]

From the Stefan-Boltzmann Law, we know that \(L = A \sigma T^4\), and therefore \(T = (\frac{L}{A\sigma})^{1/4}\). We know that the luminosity of the disk is given by \(5.7 * 10^{45} \text{ erg/s}\), and the radius of the black hole is given by \(R = \frac{2GM_{\text{BH}}}{c^2}\).  We therefore have \[T = (\frac{L_{\text{disk}}}{\frac{4}{3}π(\frac{2GM_{\text{BH}}}{c^2})^3\sigma})^{1/4}\] \[T = (\frac{5.7 * 10^45 \text{ erg/s}}{\frac{4}{3}π(\frac{2G(2 * 10^8 M_\odot)(2 * 10^{33} \frac{g}{M_\odot}}{(3*10^{10})^2})^3(5.7*10^{-5} \text{erg cm}^{-2}\text{ K}^{-4}\text{ s}^{-1})})^{1/4}\] \[T = 100 \:K\]