Blog Post 25, Worksheet 7.2, Problem 2

2. There is actually a hard upper limit to the luminosity of this system - and to the luminosity of any accreting compact object. Consider that the photons being emitted in this scenario will interact with the surrounding material (which has yet to accrete onto the black hole). These photons will undergo Thomson scattering off of electrons in this material. In detail, the electric field of the incident lightwave (i.e., the photon) will accelerate an electron, causing it to then re-emit radiation. The photons are therefore able to transfer some of their momentum to the infalling gas. 

The energy flux of these photons at a distance r from the black hole is \[F = \frac{L}{4πr^2}.\] Then, recall that the momentum of a photon of energy E is simple p = E/c Therefore, the momentum flux at r from the black hole is \(\frac{L}{4πcr^2}\). 

Finally, the rate of momentum transfer to the surrounding electrons (or the force due to photons (\(f_{\text{rad}}\))) is modulated by the Thomson cross section, \(\sigma_t\) = 6.6524 * \(10^{-25} \text{ cm}^2\) (i.e., the effective area of an electron interacting with a photon): \[f_{\text{rad}} = \sigma_t \frac{L}{4πcr^2}\]

(a) When this force from radiation pressure exceeds the force of gravity, accretion is halted and all the gas is blown away. For a black hole mass \(M_{\text{BH}}\), derive the maximum possible luminosity due to accretion. This is called the Eddington Luminosity. 

We know that the Eddington Luminosity will occur at the point where the force of gravity is equal to the force from radiation pressure, or when: \[\frac{GM_{\text{BH}}M_{\text{proton}}}{r^2} = \sigma_t \frac{L}{4πcr^2}\] If we solve for L, we get: \[L = L_{\text{Edd}} = \frac{4πcGM_{\text{BH}}M_{\text{proton}}}{\sigma_t}\]
(b) If the SMBH in Andromeda were accreting at 20% of its Eddington luminosity, how bright would it be? How does this value compare with the \(L_{\text{disk}}\) you calculated in 1(e)?

The Eddington luminosity for Andromeda's SMBH is given by \[L_{\text{Edd}} = \frac{4πcGM_{\text{BH}}M_{\text{proton}}}{\sigma_t}\] We know that \(M_\text{BH} = 2 * 10^8 M_\odot\), and \(M_{\text{proton}} = 1.7 * 10^{-24} g\). We can plug in these values to find 20% of the Eddington luminosity: \[ 0.2L_{\text{Edd}} = 0.2(\frac{4π(3 * 10^{10} \text{ cm/s})(G)(2 * 10^8 M_\odot)(2 * 10^{33} \text{g}/M_\odot)(1.7 * 10^{-24} g)}{6.6524 * 10^{-25} \text{ cm}^2})\] This gives us that the luminosity of the SMBH is \(5.2 * 10^{45}\) erg/s, which is almost exactly what we found \(L_\text{disk}\) to be in problem 1(e).