Blog #24, Worksheet #7.2, Question #2, Eddington Luminosity

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 \[F = \frac{L}{4 \pi r^2}\]

Then, recall that the momentum of a photon of energy E is simply 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 \((f_{rad})\) is modulated by the Thomson cross section, \(\sigma_t = 6.6524 \, x \, 10^{-25} \, cm^2\) (i.e., the effective area of an electron interacting with a photon): \[f_{rad} = \sigma_t \frac{L}{4 \pi c r^2}\]

(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 \(M_{BH}\), derive the maximum possible luminosity due to accretion. This is called the Eddington Luminosity.

We begin with the fact that we know the momentum flux and the force of the radiation \[Flux_{\rho} = \frac{L}{4 \pi cr^2} \qquad and \qquad f_{rad} = \sigma_t \frac{L}{4 \pi c r^2}\]Then to solve for the maximum luminosity, we set the force of the radiation equal to the force of gravity from the black hole on the protons. When the force of radiation is larger, all accretion is halted and the gas leaves the disk. \[f_{rad} = \sigma_t \frac{L}{4 \pi c r^2} = \frac{G M_{BH} M_{proton}}{r^2}\]And finally we solve for our Eddington Luminosity \[L_{max} = \frac{4 \pi c G M_{BH} M_{proton}}{\sigma_t}\]

Bonus: Express the Eddington Luminosity as a number of Solar luminosities, L , and the black hole mass in solar masses \((\frac{M_{BH}}{M_{\odot}})\), such that \[L_{Edd} = X(\frac{M_{BH}}{M_{\odot}}) \cdot L_{\odot} \]

We need our Eddison Luminosity from part a) to equal the equation on the right side. Thus, we have \[\frac{4 \pi c G M_{BH} M_{proton}}{\sigma_t} = X(\frac{M_{BH}}{M_{\odot}}) \cdot L_{\odot} \]Then we simply simplify and solve for X to give us \[X = \frac{4 \pi c G M_{proton}}{\sigma_t} \cdot \frac {M_{\odot}}{L_{\odot}} \]

(b) If the SMBH in Andromeda were accreting at 20% of its Eddington luminosity, how bright would it be? 

We know the Eddington Luminosity can be found with the equation \[L_{Edd} = \frac {4 \pi c G M_{BH} M{proton}}{\sigma_t}\]From Wikipedia we estimate that the mass of the SMBH at the center of Andromeda is about \(10^8 \, M_{\odot} \). We also know that the mass of a proton is \(1.67 x 10^{-27} \, kg\). 

Plugging in and solving for the Eddison Luminosity, we get: \[L_{Edd} = 1.257 \, x \, 10^{39} \frac{J}{s}\]However, we are given that the SMBH is operating at 20% of its Eddison Luminosity, giving us \[L = 2.515 \, x \, 10^{38} \frac{J}{s}\]

Sources Used

https://en.wikipedia.org/wiki/Supermassive_black_hole