What is the luminosity of the observable universe (excluding the blackholes)?
This is Fermi estimation problem: Find the mass of the sum, calculate the mass to energy conversion, and calculate the time needed for a photon to escape the center
The escape time is roughly a billion years, which gives a luminosity of around $10^{37}$
There are roughly $10^{10}$ stars in the universe. Assume all stars are like the sun
There are also roughly $10^{10}$ galaxies in the universe
Combining these gives that the luminosity of the universe is $10^{53}$
The point of this is that gravity is very efficient at converting mass to energy
If the black hole is moving close to the speed of light towards you (pileup of photons), then the observed luminosity can exceed this threshold
Gamma ray bursts (GRBs) are the prime example of this boosted luminosity
Imagine a EM wave propagating along the $\hat{z}$ direction. This will oscillate an electron in a direction transverse to z (fix this direction to be x)
This oscillating generates the radiation
Newton’s 2nd Law, and letting $E(z,t) = exp(i\omega t) \hat{x}$, the resulting motion is $x(t) = A exp(i\omega t)$ where $A = \frac{qE(z)}{m\omega^{2}}$
The cross section scales like $A^{2}$, which means that the cross section of the proton is roughly 1 million times smaller than the electron
Setting the gravitational force (ignoring the electron mass) equal to the radiative force (ignoring the proton cross section) gives the target luminosity (re: Eddington Luminoosity) as $L_{edd} = \frac{4\pi c G M m_{p}}{\sigma_{T}}$
Setting the mass scale to that of the sun, we get that $L_{Edd} = 1.3E38 (\frac{M}{M_{*}}) \frac{ergs}{s}$
For a black hole, $L_{edd} = 1.5E45 M_{7} \frac{erg}{s}$ where $M_{7} = \frac{M}{1E7 M_{*}}$
If we assume that the object is a perfect blackbody, we can define an effective temperature $L = 4\pi R^{2} \sigma_{B} T_{eff}^{4}$