# Can Hawking radiation be observed from the radiation (spectrum) of M87*?

Using $$M87^*$$ data from EHT observation (mass, temp of the surrounding accreting disk) and approximating area of EH by euclidean geometry $$4\pi r^2)$$, error comes around of order one or two.

One gets following result: $$T_{M87^*}\approx0K$$ and since surrounding temperature is $$6\times10^9K$$ therefore by Boltzmann formulae of blackbody radiation, $$P_{rad}\approx7.3\times10^{31}W/m^2$$, Area of EH $$\approx4.7\times10^{27}$$ therefore luminosity $$L_{haw}\approx 3.4\times10^{66}\mathrm{ergs}/s$$ which is ginormously large compared to X-ray luminosity $$L_{0.5-7\mathrm{keV}}\approx6\times10^{43}$$ which is only for a small range of energy of photon still there is a difference of order $$23$$ so it should make the Hawking radiation detectable.

Two things I am dubious of are

1. The accretion happens in a planar region so the BH is not completely surrounded by temperature gradient of $$10^9K$$.
2. The luminosity of Hawking radiation is calculated by including all the wavelengths of the spectrum whereas the second quantity is for only a selected portion of spectrum.

Edit: By $$T_{M87^*}$$ I meant the Hawking temperature of the black hole which is, neglecting the spin contribution $$\frac{1}{4GM}\approx0$$ and this black hole behaves like a black body of temp $$\frac{1}{4GM}$$ placed in the surrounding of the accreting disk.

• The outside temperature is irrelevant, you have to use the Hawking temperature of the black hole. – Javier Sep 2 at 17:29
• @Javier I don't think outside temperature is irrelevant since it's like a blackbody placed inside a room with the temp of blackbody given by Hawking temp and surrounding temp given by the accreting disk temp. – aitfel Sep 2 at 17:48