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The recent data from the EHT Consortium on the size and mass of the central black hole of M87, named M87*, are telling us that the diameter of the event horizon should be ~1.5 light-days, or stated differently, ~42 ± 3 μas (micro-arc-seconds). They also tell us that the mass is ~ $6.6\times10^9 M_{\odot}$ (billion), and that it is located at a distance of about $16.8 ± 0.8$ Mpc.

How do you calculate the diameter using those fundamentals?

Where $R$ = distance from earth, $M$ = mass of M87* and using SI units please.

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    $\begingroup$ Note that it's the emission ring that has the angular diameter of 42±3 μas, not the central black hole. If you calculate the actual diameter of the emission ring from 2*R*tan(α/2) with R=16.8 Mpc and α=42 μas, you get ~1e20 m, much larger than the Swarzchild radius of the central black hole $\endgroup$
    – Mark Beadles
    Commented Apr 11, 2019 at 15:03
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    $\begingroup$ I see. But what exactly is the definition of the emission ring in this case, and how far is it from the (ideal Schwarzchild) event ($r=2M$) or photon ($r=3M$) horizon? $\endgroup$
    – not2qubit
    Commented Apr 11, 2019 at 23:07
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    $\begingroup$ Never mind. I think I found the answer in the 5th paper. $\endgroup$
    – not2qubit
    Commented Apr 11, 2019 at 23:17
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    $\begingroup$ @MarkBeadles your physical diameter is too big by a factor of $10^6$. $\endgroup$
    – ProfRob
    Commented Nov 19, 2020 at 22:51

2 Answers 2

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The predicted radius of the photon ring is $$ r_p = \sqrt{27} \frac{GM_{\rm BH}}{c^2}, $$ for a non-spinning black hole. The result is only slightly different for a fast spinning black hole. By dividing $r_p$ by the distance to the source $D$, we have a relationship between the angular radius and the black hole mass.

This is done in equation 1 of the fifth Event Horizon Telescope paper $$ \theta_p = \frac{r_p}{D} = 18.8 \left(\frac{M_{\rm BH}}{6.2\times 10^9 M_{\odot}}\right) \left(\frac{D}{16.9\ {\rm Mpc}}\right)^{-1}\ {\rm microarcseconds}\ .$$

If the angular radius is measured to be 21 $\mu$arcsec, then this suggests a mass of 6.9 billion solar masses.

The difference between this and the final result of 6.5 billion solar masses is down to a more sophisticated modelling of the image using a radiative transfer model and a spinning black hole.

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If you are talking about the diameter of the event horizon, we can use the Schwarzschild radius to get a quick estimate. The Schwarzschild radius is given by

$$ r_s = \frac{2 G M}{c^2}$$

in SI units and using the given data (we only really need the mass of the black hole which is $M = 6.6 \times 10^9 \times 2 \times 10^{30} \approx 13.2 \times 10^{39} \text{kg}$ where the mass of the sun is approximately $2 \times 10^{30}$ kg) it is straightforward to calculate the Schwarzschild radius in SI units

$$ r_s = \frac{2 \times 6.67 \times 10^{-11} \times 13.2 \times 10^{39} }{\left(3 \times 10^8 \right) ^2} \approx 19.6 \times 10^{12} \text{m} $$

A light-day is approximately $25.9 \times 10^{12} \text{m}$ and hence you get that the diameter is approximately

$$ 19.6 \times 2 \div 25.9 \approx 1.51 \ \text{ light-days } $$

or in SI units if you like $$ 3.92 \times 10^{13} \text{ m } $$

I have taken just a few significant figures since the most crucial data (mass of the blackhole) is accurate to at most 2 significant figures.

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