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.

  • 4
    $\begingroup$ 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 Apr 11 at 15:03
  • $\begingroup$ 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 Apr 11 at 23:07
  • 2
    $\begingroup$ Never mind. I think I found the answer in the 5th paper. $\endgroup$ – not2qubit Apr 11 at 23:17

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.