By PyRsquared


2019-04-11 08:58:33 8 Comments

The latest photos of the M87 black hole capture light from around a black hole at the center of the Messier 87 galaxy, which is 16.4 Mpc ($5.06 \times 10^{20}$km) from our milky way.

Why couldn't / didn't the scientists involved take photos of black holes less distant, for example those at the center of our Milky Way or Andromeda (0.77 Mpc) or Triangulum Galaxy? Wouldn't these black holes appear larger and the photos have greater detail / resolution and be easier to capture?

My intuition would be that maybe black holes at the center of closer galaxies aren't as large, or maybe they have more matter in the way / aren't directly aligned with our view from earth making it harder to capture them, but I don't know for sure.

2 comments

@Ister 2019-04-11 21:15:49

Since this isn't covered by Rob Jeffries' answer, let me add that Sagittarius A* (the black hole in the centre of Milky Way) was considered, but as explained by Heino Falcke at press conference revealing the photo (quoted after Deccan Herald)

Sagittarius A Star is 1000 times faster and smaller. Its like a toddler who is moving constantly. In comparison, M87 is much slower, like a big bear,

On the other hand both M87 and Sgr A* were "photographed" (i.e. data required was captured) it was just "easier" task to process data for M87 so we may expect picture of Sgr A* as the next one in some time.

As stated on EHT webpage

We study supermassive black holes Sgr A* and M87 because their apparent sizes are much larger than those of stellar-mass black holes when viewed from the Earth, so they are easier to study.

@Rob Jeffries 2019-04-11 10:20:00

The "size" (Schwarzschild radius) of a black hole is directly proportional to its mass. The figure of merit that has to be considered, in order to resolve any spatial detail, is the angular diameter of the black hole as viewed from Earth. This will scale as $M/d$, where $d$ is the distance.

My understanding is that the black hole in the centre of M87 and Sgr A* at the centre of our Galaxy are the two black holes with the largest value of $M/d$.

Sgr A* : $M/d \sim 4\times 10^{6}/8 = 5\times 10^{5}\ M_{\odot}$/kpc;

M87 : $\ \ $ $M/d \sim 6\times 10^{9}/16\times 10^3 = 3.8\times 10^5\ M_{\odot}$/kpc.

Your suggested alternatives.

Andromeda : $M/d \sim 2\times 10^8/8\times 10^2 = 2.5\times 10^5\ M_{\odot}$/kpc;

Triangulum: doesn't have a known central supermassive black hole?

So Andromeda is not a crazy target. It's angular size is only 2/3 that of M87. However, another issue is how many of the 8 telescopes in the network can view Andromeda at any one time? It's impossible for the South Pole (as was M87), but also not visible for very long from Chile, so there is a reduced baseline coverage.

@PyRsquared 2019-04-11 12:39:28

Nice, I understand now. I wasn't aware of the $M/d$ angular size metric. From your figures, that makes my suggestions an order of magnitude smaller, wow! I didn't know if Triangulum had a supermassive black hole or not, my mistake

@Acccumulation 2019-04-11 15:43:40

"The "size" of a black hole is directly proportional to its mass." If by size you mean volume, that is not true. Nor is it true for solid angle. Only if you're talking about angular diameter is it true.

@J.G. 2019-04-11 21:13:11

Ah, the global coverage issue. As the old saying goes, the Earth may be small, but it blocks your view of half the universe.

@Rob Jeffries 2019-04-11 22:29:47

@J.G. Except for gravitational wave and neutrino astronomers. Lol.

@J.G. 2019-04-11 22:30:35

@RobJeffries Yes, I strictly meant electromagnetic view.

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