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I understand that radiation on planets near the galaxy’s center make life on these planets nearly impossible, and that one cannot truly “see” a black hole. However, if you could stand on the surface of a planet that orbits a star close to or within the galactic core, could you theoretically look up at the sky and see an absence of light/stars indicating the location of the central supermassive black hole?

Would it be too far away to see, obstructed by debris, or too small to notice?

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    I don't think radiation excludes life, even remotely –  Mar 28 '18 at 08:55
  • In the movie Interstellar they posit an idea that an accretion ring around a black hole might be hot enough to give off light. I don't get exactly how that works since a black hole is supposed to suck in even light, but maybe its because the disc is just outside its event horizon? Anyway, that might give you some ideas. Oh, L Dutch gives you that answer below. –  Mar 28 '18 at 20:13

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If the black hole is active, meaning it is still capturing matter from its surrounding, it will have a large accretion disk all around, which is the only way to dissipate angular momentum for the matter falling into it.

As a result of this dissipation all the matter will warm up and emit radiation. This disc will be fairly large and thus clearly visible as a bright object in the sky.

This page shows an image captured by Hubble of such a disk: disk around a black hole

Strictly speaking then, one cannot see directly the black hole, as its view will be covered by the bright emission of the disk.

However, the presence of the disk will allow to observe the black hole thanks to its gravitational effect.

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You couldn't see it as a black patch in the sky, because it's far too small. It's only 17 times the radius of our sun, which of course you can't see as a disc even from the outer reaches of our own solar system. What you could easily see is the much larger area of light and other radiation from matter falling into it.

Mike Scott
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There's a thing called gravitational lensing, which means that light coming from behind the black hole would bend towards it, and since the galactic core has a lot of stars, it might be that instead of a black spot in the sky, you'd see a great accumulation of light in and around the black hole's position.

I'm not sure how literal "lensing" is, so I don't know whether there is a focal point based on gravity and the energy of the light and whether it would matter where the planet is from this focal point.

http://www.cfhtlens.org/public/what-gravitational-lensing (google will provide more links if you search for it)

Real Subtle
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    It works like actual spherical lenses. And you can see it in Nolan's Interstellar. The black hole in the movie was designed and rendered according to the most up to date knowledge on astrophysics, with help from Kipp Thorne himself. – Geeky Guy Mar 28 '18 at 12:48
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    @Renan is that what it was? I was wondering why there was a halo around it. That question is answered, than you. :) – Real Subtle Mar 28 '18 at 13:25
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    @RealSubtle They actually "discovered" that halo thanks to their advanced rendering efforts. Nobody was really expecting it, but when they plugged all the numbers into the simulation and told it to render, there it was, and after checking the math over and over they realized it must actually appear that way in real life. We just don't have a clear enough view of one to observe it directly. – thanby Mar 28 '18 at 16:50
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Rainbow cloud.

A source of radiation from black holes is stuff spiralling into the black hole, heating up as it "fell" and released its gravitational potential energy. Again this is black body radiation but this time the regular kind: the hotter the emitters are is the shorter the wavelength. This radiation comes from next to the black hole, not out of the hole itself.

https://physics.stackexchange.com/questions/24958/how-can-a-black-hole-emit-x-rays

The X-rays come from hot gas orbiting around the black hole in an accretion disk. As the gas orbits, magnetic stresses cause it to lose energy and angular momentum, thus spiralling slowly in towards the black hole. The orbital energy is transformed into thermal energy, heating up the gas to millions of degrees, so it then emits blackbody radiation in the X-ray band.

Once the gas gets closer than a few times the horizon radius, it plunges into the black hole, so while some X-rays can still escape just before the horizon, most are emitted a fair bit outside.

Telescopes to detect black holes look for the most energetic rays, which are emitted from the hottest areas of gas nearest the hole. We cannot stand on a planet and look up and see xrays. But consider: if there is very hot gas, next to it there is less hot gas, and next to that gas that is less less hot. The cooler a blackbody is, the longer wavelength the emitted radiation is. Somewhere in that progressively cooler cloud is gas which emits radiation in the visible wavelength.

I here assert that the gradual change in temperature of this cloud as it is progressively farther from the hottest innermost gas should produce a gradual change in the frequencies emitted. The first visible light would be in the far violet, nearest the hole. This will grade through blues and greens farther away and then to red at the farthest coolest part of the cloud.

This prediction should be true not just for black holes, but for any cloud of gas heated from within. Now let me look... here we go.

rainbow nebula https://www.space.com/12051-bright-nebula-photo-supergiant-star-betelgeuse.html

The black hole rainbow cloud will be more symmetric than this one. The star is spewing this stuff out willy-nilly but the hole is sucking gas in, so it will be a symmetric spiral.

Willk
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  • The question is about supermassive black holes, which emit no detectable Hawking radiation, so none of this is relevant. – Mike Scott Mar 28 '18 at 13:55
  • @Mike Scott: Take a look at the bold heading "You could see high energy xrays / gamma rays from black hole associated gas clouds.". Maybe you missed that the first time through. – Willk Mar 28 '18 at 14:11
  • @Willik The question’s not about that either; it asks if you could see an absence of light and stars. You could edit this to actually answer the question that is asked by removing the first section and turning the second part into a negative answer, pointing out that there will be light and not an absence of light. – Mike Scott Mar 28 '18 at 14:21
  • @Mike Scott: you are right; he was asking specifically about the lack of light and not about detection more generally. Well, I learned something reading up. I will tweak this answer. – Willk Mar 28 '18 at 14:26
  • Also note that your photo of Betelgeuse was taken in the infrared band, so the colours are fake. – PM 2Ring Mar 29 '18 at 08:56
  • @PM 2Ring noooooooo! – Willk Mar 30 '18 at 00:08