According to NASA

a black hole is anything but empty space. Rather, it is a great amount of matter packed into a very small area

According to the documentary Space Unraveling The Cosmos about Black Holes

Gravity here is so strong even light cannot escape

Which leads me to believe that Black Holes basically are masses so compact but great which gives them very strong gravitational influence (Newton's Gravitational Law) that light is not allowed to go out of, hence a "Black" hole.

If my above understanding of Black Holes is correct, then why is those Spiral Galaxies pictured using Hubble Telescope is showing a big shiny ball instead of a "Black" hole ? What is the obvious clue that I am missing here ?

Here is the image of Andromeda Galaxy image taken using Hubble.

Andromeda Image taken using HST

  • 8
    $\begingroup$ At a very basic level of explanation, despite the holes being "prefectly black", there are many stars orbiting them. Centers of spiral galaxies tend to have relatively high concentrations of massive, young starts, which is why they are usually the brightest regions in such galaxies. $\endgroup$ Commented Jul 25, 2014 at 19:00
  • $\begingroup$ Related: physics.stackexchange.com/q/73705 $\endgroup$ Commented Jul 25, 2014 at 19:01
  • 1
    $\begingroup$ Check chapter 7 Black Holes Ain't So Black of the book A Brief History of Time: From the Big Bang to Black Holes. $\endgroup$
    – user5402
    Commented Jul 25, 2014 at 20:48

1 Answer 1


A typical giant galaxy, such as the one you've provided a picture of, has a radius of something like $10\;\rm kpc$ (kiloparsec - $1\;\rm pc \approx 3.2\;ly$).

A supermassive black hole hosted in such a galaxy has a mass of something like $10^6-10^9\;\rm M_\odot$ (solar mass, $1\;\rm M_\odot \approx 2\times10^{30}\; kg$). The monstrous billion solar mass black holes are really only found in particularly large ellipticals; the galaxy in your photo probably hosts one of about one to a few million solar masses. The horizon radius of such a black hole will be on the order of the Schwarzschild radius, so:

$$r_s=\frac{2GM}{c^2}\approx10^{-10}\rm\; kpc$$

So the supermassive black hole is something like 100 billion times smaller in radius than the galaxy, way way WAY smaller than a pixel in a picture like the one you show.

Furthermore, there are a lot of stars in the central region of a galaxy and many will be close (or roughly in front of) the black hole, not to mention clouds of intragalactic gas that may obscure the view to the black hole.

That said, it is becoming possible using very long baseline interferometry to take "pictures" of a couple of nearby black holes. I don't think there are any successful images yet, but we'll probably get some in the next 3 years or so using the Event Horizon Telescope. A prediction of what will be seen:

enter image description here

The formation of the image is quite complicated (the paper I link later gives a lot of the gory detail if you're interested). First, note that this is in "false colour", the colour indicates the intensity of the radiation from blue (low) to white (high). The photons come from a disk hot gas ("accretion disk") that is expected to be found near many black holes. Those in the picture are those which happen to approach the black hole, but do not enter it. Because of the curvature of spacetime, photons can orbit the hole and accumulate in these "photon orbits". The orbits occur a few Schwarzschild radii from the hole. The orbits aren't stable, so some photons eventually plunge into the hole, while others escape away - these are the ones in the picture. The strong asymmetry in the image (while you'd expect a BH to be very symmetric) is due to the fact that the source of the light (the accretion disk) is not spherically symmetric, and only approximately axially symmetric - it may be warped, have bright and dim spots, etc. One side of the image is brighter because typically one side will be relativistically beamed toward us while the other will be beamed away. This is as close to a black hole "looking black" as we're likely to get. There are photons orbiting across the "face" of the hole in the picture, but none make it to us from that direction, so the hole appears black in the image.

One paper I particularly enjoyed reading about the more theoretical aspects of these black hole images: Testing the no-hair theorem with event horizon telescope observations of Sagittarius A*. It includes more simulated images at resolutions more like what we'll realistically achieve with the EHT.

  • $\begingroup$ How does this picture change, if at all, when the black hole is surrounded by other matter? As material accelerates rapidly toward the event horizon, wouldn't radiation be generated by the collisions among similarly ill-fated particles? Could that radiation reach us, and would it occlude the "black" part in your predictive image? $\endgroup$ Commented Jul 25, 2014 at 23:16
  • 2
    $\begingroup$ @AaronNovstrup The (inner edge of the) accretion disk itself is necessarily at larger radii than the photon orbit, but that doesn't prevent the disk from lying across the hole in projection, and this could indeed obstruct the view, much like intervening stars or gas clouds. However, I'm not sure on the expected emission wavelengths for each bit and the opacities - you might be able to see the ring emission through (certain parts of?) the accretion disk while filtering out the disk itself. The emission and scattering in the vicinity of a black hole is really quite a mess. $\endgroup$
    – Kyle Oman
    Commented Jul 25, 2014 at 23:24
  • 2
    $\begingroup$ Note that google says to me that 10^-10 kpc is about 2M miles or 3M kilometers. There are cars that have driven that distance. 10^-10 kpc is about 10 light seconds (compare 100,000 light years for a diameter of a galaxy). The scale of 'kpc' can sometimes confuse or mislead a non-astronomer who has difficulty putting it into other terms. $\endgroup$
    – user20936
    Commented Jul 26, 2014 at 1:18
  • $\begingroup$ Congratulations on your 10k ;-)! $\endgroup$ Commented Apr 1, 2016 at 0:16
  • 2
    $\begingroup$ Your image was correct. :) $\endgroup$
    – fahadash
    Commented Apr 10, 2019 at 21:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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