Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Density of supermassive black holes.

  1. Mar 21, 2011 #1
    Wikipedia gives this description of supermassive black holes to distinguish then from stellar black holes:

    "The average density of a supermassive black hole (defined as the mass of the black hole divided by the volume within its Schwarzschild radius) can be much less than the density of water for very large mass black holes (the densities are similar for 108 solar mass black holes). This is because the Schwarzschild radius is directly proportional to mass, while density is inversely proportional to the volume. Since the volume of a spherical object (such as the event horizon of a non-rotating black hole) is directly proportional to the cube of the radius, average density decreases for larger black holes, being inversely proportional to the square of the mass."

    I get how the density determined by the Schwarzschild radius should be inversely proportional to the mass, but it still seems so counter-intuitive to me, with so much weight bearing down upon the center of the black hole. Could someone give me an explanation as to which forces counteract the huge gravitational forces driving the volume of the black hole towards a point, and allowing it to have a density less than that of water?

  2. jcsd
  3. Mar 22, 2011 #2


    User Avatar
    Science Advisor
    Gold Member

    The quote is talking about the AVERAGE density, which is simply the mass of the black hole divided by the volume, as determined by the Schwarzschild radius. It does not address the question of how the mass inside the black hole is distributed.
  4. Mar 22, 2011 #3

    Ken G

    User Avatar
    Gold Member

    And indeed I think you have a legitimate objection, because it's not clear that "average density" means much of anything for a black hole. It doesn't describe the gravitational field, for that treats the mass as being concentrated at the central point, and it's not even clear that "volume" has much meaning for an event horizon (the physically relevant parameter is usually surface area, and one cannot assume that connects to volume in the usual way in such a highly curved spacetime). I don't think "average density" has much physical meaning at all, but I think all the Wiki is trying to bring out is that you can't say a black hole is a black hole because it's mass is large, as any mass will be a black hole if small enough, and you can't say a black hole is a black hole because of its density, as any average density can be one if the mass is large enough. It's actually the ratio M/R, not M by itself or M/R3, that meets with fundamental limitations.
  5. Mar 23, 2011 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The density of the black hole is either zero or infinity at every point in the region they're talking about. So they average zero and infinity and get some finite value. No sweat.
  6. Mar 23, 2011 #5
    Density is defined as:

    [tex]\rho =\frac{m}{V}[/tex]

    For a sphere:

    [tex]V = \frac{4}{3}\pi r_{s}^3[/tex]

    For a black hole,


    with volume

    [tex]V=\frac{4}{3}\pi \left ( \frac{2Gm}{c^2} \right )^3= \frac{32\pi G^3m^3} {3c^6}[/tex]

    Density becomes:

    [tex]\rho =\frac{m}{V}=m\frac{1}{V}=m\frac{3c^6}{32\pi G^3m^3}=\frac{3c^6}{32 \pi G^3m^2}=\left ( \frac{3c^6}{32\pi G^3} \right ) \frac{1}{m^2}[/tex]

    Density goes down as a function of

    [tex] \frac{1}{m^2}[/tex]
  7. Mar 27, 2011 #6
    Density’s a poor way of characterizing a black hole. The event horizon, used in calculating volume, is a mathematical surface. It describes points where escape velocity exceeded the speed of light.

    The no-hair theorem


    postulates that black holes can only be described by three parameters:

    Spin (rotatition)
    Last edited: Mar 27, 2011
  8. Mar 27, 2011 #7


    User Avatar
    Science Advisor
    Gold Member
    2015 Award

    This is meaningless. So what if the average mass density inside the event horizon of a super massive black hole is about the same as water? That dense thing in the center does not care.
  9. Mar 27, 2011 #8

    A distinguishing difference would be their mass spectrum and space-time coordinates inside a galaxy. Supermassive black holes and stellar mass black holes would exist in different areas of the mass spectrum and supermassive black holes can be located inside galactic nuclei and stellar mass black holes can be located inside galactic spiral arms.
    Last edited: Mar 27, 2011
  10. Dec 2, 2012 #9
    I have similar musings. In lay terms, how can a body that has an average mass or possibly a surface density of 1, have a strong enough gravitational field to have an event horizon where the escape velocity is >c? I'm not sure if it would help, but it would be useful to know the density distribution within these supermassive black holes. It's just very hard for me to understand how the gravitational/tidal forces can be so minimal at the event horizon and yet still retard the escape of light. Can someone please explain this. You may write directly to me if you wish at jbsb250502@yahoo.com.

    JBB, M.D.
  11. Dec 8, 2012 #10
    I think the distance that the gravitational pull of the black hole, to the point where light has enough velocity to escape is not really a good measure of the size or density of the black hole.

    it says nothing about the size or density of the centre, I believe Jupiter is low enough average density to be able to float on water, but that says nothing about the density of its core!

    but interesting subject.
  12. Dec 8, 2012 #11
    i see you're using "gravitational force" and "tidal force" interchangeably, and that is the basis of your confusion. while gravity is most definitely a force, a "tidal force" is not an actual force...rather it is the gravitational force gradient that is measured over a distance. in other words, it is the difference in the gravitational force exerted by a body at a distance X and the gravitational force exerted by that same body at a distance Y. w/ a stellar mass BH (and a relatively small Schwartzchild radius), the difference in gravitational forces just above the event horizon and just below it can be incredibly significant, so much so that it would stretch out even the densest and structurally soundest of objects, like a solid lead or steel ball or instance. a supermassive BH on the other hand has such a large radius that the difference in the BH's gravitational force just outside the event horizon and just inside the event horizon is arbitrary. note that while tidal forces in the neighborhood of the event horizon of a supermassive BH may be arbitrary (or minimal as you put it), the BH's gravitational field at the event horizon is still intense enough to prevent light from escaping. i hope that makes things a bit more clear for you.
  13. Dec 8, 2012 #12


    User Avatar
    Science Advisor

    The thing is, it's the only way to characterize the size of a black hole! And of course quoting the average density doesn't give the full picture. What does? The full metric! And of course the metric tells you that there is no matter anywhere in the schwarzschild spacetime! Counter intuitive, perhaps, but that's actually the full story.
  14. Dec 8, 2012 #13


    User Avatar
    Science Advisor

    This would be the Newtonian intuition, but in GR it's actually quite opposite. There exists no 'gravitational force', and only tidal forces exist. Locally inertial coordinates will always feel no net 'force' from gravity (the equivalence principle), but since curvature causes tidal forces, these can't be gauged away in a similar manner.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?