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Does a black hole have a surface area?

  1. Aug 14, 2015 #1
    [Moderator's note: Spin-off from another thread.]

    I'm sorry, does black hole have surface area? Did you mean the sphere defined by Schwarzchild Radius?
     
    Last edited by a moderator: Aug 15, 2015
  2. jcsd
  3. Aug 14, 2015 #2

    PeterDonis

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    Yes. More precisely, the hole's horizon does. See below.

    Yes, but you're looking at it backwards. The "Schwarzschild radius" is really ##\sqrt{A / 4 \pi}##, where ##A## is the area of the horizon (which is a 2-sphere). The area is the geometric quantity; the "radius" is just a coordinate label and doesn't correspond to a physical radius.
     
  4. Aug 15, 2015 #3
    The event horizon.
    Perhaps I should state my question more clearly
    The area defined by ##4\pi {R_{schwarzshild}}^2##?
     
    Last edited: Aug 15, 2015
  5. Aug 15, 2015 #4

    PeterDonis

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    Yes.

    The answer is still the same: you are looking at it backwards. The area is not defined by ##4 \pi R_s^2##. ##R_s## is defined as ##\sqrt{A / 4 \pi}##. The area ##A##, as the geometric quantity, is logically prior to ##R_s##, which is just a coordinate label.
     
  6. Aug 15, 2015 #5
    Okay...
    So the surface area of a black hole is the surface area of a sphere which has the Schwarzschild radius IS ##4 \Pi R_s##
    While Rs itself is defined by https://en.wikipedia.org/wiki/Schwarzschild_radius#Formula
    ##R_s = \frac{2GM}{c^2}##, still Newton, right? Only instead of Ve2, we replace it with the speed of light.
     
  7. Aug 15, 2015 #6

    PeterDonis

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    No. It is the surface area of a sphere which is labeled by the radial coordinate ##R_s##. That number is not the actual geometric radius of that sphere; in fact, that sphere, the horizon, has no geometric radius in the ordinary sense. , because it has no geometric center. A black hole is not just a spherical object sitting in ordinary Euclidean space.

    That formula tells you how to calculate ##R_s##, yes. But it doesn't tell you how ##R_s## is defined, geometrically. It doesn't tell you that ##R_s## is the geometric radius of anything. It just tells you how to calculate a number.

    To see what ##R_s## actually means, you have to look at the metric; you have to look at what determines which 2-sphere is the hole's horizon; and you have to look at what geometric properties that 2-sphere actually has. When you look at all those things, you find that, as I said before, ##R_s## is not the geometric radius of the horizon; it's just a coordinate label. The geometric property that actually characterizes the horizon is its area.

    One way to see this is to find coordinates in which the horizon is labeled by a different coordinate; for example, isotropic coordinates. In these coordinates, the horizon is labeled by the radial coordinate ##R = GM / 2 c^2##, i.e., 1/4 of the value ##R_s## that labels it in Schwarzschild coordinates. But the area of the horizon is the same, because that's an invariant geometric property.

    No. A black hole is certainly not an object to which Newtonian gravity applies.
     
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