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Question about derivation of surface gravity of a black hole

  1. May 26, 2010 #1


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    The derivaton I have seen (for a Schwarschild black hole) is this

    One imagines a mass m attached to a long (well infinitely long) inextensible rope (of negligible mass!). One raises the mass a distance dl.

    A local observer (hovering above the black hole at a coordinate r) will say that the work required is [itex] m g dl [/itex] , where this g is the local acceleration given by (setting G=1)

    [tex] g= \frac{M }{r^2} (1 - \frac{2M}{r})^{-1/2} [/tex] where M is the mass of the black hole.

    On the other hand, they say that the work as measured by a distant observer is
    [tex] m g_{\infty} dl [/tex] which is taken to define the surface gravity [itex] g_{\infty}[/itex].

    Then one says that the two expressions are related by a redshift factor and this directly leads to the answer for [itex] g_{\infty}[/itex], 1/4M.

    What bothers me is that we usually say that the rest mass energy of an object with mass m located at a coordinate r is measured from the observer at infinity to be [itex] m (1-2M/r^{1/2} [/itex] On the other hand, the derivation for the surface gravity made above uses the rest mass m to write down the work done by the remote observer.

    So why is that the correct thing to do?

    Thanks in advance
  2. jcsd
  3. May 26, 2010 #2
    Would it be correct to say there's a typo in this equation and it should read [itex] m (1-2M/r)^{1/2} [/itex]? If so could you provide a source. This implies that if an object falls from rest at infinity towards an object of mass, the kinetic energy it gains on the way will not be registered at infinity once it reaches the surface of the object as [itex] m (1-2M/r)^{1/2} [/itex] will cancel out any energy gained due to the Lorentz factor. On the other hand, relative to the surface of the object, the kinetic energy would be apparent.

    Is there any way you could expand on this?
  4. May 27, 2010 #3
    this equation is in agreement with the one I give for the proper acceleration on a particle at r in the first post of this thread: https://www.physicsforums.com/showthread.php?t=402135

    and this result is in agreemment with coordinate force (as measured at infinity) acting on a particle on the surface at radius r that I gave in the same post which was:

    [tex] F = \frac{GMm}{r^2} [/tex]

    which when the particle is at the event horizon and r = 2GM/c^2 the result is:

    [tex] F = \frac{mc^2}{4} [/tex]

    Note that the mass (m) of the test particle given in the above equation is not the proper rest mass [tex]m_o[/tex] but is related by:

    [tex]m = \frac{m_o}{\sqrt{1-\frac{2GM}{rc^2}}}[/tex]

    When expressed in terms of the proper rest mass of the particle, the surface force is infinite.

    Here is one reference http://www.physics.umd.edu/grt/taj/776b/lectures.pdf See section 2.1.1 that might shed some light, but it does not go into too much detail.
    Last edited: May 27, 2010
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