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What is Feynman's excess radius for a black hole?

  1. Jun 30, 2009 #1
    In his lectures, Feynman writes on page 42-6 that the radius excess due to space-time curvature is GM/3c^2.

    How does this work out for a Schwarzschild black hole? In this case, the predicted radius should be Rp= 2GM/c^2. The observed radius should probably be 0. But the difference is not what Feynman states. What is wrong here?

  2. jcsd
  3. Jun 30, 2009 #2

    George Jones

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    of Volume II
    In typical Feynman fashion, he left out quite a bit. Feynman is not treating black holes, he is considering constant density spheres that are much larger than their Schwarzschild radii.

    If [itex]A[/itex] is the surface area of the sphere, then [itex]R[/itex], the value of the Schwarzschild r-coordinate at the surface, is defined by [itex]A = 4 \pi R^2[/itex].

    Now, dig a small tunnel from the surface to the centre of the sphere, and drop a tape measure down the tunnel. If the tape measure shows that the distance from center to surface is [itex]L[/itex], then (Feynman's sign is wrong)

    [tex]L - R = \frac{GM}{3c^2}.[/tex]

    I have had sme fun verifying this.

    The metric inside a constant density sphere is given by

    ds^{2}=A\left( r\right) dt^{2}-\frac{dr^{2}}{1-\frac{2GM}{c^{2}R^{3}}r^{2}}-r^{2}\left( d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}\right),

    where [itex]A\left( r\right)[/itex] is a (somewhat complicated) known function that doesn't come into play in this situation.

    Proper (tape measure) distance is calculated at "instant in time", and the tunnel is purely radial, so angles are constant, i.e., [itex]0 = dt = d \theta = d \phi[/itex]. Using this in the metric

    dl^2 = \frac{dr^{2}}{1-\frac{2GM}{c^{2}R^{3}}r^{2}}

    for the proper distance [itex]l[/itex]. Integrating this from centre to surface gives

    L = \int_{0}^{R}\frac{dr}{\sqrt{1-\frac{2GM}{c^{2}R^{3}}r^{2}}}.

    Making the substitution

    u = \sqrt{\frac{2GM}{c^{2}R^{3}}}r

    in the integral gives

    L = \sqrt{\frac{c^{2}R^{3}}{2GM}}\arcsin \sqrt{\frac{2GM}{c^{2}R}}.

    Now, if [itex]R[/itex] is larger than the Schwarzschild radius, then the argument of the [itex]\arcsin[/itex] is fairly small, so keep only the first two terms of a power series expansion of the [itex]\arcsin[/itex] term, giving

    L = R \left(1 + \frac{GM}{3c^{2}R} \right).
    Last edited: Jun 30, 2009
  4. Jul 1, 2009 #3
    Thank you! Is there some way that Feynman's argument can be used for black holes?
    Or can it be changed so that it is also valid for black holes?

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