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

**heinz** · Jun30-09, 04:54 AM

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?

Hz

**George Jones** · Jun30-09, 09:10 AM

Originally Posted by heinz

In his lectures, Feynman writes on page 42-6

of Volume II

Originally Posted by heinz

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?

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

is the surface area of the sphere, then

, the value of the Schwarzschild r-coordinate at the surface, is defined by $LaTeX Code: A = 4 \\pi R^2$ .

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

, then (Feynman's sign is wrong)

$LaTeX Code: L - R = \\frac{GM}{3c^2}.$

I have had sme fun verifying this.

The metric inside a constant density sphere is given by

$LaTeX Code: 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 $LaTeX Code: A\\left( r\\right)$ 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., $LaTeX Code: 0 = dt = d \\theta = d \\phi$ . Using this in the metric

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

for the proper distance

. Integrating this from centre to surface gives

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

Making the substitution

$LaTeX Code: u = \\sqrt{\\frac{2GM}{c^{2}R^{3}}}r $

in the integral gives

$LaTeX Code: L = \\sqrt{\\frac{c^{2}R^{3}}{2GM}}\\arcsin \\sqrt{\\frac{2GM}{c^{2}R}}. $

Now, if

is larger than the Schwarzschild radius, then the argument of the $LaTeX Code: \\arcsin$ is fairly small, so keep only the first two terms of a power series expansion of the $LaTeX Code: \\arcsin$ term, giving

$LaTeX Code: L = R \\left(1 + \\frac{GM}{3c^{2}R} \\right). $

**heinz** · Jul1-09, 01:03 AM

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?

Hz