# Calculating the area of a sphere

## Main Question or Discussion Point

How do you calculate the area of a sphere? for example in schwarzschild geometry and considering only the three dimensional spatial metric, what would be the area (and how do u calculate it) of a sphere of radius 2M?

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bcrowell
Staff Emeritus
Gold Member
It depends on how you define the radius. In Schwarzschild coordinates, the radial coordinate r is simply defined so that the surface area is $4\pi r^2$ (or, equivalently, so that the circumference is $2\pi r$).

The area is 4 pi M, approximately 12.56637062*M

For completeness sake:

The radial distance for an stationary observer between r1 and r2 is:

$$\sqrt {r_{{2}} \left( r_{{2}}-2\,M \right) }-\sqrt {r_{{1}} \left( r_{ {1}}-2\,M \right) }+2\,M\ln \left( {\frac {\sqrt {r_{{2}}}+\sqrt {r_{ {2}}-2\,M}}{\sqrt {r_{{1}}}+\sqrt {r_{{1}}-2\,M}}} \right)$$

The volume between r1 and r2 is :

$$4/3\,\pi \,\sqrt {{r_{{2}}}^{5} \left( r_{{2}}-2\,M \right) }+10/3\, \pi \,M\sqrt {{r_{{2}}}^{3} \left( r_{{2}}-2\,M \right) }+10\,\pi \,{M }^{2}\sqrt {r_{{2}} \left( r_{{2}}-2\,M \right) }+20\,\pi \,{M}^{3} \ln \left( 1/2\,{\frac {r_{{2}}}{M}}+1/2\,\sqrt {2\,{\frac {r_{{2}}}{ M}}-4} \right) -4/3\,\pi \,\sqrt {{r_{{1}}}^{5} \left( r_{{1}}-2\,M \right) }-10/3\,\pi \,M\sqrt {{r_{{1}}}^{3} \left( r_{{1}}-2\,M \right) }-10\,\pi \,{M}^{2}\sqrt {r_{{1}} \left( r_{{1}}-2\,M \right) }-20\,\pi \,{M}^{3}\ln \left( 1/2\,{\frac {r_{{1}}}{M}}+1/2 \,\sqrt {2\,{\frac {r_{{1}}}{M}}-4} \right)$$

The Latex is cut off, but if you click on it you get the complete formula. I do not know how to use line breaks as the standard \\ does not seem to work, perhaps a moderator could help.

These formulas work up to and including the EH. There are other formulas that work passed the EH but only up to and not including r=0.

Last edited:
bcrowell
Staff Emeritus
Gold Member
Passionflower, you're assuming what you want to prove, because the r coordinate you're using was defined so that the result was true.

See, for example, Hawking and Ellis, p. 149: "The coordinate r in this metric form is intrinsically defined by the requirement that $4\pi r^2$ is the area of these surfaces of transitivity."

Passionflower, you're assuming what you want to prove, because the r coordinate you're using was defined so that the result was true.

See, for example, Hawking and Ellis, p. 149: "The coordinate r in this metric form is intrinsically defined by the requirement that $4\pi r^2$ is the area of these surfaces of transitivity."
I don't want to prove anything I simply gave the poster the formulas to calculate area, distance and volume using Schwarzschild coordinates.

If there is something wrong with the formulas please tell if not I do not know what your problem is.

So is it 4 pi M or 16 pi M. Because if the area is 4*pi*r^2 and r = 2M it should be 16*pi*M right?

...the area is 4*pi*r^2 and r = 2M it should be 16*pi*M right?
Right!

Or if you want to be complete:

A = 16 pi G2 m2/c4.​

You can go directly from the mass to the area, there is no need to actually consider the radius.