Calculating the area of a sphere

  • #1

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?
 

Answers and Replies

  • #2
bcrowell
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It depends on how you define the radius. In Schwarzschild coordinates, the radial coordinate r is simply defined so that the surface area is [itex]4\pi r^2[/itex] (or, equivalently, so that the circumference is [itex]2\pi r[/itex]).
 
  • #3
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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:

[tex]
\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)
[/tex]

The volume between r1 and r2 is :

[tex]
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)
[/tex]

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:
  • #4
bcrowell
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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 [itex]4\pi r^2[/itex] is the area of these surfaces of transitivity."
 
  • #5
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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 [itex]4\pi r^2[/itex] 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.
 
  • #6
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?
 
  • #7
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...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.
 

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