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tommac
2010-Mar-17, 06:41 PM
How do I calculate the volume of a sphere in curved 3d space? I would assume that i need to use the Ricci Tensor ... would I need to take the integral of the Ricci tensor or something like that?

Regards,
Tom

JohnD
2010-Mar-17, 06:44 PM
(4PiR^3)/3 just like a sphere in flat 3D space.

John

tommac
2010-Mar-17, 06:49 PM
(4PiR^3)/3 just like a sphere in flat 3D space.

John
????

really? but what woudnt R be warped around warped space depending on what direction you took it in?

For example lets say you drew a triangle inside the sphere taking 3 points on the surface of the sphere and connecting the lines on the inside of the sphere. That triangle would probably not have angles summing 180. If that is not true then how could the 4piR^3/3 hold true?

JohnD
2010-Mar-17, 07:02 PM
Triangles, schmiangles! Space is warped, not the volume in it.

John

korjik
2010-Mar-17, 07:39 PM
????

really? but what woudnt R be warped around warped space depending on what direction you took it in?

For example lets say you drew a triangle inside the sphere taking 3 points on the surface of the sphere and connecting the lines on the inside of the sphere. That triangle would probably not have angles summing 180. If that is not true then how could the 4piR^3/3 hold true?

Should be:

Integral(4(pi)r^2)dr

Where r and dr are functions of the curvature of space.

At that point, you need to know the curvature

tommac
2010-Mar-17, 08:26 PM
Triangles, schmiangles! Space is warped, not the volume in it.

John

What is the area of a warped triangle?

tommac
2010-Mar-17, 08:27 PM
Should be:

Integral(4(pi)r^2)dr

Where r and dr are functions of the curvature of space.

At that point, you need to know the curvature

Yes so to get the curvature you can have the ricci tensor (field ??? ) for each point in the sphere or what?

chornedsnorkack
2010-Mar-17, 10:21 PM
What is the area of a circle on a sphere?

Tensor
2010-Mar-18, 12:40 AM
Triangles, schmiangles! Space is warped, not the volume in it.

John

I would suggest the study differential geometry would show you the error of this statement. The Ricci tensor can tell you how much of a change, in the volume of a sphere, the curvature will cause.

Jens
2010-Mar-18, 03:29 AM
How do I calculate the volume of a sphere in curved 3d space? I would assume that i need to use the Ricci Tensor ... would I need to take the integral of the Ricci tensor or something like that?

I also assume it would require the use of pi. Speaking of which, congratulations on your induction.

tommac
2010-Mar-18, 02:27 PM
I also assume it would require the use of pi. Speaking of which, congratulations on your induction.

Thank you.

tommac
2010-Mar-18, 02:32 PM
Is there a ricci tensor for each point within the volume?

Lets take a simple example of a neutron star tightly orbiting a black hole with nothing else in the universe. How would I start to calculate the volume of the neutron star?

chornedsnorkack
2010-Mar-18, 06:28 PM
The curvature of space may be a tensor, but it may also be given by a single scalar.

On a sphere with radius R, if you draw a circle with radius r measured along surface (great circles) of the sphere - if the radius of circle r=(π/2)*R then the circle is great circle, with circumference c=2*π*R=4*r, and area S=2*π*R²=(8/π)*r². If the radius of the circle r=π*R then the circle is the opposite pole, with zero circumference and area S=4*π*R²=(4/π)*r². The expression for the area of any other circle should, I think, be derivable through trigonometric functions.

tommac
2010-Mar-18, 06:42 PM
The curvature of space may be a tensor, but it may also be given by a single scalar.

On a sphere with radius R, if you draw a circle with radius r measured along surface (great circles) of the sphere - if the radius of circle r=(π/2)*R then the circle is great circle, with circumference c=2*π*R=4*r, and area S=2*π*R²=(8/π)*r². If the radius of the circle r=π*R then the circle is the opposite pole, with zero circumference and area S=4*π*R²=(4/π)*r². The expression for the area of any other circle should, I think, be derivable through trigonometric functions.

Lets draw a circle on the surface of a sphere ... The area of that circle depends on the radius as a determinant of its curvature ... The circumference will not equal pi r^2 ( with r being the point from the center of the circle ( on the sphere ) to any point of the line. This r would be longer than the real r which would be the distance from the center point of the circle inside the sphere. So the area of the circle would be greater than pi r ^2. We can see that if we add another dimension, lets even say a non curved space-time, to make a warped cylinder you can quickly see how taking the area of the circle in curved space time and multiplying it by a non curved legnth will give you a volume that is greater than: pi r^2 h.

chornedsnorkack
2010-Mar-18, 09:32 PM
The circumference will not equal pi r^2 ( with r being the point from the center of the circle ( on the sphere ) to any point of the line.

Circumference never contains the square of radius.

The circumference of a circle drawn on a sphere is 2*π*R*sin(r/R).

tommac
2010-Mar-18, 11:42 PM
Circumference never contains the square of radius.

The circumference of a circle drawn on a sphere is 2*π*R*sin(r/R).

Yes ... typo meant 2 pi r ...

chornedsnorkack
2010-Mar-19, 05:01 PM
The volume of 3-sphere V=2*π²*R³.

So the volume of a great sphere, that is the volume of a plane, radius r=(π/2)*R, is V=π²*R³=(8/π)*r³.

tommac
2010-Mar-19, 07:58 PM
The volume of 3-sphere V=2*π²*R³.

So the volume of a great sphere, that is the volume of a plane, radius r=(π/2)*R, is V=π²*R³=(8/π)*r³.

But that does not include curvature within the sphere right?

Celestial Mechanic
2010-Mar-21, 04:05 AM
Volume of a Sphere in Schwarzschild Coordinates

Isn't everyone forgetting something? Like the metric? This is of far greater relevance to tommac's question than the Ricci tensor.

We are going to consider one of the simplest of volume problems in general relativity: calculate the volume contained between two spheres of radius r1 < r2 external to a spherically-symmetric distribution of matter, such as (approximately) the Sun or the Earth, or a non-rotating uncharged black hole.

We have to restrict ourselves to the exterior of the mass distribution because to say anything about the volume of the interior we have to know some detail of the distribution of mass/energy. Outside of a spherically-symmetric distribution of mass/energy the metric reduces to the Schwarzschild metric:

ds2 = (1-2M/r)dt2 - (1-2M/r)-1dr2 - r2dtheta2 - r2sin2(theta)dphi2. (1)

Here we are using units where G = c = 1. In SI units 2M/r would be written 2GM/c2/r.

The element of area for a constant r is:

dA = r*dtheta * r*sin(theta)*dphi = r2sin(theta) dtheta dphi (2),

just as it is in ordinary Euclidean space. Thus the area of a sphere of radius r is 4*pi*r2. In fact, we may consider this as the definition of the Schwarzschild radial coordinate, r. It's not a practical one, however. :D

But now consider the volume element:

dV = (1-2M/r)-1/2dr * r*dtheta * r*sin(theta)*dphi

= (1-2M/r)-1/2r2sin(theta) dr dtheta dphi

= (1-2M/r)-1/2 dr dA (3).

Note the extra factor of (1-2M/r)-1/2. That's what's going to make this problem difficult.

The volume contained between the two radii r1 and r2 is:

V = (4*pi) * Integral(r1, r2) {r2*(1-2M/r)-1/2*dr} (4).

We make a trigonometric change of variables r=(2M)*sec2(psi), dr=2(2M)sec2(psi)tan(psi)dpsi, so that the integral becomes:

V = (8*pi) * (2M)3 Integral(psi1, psi2) {sec7(psi)dpsi} (5).

Doesn't that look better? Three integrations by parts (or a quick lookup in Gradshteyn and Rizhek) we find the antiderivative to be:

(8*pi) * (2M)3 * [(1/6)sec5(psi)tan(psi) + (5/24)sec3(psi)tan(psi) + (5/16)sec(psi)tan(psi) +(5/16)ln|sec(psi)+tan(psi)|] (6).

Going back to r, this becomes:

(4*pi/3)sqrt(r5(r-2M)) + (5*pi/3)(2M)sqrt(r3(r-2M)) + (5*pi/2)(2M)2sqrt(r(r-2M)) + (5*pi/2)*(2M)3ln|r/(2M)+sqrt((r/2M)-1)| (7).

Evaluate (7) at r2 and subtract (7) evaluated at r1. Whew!

Of course if 2M/r is a small enough quantity that we may neglect its square and higher powers the formula simplifies considerably:

(4*pi/3)(r23 - r13) + pi*(2M)*(r22 - r12), 2M << r, (8).

The conclusion is this: in the Schwarzschild metric the volume between two concentric spheres in the exterior centered on the distribution of mass will be greater than the Euclidean value.

Also, note that for a black hole (7) is zero at r=2M, so we also have this theorem:

The volume between the event horizon and a sphere centered on the black hole is equal to (7), or to first order in (2M):

(4*pi/3)*r3 + pi*(2M)*r2, (2M)<<r. (9)

Again, this volume is greater than expected. Of course this is all very academic because it is highly impractical to measure areas and volumes on an astronomical scale. What about ordinary everyday-sized spheres, located somewhere in a gravity field, not concentric with the central mass distribution? I'll see what I can do with this one. I expect that for a sphere of radius r located at a distance of R in a gravity field of mass M the volume should be the Euclidean value times 1 plus or minus some multiple of (2M/R). The question is how much and what sign. Stay tuned!

tommac
2010-Mar-21, 04:39 AM
Volume of a Sphere in Schwarzschild Coordinates

Isn't everyone forgetting something? Like the metric? This is of far greater relevance to tommac's question than the Ricci tensor.

We are going to consider one of the simplest of volume problems in general relativity: calculate the volume contained between two spheres of radius r1 < r2 external to a spherically-symmetric distribution of matter, such as (approximately) the Sun or the Earth, or a non-rotating uncharged black hole.

We have to restrict ourselves to the exterior of the mass distribution because to say anything about the volume of the interior we have to know some detail of the distribution of mass/energy. Outside of a spherically-symmetric distribution of mass/energy the metric reduces to the Schwarzschild metric:

ds2 = (1-2M/r)dt2 - (1-2M/r)-1dr2 - r2dtheta2 - r2sin2(theta)dphi2. (1)

Can we slow down this answer for a layman such as myself ... how did you get this equation from the Schwarzschild metric?

Sorry I am just in the beginning phases of trying to figure out this stuff.

Celestial Mechanic
2010-Mar-22, 05:06 AM
Can we slow down this answer for a layman such as myself ... how did you get this equation from the Schwarzschild metric? {Snip!}
Equation (1) is the Schwarzschild metric. I've spelled out "theta" and "phi" rather than used the Greek letters for them because of the limitations of my plaintext editor.

Edited to add: Thank you for asking the question in the OP. Answering the question gives us all an opportunity to appreciate some of the lesser-known and somewhat unexpected aspects of general relativity.

Ken G
2010-Mar-23, 01:55 PM
Thanks for that effort Celestial Mechanic, it's nice to see the calculation worked out so clearly. Note that another way to frame your asymptotic result is that the Schwarzschild radius is rs = 2M, so you could also put that for r >> rs, you have
V(r) = 4*pi*r2 * (r/3 + rs/4).
That means if we think of the volume of a sphere as its surface area times its "effective radius", then the mass in the center extends the effective radius by rs/4.

mugaliens
2010-Mar-25, 09:00 AM
(4PiR^3)/3 just like a sphere in flat 3D space.

John

No. I believe the operative word in the OP is "curved."

Thanks for that effort Celestial Mechanic, it's nice to see the calculation worked out so clearly. Note that another way to frame your asymptotic result is that the Schwarzschild radius is rs = 2M, so you could also put that for r >> rs, you have
V(r) = 4*pi*r2 * (r/3 + rs/4).
That means if we think of the volume of a sphere as its surface area times its "effective radius", then the mass in the center extends the effective radius by rs/4.

Thank you, Gen G, and CM.