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If we assume space is flat the volume of Earth is:

$$ V = \frac{4 \pi R^3}{3} = \frac{4 \pi (6378.1 km)^3}{3} = 1.086 \times 10^{21} m^3 $$

The Einstein field equations, however, predict that the presence of matter warps space-time. As I understand it, the hyperbolic geometry means that the actual volume will be smaller than that predicted by the radius with the above equation, because the space in the center of the Earth (and around the Earth) is "dimpled". How much will these two volumes differ by?

There would be different ways to answer this question, depending on how you define "radius". There is one sense, where you could drop a measuring tape from the surface to the center of the Earth, and I think this would be the "proper" length. This is my preferred definition, but I think it would be the most complicated to answer. Otherwise, you could use the universal coordinates from someone far from Earth, which would give a smaller measure of radius. Another, more meaningful, way might be to take radius from the circumference as measured by us on Earth $R=C/(2 \pi)$, in which case the real volume would be greater than the naive $4 \pi R^3/3$ number. I welcome correction on any of these predictions.

I thought it could be done easily after discovering Flamm's paraboloid, but my attempts have so far failed. For reference, this shape describes the shape of space if the "extra" space were extended out into another dimension, $w$.

the space curvature

$$ w = 2 \sqrt{r_{s} \left( r - r_{s} \right)}. $$

Specific to Earth:

$$ r_s = \frac{2 GM}{c^2} = 8.868 \text{mm} $$

This shape, however, is only valid past $R$, and some kind of clever corrections would be needed to get around this fact. I originally wanted to ask what the added volume would be over the entire Schwarzschild metric compared to flat space (meaning from $r=R$ to $r=\infty$). It looks like that would be infinity, so I didn't ask that question.

So in short, how much does general relativity math change the volume by Earth by? Is it really tiny? Is it huge? This is exciting to know!

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Love the question! My first thought is that an accurate answer to this would require a determination of the metric inside the Earth, which is not Schwarzschild; it would have to be a non-vacuum solution to the Einstein equation. I'm not sure how much is known about such solutions. But since the curvature is so low it might (should?) be possible to make a perturbative estimate of the factor you're asking for without knowing the exact metric. –  David Z Sep 13 '12 at 17:25
    
@DavidZaslavsky I don't know if you can easily turn this into the field equations, but the obvious Newtonian form for gravity in Earth is $\propto r$. Funny enough, due to the crazy distribution of Earth's matter, gravity actually increases as you go down into the crust. Obviously, I think we should seek the homogenous sphere answer. IMO, first thing I think someone should do is trash $w(r)$ and only use $dw/dr$, which I hope would be unaffected by the symmetrically distributed matter everywhere $>r$. But that's part of my problem, I don't know what rules still apply. –  AlanSE Sep 13 '12 at 18:33
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1 Answer

To start out, we can quickly simplify the equation for the surface. The $r_s$ metric is laughably small compared to Earth-scale dimension, so when we add those two it can be simply ignored.

$$ w(r) = 2\sqrt{ r_s (r+r_s) } \approx 2\sqrt{ r_s r}$$

We would like to apply this to the interior of the Earth but obviously we can't. GR only gives the curvature of space, the equation above is a constructed form of the surface that satisfies that curvature. So I'll work backwards here to get an equation for the curvature. That's not all though. If we're looking at some point in the interior of the Earth, only the matter below that point contributes to the gravitational field, and thus the curvature.

$$ \frac{dw}{dr} = \frac{\sqrt{r_s}}{2 \sqrt{r}} $$

$$ r_s = \frac{ 2 G M(r) }{ c^2 } $$

$$ M(r) = \rho \frac{4}{3} \pi r^3 $$

$$\frac{dw}{dr} = \sqrt{\frac{2G \pi \rho}{3 c^2} } r = \alpha r = (2.9 \times 10^{-12} 1/m ) r $$

Here, $\alpha$ comes out to a physical constant that is easy to calculate. This metric is for curvature. It is the rise to run for the surface in the extra dimension.

$$ ds = \sqrt{1+\left( \frac{dw}{dr} \right)^2 } dr \approx \left( 1 + \frac{1}{2} \left( \frac{dw}{dr} \right)^2 \right) dr = \left( 1 + \frac{\alpha^2}{2}r^2 \right) dr$$

The Volume of a normal sphere is:

$$ V = \int_0^R 4 \pi r^2 dr = \frac{4 \pi R^3 }{3}$$

In doing the correct integral with GR we integrate about $ds$. I should note, the radius is a difficult value to pin down, as I discussed in the question. I'll use $R'$ notation, but I'm really going to continue from this point with $R'=R$. If you disagree with this, this would be the point you'd want to pick the calcs up. My position for making these radii the same is to get the circumference the same. Keep in mind, the w-axis offset of $w(r)$ is arbitrary, so I can claim it intersects the r-axis at the surface of Earth. In this way, $R$ is a straight line in 4-dimension space (3 space plus the fictitious buckling dimension) from the surface to "under" the center of the Earth. The thing that is preserved with this method is the circumference and the surface area of Earth, which are the only values Earthlings have direct measurement of. I also made an illustration to convince you of this (I needed it myself).

The Earth is a bowl

$$ V' = \int_0^{R'} 4 \pi r^2 ds = \int_0^{R} 4 \pi r^2 \left( 1 + \frac{\alpha^2}{2}r^2 \right) dr $$

Subtract the normal volume to find the difference.

$$ \Delta V = V' -V = \int_0^R 4 \pi r^2 \frac{\alpha^2}{2}r^2 dr =\frac{4 \pi \alpha^2}{2} \int_0^R r^4 dr $$

$$ \Delta V = \frac{4 \pi \alpha^2 R^5}{10} = \frac{4 G \pi^2 \rho R^5}{15 c^2} = \frac{ G M \pi R^2}{5 c^2} = 113 km^3 $$

This comes out to 113.33 km^3, or 1000 Strategic Petroleum Reserves (I stand corrected on this point). This is a common size for lakes around the world.

Coincidentally, the same calculation for the sun yields about half the volume of the Earth.

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The SPR contains 110,640,000 m^3 ~= 0.11 km^3. But there are lots of things that must remain hidden... :P –  mmc Sep 14 '12 at 15:46
    
Your comment 'laughably small' made me look up the Schwarzschild Radius of the Earth. Turns out $r_s = 8.87*10^{-3} = 9mm$ –  Ehryk Sep 14 '12 at 15:52
    
The formula for the integral should be $\int \epsilon_{abcd}\sqrt{|det(g)|}t^{a}dr^{a}d\theta^{b}d\phi^{c}$, where $t^{a}$ is the unit timelike normal. For most of the Schwarzschild-y coordinate systems, det(g) = -1, so this will come out to $\int \frac{4\pi r^{2}}{\sqrt{|g_{tt}|}}dr$ after you use the spherical symmetry to your advantage, which differs from the expression you use. –  Jerry Schirmer Dec 13 '12 at 21:48
    
@JerrySchirmer I was returning to the basics of Einstein notation to get what that means in terms of $r$, but it's still beyond me. What is $g_{t,t}$ in that expression? –  AlanSE Feb 3 at 2:27
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