# What is the general dependence of volume change from mass change due to general relativity?

The volume excess of earth due to general relativity in comparison to euclidean space has been calculated to

$$\Delta V = \frac{ G M \pi R^2}{5 c^2} = 113 km^3$$

(this is done in this physics.stackexchange.com - question:How much does the curvature of space change the volume of Earth by?)

That means $$\Delta V$$ increases linearily with $$M$$. Now, I'm wondering whether this is valid in general or only regarding the earth? Is that ("$$\Delta V$$ increases linearily with $$M$$") an inference out of Einstein's field equations? Is it true for any metric?

That means ΔV increases linearly with M. Now, I'm wondering whether this is valid in general or only regarding the earth? Is that ("ΔV increases linearily with M") an inference out of Einstein's field equations? Is it true for any metric?

No, it depends on metric. In spherically symmetric spacetime the proper volume is defined as: $$$$V_{prop}=4 \pi \int_{0}^{R} r^2~g_{rr}(r,\alpha)^{1/2} dr, ~~~~~~\alpha\equiv r_{S}/R=(2GM/c^2)/R).$$$$ For example, in case of constant density sphere (interior Schwarzschild solution) with metric component $$g_{rr}=\frac{1}{1-\alpha/R^2~r^2}$$ the proper volume results in $$$$V_{prop}=V_{flat}~\frac{3}{2}~\Bigg[\frac{\arcsin{\sqrt{\alpha}}-\sqrt{\alpha (1-\alpha)}}{\alpha^{3/2}}\Bigg].$$$$ Only for $$\alpha\ll 1$$ the proper volume is proportional to $$\alpha$$ and thus to $$M$$: $$V_{prop}= V_{flat}\cdot 3/10\cdot\alpha~\propto~M$$.

• Please, how do you get from your initial formula for $V_{prop}$ and the special $g_{rr}$ to your final $V_{prop}$? Especially the origin of the arcsin I do not see.
– user336075
May 21 at 15:59
• And, sorry for asking stupid questions, is the $R$ in the formula for $g_{rr}$ the same $R$ as in the upper limit of the integral in the formula for the initial $V_{prop}$?
– user336075
May 21 at 16:03
• Yes, and you can use Wolfram online integrator for that purpose. May 21 at 16:08