# Assuming a fixed total mass, will the spacetime geometry outside a spherical mass distribution depend on the shape (of the distribution)?

Consider two independent spheres of equal masses but of different radius and in different spacetimes. The first sphere is less dense than the second one, i.e., it has a larger radius. For example, if the first sphere is considered to be size of the Sun, the second one will be the size of a golfball.

Now my question is, will geometry of spacetime curvature be similar outside these two spheres or different? If different, why?

Note: the two masses are nowhere near each other so there is no influence between them.

• Related: physics.stackexchange.com/q/21705/2451 and links therein. Dec 10 '17 at 14:40
• The geometry of spacetime outside both spheres is the Schwarzchild geometry, so outside the larger of the two spheres, the spacetime geometry is exactly the same. Dec 10 '17 at 14:46
• @PeterShor are you saying that after a certain distance (equal to radius of larger sphere) from the center of two spheres the spacetime geometry will be same for two sphere Dec 10 '17 at 15:30

Thanks to Birkhoff's theorem, we know that the field outside a spherically symmetric isolated mass will always be the Schwarzschild metric. In other words, the metric will look like this just outside the surface of the object $$d s^2 = -\left(1 - \frac{2 M}{r}\right) d t^2 + \frac{1}{1 - 2M/r} dr^2 + r^2 d \Omega^2$$ where $d\Omega^2 = d\vartheta^2 + \sin^2\vartheta d \varphi^2$ is the surface element of a unit sphere. So in some sense the geometry looks the same outside spherical objects in relativity, one only changes the position of the surface and the value of $M$.
However, you will find it surprising that even when you take the same number of particles (say protons and electrons) of a fixed total rest mass $M_0$ and squeeze it into a body of different radius, the resulting value of the parameter $M$ in the metric can be somewhat different.
For instance, when we are talking about a body made out of a perfect fluid, we can use the analysis of Tolmann, Oppenheimer and Volkoff to see that the gravitating mass can be understood to consist of three contributions $$M = M_0 + \delta M_\mathrm{thermo} + \delta M_\mathrm{binding}$$ $\delta M_\mathrm{thermo}$ corresponds to the internal thermodynamical energy of the gas, and $\delta M_\mathrm{binding}$ is the gravitational binding energy. When we are close to a Newtonian regime, the binding energy can be expressed simply as the Newtonian binding energy (divided by $c^2$) $$\delta M_\mathrm{binding} = -\int_0^\mathrm{surf.} \frac{G m_0(r) \rho_0 }{r c^2} 4 \pi r^2 dr$$ where $\rho_0$ is the rest-mass density, and $m_0(r) = \int_0^r \rho_0(r') 4\pi r'^2dr'$ is the rest mass contained in the sphere of radius $r$.