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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.

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  • $\begingroup$ Related: physics.stackexchange.com/q/21705/2451 and links therein. $\endgroup$ – Qmechanic Dec 10 '17 at 14:40
  • $\begingroup$ 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. $\endgroup$ – Peter Shor Dec 10 '17 at 14:46
  • $\begingroup$ @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 $\endgroup$ – Darkray5 Dec 10 '17 at 15:30
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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 $r$.

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