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An answer to What would be the rate of acceleration from gravity in a hollow sphere? states "that according to General Relativity time passes more slowly inside a hollow massive sphere than it does outside".

How much does time slow down by? Is it enough to require adjustments for, say, predicting how Earth's molten core behaves or nuclear reactions in the heart of the Sun?

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CJ Dennis
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That discussion is about a spherical shell and does not apply to the Earth nor to the sun. In the latter situations assuming uniform mass distribution we know that time is dilated by $t(\infty)/t(r)=\sqrt{g_{00}}=\sqrt{1 - \frac{2GM}{rc^2}}$, where $r$ is the distance from the center.

For a spherical SHELL however we have to go back and forth between the GR and its Newtonian limit. Since the Earth's and Sun's gravity is weak we can approximate $\sqrt{g_{00}} = \sqrt{1 - 2\Phi(r)}$, now the potential $\Phi(r)$ outside the shell is the same as before $\Phi_{\text{out}}(r)=GM/rc^2$, however inside the shell it is $\Phi_{\text{in}}(r) = GM/Rc^2 = $const.

There fore the different in passage of time between a position $r$ from the center and infinitely far away is $$ t(\infty)/t(r) = \sqrt{1 - 2\Phi(r)}=\left\{\begin{array}{cc} \sqrt{1 - \frac{2GM}{Rc^2}} =\text{const}\qquad & \forall r\leq R \\ \sqrt{1 - \frac{2GM}{rc^2}} \qquad & \forall r\geq R \end{array}\right. $$

Ali Moh
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  • The way I understand the question I referenced is using the rubber sheet model. Consider a hollow sphere to be two-dimensionally like a ring, much like a perfectly circular biscuit cutter. Press the ring into the rubber sheet. It will deform but the interior will be perfectly level. Even though there is no difference in gravitational potential there is still a gravitational field so time dilation applies. A solid sphere is simply a sphere, so it deforms the rubber sheet more in the middle than the edge. There is a gravitational field again (uneven this time) so there should be time dilation. – CJ Dennis May 23 '15 at 05:22
  • "That discussion is about a spherical shell and does not apply to the Earth nor to the sun." Aren't the Earth and Sun easily modeled as series of concentric hollow shells? – Asher May 23 '15 at 06:57
  • The GM/r solution isn't valid inside of a solid sphere either; you can see that it breaks down by noticing that it tries to divide by r=0 at the center of the object! In fact you actually have GM(3R^2-r^2)/2R^3 inside the sphere which is something completely different, and works out to 3GM/2R at the center. – hobbs May 23 '15 at 07:27
  • True in the first paragraph i'm talking about being outside the solid sphere – Ali Moh May 27 '15 at 17:33