1
$\begingroup$

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?

Improve this question
$\endgroup$
3
  • $\begingroup$ The Earth is only a hollow sphere in Jules Verne. $\endgroup$
    – hobbs
    May 23 '15 at 4:33
  • 1
    $\begingroup$ related:how do we age if we tunneled to Earth core $\endgroup$
    – user6760
    May 23 '15 at 6:25
  • $\begingroup$ @hobbs In real life it is an infinite series of concentric hollow spheres, which each follow the same rules. $\endgroup$
    – Asher
    May 23 '15 at 6:54
0
$\begingroup$

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 $$ t(\infty)/t(r) = \sqrt{1 - 2\Phi(r)}=\left\{\begin{array}{cc} \sqrt{1 - \frac{2GM}{Rc^2}} =\text{const}\qquad & \forall r\leq R \\ \sqrt{1 - \frac{2GM}{rc^2}} \qquad & \forall r\geq R \end{array}\right. $$

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – CJ Dennis
    May 23 '15 at 5:22
  • $\begingroup$ "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? $\endgroup$
    – Asher
    May 23 '15 at 6:57
  • $\begingroup$ 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. $\endgroup$
    – hobbs
    May 23 '15 at 7:27
  • $\begingroup$ True in the first paragraph i'm talking about being outside the solid sphere $\endgroup$
    – Ali Moh
    May 27 '15 at 17:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.