I know the time dilation (relative to an observer at infinity) outside a non-rotating sphere can be found from the Schwarzschild metric to be $$t_{0}=t_{f}\sqrt{1-\frac{r_{s}}{r}}$$ where $r_{s}$ is the Schwarschild radius. I also know there is a similar expression for the time dilation inside a non-rotating sphere using the interior Schwarzchild metric. The calculation is done here.

I would like to find similar relations for a non-uniform sphere. For example, a neutron star is much less dense on the surface than the core. What would a good assumption of density vs. distance from center be in this case? From here, what is the relative time dilation for inside and outside the sphere?

  • $\begingroup$ According to Birkhoff's theorem, outside the sphere, the metric will be the same as the Schwarzschild metric since the mass distribution is spherically symmetric. If you want to see what happens inside the sphere, you should start form the general spherical symmetric metric and solve Einstein's field equations for the mass distribution you are considering and apply a continuity condition at the surface of the sphere. $\endgroup$ – Panos C. Apr 13 '18 at 14:14

I'm afraid there isn't a nice simple solution for this. The problem is that the result depends on the equation of state of the stuff inside your sphere. The pressure inside the sphere compresses the stuff in the sphere and that increases the density which in turn affects the gravity and therefore pressure. How much the stuff compresses depends on its compressibility, and that is a property of whateverthe stuff is. There may also be phase transitions e.g. if you take a big enough ball of water the pressure will cause the water to solidify to ice at some depth then further transition to the many different crystal phases that ice has at different pressures.

Wald discusses the interior metric in section 6.2 of his book General Relativity, but even there he treats only the case of a perfect fluid. The derivation takes many pages so I won't reproduce it here. I'll give only the final result, which is:

$$ ds^2 = -e^{2\phi}dt^2 + \left(1 -\frac{2m(r)}{r}\right)^{-1}dr^2 + d\Omega^2 $$

The mass function $m(r)$ is the mass within the radius $r$ and is given by:

$$ m(r) = 4\pi \int_0^r \rho(r')r'^2 \,dr' $$

The time function $\phi$ is given by:

$$ \frac{d\phi}{dr} = \frac{m(r) + 4\pi r^3 P}{r(r - 2m(r))} $$

To calculate the time dilation you would need to specify the properties of your matter and use those properties to (probably numerically) calculate $\phi$.

You mention the specific case of a neutron star, but I have never worked in that area so I can't comment beyond saying that it's a complicated problem and there is a huge volume of literature on the subject. The simplified case is the Tolman, Oppenheimer, Volkoff equation, though simple is a misleading word to use for it!

  • $\begingroup$ I have found an article dealing with the EOS at link.springer.com/article/10.1007/s10509-016-2941-7. Do you think this could help me solve this problem? They have a nice graph of mass vs. radius, but I don't see one for pressure. To be honest, my mathematical knowledge is lacking, so I am unsure how to utilize their results. $\endgroup$ – Java Newbie Apr 15 '18 at 12:58

Gravitational time dilation requires clocks at stationary r-coordinates. This is possible outside the event horizon, because here clocks can hover with r = constant. Inside the event horizon the spacetime is not static anymore however, because here the r-coordinate changes from spacelike to timelike. Simply said the r-coordinate "behaves" like a time coordinate, it knows only one direction and that is downwards to the singularity. Therefore nothing can hover at constant r-coordinate inside the event horizon and hence the formula which you show is not applicable.

  • 1
    $\begingroup$ The question is about massive bodies, not the interior of a black hole. $\endgroup$ – Javier Apr 13 '18 at 14:30

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.