general relativity - Gravitational time dilation at the earth's center

I would like to know what happens with time dilation (relative to surface) at earth's center .

There is a way to calculate it?

Is time going faster at center of earth?

I've made other questions about this matter and the answers refers to:

$\Delta\Phi$ (difference in Newtonian gravitational potential between the locations) as directly related, but I think those equation can't be applied to this because were derived for the vecinity of a mass but not inside it.

Any clues? Thanks

Answer

The rule I mentioned in another question, that the time dilation factor is $1+\Delta\Phi/c^2$, applies here. The derivation (found in various textbooks) depends only on the assumptions that fields are weak and matter is nonrelativistic, both of which are true for the Earth.

Modeling the Earth as a uniform-density sphere (not true, of course, but I don't care), we find that $g(r)=GMr/R^3$ where $R$ is the radius of the Earth. So $$ \Delta\Phi={GM\over R^3}\int_0^Rr\,dr={GM\over 2R}. $$ That means that $$ {\Delta\Phi\over c^2}={GM\over 2Rc^2}={1\over 4}{R_s\over R}. $$ Here $R_s=2GM/c^2$ is the Schwarzschild radius corresponding to the Earth's mass. Numerically, $R_s$ is about 9 mm, and $R$ is about 6400 km, so $\Delta\Phi/c^2=3\times 10^{-10}$.

The sign of the effect is that clocks tick slower when they're deeper in the potential well. That is, a clock at the Earth's surface ticks 1.0000000003 times faster than one at the center.