A.T. said:
That is just a very complicated way to say that the proper acceleration of a clock resting in the center is zero. This is less than the proper acceleration of a clock resting on the surface. Yet the center clock runs slower than the surface clock.
Start with the Schwarzschild solution in the weak field approximation:
(cd\tau)^2=(1-\frac{2\Phi}{c^2})(cdt)^2+(1-\frac{2\Phi}{c^2})^{-1}(dr)^2+...
For the case dr=d\theta=d\phi=0 you get the well known relationship:
d\tau=\sqrt{1-\frac{2\Phi}{c^2}}dt
Writing the above for two different gravitational potentials \Phi_1 and \Phi_2 you obtain the well-known time dilation relationship:
\frac{d\tau_1}{d\tau_2}=\sqrt{\frac{1-\frac{2\Phi_1}{c^2}}{1-\frac{2\Phi_2}{c^2}}}
At the Earth surface :
\Phi_1=-\frac{GM}{R}
At the Earth center:
\Phi_2=-3/2\frac{GM}{R}
Now, due to the fact that \frac{\Phi}{c^2}<<1 you can obtain the approximation:
\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}<1
So, f_1>f_2 where f_1 is the clock frequency on the Earth crust and f_2 is the frequency of the clock at the center of the Earth.
In addition, the time dilation depends on the difference in the gravitational field \Phi_1-\Phi_2.
Generalization:
At a distance r<R from the center of the sphere, inside the sphere, the gravitational potential is:
\Phi_2(r)=-\frac{GM}{R}(\frac{3}{2}-\frac{r^2}{2R^2})
The above gives:
\frac{d\tau_1}{d\tau_2}=1-\frac{\Phi_1-\Phi_2}{c^2}=1-\frac{GM}{2Rc^2}(1-\frac{r^2}{R^2})<1
For r=0 (clock2 at the center of the Earth) you recover the results from above.
For r=R you get the expected:
\frac{d\tau_1}{d\tau_2}=1