Time Dilation $\gamma(x)$

Since,

$g=-\cfrac { 1 }{ 2 } .\cfrac { d{ v }_{ t }^{ 2 } }{ dx }$

and

$\cfrac{{v}_{t}^{2}}{{c}^{2}}=\gamma(x)^{2}$

$g=-\cfrac {{c}^{2} }{ 2 } \cfrac { d\gamma(x)^{ 2 } }{ dx }$

because gravity is given by,

$g=-{g}_{o}{e}^{-\cfrac{{g}_{o}{r}_{e}}{{G}_{o}}(x)}$

We have an expression for time dilation $\gamma$,

$\cfrac { d\gamma(x)^{ 2 } }{ dx }=\cfrac{2{g}_{o}}{{c}^{2}}{e}^{-\cfrac{{g}_{o}{r}_{e}}{{G}_{o}}(x)}$

$\gamma(x)^{ 2 }=\int {d \gamma(x)^{ 2 }}=\int{\cfrac{2{g}_{o}}{{c}^{2}}{e}^{-\cfrac{{g}_{o}{r}_{e}}{{G}_{o}}(x)}dx}$

$\gamma(x)^{ 2 }=A-\cfrac{2{G}_{o}}{{c}^{2}{r}_{e}}{e}^{-\cfrac{{g}_{o}{r}_{e}}{{G}_{o}}(x)}$

When $x\rightarrow\infty$, $\gamma(x)^2\rightarrow1$, because ${v}_{t}=c$ when space is normal.

$\gamma(x)^{ 2 }=1-\cfrac{2{G}_{o}}{{c}^{2}{r}_{e}}{e}^{-\cfrac{{g}_{o}{r}_{e}}{{G}_{o}}(x)}$

As such a gamma field, the change in $\gamma=\cfrac{{v}_{t}}{c}$ over $x$ is given by

$\gamma(x)=\sqrt{1-\cfrac{2{G}_{o}}{{c}^{2}{r}_{e}}{e}^{-\cfrac{{g}_{o}{r}_{e}}{{G}_{o}}(x)}}$

This is an expression for time dilation over a gravitational field.