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.
