Consider energy conservation equation again.
\({v}^{2}_{t} +{v}^{2}_{s} = {c}^{2}\)
Differentiating with respect to time,
\(\frac{d{v}^{2}_{t}}{dt}+2{v}_{s}.\frac{d{v}_{s}}{dt} = 0\)
\(-2{v}_{s}.\frac{{v}_{s}}{dt} = \frac{d{v}^{2}_{t}}{dx}.\frac{dx}{dt}\)
since \(\frac{{v}_{s}}{dt}= g\), \(\frac{dx}{dt}= {v}_{s}\) and \({\gamma (x)}^{2}=\frac{{v}_{t}}{{c}^{2}}\)
\(g = -\frac{1}{2} \frac{d{v}^{2}_{t}}{dx} = -\frac{{c}^{2}}{2}\frac{d{\gamma (x)}^{2}}{dx}\)
but
\(g = D\frac{d({d}_{s}(x))}{dx}\)
(This is an assumption equivalent to assuming that time speed is a terminal velocity in free space related to space density by \({v}_{t}^{2}=F-D{ d }_{ s }(x)\) where \(F, D\) are constants and \({ d }_{ s }(x)\) is a decreasing exponential with a value of \({d}_{n}\) at \(x\rightarrow\infty\), the normal density of free space. Refer to post "Technically, the Moon")
\(D\frac{d({d}_{s}(x))}{dx}=-\frac{{c}^{2}}{2}\frac{d{\gamma (x)}^{2}}{dx}\)
Integrating both sides,
\({\gamma (x)}^{2} = -\frac{2D}{{c}^{2}}{d}_{s}(x)+A\)
Since \(x \rightarrow \infty\), \({d}_{s}(x)\rightarrow{d}_{n}\) and \(\gamma (x)\rightarrow 1\)
\(A =1+ \frac{2D}{{c}^{2}}{d}_{n}\)
\({\gamma (x)}^{2} =1-\frac{2D}{{c}^{2}}({d}_{s}(x)-{d}_{n})\)
In a black hole \(\gamma (x)\) = 0,
\(\frac{2D}{{c}^{2}}({d}_{B}-{d}_{n}) = 1\)
\(\therefore \frac{2D}{{c}^{2}}=\frac{1}{({d}_{B}-{d}_{n})}\)
So,
\({\gamma (x)}^{2} =1-\frac{{d}_{s}(x)-{d}_{n}}{{d}_{B}-{d}_{n}}\)
\({\gamma (x)}^{2} =1-\frac{{d}_{s}(x)-{d}_{n}}{{d}_{e}-{d}_{n}}\frac{{{d}_{e}-{d}_{n}}}{{d}_{B}-{d}_{n}}\)
\(\because \frac{{ d }_{ B }-{ d }_{ n }}{{ d }_{ e }-{ d }_{ n }}\) = 7.191e8
\({\gamma (x)} =\sqrt{1-\frac{1}{7.191e8}\frac{{d}_{s}(x)-{d}_{n}}{{d}_{e}-{d}_{n}}}\)
\({\gamma (x)} = \sqrt{1-1.3906e(-9)\frac{{d}_{s}(x)-{d}_{n}}{{d}_{e}-{d}_{n}}}\)
When space density compression goes beyond 7.191e8, \(\gamma (x)\) is complex.
