Yesterday Once More

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.