Space Density Compression Ratio

Consider a space density function of the form,

${ d }_{ s }(x)=A{ e }^{ -bx }+B$  where A, B and b are constant to be determined.

At  ${ d }_{ s }(0)={ d }_{ e }$ space is compressed, its space density is ${d}_{s}$, on surface of earth. And ${ d }_{ s }(x\rightarrow \infty )={ d }_{ n }$ where at long distance, space is relaxed and the corresponding space density is ${d}_{n}$.

We have,

${ d }_{ s }(x)={ e }^{ -bx }({ d }_{ e }{ -d }_{ n })+{ d }_{ n }$

Then we let the inverse relationship between time speed squared ${v}^{2}_{t}$and space density, ${ d }_{ s }(x)$ to be,

${v}_{t}^{2}=C-D{ d }_{ s }(x)$ where $C, D$ are to be determined.

We know that as $x\rightarrow \infty, { d }_{ s }(x)\rightarrow{ d }_{ n }$where space is relaxed and time speed is $c$,

${v}_{t}^{2}={c}^{2}$, ${ d }_{ s }(x)={ d }_{ n }$

$C={c}^{2}+D{d}_{n}$ then,

${v}_{t}^{2}={c}^{2}-D({ d }_{ s }(x)-{d}_{n})$

If we then consider the time dilation equation from previously, (The use of this equation is still valid because a linear approach can be consider as a first approximate to an exponential one.  Consider the power series expansion of $e$.)

${ v }_{ t }= \sqrt {{ c }^{ 2 }-\cfrac { 2{ G }_{ o } }{ (x+r_{ e }) }}$

when $x=0$

${ v }^{2}_{ t }={ c }^{ 2 }-\cfrac { 2{ G }_{ o } }{ (r_{ e }) }$

 ${v}_{t}^{2}={c}^{2}-D({ d }_{ e } -{d}_{n})$ where ${ d }_{ s }(0)={ d }_{ e }$

we have,

$\cfrac { 2{ G }_{ o } }{ (r_{ e })}=D({ d }_{ e } -{d}_{n})$

therefore,

$D=\cfrac{ 2{ G }_{ o } }{r_{ e }({ d }_{ e } -{d}_{n})}$

Hence,

${v}_{t}^{2}={c}^{2}-\cfrac{ 2{ G }_{ o }({ d }_{ s }(x)-{d}_{n}) }{r_{ e }({ d }_{ e } -{d}_{n})}$

So,

$\gamma (x)^{ 2 }=1-\cfrac{ 2{ G }_{ o }({ d }_{ s }(x)-{d}_{n}) }{{c}^{2}r_{ e }({ d }_{ e } -{d}_{n})}$

Comparing this with the equation obtained from gravity,

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

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

But, ${ d }_{ s }(x)={ e }^{ -bx }({ d }_{ e }{ -d }_{ n })+{ d }_{ n }$

${ e }^{ -\cfrac { { g }_{ o }{ r }_{ e } }{ { G }_{ o } } }={ e }^{ -b }$

So, $b=\cfrac { { g }_{ o }{ r }_{ e } }{ { G }_{ o } }=\cfrac{1}{{r}_{e}}$

because ${g}_{o}=\cfrac{{G}_{o}}{{r}^{2}_{e}}$

We have for the case of earth,

${ d }_{ s }(x)={ e }^{ -\cfrac { 1 }{ { r }_{e } } x }({ d }_{ e }{ -d }_{ n })+{ d }_{ n }$

Similarly, for black hole

 ${ d }_{ sB }(x)={ e }^{ -\cfrac {1 }{ { r }_{eo } } x }({ d }_{ B }{ -d }_{ n })+{ d }_{ n }$

Differentiating both expressions

$\cfrac{d({ d }_{ s }(x))}{dx} =  -\cfrac { 1 }{ { r }_{e } }{ e }^{ -\cfrac { 1 }{ { r }_{e } } x }({ d }_{ e }{ -d }_{ n })$

$\cfrac{d({ d }_{ sB }(x))}{dx} =  -\cfrac { 1 }{ { r }_{ eo } }{ e }^{ -\cfrac { 1 }{ { r }_{eo } } x }({ d }_{ B }{ -d }_{ n })$

Since both are proportional to gravity,

$\cfrac{{g}_{e}(x)}{{g}_{B}(x)} = \cfrac { { r }_{ eo } }{ { r }_{e } }\cfrac{{ e }^{ -\cfrac { 1 }{ { r }_{e } } x }({ d }_{ e }{ -d }_{ n })}{{ e }^{ -\cfrac { 1 }{ { r }_{eo } } x }({ d }_{ B }{ -d }_{ n })}$

and so,

$\cfrac{{ d }_{ B }-{ d }_{ n }}{{ d }_{ e }-{ d }_{ n }} =\cfrac{{g}_{B}(x)}{{g}_{e}(x)}\cfrac { { r }_{ eo } }{ { r }_{e } } { e }^{ (\cfrac { 1 }{ { r }_{eo }}-\cfrac { 1 }{ { r }_{e } }) x}$

when $x$ = 0,

$\cfrac{{ d }_{ B }-{ d }_{ n }}{{ d }_{ e }-{ d }_{ n }} =\cfrac{{g}_{B}(0)}{{g}_{e}(0)}\cfrac { { r }_{ eo } }{ { r }_{e } }.1$

Therefore, the relative amount of space compression around a black hole is,

$\cfrac{{ d }_{ B }-{ d }_{ n }}{{ d }_{ e }-{ d }_{ n }}$ = (5.07071e18/9.80665)*(0.00886/6371000) = 7.191e8