\( { v }_{ t }^{ 2 }=\cfrac { A }{ { d }_{ s } }\) where A is a constant
that shows time speed is inversely proportional to space density. So
\( { c }^{ 2 }=\cfrac { A }{ { d }_{ normal} }\)
ie. \(A={c}^{2}{d}_{normal}\)
with the understanding that at normal space density time \({d}_{normal}\) speed is c. And
\({ d }_{ s }=-a(x+{ r }_{ 0 })+{ a }_{ 0 }\) ----(*)
that gives linear space density change. And
\(g=-\cfrac { 1 }{ 2 } .\cfrac { d{ v }_{ t }^{ 2 } }{ dx } =-\cfrac { 1 }{ 2 } .\cfrac { d }{ dx } (\cfrac { A }{ { d }_{ s } } )=\cfrac { A }{ 2 } .\cfrac { 1 }{ { d }_{ s }^{ 2 } } .\cfrac { d{ d }_{ s } }{ dx }\)
\(g=-G.\cfrac { 1 }{ { (-a(x+{ r }_{ o })+{ a }_{ 0 }) }^{ 2 } } =-\cfrac { G }{ { a }^{ 2 } } \cfrac { 1 }{ { (x+\cfrac { { a{ r }_{ o }-a }_{ o } }{ a } ) }^{ 2 } } =-\cfrac { G }{ { a }^{ 2 } } \cfrac { 1 }{ { (x+{ r }_{ e }) }^{ 2 } }=-{ G }_{ 0 }.\cfrac { 1 }{ { (x+{ r }_{ e }) }^{ 2 } }\)
that expresses gravity derived from considering the energy conservation equation and changing space density affecting time speed. Looking at \({G}_{o}\)
\(G=\cfrac{Aa}{2}\) and so,
\({G}_{o}=\cfrac{G}{{a}^{2}}=\cfrac{A}{2a}\)
Therefore
\(a=\cfrac{{c}^{2}{d}_{normal}}{2{G}_{o}}=\cfrac{{d}_{normal}}{{r}_{eo}}\), since \({r}_{eo}=\cfrac{2{G}_{o}}{{c}^{2}}\)
When \(x=0\) the space density equation gives
\({d}_{s}={d}_{earth}=-\cfrac{{d}_{normal}}{{r}_{eo}}({ r }_{ 0 })+{ a }_{ 0 }\)
so,
\({ a }_{ 0 }={d}_{earth}+\cfrac{{d}_{normal}}{{r}_{eo}}{ r }_{ 0 }\)
and so (*) becomes
\({d}_{s}(x)=-\cfrac{{d}_{normal}}{{r}_{eo}}(x+{ r }_{ 0 })+{d}_{earth}+\cfrac{{d}_{normal}}{{r}_{eo}}{ r }_{ 0 }\)
\({d}_{s}(x)=-\cfrac{{d}_{normal}}{{r}_{eo}}(x)+{d}_{earth}\) or
\({d}_{s}(x)=-\cfrac{{c}^{2}{d}_{normal}}{2{G}_{o}}(x)+{d}_{earth}\)
We see clearly the absurdity of this formulation but as a first linear approximation it is OK. The factor
\(\cfrac { 1 }{ { d }_{ s }^{ 2 } } .\cfrac { d{ d }_{ s } }{ dx }\)
needs further consideration, in order that the space density equation make more sense. Consider, the expansion.
\({e}^{-b(x+{r}_{e})}={e}^{-b{r}_{e}}(1-bx+.....)=(-{e}^{-b{r}_{e}}bx+{e}^{-b{r}_{e}}....).\)
then \({d}_{s}(x)={e}^{-b(x)}{e}^{-b{r}_{e}}\), where \({e}^{-b{r}_{e}}={d}_{earth}\), \( b=\cfrac { { c }^{ 2 }{ d }_{ normal } }{ 2{ G }_{ o }{d}_{earth} }\) and \({ e }^{ -\cfrac { { c }^{ 2 }{ d }_{ normal } }{ 2{ G }_{ o }{d}_{earth} } { r }_{ e } }={ d }_{ earth }\)
where \(a, b\) are constants and \({r}_{e}\) is defined such that x=0 is on the surface of earth where space is most compressed. By comparing coefficients, we have a consistent expression,
\({ d }_{ s }(x)={ e }^{ -\cfrac { { c }^{ 2 }{ d }_{ normal } }{ 2{ G }_{ o }{d}_{earth}} (x) }{ e }^{ -\cfrac { { c }^{ 2 }{ d }_{ normal } }{ 2{ G }_{ o }{d}_{earth} } { r }_{ e } }+{d}_{normal}\)
The last term is added because we know that \(x\rightarrow \infty \), \({d}_{s}\rightarrow{d}_{normal}\) and the term constant term \(e^{-b{r}_{e}}\) is modified to
\({ e }^{ -\cfrac { { c }^{ 2 }{ d }_{ normal } }{ 2{ G }_{ o }{d}_{earth} } { r }_{ e } }={ d }_{ earth }-{d}_{normal}\).
\({ d }_{ s }(x)={ e }^{ -\cfrac { { c }^{ 2 }{ d }_{ normal } }{ 2{ G }_{ o }{d}_{earth}} (x) }({ d }_{ earth }-{d}_{normal})+{d}_{normal}\)
This derivation however, will lead to unfamiliar expressions, although we have seen the first approximation to be the same as textbook derivation. Have a nice day.
\({G}_{o}=\cfrac{G}{{a}^{2}}=\cfrac{A}{2a}\)
Therefore
\(a=\cfrac{{c}^{2}{d}_{normal}}{2{G}_{o}}=\cfrac{{d}_{normal}}{{r}_{eo}}\), since \({r}_{eo}=\cfrac{2{G}_{o}}{{c}^{2}}\)
When \(x=0\) the space density equation gives
\({d}_{s}={d}_{earth}=-\cfrac{{d}_{normal}}{{r}_{eo}}({ r }_{ 0 })+{ a }_{ 0 }\)
so,
\({ a }_{ 0 }={d}_{earth}+\cfrac{{d}_{normal}}{{r}_{eo}}{ r }_{ 0 }\)
and so (*) becomes
\({d}_{s}(x)=-\cfrac{{d}_{normal}}{{r}_{eo}}(x+{ r }_{ 0 })+{d}_{earth}+\cfrac{{d}_{normal}}{{r}_{eo}}{ r }_{ 0 }\)
\({d}_{s}(x)=-\cfrac{{d}_{normal}}{{r}_{eo}}(x)+{d}_{earth}\) or
\({d}_{s}(x)=-\cfrac{{c}^{2}{d}_{normal}}{2{G}_{o}}(x)+{d}_{earth}\)
We see clearly the absurdity of this formulation but as a first linear approximation it is OK. The factor
\(\cfrac { 1 }{ { d }_{ s }^{ 2 } } .\cfrac { d{ d }_{ s } }{ dx }\)
needs further consideration, in order that the space density equation make more sense. Consider, the expansion.
\({e}^{-b(x+{r}_{e})}={e}^{-b{r}_{e}}(1-bx+.....)=(-{e}^{-b{r}_{e}}bx+{e}^{-b{r}_{e}}....).\)
then \({d}_{s}(x)={e}^{-b(x)}{e}^{-b{r}_{e}}\), where \({e}^{-b{r}_{e}}={d}_{earth}\), \( b=\cfrac { { c }^{ 2 }{ d }_{ normal } }{ 2{ G }_{ o }{d}_{earth} }\) and \({ e }^{ -\cfrac { { c }^{ 2 }{ d }_{ normal } }{ 2{ G }_{ o }{d}_{earth} } { r }_{ e } }={ d }_{ earth }\)
where \(a, b\) are constants and \({r}_{e}\) is defined such that x=0 is on the surface of earth where space is most compressed. By comparing coefficients, we have a consistent expression,
\({ d }_{ s }(x)={ e }^{ -\cfrac { { c }^{ 2 }{ d }_{ normal } }{ 2{ G }_{ o }{d}_{earth}} (x) }{ e }^{ -\cfrac { { c }^{ 2 }{ d }_{ normal } }{ 2{ G }_{ o }{d}_{earth} } { r }_{ e } }+{d}_{normal}\)
The last term is added because we know that \(x\rightarrow \infty \), \({d}_{s}\rightarrow{d}_{normal}\) and the term constant term \(e^{-b{r}_{e}}\) is modified to
\({ e }^{ -\cfrac { { c }^{ 2 }{ d }_{ normal } }{ 2{ G }_{ o }{d}_{earth} } { r }_{ e } }={ d }_{ earth }-{d}_{normal}\).
\({ d }_{ s }(x)={ e }^{ -\cfrac { { c }^{ 2 }{ d }_{ normal } }{ 2{ G }_{ o }{d}_{earth}} (x) }({ d }_{ earth }-{d}_{normal})+{d}_{normal}\)
This derivation however, will lead to unfamiliar expressions, although we have seen the first approximation to be the same as textbook derivation. Have a nice day.
