\(\cfrac{F}{m}\cfrac { \partial ^{ 2 }\, \psi }{ \partial \, x^{ 2 } } =-i c^{ 2 } (\cfrac { \partial ^{ 2 }F }{ \partial \, x^{ 2 } } )\)
but \( \cfrac { \partial ^{ 2 }\, \psi }{ \partial \, x^{ 2 } } =\cfrac { \partial F }{ \partial \, x } \)
So,
\(F\cfrac { \partial \, F }{ \partial \, x } =-i mc^{ 2 } (\cfrac { \partial ^{ 2 }F }{ \partial \, x^{ 2 } } )\)
\( \cfrac { 1 }{ 2 } \cfrac { \partial \, F^{ 2 } }{ \partial \, x } =-imc^{ 2 }(\cfrac { \partial ^{ 2 }\, F }{ \partial \, x^{ 2 } } )\) --- (*)
If we consider a constant term after the first integration,
\(F^{ 2 }-A=-i2{ mc^{ 2 } }\cfrac { \partial F }{ \partial \, x } \)
This term corresponds to loss in \(F\) as it progresses along \(x\). This is equivalent to \(F\) being in a medium with mass, ie. space has inertia.
\( \int { 1 } d\, x=-i2{ mc^{ 2 } }\int { \cfrac { 1 }{ F^{ 2 }-A } } dF\)
\(x=i2{ mc^{ 2 } }\cfrac { 1 }{ \sqrt { A } } archtanh(\cfrac { F }{ \sqrt { A } } )+x_z\)
\(-i\cfrac { \sqrt { A } (x-x_z) }{ 2{ mc^{ 2 } } } =archtanh(\cfrac { F }{ \sqrt { A } } )\)
\(F=\sqrt { A } tanh\left( \cfrac { {- i\sqrt { A } } }{ 2mc^{ 2 } }( x-x_z) \right) \)
From (*) it is possible that
\(F^{ 2 }=-2i{ mc^{ 2 } }\left(\cfrac { \partial F }{ \partial \, x }+D^2\right) \)
where \(D^2\) is the constant of integration,
\(F^{ 2 }+i2{ mc^{ 2 }D^2 }=-i2{ mc^{ 2 } }\cfrac { \partial F }{ \partial \, x } \)
and so,
\( A=-i2{ mc^{ 2 } }D^2=e^{i3\pi/2}.2{ mc^{ 2 } }D^2\),
\( \sqrt{A}=e^{i3\pi/4}D\sqrt{2{ mc^{ 2 }}}\),
\(-i\sqrt{A}=e^{-i\pi/2}.e^{i3\pi/4}D\sqrt{2{ mc^{ 2 }}}=e^{i\pi/4}D\sqrt{2{ mc^{ 2 }}}\)
Therefore,
\(F=e^{ i3\pi /4 }D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } } }( x-x_o).e^{ i\pi /4 } \right) \)
If \(G=D.e^{ i\pi /4 }\) is real then
\(F=e^{i\pi/2}\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } (x-x_z) \right) \)
\(F=i\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } (x-x_z) \right) \)
where \(e^{i\pi/2}=i\) is interpreted as a phase lag in time. An illustrative plot of \(-F\), which is often the case where minus gravity, \(-g\) is shown,
This solution suggests that there is a zero gravity value in the space around a body at some point \(x_z\), beyond this point gravity is positive and pushes other mass away.
The series expansion of \(\tanh(x)\) is given by,
\(tanh(x)=x-\frac { 1 }{ 2 } x^{ 3 }+\frac { 2 }{ 15 } x^{ 5 }-\frac { 17 }{ 315 } x^{ 7 }...\left| x \right| \)
\(tanh(x)\approx x\). This is a different result from when \(A=0\).
This force, \(F\) was derived from energy density, as such it is a force per unit volume, ie Nm-3. To obtain the normal Newtonian formulation, \(F_g\),
\(F_g(x)=F\cfrac{4}{3}\pi x^3\cfrac{1}{4\pi x^2}\cfrac{1}{4\pi x^2}=F\cfrac{1}{12\pi x}\)
Note: \(arctan\left( x \right) =i\frac { 1 }{ 2 } ln\left( \cfrac { 1-ix }{ 1+ix } \right) \)
