\(v^{ 2 }_{ { max } }=\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\} e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }\) --- (*)
where \(\psi(x)=-ln(cosh(x-\cfrac{1}{2}x_a)+C\) without scaling both \(x\) and \(y\). With proper scaling,
\(\psi=-i{ 2{ mc^{ 2 } } }\,ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }(x-x_z)))+c\)
which is the case for the post "Twirl Plus SHM, Spinning Coin" dated 17 Jul 2015. \(-i\) denotes the direction of \(\psi\) and can be ignored as we deal with magnitude here. From that post,
\(\cfrac { \psi _{ d } }{ m } =2{ c^{ 2 } }cos(\theta )ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } x_z))\)
where \(\psi_d=\psi_{max}-\psi_n\) is to oscillate on the surface of \(\psi_{max=77}\) at \(x=x_z\) where
\(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } x_z=3.135009\) and so that,
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } x_z))=2.4438\)
for one charge or one particle.
When we rearrange (*)
\(v^{ 2 }_{ { max } }=-\cfrac { \psi _{ n }-\psi _{ max }}{ m }. \cfrac { \psi _{ n } }{ \psi _{ max } }\ e^{ \psi _{ max } }.\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }\)
\(v^{ 2 }_{ { max } }=-\cfrac { \psi _{ d }}{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\ e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }\)
and we set, \(\psi_{max}=\psi(x_z)=ln(cosh(3.135009)=2.4438\), and in both solutions \(n=1\) and \(max=77\)
\(\cfrac { \psi _{ n } }{ \psi _{ max } }=\cfrac{1}{4}.\cfrac{1}{2.4438}\) a constant irrespective of the scaling needed to the \(y-axis\).
We have,
\(v^{ 2 }_{ { max } }=-2{ c^{ 2 } }cos(\theta )*2.4438*\cfrac{1}{4}*\cfrac{1}{2.4438}.e\left( { e^{ 2 }-1 } \right) ^{ 1/2 }\)
because
\(e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }=e\left( { e^{ 2 }-1 } \right) ^{ 1/2 }\)
as we set \(\psi _{ max }=1\); \(\psi_{max}\) to be one particle.
Therefore,
where the negative sign provides a \(i\) that indicates \(\psi_{n=1}\) escape perpendicular to the radial distance \(x\).
If we apply this to water molecules, assuming that they are sufficiently spherical,
\(cos(\theta )\approx 1\) as \(\theta\approx0\)
The density of water is 1000 kg m-3, and there are a total of ten particles of each type (counting total atomic numbers) per water molecule,
\(\cfrac{1}{2}m_{\rho}v^{ 2 }_{ { max } }=E_{input/proton}=\cfrac{1}{2}c^{ 2 }*3.4354*1000*\cfrac{1}{10}\)
where \(E_{input/proton}\) is the energy density input per particle in a unit volume that would eject a basic particle \(\psi_{n=1}\) or \(\psi_{c}\). We have,
\(E_{input/proton}=\cfrac{1}{2}(one\,\,particle)*c^{ 2 }*(one\,\,unit\,\,volume)*343.54\)
since the particles are already at light speed, \(c\) in the unit volume. We would need to move in \(343.54\) such volume per second. ie at a speed of \(343.54 ms^{-1}\)
For this reason, a sonic boom at the speed of \(343.54 ms^{-1}\).
Don't take this too seriously...
Note: Missing post remains missing. This hopefully replaces the lost post. Each type of particles does not interact with another different type, eg. gravity particles do not interact with protons.
With
\(\theta_{\pi}=3.135009\)
\(\psi_{max}=\psi(x_z)=2mc^2ln(cosh(\theta_{\pi})\) was set to one, ie
\(2mc^2ln(cosh(\theta_{\pi}))=1\)
that corresponds to the definition/observation that \(\psi_{max}\), \(n=77\) is one particle. When the scaling factor \(2mc^2\) is also considered; as we call \(\psi_{max}\) one particle,
\(m=\cfrac{1}{2c^2}*\cfrac{1}{2.4438}=2.2765e-18\)
we have also defined its mass density.
