Sonic Boom

From the post "No Solution But Exit Velocity Anyway" dated 14 Jul 2015,

$v^{ 2 }_{ { max } }=\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\}  e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }$ --- (*)

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,

$v^{ 2 }_{ { max } }=-c^{ 2 }cos(\theta )*3.4354$

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.