Equating To A Indefinite Integral

This post is invalid because the post "Into A Pile Of Deep Shit" dated 09 Jun 2016, has been corrected.

$q=\psi$

although the discussion of the Lagrangian as Leftover energy is valid.

From the previous post "Into A Pile Of Deep Shit" dated 09 Jun 2016,

${ q_{a_{\psi}} } =2\dot { x }F_{\rho}|_{a_{\psi}}$

$F_{ \rho  }=i\sqrt { 2{ mc^{ 2 } } } \, G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{ z }) \right)$ --- (1)

For simplicity, $x_z=0$ and

$\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (a_{\psi})=\pi$

and

$tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (a_{\psi}) \right)=1$ per unit volume

and so,

${ q_{a_{\psi}} } =i4\pi { \cfrac { \dot { x }  }{ a_{ \psi  } } mc^{ 2 } }$

where the imaginary sign, $i$ has been returned and $i\dot{x}$ is in the direction perpendicular to the radial line along $a_{\psi}$.  $i\dot{x}$ is going around a circle with radius along a radial line, in this case the radius is $x=a_{\psi}$.

$F_{\rho}$ is force per unit volume, so the term,

$i\sqrt { 2{ mc^{ 2 } } } \cfrac{\pi}{a_{\psi}} \sqrt { 2{ mc^{ 2 } } }=i2\pi { \cfrac { 1 }{ a_{ \psi  } } mc^{ 2 } }$

the right hand side of expression (1), is also force per unit volume.  And so,

${ q_{a_{\psi}} } =2\dot { x }F_{\rho}|_{a_{\psi}}=i4\pi { \cfrac { \dot { x }  }{ a_{ \psi  } } mc^{ 2 } }$

is power per unit volume.  This is the total power from an conceptual unit sphere centered at the particle.  This power is divided over the surface area of the particle of radius $a_{\psi}$; this is actually intensity,

$\cfrac { q_{ a_{ \psi  } } }{ 4\pi \varepsilon _{ o }a^{ 2 }_{ \psi  } } =\cfrac { 1 }{ 4\pi \varepsilon _{ o }a^{ 2 }_{ \psi  } } i4\pi { \cfrac { \dot { x }  }{ a_{ \psi  } } mc^{ 2 } }=i{ \cfrac { \dot { x }  }{ a^{ 3 }_{ \psi  } } \cfrac { mc^{ 2 } }{ \varepsilon _{ o } }  }$

the power along a radial line when the source is spherical or power per unit area.

All forces distributed on the uniform sphere except the radial component will cancel; only the radial component of the force associated with this power remains.  Alternatively, given two such power spheres at $\Delta x$ distance apart at the surface, (ie radii differ by ($\Delta x$), a force develops from the higher power sphere to the low power sphere along a radial line.

$\cfrac{\partial\,F}{\partial\,t}=-\cfrac{\partial\,P}{\partial\,x}$

This is where we have an extra dimension of "per second", compared to previously when

$F=\cfrac { q_{a_{\psi}} }{4\pi \varepsilon_o a^2_{\psi} }$

In the latter, there is no natural resonance frequency and the associated phenomenon of  a body gaining potential energy while in resonance in a field cannot be accounted for.

What happens when we equate,

$q=4\pi { \cfrac { \dot { x }  }{ a_{ \psi  } } mc^{ 2 } }=4\pi a^2_{\psi}m$ --- (1)

??

When we integrate,

$F=\int{F_{\rho}}dx$

where $F_{\rho}$ is given by (1).

$F=i2{ mc^{ 2 } }.ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{ z }))$

with $x_z=0$ for simplicity.

$F$ is the force radiating from a sphere per unit area with radius $x$, centered at the particle.  This force is along a radial line $x$.  $F$ can be interpreted as either force per unit area, or energy (work done) along $dx$ per unit volume, as $dx\rightarrow 0$.

When we formulate expression (1),

we are equate the energy change while crossing the spherical boundary of radius $a_{ \psi}$ with the velocity $i\dot{x}$, $\dot{x}$ being along a radius line over an infinitesimal displacement of  $dx$.  Because, at $x=a_{\psi}$,

$\Delta E=\int_{a_{\psi}-\Delta x/2}^{a_{\psi}+\Delta x/2}{F_{\rho}}dx$

$\dot{x}=\cfrac{dx}{dt}$

$\Delta E=\dot{x}\int_{a_{\psi}-\Delta x/2}^{a_{\psi}+\Delta x/2}{F_{\rho}}\cfrac{dx}{\dot{x}}$

$\Delta E=\dot{x}\int_{-\Delta t/2}^{+\Delta t/2}{F_{\rho}}dt$

$\Delta E=\dot{x}\left(F_{\rho1}\cfrac{\Delta t}{2}-F_{\rho2}(-\cfrac{\Delta t}{2}) \right)$

where $F_{\rho1}$ is $F_{\rho}$ at $\cfrac{\Delta t}{2}$ and $F_{\rho2}$ is $F_{\rho}$ at $-\cfrac{\Delta t}{2}$.  As $\Delta t\rightarrow 0$

$F_{\rho1}\rightarrow F_{\rho2}\rightarrow F_{\rho}$

$\Delta E=\dot{x}F_{\rho}\Delta t$

$\cfrac{\Delta E}{\Delta t}=P=\dot{x}F_{\rho}$

as $\Delta t\rightarrow 0$

Which leads us back to the power, $P$ per unit volume along $x$ about $x=a_{\psi}$.

This invalidates the expression,

$q=4\pi { \cfrac { \dot { x }  }{ a_{ \psi  } } mc^{ 2 } }=4\pi a^2_{\psi}m$

because we equated to an indefinite integral that should have been an definite integral along $x$ between $x=a_{\psi}-\Delta x/2$ and $x=a_{\psi}+\Delta t/2$.

$4\pi { \cfrac { \dot { x }  }{ a_{ \psi  } } mc^{ 2 } }\ne4\pi a^2_{\psi}m$

but if $\psi$ of  the system (eg. an uniformly distributed charges over the surface of a conductor) is conserved.

$i4\pi { \cfrac { \dot { x }  }{ a_{ \psi  } } mc^{ 2 } }=q=constant$

When we replace $c$ with $i\dot{x}$, where we had previously assume light speed,

$i4\pi { \cfrac { \dot { x } ^3 }{ a_{ \psi  } } m }=-q$

which suggests a normally negative charge associated with a particle.

$i\dot { x }^3=-\cfrac{1}{4\pi  }\cfrac{q}{m} a_{ \psi  }$

$T=\cfrac{1}{2}m \dot { x }^2$  Is this true for a wave??

$\cfrac{\partial\,T}{\partial\,x}=\cfrac{1}{2}m 2.\dot { x }\cfrac{\partial\,\dot{x}}{\partial\,x}$

$\cfrac{\partial\,T}{\partial\,x}=m \dot { x }\cfrac{\partial\,\dot{x}}{\partial\,x}$

$\cfrac{\partial\,T}{\partial\,x}$ is dependent on $\dot { x }$.

Normally,

$\cfrac{\partial\,\dot{x}}{\partial\,x}=0$

but, $i\dot{x}$ is perpendicular to $x$, the change in $ix$ need not be equal to the change in $x$ per change in time.  ie,

$\cfrac{\partial\,i\dot{x}}{\partial\,x}=\dot{\left(\cfrac{\partial\,ix}{\partial\,x}\right)}=\cfrac{\partial}{\partial\,t}\cfrac{\partial\,ix}{\partial\,x}$

but

$\cfrac{\partial\,ix}{\partial\,x}\ne 1$

So,

$\cfrac{\partial}{\partial\,t}\cfrac{\partial\,ix}{\partial\,x}\ne0$

this allows the wave $\psi$ to slow at the boundary,

$\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (a_{\psi})=\pi$

such that,

${ q_{ a_{ \psi  } } } =3k-2\cfrac { \partial \, T }{ \partial \, x }|_{a_{\psi}}$

can take on a different sign.  This expression is from the post "Deep Blue Deeper" dated 01 Jun 2016 where,

$\cfrac{\partial\psi}{\partial x}=\cfrac{\partial V}{\partial x}+\cfrac{\partial T}{\partial x}=k$

at $x=a_{\psi}$

and the post "Why A Positron And Deep Blue..." dated 01 Jun 2016

$q=\left[3 \cfrac { \partial V\,  }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi}}$

Have a nice day.

Note:  The wave $\psi$ is a also particle, with kinetic and potential energy.  When such a particle enters a field with initial kinetic energy $T$, all its potential energy is derived from this initial energy $T$.  What's left of its kinetic energy is,

$L=T-U$ or $L=T-V$

the Lagrangian.

From this $L$eftover energy we derive the Lagrange equations of motion from Hamilton’s Principle, which is Newton's law at some level.  $T$, kinetic energy is the part of the particle's energy in space.   Potential energy is the other part of the particle's total energy in the respective time dimension, $t_c$, $t_g$ or $t_T$.  The particle as a wave has energy oscillating between one space dimension and one of three time dimensions. So,

$T=\cfrac{1}{2}m \dot { x }^2$

is true.  What about light speed being constant and light being a wave?