High Frequency EMW From Laser

When we reverse the direction of circular motion of the particles, in the post "It's All Fluorescence Outside, Inside" dated 29 Jul 2015,

The $E$ field lines looking into the direction of of propagation, along $OP$,

This however, is an EMW, where

$B=-i\cfrac{\partial\,E}{\partial x}$

If both directions of rotation are equally probable then, a laser should also be radiating high frequency EMW.  Both light and EMW outputs share the input power.

Transverse Mode And Pretty Flowers

From the post "It's All Fluorescence Outside, Inside" dated 29 Jul 2015,

the rotating $E$ fields generated by $g^{+}$ and $T^{+}$ particles interact and give rise to the Transverse Modes in a laser.

An aperture that restrict the spread of this field in the plane perpendicular to $OP$, prevents such interactions and "destroys" the transverse modes.  A conductive wire within the resonance cavity of the laser would also short the $E$ fields and prevent their interaction in certain directions and so produce different interference patterns.

Have a nice day.

Pole Dancing

From the previous post "Clock Face And Life, Two Pies And Habits" dated 30 Jul 2015,

$\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta  }) }=-\int^{\infty}_{-\infty}{it.g(\omega t)e^{-i\omega t}}d\,\omega=-it.G(t)$

if instead,

$\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta  }) }=-\int^{\infty}_{-\infty}{i\omega.g(\omega t)e^{-i\omega t}}d\,t=-i\omega.G(\omega)$

which is exactly Fourier.  However,

since,  $\theta=\omega t$,   $\omega=2\pi.\cfrac{1}{T}$

$t=\cfrac{T}{2\pi}\theta$

and

$\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta  }) }=-i\cfrac{T}{2\pi}\theta.G(\cfrac{T}{2\pi}\theta)$

If I go fishing,

$\int ^{\infty}_{-\infty}{\cfrac{h(z)}{(z-z_o)}d(e^{ -i\theta  }+z_o) }=-i\cfrac{T}{2\pi}\theta.G(\cfrac{T}{2\pi}\theta)$ --- (*)

where $f(z)=\cfrac{h(z)}{(z-z_o)}$, $z_o$ a complex constant.

from the post "What Holomorphic?" dated 30 Jul 2015.  When we consider $-\infty$  to  $\infty$ as the biggest close loop possible.  From Cauchy's integral formula,

$h(z_o)=\cfrac{1}{2\pi i}\oint_c{\cfrac{h(z)}{(z-z_o)}}dz$

Unfortunately on the complex plane, $z_o$ cannot just disappear; is (*) still valid?  If so,

$\oint_c{\cfrac{h(z)}{(z-z_o)}}dz=\int ^{\infty}_{-\infty}{\cfrac{h(z)}{(z-z_o)}d(e^{ -i\theta  }+z_o) }$

$2\pi ih(z_o)=i\cfrac{T}{2\pi}\theta_{z_o}. g\left(\cfrac{T }{2\pi}\theta_{z_o} \right)$

$h(z_o)=\cfrac{T}{(2\pi)^2}\theta_{z_o}. g\left(\cfrac{T }{2\pi}\theta_{z_o} \right)$

where $z_o=R_{\theta_{zo}}e^{i\theta_{zo}}$

$h(z_o)$ depends on $\theta_{z_o}$ only.  Which is very odd, I loop around a pole with $e^{-i\theta}$ but it does not matter how long my reach is.

This must be wrong.  Previously, the dependence on $R_\theta$ disappeared because I am always on a unit circle.  If we change $e^{-i\theta}\to R_{zo}e^{-i\theta}$ such that we loop around $z_o$,

$\int ^{\infty}_{-\infty}{\cfrac{h(z)}{(z-z_o)}d(R_{zo}e^{- i\theta  }) }=R_{zo}.i\cfrac{T}{2\pi}\theta. g\left(\cfrac{T }{2\pi}\theta \right)$

then,

$h(z_o)=R_{zo}.\cfrac{T}{(2\pi)^2}\theta_{z_o}. g\left(\cfrac{T }{2\pi}\theta_{z_o} \right)$

provided $z_o$ is the only pole within the circle $R_{zo}e^{-i\theta}$.  Still,

$\oint_c{\cfrac{h(z)}{(z-z_o)}}dz\ne \int ^{\infty}_{-\infty}{\cfrac{h(z)}{(z-z_o)}d(R_{zo}e^{- i\theta  }) }$

And this is all wrong.  Unless $\oint_c{...}dz$ allows us to go around in circle specifically.

$\oint_c{\cfrac{h(z)}{(z-z_o)}}dz=\oint_{cR}{\cfrac{h(z)}{(z-z_o)}}d(R_{zo}e^{- i\theta  })=R_{zo} \int ^{\infty}_{-\infty}{\cfrac{h(z)}{(z-z_o)}d(e^{ -i\theta  }) }$

which is unfortunately valid for any $R\gt R_{zo}$.  Which leads us to strict $R_{zo}\lt R\lt R_{zo}-\Delta R$.
...

All these suggest that with,

$\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta  }) }=F(\omega)$

where $\theta=\omega t$,

we have just extracted the argument part of the function $f(z)$.

$\cfrac{d F(w)}{d(e^{-i\theta})}=f(z)$

put back the argument part of the function.

Mean And Variance Polarization

Cauchy Distribution does not have a mean, but

$\int^\infty_0{x.f(x)}dx=\int^\infty_0{\cfrac{2x^2}{\pi(1+x^4)}}dx=\cfrac{1}{\sqrt{2}}$

when we consider both $-\theta$ and $\theta$

$E\left\{x\right\}=2\cfrac{1}{\sqrt{2}}=\sqrt{2}$

The emitted photons, assuming that $\theta$ is uniformly distributed, $-\infty\lt\theta\lt\infty$, has an expected value of $\sqrt{2}$.

But,

$\int{x^2.f(x)}dx=\int{\cfrac{2x^3}{\pi(1+x^4)}}dx=\cfrac{1}{2\pi}ln(x^4+1)$

does not converge for all values of $\theta$, however for the range $0\le x\le \cfrac{\pi}{2}$,

$E\left\{x^2\right\}=0.31169$

in which case, over the same range,

$E\left\{x\right\}=0.314044$

and the variance is,

$Var\left\{x\right\}=2*0.31169-(2*0.314044)^2$

$Var\left\{x\right\}=0.22889$,   $\sigma=\sqrt{Var\left\{x\right\}}=0.4784$

Polarization mean and variance has little meaning, compared to mode at $\cfrac{1}{\sqrt[4]{3}}$.

What Holomorphic?

From the previous post "Clock Face And Life, Two Pies And Habits" dated 30 Jul 2015,

$\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta  }) }=-\int^{\infty}_{-\infty}{it.g(\omega t)e^{-i\omega t}}d\,\omega=-it.G(t)$

$\theta=\omega t$ and $z=R_\theta e^{i\theta}$

$R_\theta$ being a function of $\theta$.

$\int { f(z)d(e^{ -i\theta  }) }$

takes a free ride with Fourier.  Must it still be holomorphic?

$f(z)=g(\omega t)$

as long as the Fourier tansform of $g(\omega t)$ exist,  $\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta  }) }$ is valid.  In the case of simple periodicity, $T$ a constant,

$\int ^{\infty}_{-\infty}{ f(z)d(e^{ -i\theta  }) }=\int ^{T}_{0}{ f(z)d(e^{ -i\theta  }) }=-it.G(t)|^{T}_0=-i.TG(T)$

where $G(t)$ is the Fourier transform of $f(z)$ with $z$ replaced by $\omega$.  $-i$ is in the direction of $\theta=0$.

Most liberating.

Note: $-i.tG(t)$ is a complex function ($\theta=\omega t$ being real), with a dual $f(z)$, running around in circles in the complex plane.

$e^{i\theta}$ being replaced with $e^{-i\theta}$.

Clock Face And Life, Two Pies And Habits

Instead of,

$\cfrac{df(z)}{dz}$  or  $\int{f(z)}dz$

we do,

$\cfrac{df(z)}{d(e^{-i\theta})}$  or  $\int{f(z)}d(e^{-i\theta})$

or equivalently,

$\cfrac{df(z)}{d(-\hat{z})}$  or  $\int{f(z)}d(-\hat{z})$

where $-\hat{z}$ denote the unit vector opposite to the $z$ direction.

This way $-d\hat{z}$ or $d(e^{-i\theta})$ is opposite to the direction of $z$; at an angle $-\theta$ to the $x$ axis on the $i-x$ plane.  Furthermore,

$\int { f(z)d(e^{ -i\theta  }) }=- i\int { f(z)e^{ -i\theta  } } d\theta$

and

$\cfrac { df(z) }{ de^{- i\theta  } } =-\cfrac { 1 }{ ie^{ -i\theta  } } \cfrac { df(z) }{ d\theta  }$

In the famous counter example to holomorphic function, $f(z)=\bar z$

$-\cfrac { 1 }{ ie^{ -i\theta  } } \cfrac { d\bar { z }  }{ d\theta  } =-\cfrac { 1 }{ ie^{ -i\theta  } }\left\{R\cfrac { de^{ -i\theta  } }{ d\theta  }+e^{ -i\theta  }\cfrac { d\,R}{ d\theta  } \right\}  =-\cfrac { 1 }{ ie^{- i\theta  } } \left\{Re^{ -i\theta  }.(-i)+e^{ -i\theta  }\cfrac { d\,R}{ d\theta  }\right\}=R+i\cfrac{d\,R}{d\theta}$

this result is independent of $\theta$ when $R$ is a constant (a circle) or has a linear dependence on $\theta$ (ie $\small \cfrac{d\,R}{d\theta}=constant$, a linear spiral).

Furthermore,

$\int { f(z)d(e^{- i\theta  }) }= f(z)e^{-i\theta}-\int{e^{-i\theta}}df(z)=- i\int { f(z)e^{ -i\theta  } } d\theta$

when we have,

$\theta=\omega t$   and  $z=R(\omega t)e^{-i\omega t}$

$f(z)=g(\omega t)$

$\int { f(z)d(e^{ -i\theta  }) }=  -i\int { g(\omega t)e^{- i\theta  } } d\theta=-i\int{g(\omega t)e^{-i\omega t}}.td\omega$

In fact,

$\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta  }) }=-it\int^{\infty}_{-\infty}{g(\omega t)e^{-i\omega t}}d\omega=-it.G(t)$

we have Fourier Transform on the RHS of the expression,

$-it.G(t)$

$t$ being independent of $\omega$, $G(t)$ is the Fourier transform of $g(\omega t)$ or $f(z)$ with parameter $t$, and the $-i$ factor returns us to the positive $\theta=0$ direction.

On the LHS, we have $f(z)$ integrated around a unit circle $e^{-i\theta}$, for all values of $\theta$ from $-\infty$ to $\infty$.

As we go around in circles with $f(z)$, we travel linearly through time $t$ by $G(t)$.

$f(z)$ should be called the clock function, and $G(t)$ our life function.

$g(\omega t)$ is our habits, the sum of which in all magnitudes of frequencies is our life, $G(t)$.  $g(\omega t)$ is the habit function.

Note:  After the correction $e^{i\theta}\to e^{-i\theta}$, life is no longer negative.

Delay, Once Only

From,

$E_p=\hbar\left\{1-\sqrt{tan(\theta_v)}\right\}\cfrac{c}{x_c}$

for $0\lt\theta\lt45^o$,

$E_p\gt0$

emission is followed by absorption.

For $45^o\lt\theta\lt90^o$,

$E_p\lt0$

absorption is followed by emission, as a result there is a small delay before emission.

At $\theta=45^o$,

$E_p=0$

there is no emission, absorption occurs once of the initial excitation photon.

At $\theta=90^o$,

$E_p\to\infty$

Not probable, but $\small{\theta}$ can come close to ninety degrees.

It's All Fluorescence Outside, Inside

In the case of a laser with internal mirror both peaks in the fluorescence response are set oscillating.  They constitute the two orthogonal modes of the laser.

In the case of a laser with external mirror, only the higher intensity left emission peak on the fluorescence response (right arm on the $v^2$ vs $x$ curve) is set oscillating.  The low intensity right emission peak decayed.  Only one mode is in the output of the laser.

This suggests $\theta$ is polarization.

with,

$-\theta\equiv\theta$

negative $\theta$ is equivalent to positive $\theta$.

As the particle performs circular motion, $x_d$ rotates along the perimeter of the circle.  When we hold $x_d$ fixed,

the particle performs circular motion in a rotating circle whose diameter traces the slant surface of a cone.  The particle is always on $x_d$; the circular path rotates at an incline $\theta$ to the direction of $x_d$, $PO$.  $PO$ is the direction of incident.

This way,

$-\theta\equiv\theta$.

If particle has energy oscillating in the orthogonal dimension, $t_c$ (eg. $g^{+}$,  $T^{+}$), the particle in circular motion creates an $E$ field given by the right hand screw rule.  This field rotates as the circle rotates, and always makes an angle of $\small{90^o-\theta}$ with $PO$.  This rotating $E$ field does not mean circular polarization.  Polarization as detected by rotating a polarizer in front of a laser is $\theta$.

This rotating $E$ field is consistent with the rotating dipole model from which we have,

$B=i\cfrac{\partial\,E}{\partial x}$

from the post "Whacko and the Free Photons" dated 30 May 2014.

The $E$ field is always rotating given non zero $\theta$.  When $\theta=90^o$, $E$ is parallel to $PO$.  From,

$E_p=\hbar\left\{\sqrt{tan(\theta_v)}-1\right\}\cfrac{c}{x_c}$

$E_p\to\infty$

In this case, the particle travels along $PO$ with circular motion.  At the center $O$, the particle is in circular motion but with $v_{shm}\approx 0$.  The particle no longer oscillate.  Emission at $\small{E_p\to\infty}$ occurs only once.

We are assuming that all forces on the particle are finite as such,

$F_{cir}=m\cfrac{v^2_{cir}}{x_c}$

$x_c\to 0$,    $v_{cir}\to 0$  as  $\theta\to90^o$, for $F_{cir}$ finite.

Squeeze Laser

From the posts "A Pump!"  and  "Polarization And Invisibility" dated 25 Jul 2015, when

$\theta=45^o$

no photons are emitted, the particle continues to oscillate.  We do not have resonance, as the amplitude of oscillation cannot increase further, but we have no loss from the oscillation.

When

$\theta=\cfrac{1}{\sqrt[4]{3}} \small{rad}=43.535^o$

we have maximum intensity.

If it is possible, to negate the mere $\small{\Delta \theta=1.4646^o}$  difference, it would be possible to squeeze the the energy stored at $\small{\theta=45^o}$ and releases them at maximum intensity.

At $\small{\theta=45^o}$,

$x_d=\cfrac{x_v}{cos(45^o)}=x_v\sqrt{2}$

At  $\small{\Delta \theta=1.4646^o}$,

$x_d=\cfrac{x_v}{cos(\cfrac{1}{\sqrt[4]{3}})}=1.3794x_v$

A squeeze factor of $0.9754$.

Light amplification squeeze emission radiation!  Almost pornographic.

This photons released when the material (eg. iron) is hit and deformed could account for the flashes seen.

Not Exactly A Fluorescence Polarizer

$tan (\theta)$ has a Cauchy distribution,

$f(x)=\cfrac{1}{\pi(1+x^2)}$

for $-\cfrac{\pi}{2}\lt\theta\lt\cfrac{\pi}{2}$

then $\sqrt{|tan(\theta)|}$ has a distribution

$f(x)=\cfrac{1}{\pi(1+x^4)}|2x|$

for $-\cfrac{\pi}{2}\lt\theta\lt\cfrac{\pi}{2}$  in blue below.

These set of peaks for the probability density of $\sqrt{tan(\theta)}$ are not at $\cfrac{\pi}{2}$ apart but $\cfrac{2}{\sqrt[4]{3}}$ apart.

$\cfrac{2}{\sqrt[4]{3}}=1.5197\,\, \small{rad}=87.071^o\ne90^o$

From,

$\cfrac{E_{p\,v}}{\hbar}=\left\{1-\sqrt{tan(\theta_v)}\right\}\cfrac{c}{x_v}$

The probability density of $1-\sqrt{tan(\theta_v)}$ has been right shifted by $1$.  The peak intensity of $E_p$ occurs not at $\theta=0\,\,\small{rad}$ but at $\theta=1\,\, \small{rad}=57.296^o$.

$a=\cfrac{c}{x_v}$

is assumed constant.

Note:  Also,

$\cfrac{E_{p\,v}}{\hbar}=\left\{\sqrt{tan(\theta_v)}-1\right\}\cfrac{c}{x_v}$

in which case two peaks occurs centered about $\theta=-1\,\, \small{rad}$,  $\theta=1.5197\,\, \small{rad}$ apart.  A notch occurs at $\theta=-1\,\, \small{rad}$.

The Thin And Fat Issue

Why is the variance of the absorbed photons greater than the emitted photons of either peaks?

The variance around the absorption peak is due to the spread of particle velocities around mode/mean $v_m$.

The variance at the two fluorescence peaks are due to the variance in $\theta$ as both $x_p$ and $x_v$ in ratio, distribute themselves around their respective mean values.  $c^2$ is a constant.  Although, from the post "Emmy Noether" dated 6 Jul 2015,

$299792458=\cfrac{1}{2}(\cfrac{\bar\alpha N_p}{\bar\alpha N_p-1})=c$

where $\bar\alpha$ is the average fraction energy shared, and $N_p$ total number of particles in entanglement.  Both $\bar\alpha$ and $N_p$ has a mean and a variance, for a specific instance of a particle.

Bullshit is otherwise fiction, if not for bulls.

It's Fluorescence

The emitted photons frequencies are less than the absorbed frequencies, emitted after a time delay; this is fluorescence.

Stokes shift is the direct result of,

$E_p=h.\Delta f_{cir}$

as

$\Delta f\lt f_{cir}$

And the spectra overlap occurs exactly at the cross point,

 $f_{cross}=f_{max}+\cfrac{1}{2}\Delta f$.

The second lower hump on the left corresponds to secondary photons absorbed that drive the particle beyond $v^2_{min}$ up the two arms of the $v^2$ vs $x$ curve.  That is the Delta Spectrum.

The graph above allow us to obtain $\theta$,

$\Delta f=f_{blue\,max}-f_{red\,max}=\left\{1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right\}.f_{cir}=\left\{1-\sqrt{{tan(\theta)}}\right\}.f_{cir}$

Maximum absorption corresponds to an energy level of $\cfrac{1}{2}mc^2$, where the particle is orbiting at speed $c$ afterward (from the post "Initial High Energy Photon" dated 26 Jul 2015).  The majority of the particle is at a nominal energy state, $v^2_m\lt c^2$ given temperature, after absorption of photons (at  $\cfrac{1}{2}m(c^2-v^2_m)$) most particle will be in circular motion at speed $c$.

So,

$f_{cir}=\cfrac{c}{2\pi x_r}$

$\Delta f=f_{blue\,max}-f_{red\,max}=\left\{1-\sqrt{{tan(\theta)}}\right\}.\cfrac{c}{2\pi x_r}$

where,

$x_r=x_v$  or  $x_c$

Since other particles along the two arms of the $v^2$ vs $x$ curve around $v^2_{min}$ can absorb smaller packets of energy (which results in the Delta Spectrum) and reach $c^2$, the emitted peak is higher than the absorption peak.  There is more particle with circular energy $\cfrac{1}{2}mc^2$ than those that absorbed an energy packet $\cfrac{1}{2}m(c^2-v^2_m)$ directly.

When particles can travel along both the left and right arms of the $v^2$ vs $x$ curve, the emission spectrum will have twin peaks as shown,

When the particles received higher energy photons and is moved to orbit of light speed, $c$, from the post "Initial High Energy Photon",

$cos(\theta_v)=\cfrac{x_p}{x_r}$

where $x_p$ is the initial orbital radius and $x_r$ is the orbital radius when the particle has speed $c$ according to the $v^2$ vs $x$ curve,

with

$x_v\lt x_c$

$cos(\theta_v)=\cfrac{x_p}{x_v}\gt cos(\theta_c)=\cfrac{x_p}{x_c}$

The change in frequency that results in photon emission is,

 $\cfrac{E_{p\,v}}{\hbar}=\left\{1-\sqrt{tan(\theta_v)}\right\}\cfrac{c}{x_v}=\left\{ 1-\sqrt { \frac { \sqrt { \left( \cfrac { x_{ p } }{ x_{ v } }  \right) ^{ 2 }-1 }  }{ \cfrac { x_{ p } }{ x_{ v } }  }  }  \right\} \cfrac { c}{ x_{ v } }$

$= \left\{ 1-\sqrt { \cfrac { x_{ v } }{ x_{ p } } \sqrt { \left( \cfrac { x_{ p } }{ x_{ v } }  \right) ^{ 2 }-1 }  }  \right\} \cfrac { c }{ x_{ v } }$

$=\left\{ 1-\sqrt [4]{ 1-\left( \cfrac { x_{ v } }{ x_{ p } }  \right) ^{ 2 } }  \right\} \cfrac { c}{ x_{ v } }$

as $x_p\gt x_c\gt x_v$

$\cfrac{x_v}{x_p}\lt \cfrac{x_c}{x_p}\lt 1$

From the following plot,

when $x_v\lt x_c$

$\left\{ 1-\sqrt [4]{ 1-\left( \cfrac { x_{ v } }{ x_{ p } }  \right) ^{ 2 } }  \right\} \cfrac { c }{ x_{ v } } \lt\left\{ 1-\sqrt [4]{ 1-\left( \cfrac { x_{ c } }{ x_{ p } }  \right) ^{ 2 } }  \right\} \cfrac { c }{ x_{ c } }$

$E_{p\,v}\lt E_{p\,c}$

Photons emission from the left arm has less energy than photon from the right arm.  Particles along the right arm result in energy emission closer to the absorption spectrum, at energy higher than the emission due to particles moving along the left arm.  In the case when the left arm is truncated,

less particles on the left arm absorb photons to the elevated state.  At the same time, since

$v^2_l\lt c^2$

$v_l\lt c$

$E_{p\,v}\lt\lt E_{p\,c}$

The second emission peak due to the left arm is lower in fluorescence than the right arm, and it is also of greater wavelength.

Sustained Oscillation?

But,

 $v^2(x_v)\ne v^2(x_v\sqrt{2})$ in general,

$v^2_{cir}\ne v^2_{shm}$

so,

$v_{cir}\ne v_{shm}$

However,

specifically there can be $v^2$ at $x$ such that $x\sqrt{2}$ is the other intersection on the $v^2$ vs $x$ graph along a horizontal $v^2$.  In stating,

$cos(\theta)=\cfrac{1}{\sqrt{2}}$

we have also specified the value for $v^2$ for this to occur.  So,

 $v^2(x_v)=v^2(x_v\sqrt{2})$

only for a specific value of $v$ if the solution exist.  If the solution does exist then the non emitting/absorbing mode of oscillation exist.  Whether the oscillation is sustained is another question.

How to detect such oscillation?  If this mode of oscillating leaks in any way, it is then not sustained.

It is even oscillating?!

As $v_{cir}$ and $v_{shm}$ changes along its path $\theta$ varies.  Because of the $sin(\theta)$ and $cos(\theta)$ terms, $v_{cir}$ leads $v_{shm}$.  At $x=0$, the center however, $v_{cir}=0$, $v_{shm}$ is also zero due to the common factor that they both share, as plotted above.  When $v_{cir}$ is maximum at $a$, $v_{shm}$ is zero due to their phase difference.

Although both $v_{cir}$ and $v_{shm}$ share the same graph, they are at 90^o phase difference.  When one is at the extrema the other is at zero and they are both zero at $x=0$.

And silent is the night.

Initial High Energy Photon

From the previous post,

$E_{p\,e}=h.\left\{1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right\}.f_{cir}$

$E_{p\,e}=-E_{p\,a}$

The energy of the absorbed photon that set off the oscillation about the center has less energy than the particle in circular motion.  As the energy of the absorbed photon decrease, the term

$cos(\theta)\to\cfrac{1}{\sqrt{2}}$

subsequent emitted photons has progressively less energy.  At,

$cos(\theta)=\cfrac{1}{\sqrt{2}}$

no photons are emitted nor absorbed as the particle oscillates.  Beyond this point the particle first absorb energy then emit energy during its oscillation.

What happens when the initially absorbed photon has higher energy than the particle in circular motion?

$x_v\gt x_1$

It is impossible to displace the circular motion in the direction perpendicular to its plane.  It is possible that in the first instant, the particle switches to circular motion with radius $x=x_1$ and is displaced along a radial line by $x=x_v$.  The particle oscillate along this radial line.  So,

$cos(\theta)=\cfrac{x_1}{x_v}$

and the particle emit/absorb energy as previously.

Active Absorption

Obviously,

$E_{p\,e}=h.\left\{1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right\}.f_{cir}$

$E_{p\,a}=h.\left\{1-\sqrt{\cfrac{cos(\theta)}{sin(\theta)}}\right\}.\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}f_{cir}=h.\left\{\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}-1\right\}f_{cir}=-E_{p\,e}$

The emitted photon is exactly the absorbed photon.

A Pump!

Consider,

a particle in orbit about the center of a $\psi$ cloud at light speed $c$.  On impact of a photon, this particle is displaced from its initial position.  Since it is at the speed limit, $c$, its speed remained unchanged at the new position and so neither does its radius of circular motion change.

From the post "Nature Has Orbits" dated 19 Jul 2015,

$v^2=-\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

and the energy stored is,

$E_{xd}=\cfrac{1}{2}mv^2=\cfrac{1}{2}m\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

So, given the photon energy as $E_i$,

$E_i=E_{xd}=\cfrac { 1 }{ 2 }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }|_{x_d}$

and the particle is orbiting at speed $c$,

$c^2=\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }|_{x_v}$

and so,

$cos(\theta)=\cfrac{x_v}{x_d}$

$x_v$ is fixed given light speed, but $x_d$ changes with photon energy.  Because $v^2$ has a minimum value, there is also a minimum photon energy below which the particle does not move on impact.

$E_{i\,min}=\cfrac{1}{2}mv^2_{min}$

Above this threshold all energy from the photon is emitted EXCEPT when,

$c^2=\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }|_{x_c}$

at the other solution of $v^2=c^2$.  The particle at point this will continue to orbit around the center with this higher value radius and does not oscillate  It is not possible to excite the particle beyond this radius because of the light speed limit.  At values less than this upper limit, the lower solution of $x$ to a given $v^2$ is occupied first than upon further absorption of photons the next higher solution above the minimum point is occupied.

It is counter intuitive that a smaller packet of energy at $v^2_{min}$ can cause the particle to go further out along the radial line.  In this situation, it is not the greater energy going the furthest distance, but the appropriate energy state at the appropriate location in a $\psi$ cloud.

The range of energy over which the photon is absorbed and emitted is delimited between $[x_v,\,x_c)$, not including $x_c$.

Since,

 $x_c\gt x_d\gt x_v$

$cos(\theta)=\cfrac{x_v}{x_d}\lt 1$

It is then possible that when,

$\sqrt{2}.x_v=x_d$

$cos(\theta)=\cfrac{1}{\sqrt{2}}$

$\theta=45^o$  at this point no photons are emitted (from the post "Polarization And Invisibility" dated 25 Jul 2015), the particle continues to oscillate, otherwise the oscillation is damped with emission of photons.  At $\theta=45^o$, we do not have resonance, as the amplitude of oscillation cannot increase further, but we have no loss from the oscillation.  It could be that the particle begins to perform circular motion around a second axis.

The range of energy that the particle can absorb and emit is,

$\cfrac{1}{2}mv^2_{min}\lt E_p\lt\cfrac{1}{2}mc^2$

except a sharp bandwidth, when $v^2(x=x_v\sqrt{2})$ at which $v^2=v^2_{v}$, absorption of this photon is not emitted.

The diagram above shows three issues.  A particle at $c^2$ receive a photon $E_p$ and transit to higher radius.  It is still in circular motion at $c$.  This transition is a gain of energy by $E_p$ over and above $\cfrac{1}{2}mc^2$.  The other solution of $x$ also at an energy level of $E_p$ is accessible only through $v^2_{min}$ upon a further absorption of energy at $v^2_{min}$.  When the photon is at $\cfrac{1}{2}mv^2_v$, which drive the particle to the point $x=x_v\sqrt{2}$, this photon is not emitted.

The particle absorb a photon and finds its appropriate location in the $\psi$ cloud.  Absorbing a photon with less energy can propel the particle to a further location.  The minimum energy absorbed is $\cfrac{1}{2}mv^2_{min}$.  Further absorption at any position on the curve below $c^2$, pushes the particle further along the curve.  This results in a Delta Spectrum of lower frequencies, that is the difference of the energy between the final and initial positions on the curve. (Remember the little hump on the left).

The initial $E_{p}$ finds the position $x$ on the $v^2$ vs $x$ curve, subsequent absorption of lower energy photons move the particle along the curve.

While on the curve, the particle is in circular motion at light speed $c$.  Once displaced the particle is no longer on the curve above, it is in oscillation about the center of the $\psi$ cloud (from the post "Twirl Plus SHM, Spinning Coin" dated 17 Jul 2015.

The particle is oscillating between point $a$ and $b$.  This graph is plotted with the requirement that $x=0$, $v^2=0$.  As the particle move along the radial line, it is at zero radius with respect to its circular motion, when $x=0$, this requires that $v=0$, thus $v^2=0$.  The graph above is both $v^2_{cir}$ and $v^2_{shm}$; these two differ by a constant factor of $\cos(\theta)$ and $sin(\theta)$, respectively.

As the particle passes through $x=0$, it emits a photon.  This lost of energy, damps the oscillation.
The photon emitted is given by,

$E_{p}=h.\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|.f_{cir}$

from the post "Color!" dated 25 Jul 2015.  The next emission as the particle passes through $x=0$ again is,

$E_{p}=h.\left|1-\sqrt{\cfrac{cos(\theta)}{sin(\theta)}}\right|.f_{shm}=h.\left|1-\sqrt{\cfrac{cos(\theta)}{sin(\theta)}}\right|.\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}f_{cir}$

and the next emission,

$E_{p}=h.\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|.f_{cir}$

and

$E_{p}=h.\left|1-\sqrt{\cfrac{cos(\theta)}{sin(\theta)}}\right|.\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}f_{cir}$

...

Here we have a problem that was set aside when we have the expression,

$\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}$

It is unbounded, $cos(\theta)=0$, and is not guaranteed to be smaller than one.  But from the four expressions above for $E_P$, two will result in a positive emission (ie. lost in energy).

What is then the physical interpretation of the the remaining pair?   In this case, photon absorption is not a passive process but an active one, where energy from the surrounding $\psi$ is absorbed at a determined place $x=0$, in an determined amount, $E_p$.

One set of expressions for $E_p$ emits a photon.  The reciprocal set of expressions for $E_p$ absorb packets of energy from the $\psi$ cloud.  In this case, $E_p$ is best express as,

$E_{p}=h.\left\{1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right\}.f_{cir}$

and

$E_{p}=h.\left\{1-\sqrt{\cfrac{cos(\theta)}{sin(\theta)}}\right\}.\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}f_{cir}$

where a negative $E_p$ is an absorption of a photon. And we get rid of the absolute value bar by allowing for active spontaneous photon absorption from the surrounding $\psi$.

A pump!

Note:  We still need a constraint on $\theta$.

Polarization And Invisibility

On closer inspection,

only solutions with $v^2=0$ at $x=x_z$ are admissible in the post "Twirl Plus SHM, Spinning Coin" date 17 Jul 2015.

$\theta$ could be polarization measured from the vertical.  Photons emission/absorption occurs at $x=x_z$, at the center of the $\psi$ cloud.

What if $\theta$ can be tuned, at 45^o, the factor,

$1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}=0$   and

$E_p=0$

when $\theta=45^o$,

does the particle also vanishes?  It is no longer emitting nor absorbing photons.

Color!

From the post "Twirl Plus SHM, Spinning Coin" dated 17 Jul 2015,

$v^{ 2 }_{ cir }=cos(\theta )(x+x_{ z })\left\{ -2{ c^{ 2 } }cos(\theta )ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } x))+\cfrac { \psi _{ d } }{ m }  \right\}$

and

$v^{ 2 }_{ shm }=sin(\theta )(x+x_{ z })\left\{ -2{ c^{ 2 } }cos(\theta )ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } x))+\cfrac { \psi _{ d } }{ m }  \right\}$

where a particle is both in circular motion and oscillating along a radial line.  A change in sign of $v^2_{shm}$ occurs at $v_{shm}=0$.

When both $v^2_{cir}$ and $v^2_{shm}$ are negative (ie. $v_{cir}$ and $v_{shm}$ are complex), their role switches.

$v_{cir}\to v_{shm}$   and

$v_{shm}\to v_{cir}$

We have a change in the radius of the circular motion and a change in the velocity along the radial line at the same time.  This is so because both expressions for $v_{cir}$ and $v_{shm}$ instantaneously attains zero at the same time.  $KE$ is still conserved, but the frequency of oscillation changes abruptly.  The latter result in a abrupt change in $PE$, from $E=h.f$.

This could be the mechanism by which a particle releases or gains packets of energy.

When the switch occurs, $v^2_{cir}$ changes by a factor,

$\cfrac{sin(\theta)}{cos(\theta)}$

and the corresponding change in frequency is by a factor,

$\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}$

If $\Delta E= h. \Delta f$, then, the photon released is,

$E_{p}=h.\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|.f_{cir}$

since $v^2_{cir}$ is changing, we have only a specific $v^2_{min}$ and expect the frequency, $f_{cir}$ to spread to higher values.  So,

$E_{p}=h.\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|\cfrac{v_{min}}{2\pi x_{min}}=\hbar.\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|\cfrac{v_{min}}{x_{min}}$

where $x_{min}$ is the value of $x$ at which $v^2$ is minimum and is also the radius of the circular motion around the center of the $\psi$ cloud.  From the post "Two Quantum Wells, Quantum Tunneling, $v_{min}$" dated 19 Jul 2015,

$v^2_{min}=-\cfrac{\psi^2_c}{m}.tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) }$

Considering only magnitude,

$E_{p}=\hbar\cfrac{\psi_c}{\sqrt{m}}\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|.\sqrt{tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) }}.\cfrac{1}{x}$

We have an expression for the range of energy for the photon emitted/absorbed by a particle orbiting at a radius $x$.

The particle is in motion by an initial excitation, afterwards the emission and absorption of such packets of energy occurs naturally when both $v_{cir}$ and $v_{shm}$ attains zero instantaneously.

In the derivation for $v^2_{cir}$, $\theta$ is the angle between the plane of the circular motion and the radial line joining the particle with the center of the $\psi$ cloud.  The particle in effect move on the surface of two cones pointed tip to tip at $x=x_z$ , with $\theta$ held constant.

Now that $\theta$ has gained greater significance, what is $\theta$?

When the particle is at light speed,

$E_{p}=\hbar.\left|1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right|\cfrac{c}{x_{orbit}}$

If $\theta$ is a constant, then this is a sharp photon emission/absorption from the particle in orbit.

What is $\theta$?

Particle Pairing

Electron pairing or particle pairing occurs the positive and negative solutions to $v^2$.  We take the negative root for the negative solution to make the final solution still in circular motion about the center (ie. $v$ complex).  We can visualize the second solution to be in counter motion to the first.(anti-clockwise vs clockwise)

The rest of the solutions are velocities along $x$, the radial line.

Rules Of Numbers

If a configuration of orbiting particles are stable with respect to sharing/exchange of particles with other configurations (ie. chemical reactions) when all possible solutions provided by the outermost set of curves are occupied (complete solution set criteria), then

The need to fully occupy all solutions provided by the outermost set of curves will give the Octet Rule.

When the outermost curves overlaps with a inner curve, then, because two solutions on the inner have to be included before the last pair of solutions on the outermost curves can be included, the remaining  two solutions on the inner curve must also be filled, in order to satisfy the complete solution criteria.  This makes twelve solutions for a stable configuration; the Duodectet rule

However, as we move further in, there is a 16 particle rule, then a 20 particle rules, but no 18 particle rule, as each loop pair of the curves provides 4 solutions.

Furthermore at the other end of the line, Rule of 2 is possible if we prohibit the lower solutions as being too close to the center.  In reality, the particle at the center extends a finite distance $x_a$.

And the Rule of 4 will be filling one loop pair of the set of curves.  This is a stability criteria only if this loop pair is the outermost pair, in which case the whole solution curves are being filled.

Have a nice day.

Electronic Shells By The $\psi$ Shore

Since the energy density of both orbiting and the $\psi$ cloud are of the same nature, we may consider them as a whole, a particle of  greater $\psi$.  This big particle can in turn have particles orbiting around it; which in turn can be considered as a whole with more particles orbit around it.  We can then expect a solution to $v^2$ that is the sum of many curves further and further from the center of the $\psi$ cloud.

In cases when the $4s$ shell are filled up first, it is because the last solution overlaps with the last but one.

In which case $4s$ is physically closer to the center than the four solutions on the last but one set of curves.  Such overlaps results in the higher shell being filled up first.

For the diagram above, when fully filled, it is the element Zinc (Zn).  The following diagram shows the furthest curve overlapping by one pair of solution with the last but one curve.  When filled up to the $4s$ shell, this is titanium (Ti).

Although $4s$ is a solution closer to the center than the last pair of solutions for the $3p$ shells, it is a solution based on all particles before it, taken as a whole, around which the next particle orbits with velocity $v^2$.  Each set of curves provides for the addition of four possible particles. This means for the case of electrons around a positive nucleus a possible charge discrepancy from $+7$ (after losing seven electrons) to $-7$ (after gaining seven electrons) due to a single set of curves. ($+3$ to $-3$ due to a single loop pair of the curve.) Higher discrepancy are possible when the curves overlap.

The first set of curves may provide 2, 4, 6 or 8 solutions.  Beyond the first set of curves, the next solutions come in 4 pairs (total 8).  These curves unfortunately does not fit well with the boundaries of electronic shells (1s, 2s, 2p, 3s...).  Grouping the electron into shells based on their relative distance from the center does not fit the solutions provide for by considering $\psi$, energy density.

The last pair of solutions is still the furthest.  And we can make the correspondent that a closed shell is a single solution with a set of curves of eight solutions.

More Force Density

$F=-\psi$

$F_{\rho}=-\cfrac{\partial\,\psi}{\partial x}$

but,

$F=\int_V{F_{\rho}}dV$

So,

$F=\int_V{-\cfrac{\partial\,\psi}{\partial x}}dV$

$F=\int_A{-\psi}dA$

that means,

$\psi=\int_A{\psi}dA$???

This is because,

$F\ne\int_V{F_{\rho}}dV$

but,

$F=\cfrac{1}{4\pi x^2}\int_V{F_{\rho}}dV$

where $x$ is the radius of the volume integrated over with,  $F$ is the force along a radial line $x$ and it is the force distributed over the surface of the volume, $V$.

$\int_V{F_{\rho}}dV$

is the sum of  all $F$s over the area of the sphere.

$\int_V{F_{\rho}}dV=4\pi x^2.F$

And so,

$\because dV=4\pi x^2.dx$ at $x$

$F=\cfrac{1}{4\pi x^2}\int_0^x{-\cfrac{\partial\,\psi}{\partial x}}4\pi x^2.dx$

$F=\cfrac{1}{4\pi x^2}\int_0^x{-4\pi x^2}d\psi$

as $x$ is the independent variable.
$\require{cancel}$
$F=\cfrac{1}{\cancel{4\pi x^2}}\cancel{4\pi x^2}.\int_0^x{-1}.d\psi$

$F=-\psi$ at $x$

What happen then to

$\cfrac{\partial\,L}{\partial x}=-\cfrac{\partial}{\partial x}\left\{\int_V\psi dV\right\}=\int_V{-\cfrac{\partial\,\psi}{\partial x}}dV=\int_V{F_{\rho}}dV=F$

from the post "On Second Thought... Potential Density" dated 24 Jul 2015?

$F$ here is the force on the particle not the force due to the particle's $\psi$ distributed around it.  This force is more accurately expression as,

$\psi_n-\psi(x)$

$F=F_n-F(x)$

where $F(x)$ is the force due to the containing particle.  The force $F$ on the orbiting particle is the result of the interaction between its energy density, $\psi_n$ and the surrounding $\psi(x)$ due to the containing particle.

Have a nice day.

On Second Thought... Potential Density

From the post " $PE$ Ain't Physical Education" dated 22 Jul 2015,

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }-\psi _{ n }$

$\cfrac{d}{dt}\left\{\cfrac{\partial\,L}{\partial\dot{x}}\right\}=\cfrac{d}{dt}\left\{\cfrac{1}{2}m2v\right\}=ma$

since,

$v^2=-\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

And

$\cfrac{\partial\,L}{\partial x}=-\cfrac{\partial\,\psi_n}{\partial x}=F_{\rho}$

And we have,

$F_{\rho}=ma$

Wrong!  This is not Newton's Law Of Motion.  Which means $\psi_n$ is not potential but potential density.  In which case,

$\cfrac{\partial\,L}{\partial x}=-\cfrac{\partial}{\partial x}\left\{\int_V\psi dV\right\}=\int_V{-\cfrac{\partial\,\psi}{\partial x}}dV=\int_V{F_{\rho}}dV=F$

where $F$ is the total force on the point particle and is in the direction of lesser $\psi$ for a positive particle.

And we have,

$F=ma$

But, more importantly, should

$\psi_n$  be  $\psi_n-\psi(x)$

instead?  In which case,

$F=F_n-F(x)$

where the resultant force $F$ on the particle is $F_n-F(x)$ in the direction of lesser $\psi$ for a positive particle and reversed for a negative particle.  $F(x)$ being due to the containing particle.

Good  Morning.

Theoretical Absorption Spectrum

The convolution plot of a Gaussian distribution with $v^2$ and circular standing wave resonance,

A very tight Gaussian with $\mu=1$ and $\sigma=0.25$,

$v^2$ with Circular Standing Wave Resonance at three discrete points,

The result of convolution with distinct peaks corresponding to CSWRs.

The result of convolution inverted, and is valid in the region between $a$ and $b$.

Given a good plotting software everything can look like anything else.  Nonetheless, a theoretical Absorption Spectrum between $a$ and $b$.

Have a nice day.

Magic Numbers, Get Them Here!

With the light speed limit, the orbiting particles cannot be at greater speed.  When the particle is at light speed, depending on $x_z$ there are limited number of possible valid orbits,

 When $x_z$ is small, only one of the Quantum Wells is available for occupancy.  It is possible that only the solution of higher $x$ are possible, the other being too close to the containing particle.  If this is so,  only two particles are in orbit.  Otherwise there are four particles.

It is likely that, as in the case of the hydrogen atom, only the two higher orbit can be occupied, the lower orbits being too close to the nucleus.

As $x_z$ get greater,

and greater,

These are single centered $\psi$ clouds.  Apart from being a container for magic numbers, non sequitur.

$\psi$ As A Force, Force Shield

We also have,

$F=-\psi$

$F$ is in Newton.

$\psi$ is also the Newtonian force.

As we started $\psi$ is energy oscillating between two orthogonal directions, then it become energy oscillating between two general dimensions.

Both force and potential energy, that is $\psi$, is then also oscillating.

There is then another dimension to force and potential, its oscillating frequency.  In the case of a particle as big as Earth, this frequency is low at 7.489 Hz.  We can counter gravity with a pulse at 7.489 Hz.  At the particle level, very small particle, the associated frequency is very high, at the infra red, optical range and beyond.

An associated frequency means we can counter another force or $\psi$ at an appropriate frequency:  a $\psi$ shield or force shield.

The key is an appropriate frequency.

...To be continued...

$\psi$ As Potential Energy

Could it be that $-\psi$ exist but is not accessible to us?

We have seen its shadow in the interaction of particles through their energy density $\psi$, and the manifestation of heavy particle in orbit around other particle.  In the former case, the forces between particle is fully accounted for by the interactions of both positive and negative $\psi$s from the participating particles.  In the latter, $\psi$ was made negative relative to its surrounding.

We know,

$PE=\psi$

and just like potential energy, $-\psi$ is made available to us when we made ourselves of $-\psi$.

A lower potential we fall into, a higher potential does work for us.

Best Of Both Worlds

Any velocity along $x$, a radial line joining the particle and the center of the $\psi$ cloud will result in oscillations.

So all roots, positive and negative, to the factor,

$e^u(e^{2u}-1)^{1/2}$

are admissible.  And we have both circular motion and oscillation where the two velocities, circular motion velocity and instantaneous oscillation velocity, are perpendicular at all times.

Good night.

Orbiting Heavy Particle

The problem with heavy particle is, given,

$v^2=-\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

and,

$0\gt \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\gt-2$ --- (*)

this implies that the velocity is once again along $x$ not perpendicular to $x$.

Also,

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\left\{2-\left|  \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\right| \right\} -\psi _{ max }$

is not indicative of the type of motion the particle is in.  Furthermore even with (*) and $\psi_c\lt0$, the orbiting particle is at most at ground state, $\psi_o$, its energy density is not negative.

However, if we make $\psi_n=\psi_{max}-\delta$, where $\delta\gt0$ then from,

$v^2=-\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

we have,

$v^2=-\cfrac { 1 }{ m }  .(-\delta)\cfrac { -\delta+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

$v^2=+\cfrac { 1 }{ m } \delta \cfrac { -\delta+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

But we take the negative root instead,

$v^2=-\cfrac { 1 }{ m } \delta \cfrac { -\delta+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

and we let $\delta=-\psi$,

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

the particle is then still in orbit about $x=x_z$, but with "$-\psi$".  The factor,

$\cfrac { \psi+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

is dimensionless.  The change in sign for $v^2$ is adjusted by taking the negative root.  Graphically,

the particle have "$-\psi$" and that results in a heavy particle.  This happen when the transition to a lower energy state upon the release of a photon go past $\psi_{max}$ of the containing particle.  In effect the orbiting particle given its velocity $v$ and location $x$, is squeezed by the surrounding $\psi$ of the containing particle.

Must the orbiting particle passes beyond its ground state $\psi_o$, and be in the absolute $-\psi$ region, for it to manifest heavy mass?  As far as the expression (**) is concerned, no.  The particle is in "$-\psi$" when it transits pass $\psi_{max}$, it may actually be above $\psi_o$.  Once we obtained (**) we may write,

$c^2\int_V{m_{\rho\,particle}} dV =c^2\int_V{m_\rho} dV-\int_V(-{\psi})dV=c^2\int_V{m_\rho} dV+\int_V{\psi}dV$

 Mathematically, $-\psi$ is just as valid and the particle just manifest more mass.  But is taking the negative root justifiable?

Fiction is my game.  In my dreams, yes I can!

$PE$ Ain't Physical Education

If,

$KE_{min}=\cfrac{1}{2}mv^2_{min}$

$KE=\cfrac{1}{2}mv^2$

what is $PE$?  Obviously,

$PE=\psi_n$

And we form the Lagrangian,

$L=\cfrac{1}{2}mv^2-\psi_n$

From the post "Nature Has Orbits" dated 19 Jul 2015,

$v^2=-\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

where $u=ln(cosh(x- x_{ z }))$

$v$ being complex being perpendicular to the radial line along $x$.

Therefore,

$L=-\cfrac{1}{2}m.\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }-\psi_n$

We see that $m$ cancels off immediately,  which is a good thing as this equation is for all types of $\psi$, in $t_c$, $t_g$ and $t_T$.

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }-\psi _{ n }$

Given,

$\psi_c=\psi _{ n }-\psi_{max}$

$\psi_n=\psi _{ photon }+\psi_{o}$

$\psi_n$ is the elevated energy level of the particle from its ground state $\psi_o$ after receiving the photon $\psi_{photon}$.  $\psi_{max}$ is the maximum $\psi$ of the confining particle cloud.

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }-\psi _{ c }-\psi _{ max }$

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\left\{ \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }+2 \right\} -\psi _{ max }$

This is the Lagrangian of a particle being in the $\psi$ cloud of another particle of the same type, either positive or negative.  It is possible for both of them to be positive or negative, and for them to have different sign.

The relation,

$\cfrac{d}{dt}\left\{\cfrac{\partial\,L}{\partial\dot{x}}\right\}=\cfrac{\partial\,L}{\partial x}$

will lead us back to the Newtonian force but without the involvement of mass, $m$.  We note that

$F_{\rho}=-\cfrac{d\,\psi}{dx}$

where $F_{\rho}$ is the force density.

From the post "Opps! Lucky Me" dated 25 May 2015, the Newtonian force $F$,

$F=-\psi$

where $\psi$ is energy density and $F$ is in $N$, the newton.  This relation give us the resultant of two objects with interacting $\psi$,

$F_{1\,2}=\psi_1-\psi_2$  and

$F_{2\,1}=\psi_2-\psi_1$

irrespective of the notation, we have,

$F_{1\,2}=-F_{2\,1}$

Newton's Third Law of motion, equal opposite action and reaction.

Furthermore, in the Lagrangian,

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\left\{ \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }+2 \right\} -\psi _{ max }$

the negative sign before the $T$ or $KE$ term allows for a negative sign without making mass negative forcefully.  And the $+2$ constant multiplied to $\psi_c$ allows for the term,

$\cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

to dip two below zero, without introducing negative energy.

In the latter point, a negative $\psi$ results in, from the post "We Still Have A Problem" dated 23 Nov 2014,

$m_{\rho\,particle} c^2=m_\rho c^2-\int^{x_a}_{0}{\psi}dV$

where it was proposed the particle's total $\psi$ and its mass, $m_{\rho\,particle}$ in space add to give the mass, $m_\rho$ along the time dimension in which particle exist.

If $\psi\to-\psi$,

$m_{\rho\,particle} c^2=m_\rho c^2-\int^{x_a}_{0}{(-\psi)}dV=m_\rho c^2+\int^{x_a}_{0}{\psi}dV$

The particle gained mass density, above it inertia density (from $m_{\rho}c^2$), along the respective time dimension that it exist.

(We would want $-\psi$ to see a gain in mass, for example. Now the negative sign can come from (*) below.)

This could be the mechanism by which photon gain mass.

$0\gt \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\gt-2$  -- (*)

and still have the same Lagrangian (the same physic behavior like a photon).  From, the velocity equation,

$v^2=-\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }=-\cfrac { 1 }{ m }(-\psi _{ c })\left|  \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } \right|$

and so,

$L=-\cfrac { 1 }{ 2 } \psi _{ c }\left\{2-\left|  \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\right| \right\} -\psi _{ max }$

with condition (*), the Lagrangian is still of the same form, which implies that the particle behave the same as far as the principle of least action is concerned.

Since,

$\psi_c=\psi _{ n }-\psi_{max}$

when $\psi_{max}\gt\psi_{n}$ as we superimpose more particles of the same as the containing particle or remove the previously gained photon of a orbiting particle at its $v_{min}$.  We can effect (*), where the particles in orbit gain mass.  And because of the upper bound in (*), this occurs when $\psi_c$ is small and the orbiting particle is close to $x=x_z$ at low speed.

This is how we make heavy particles, lest they are still orbiting particles.

Have a nice day.

Note:

$m_{\rho\,particle} c^2=m_\rho c^2-\int^{x_a}_{0}{\psi}dV$

should be more accurately written as,

$c^2\int_V{m_{\rho\,particle}} dV =c^2\int_V{m_\rho} dV-\int_V{\psi}dV$

as volume integrals over the extend of the particle.

The Sharks Are Coming

A convolution plot of $v^2_{min}$,

$v^2_{min}=tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) }$


with a Gaussian distribution of mean, $\mu=1$ and spread $\sigma=0.25$

$v^2_{min}$,

and the result of convolution,

when inverted,

where the valid region is over the extend of $v^2_{min}$ between lines $a$ and $b$.

This is the distribution of energy, $KE$ of a population of particle confined in another particle excited by a non zero $\psi_c$.

Another flat top.

Going This Way?

Here is a more accurate graph of $v^2$,

A negative particle experiences a reversed force by the way its time axes are arranged and when aligned with that of a positive particle, points its $x$ axis  at the negative $x$ direction of the positive particle.  The energy density $\psi$, a scalar, is the same type on both positive and negative particle.

Notice the abrupt change in gradient on the factor $e^u(e^{2u}-1)^{1/2}$ at the minimum $x=x_z$.  This is the reason for the $-i$ factor when $x\to x_z^{-}$, which switches the direction of $v$ to be along a radial line $x$.

Quantum Tunneling! But only in one direction.

Bolding Cone Head

And if we add Circular Standing Wave Resonance to the minimum $KE$ from the post "Two Quantum Wells, Quantum Tunneling, $v_{min}$" dated !9 Jul 2015.

This standing wave resonance occurs when the frequency of oscillation $f_{SHM}$ (SHM along a radial line), is an integer multiple of the angular frequency, $f_{cir}$.  The results of resonance are sharp peaks at discrete locations along $x$.

When we flip this graph vertically,

which is the absorption spectrum below a threshold intensity.

$f_{cir,\,1}=\cfrac{v_{min,\,1}}{2\pi x_1}$

$f_{cir,\,2}=\cfrac{v_{min,\,2}}{2\pi x_2}$

But,

$\cfrac{f_{SHM}}{f_{cir,\,1}}=N$

$\cfrac{f_{SHM}}{f_{cir,\,2}}=M$

which leads to the proportion identity,

$\cfrac{x_1}{x_2}\cfrac{v_{min,\,2}}{v_{min,\,1}}=\cfrac{N}{M}$

For three dip points on the graph, $X_1,\,\,Y_1$, $X_2,\,\,Y_2$ and $X_3,\,\,Y_3$,

$\cfrac{X_1}{X_2}\sqrt{\cfrac{Y_2}{Y_1}}=\cfrac{N}{M}$

$\cfrac{X_1}{X_3}\sqrt{\cfrac{Y_3}{Y_1}}=\cfrac{N}{O}$

$\cfrac{X_3}{X_2}\sqrt{\cfrac{Y_2}{Y_3}}=\cfrac{O}{M}$

From which $f_{SHM}$ can be calculated, once $M$, $N$ or $O$ has been determined graphically.

You are not going to get a graph of $v^2_{min}$ versus $x$, of course.

But in my dream...

My Own Acronym: CSWR

For a particle oscillating in the $\psi$ cloud of another, it is in circular motion over a spread of velocities,  and at the same time oscillating perpendicular to its circular path.

Just like in Bohr's Model where waves are restricted to multiple whole wavelengths around fixed orbits, the particle display resonance when its circular frequency $f_{cir}$ and its oscillation frequency, $f_{SHM}$ are related by,

$\cfrac{f_{SHM}}{f_{cir}}=n$

$n=1,2,3$

In this case, the particle need not be in resonance, but if in resonance it will absorb high amount energy from its environment and the standing wave gain amplitude rapidly. (Circular Standing Wave Resonance).

The spread of velocities such particle can have, results in a board absorption spectrum.  A minimum velocity requirement truncates the spectrum at the lower energy side.  Lower velocity particles on both sides of the center $\psi_{max}$, give the lower energy spectrum a burgle.  Higher velocity particles being restricted to above $\psi_{max}$ only thins out rapidly but are not restricted from attaining higher velocitis.  Resonance due to circular standing wave causes sharp dips in the absorption spectrum as such waves at resonance absorb energy rapidly.  When this energy at specific frequencies are emitted they form the emission spectrum.

Spectra lines still but in a fuzz, all frequency are absorbed but more so at specific circular standing wave resonance (CSWR) frequencies.

Scilab...

Two Quantum Wells, Quantum Tunneling, $v_{min}$

From the post "Nature Has Orbits"

$v^2=-\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$ --- (1)

$v$ is complex.  Where,

$u=ln(cosh(x- x_{ z }))$,  $x_z=\cfrac{1}{2}x_a$  and

$\psi_c=\psi_n-\psi_{max}$

A plot of which,

The scaling factor $\cfrac{\psi_c}{m}$ plays a minor role in the shape of $v^2$ vs $x$.  $v^2$ and so the orbiting particle $KE$, is confined to a narrow range about $\psi_{max}$, at $x=x_z$ of the particle.  At values of $x$ closer $x=x_z$, $v^2$ has a minimum, $v^2_{min}$.

On approach to $x_z$ from higher values of $x$, since,

$\lim\limits_{x\to x_z^{+}}\cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\to\infty$

This means $v$ increases rapidly as the particle fall from higher orbit but because of the speed limit $c$, at light speed, the particle does not reach $x=x_z$.

Also, on approach to $x_z$ from lower values of $x$,

$\lim\limits_{x\to x_z^{-}}\cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\to(-i)\infty$

As $v$ is already complex, $-i$ rotates $v$ back along the radial line and the particle is expelled beyond $x=x_z$ with great velocity.  The particle jumps to higher orbit, $x\gt x_z$ at light speed $c$ (assuming that it has reach light speed, $c$ before passing through $x=x_z$).  A plot of $v^2$ around $x_z$ is shown below,

To obtain $v_{min}$ consider this, when $x\to x_z\pm\delta$, where $\delta=small$,

$\lim\limits_{x\to x_z}\cfrac{d(v^2)}{dx}=\lim\limits_{x\to x_z}\cfrac{d(v^2)}{du}\cfrac{du}{dx}=\epsilon$

as

 $\delta\to small$, not necessarily zero,  $\epsilon\to0$.  Substitute equation (1) for $v^2$,

$\cfrac{d(v^2)}{dx}=-\cfrac{\psi_c}{m}\cfrac{d}{du}\left\{\cfrac{1}{u}\left(\psi_c+u\right)e^{u}\left(e^{2u}-1\right)^{1/2}\right\}\cfrac{du}{dx}=\epsilon$

Consider,

$\cfrac { e^{ u }\left\{\psi _{ c }(-u+e^{ 2u }(2u-1)+1)+u^{ 2 }(2e^{ 2u }-1)\right\} }{ u^2\sqrt { e^{ 2u }-1 }  } \cfrac{du}{dx}=\epsilon$

$\cfrac { e^{ u }\left\{u^{ 2 }(2e^{ 2u }-1)+\psi _{ c }(u(2e^{ 2u }-1)-(e^{ 2u }-1))\right\} }{ u^{ 2 }\sqrt { e^{ 2u }-1 }  } \cfrac { du }{ dx } =\epsilon$

$\cfrac { e^{ u }\left\{u(\psi _{ c }+u)(2e^{ 2u }-1)-\psi _{ c }(e^{ 2u }-1)\right\} }{ u^2\sqrt { e^{ 2u }-1 }  } \cfrac { du }{ dx } =\epsilon$

$e^{ u }\left\{ \cfrac { (\psi _{ c }+u) }{ u } e^{ u }-\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1) }{ u^{ 2 }(2e^{ 2u }-1) }  \right\} \cfrac { du }{ dx } =\cfrac { e^{ u }(e^{ 2u }-1)^{ 1/2 } }{ (2e^{ 2u }-1) } \epsilon$

$\cfrac { (\psi _{ c }+u) }{ u } e^{ 2u }=\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1)\cfrac { du }{ dx }  } \epsilon +\psi _{ c }\cfrac { e^{ 2u }(e^{ 2u }-1) }{ u^{ 2 }(2e^{ 2u }-1) }$

$\because \cfrac{\epsilon}{\cfrac { du }{ dx }}=\cfrac { dv^{ 2 } }{ du }$

$\cfrac { (\psi _{ c }+u) }{ u } e^{ 2u }\cfrac { du }{ dx } =\cfrac { e^{ u }(e^{ 2u }-1) ^{1/2}}{ (2e^{ 2u }-1) } \cfrac { du }{ dx } \cfrac { dv^{ 2 } }{ du } +\psi _{ c }\cfrac { e^{ 2u }(e^{ 2u }-1)}{ u^{ 2 }(2e^{ 2u }-1) } \cfrac { du }{ dx }$

$\cfrac { du }{ dx }=tanh(x)=\cfrac{(e^{2u}-1)^{1/2}}{e^u}$,

$\cfrac { (\psi _{ c }+u) }{ u } e^{ 2u }\cfrac { (e^{ 2u }-1)^{ 1/2 } }{ e^{ u } } =\cfrac { e^{ u }(e^{ 2u }-1) ^{1/2}}{ (2e^{ 2u }-1) }  \cfrac { dv^{ 2 } }{ dx } +\psi _{ c }\cfrac { e^{ 2u }(e^{ 2u }-1) }{ u^{ 2 }(2e^{ 2u }-1) } \cfrac { (e^{ 2u }-1)^{ 1/2 } }{ e^{ u } }$

$\cfrac { (\psi _{ c }+u) }{ u } e^{ u }(e^{ 2u }-1)^{ 1/2 }=\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1) }  \cfrac { dv^{ 2 } }{ dx } +\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) }$

Wrong! Never divide by zero or the possibility of zero.  But $\cfrac { du }{ dx }$ was not actually divided with! 

So,

$v^2=-\cfrac{\psi_c}{m}\left\{\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1) }  \cfrac { dv^{ 2 } }{ dx } +\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) } \right\}$

So, when $\delta$ is sufficiently small, from the diagram above, the velocity is minimum $v=v_{min}$ when,

$\cfrac{d(v^2)}{dx}=0$

which occurs before $x=x_z$.   This leads to,
$\require{cancel}$
$v^2_{min}=-\cfrac{\psi_c}{m}\left\{\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1) }  \cancelto{0}{\cfrac { dv^{ 2 } }{ dx }} +\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1) ^{3/2}}{ u^{ 2 }(2e^{ 2u }-1) } \right\}_{x=x_{min}}$

$v^2_{min}=-\cfrac{\psi^2_c}{m}\left\{\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) } \right\}_{x=x_{min}}$

Since,

$u=log(cosh(x-x_z))$

$\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) }=\cfrac { cosh(x-x_z)sinh^{ 3 }(x-x_z) }{ ln^{ 2 }(cosh(x-x_z))(2cosh^{ 2 }(x-x_z)-1) } =\cfrac { sinh(2(x-x_z))sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z))cosh(2(x-x_z)) } =tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) }$

All $v^2_{min}$ for $\psi_c\gt0$ lie on this line,

$y=tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) }$

For each elevation $\psi_c$ in energy, the particle orbits at different radius with different velocity.  The minimum velocities squared of all particles elevated to different $\psi_c$s are given by the curve above.  If energy above this minimum $KE$ represented by the curve is emitted, then this curve is the baseline of the emission spectrum.  This is emission across a continuous spectrum not in packets of discrete frequencies.

How??  In packets of $\psi$, at low speed, as the particle's photon equivalent, but across a continuous spectrum as v^2 is continuous.

The particle is in circular motion around with velocity, $v_{min}$ in a direction perpendicular to $x$.  Only when $x\to x_z^{-}$ does the particle switch direction and travel along $x$ with great velocity.

When $x\to  x_z^{+}$, the particle is in circular motion with radius, $x_{radius}=x$.  When $x\to x_z^{-}$, the particle travels along the radial line, $x$.

The steep values of $v$ on either side of $x=x_z$ restrict the location of the particle to small values around $x=x_z$.  This is a Quantum Well except that $x=x_z$ is also forbidden.  They are two Quantum Wells side by side.  It is however possible to escape from the lower Quantum Well in the region $x\lt x_z$ by approaching $x=x_z$.  At that point $iv$ is rotated by the $-i$ factor and the particle speed beyond $x=x_z$ along the radial line, in the direction of $x$ into the adjacent Quantum Well at very high speed.  This is quantum tunneling, but it is possible only from the well at lower region $x\lt x_z$, to the well at higher region $x\gt x_z$.

How is $iv$ applied as the particle approachs $x_z^{-}$?  Does the particle turn slowly transcribing a spiral or does it turn abruptly near $x_z^{-}$.

Note:

Do not try to take the square root of $-i$,  rotate $v$ by $i$ for the first negative sign, then rotate $v$ by $-i$ as the result of taking the limit to approach $x_z^{-}$.