Anti Matter Existence Along Negative Time Axis

If  Thermal Gravity is anti-gravity, and gravity is along the positive time axis, then thermal gravity is on the negative time axis.

And we experience anti-matter as heat, or temperature;  just as a proton and an electron feel each other hot when they collide.

Temperature can still be spins on the time axis.  We are experiencing the effects of anti-matter not the anti-matter itself.

Consider the change in rotational energy,

$\Delta E_{r}=\cfrac{1}{2}mv^2_f-\cfrac{1}{2}mv^2$

Lost in KE along the time axis equals change in rotational energy for one particle,

$mc^2=\Delta E_{\omega}=\cfrac{1}{2}m(v^2_f-v^2)$

$c^2=\cfrac{1}{2}(v^2_f-v^2)$

$v^2_f=2c^2+v^2$,    $v^2_f=(\sqrt{2}c)^2+v^2$

If we expand the Big Band to include the time axis as well,  this is consistent with the fact that the first kinetic energy lost is the result of collision of particles with speed  $\sqrt{2}c$, from the post "If The Universe Is A Mochi".  It is not surprising,  for both cases started with  $E=m^2$.  So, the result of the very first temperate increment is,

$v^2_i=v^2_f=(\sqrt{2}c)^2+0$

$v_i=\sqrt{2}c$

If temperature is defined as energy, and

$m = 2m_p$

where  $m_p$  is the particle when the universe reaches maximum entropy, and  $2m_p$ is after the first coalescence,  after which we have masses at zero velocity, common masses we interact with day to day.  The rotation KE for a mass  $2m_p$ about the time axis,

$\cfrac{1}{2}.2m_p.v_i^2=2m_pc^2$

This is consistent with an matter/anti-matter collision of  2 particles of mass  $m_p$  on the time axis.  So, the first temperature increment is given by,

$\Delta T_1=2m_pc^2=2.\cfrac{1}{2}.2m_pc^2$

And each particle has a rotational kinetic energy of , temperature of,

 $T_i=KE_r=\cfrac{1}{2}.2m_pc^2$

this is the smallest temperature increment possible.

What?  Point mass has not rotational concerns?  That is the point,  point mass does rotate and it its temperature.  Intuitively, when the distance from a rotational axis collapses,  $r\rightarrow0$ (from the  $I=r^2m$ fame),  space collapses and that leaves the particle on the time axis.  Rotational energy is still defined but rotational moment is at this moment, not.

Matter/anti-matter interaction, is a head on collision on the time axis, the energy released is the total loss in translational kinetic energy on the time axis,

$E=mc^2$

where  $m$  is the total mass $m=m_m+m_{anti-m}=2m_m$,

$E$  is the gain in rotational energy about the time axis of the two particles, now stationary on the time axis, and so, disappeared from our existence as we move forward in time.  We experience this rotational KE as temperature.  This suggests that the decay as a result of matter/antimatter interaction must be slow.  The remaining mass of the interaction over time, carries the temperature with it and we experience that as plasma.  Mass that interacted has time speed zero, falls back in time and disappear from our time.  It is as if mass has been converted to energy.  If the interaction is instantaneous then both mass and temperature would have disappeared instantaneously.

Anti-matter then, is mass travelling in the negative time direction.  A proton is an anti-matter.  We are in the same direction as an electron on the time axis.  An charged anti matter can be captured by a very fast rotating opposite charge and be brought into our reality, just as electrons in orbits around a proton nucleus.

What's the energy need for containment?  Consider the case of a hydrogen atom...

$\tau_o$ The New Guy On The Block

We would now formulate an expression for the repulsion between hot particles.  The simplest of which is to consider conservation of flux, as Gauss did

$F_T=\cfrac{T_aT_b}{4\pi\tau_or^2}$

where  $F_T$  is the repulsive thermal force between hot particles,  $\tau$ is a measure of the resistance in establishing a thermal gradient between the two particles, and  $r$  the distance between the hot particles and,  $T_a$  and  $T_b$  are temperature on particle  $a$  and  $b$.

Not bad for a first guess.

And so, for the post "Not This Way",

$F_{ h }=Fsin(\theta )=\cfrac { q_{ p }q }{ 4\pi \varepsilon _{ o }r^{ 2 } } sin(\theta )$

can instead be,

$F_{ h }=Fsin(\theta )=\cfrac { T_{ p }T_e }{ 4\pi \tau _{ o }r^{ 2 } } sin(\theta )$

and we are fine with diffraction on a straight flat edge.   $T_p$ is the temperature on the photon and  $T_e$  is the temperature on the electron.

Hurra!  More importantly, the Universe now has a counter force to balance gravity.  Hot Banana Mochi safe.

No Charge But Thermal Gravity To The Rescue

The problem with an expression like,

$F_{ h }=Fsin(\theta )=\cfrac { q_{ p }q }{ 4\pi \varepsilon _{ o }r^{ 2 } } sin(\theta )$

from the post "Not This Way", is that  $q_p$ don't exist.

Even if the photon is a dipole, its net charge from afar is zero.

There is one other possibility, that is for the matter-antimatter annihilation process to be slow.  That popular literature on explosive matter/anti-matter reaction may not be true.  Take the case of a hydrogen atom,  when the electron collide into the proton nucleus,

the rate of annihilation along the charge-time line is,

$E_a=\cfrac { d\, q }{ dt } =\cfrac { \, d(q_{ pr }+q_{ e }) }{ dt } =\cfrac { dm }{ dt } =\cfrac { \, d(m_{ pr }+m_{ e }) }{ dt }$

And  $E_a$  is slow.  The reside charge on each of the particles forms a dipole and the energy released from  $E=mc^2$ make this both a very hot particle and a electric dipole.  This is a  $H$ plasma particle,  $p_p$,  that is experiencing a observable decay.  What would its decay half life,  $p_pT$ be?  This hot dipole is a likely candidate for photon, both mechanisms for acceleration to light speed/terminal speed (as a dipole or hot particle) can apply to this particle.

Moreover since,

$\cfrac { \, d(q_{ pr }+q_{ e }) }{ dt } =\cfrac { \, d(m_{ pr }+m_{ e }) }{ dt }$

we have a charge mass equivalence,

$\int^0_{q_{ pr }+q_{ e }}{1.}d\,q=\int^{m_{ pr }-m_{ e }}_{m_{ pr }+m_{ e }}{1.}d\,m$

$-(q_{pr}+q_{e})=-2m_{e}$  and  $q_{pr}=q_{e}$

we should have,

$q_{e}=m_{e}$

the resulting neutral hot particle has mass  $m_{pr}-m_{e}$ , likely a neutron.  In this instance, both mass and charge are treated as inertia whether they are on the positive or negative time line.  We also have,

$q_{pr}=m_{e}$

that all the positive charge in a proton is from a mass of  $m_e$.

It is more likely,

$q_{e}=M_cm_{e}$  where  $M_c$  is a scaling factor that also adjust for unit dimension, ie  charge per unit mass, C kg^-1.

Hot Photons And Hot Electrons And Light Speed

Photons and electron interact at a distance and are mutually repulsive but their interaction may not be electrostatic.  Consider both force/motion equations that keep the particles in circular motion,

$\cfrac { m_{ e }v^{ 2 } }{ r_{ e } } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ e } } +Av^{ 2 }$    for electrons

and

$\cfrac{m_pc^2}{r_p}=Ac^2$    for photon

where  $A$  is the drag factor of space at high speed.

We see that if  $A$  were to decrease both  $r_p$  and  $r_e$  can increase.  It was postulated that high temperature thin out space reducing space density and that gravity is the result of differing space density.  Gravity points from region of less dense space to more dense space and so, a hot region of low space density has a gravity component pointing outwards.  The drag factor is inversely proportion to space density.  Therefore a consistent view is that the drag factor decreases with high temperature, and there is a thermal gravitational force pointing away from low drag factor region.

What if the photon is very hot?  What if the electron is also very hot?

As the two particles approach each other a hot region develops between them, the drag factor  $A$ decreases at the same time a thermal gravitational field develops pointing outwards from this hot region.  Both factors serve to increase $r_p$  and  decreases  $r_e$,  the respective radii of the circular path of photon and electron.

The two particles are pushed away from each other and it seems, their interaction is repulsive; this is not electrostatic.

Given that a hot region has a lower space density, then gravitational forces pull a hot particle outwards due the region of less density space around it.  Under normal circumstances the forces are symmetrical around the particle and there is no net acceleration.  However when the particle is in motion in space with a high drag factor, then it is reasonable to imagine that a region of denser space builds up before the the particle in the direction of travel.

And a net acceleration develops in the direction of travel.  Given low mass and high temperature,  it is possible that the particle is self propelling.   Once again we have light speed!  This would also explain why such hot particles are always in motion; the slightest velocity sends the particle accelerating to light speed.

Have a nice day.

Another Broken Glass.

It is likely that Alice broke the looking glass this way,

Both cases theoretical.

This Way...But Short, Very Short.

From the previous post "Not This Way", we have shown the deflection of a photon by an electron, where the deflection angle is given by,

$tan(\theta)=\cfrac{x_{d1}+x_{d2}}{D+D_o}$

Now consider a short array of such valence electrons on the surface of a material, that is part of the material structural lattice.

The optical path difference between two adjacent deflections is given by

$\Delta OP=a_icos(\theta)$

where  $a_i$  is the atomic distance of the material.  If  the deflected paths are out of phase,

$\Delta OP=a_icos(\theta_d)=(2n+1)\cfrac{\lambda}{2}$

$cos(\theta_d)=(2n+1)\cfrac{\lambda}{2a_i}$

$\theta_{d0}=cos^{-1}(\cfrac{\lambda}{2a_i})$,  $\theta_{d1}=cos^{-1}(\cfrac{3\lambda}{2a_i})$,  $\theta_{d2}=cos^{-1}(\cfrac{5\lambda}{2a_i})$...

we will see destructive interference on a screen in the path of the deflection.

If the deflected paths are in phase,

$\Delta OP=a_icos(\theta)=n\lambda$

$cos(\theta)=\cfrac{n\lambda}{a_i}$

$\theta_{c0}=cos^{-1}(\cfrac{\lambda}{a_i})$,  $\theta_{c1}=cos^{-1}(\cfrac{2\lambda}{a_i})$,  $\theta_{c2}=cos^{-1}(\cfrac{3\lambda}{a_i})$...

If  $\lambda\approx a_i$  then ,

$\theta_{d0}=1.0472\,r = 60^o$

and

$\theta_{c0}=0^o$

The two angles are sufficiently wide apart and will project as distinctive bright and dark bands.

In order to achieved a uniformly aligned array of electrons, the material dimension must be very short to be straight and flat along the path of the photons, otherwise other interference from different angles as a resulting of material non-uniformity will over lap and render a indistinguishable image.

So, what about diffraction?  Diffraction is the above interference over a short dimension that is the thickness of a thin material, where photons run parallel along the short surface.  This can happen at a flat sharp edge or over the thickness of a thin material, smooth and flat.

Not This Way...

The interaction between electron and photon is repulsive,

Consider a photon on approach to a electron, at a horizontal distance  $d$  away,  what is the deflection?  We break the path of the photon into 2 parts, before it is level horizontally with the electron and after it is level with the electron.

Deflection before being level with the charge,

$r=\sqrt { d^{ 2 }+(D_{ o }-ct)^{ 2 } }$

$sin(\theta )=\cfrac { d }{ r } =\cfrac { d }{ \sqrt { d^{ 2 }+(D_{ o }-ct)^{ 2 } }  }$

$F_{ h }=Fsin(\theta )=\cfrac { q_{ p }q }{ 4\pi \varepsilon _{ o }r^{ 2 } } sin(\theta )=\cfrac { q_{ p }q }{ 4\pi \varepsilon _{ o }(d^{ 2 }+(D_{ o }-ct)^{ 2 }) } \cfrac { d }{ \sqrt { d^{ 2 }+(D_{ o }-ct)^{ 2 } }  }$

$x_{ d }=\int _{ 0 }^{ t=D_{ o }{ / }c }{ \cfrac { F_{ h } }{ m_{ p } }  } dt=\int _{ 0 }^{ t=D_{ o }c }{ \cfrac { q_{ p }q }{ 4\pi m_{ p }\varepsilon _{ o } } d(d^{ 2 }+(D_{ o }-ct)^{ 2 })^{ -3/2 } } dt=\cfrac { q_{ p }q }{ 4\pi m_{ p }\varepsilon _{ o } } \left\{ \cfrac { (ct-D_{ o }) }{ cd\sqrt { d^{ 2 }+(D_{ o }-ct)^{ 2 } }  }  \right\} ^{ D_{ o }/c }_{ 0 }$

$x_{ d1 }=\cfrac { q_{ p }q }{ 4\pi  m_{ p }\varepsilon _{ o }c } \left\{ \cfrac { D_{ o } }{ d\sqrt{d^2+D^2_o} }  \right\}$

Deflection after being level with the electron,

$r=\sqrt { d^{ 2 }_1+(ct)^{ 2 } }$

where  $d_1=x_{d1}+d$

$sin(\theta )=\cfrac { d_1 }{ r } =\cfrac { d_1 }{ \sqrt { d^{ 2 }_1+(ct)^{ 2 } }  }$

$F_{ h }=Fsin(\theta )=\cfrac { q_{ p }q }{ 4\pi \varepsilon _{ o }r^{ 2 } } sin(\theta )=\cfrac { q_{ p }q }{ 4\pi \varepsilon _{ o }(d^{ 2 }_1+(ct)^{ 2 }) } \cfrac { d_1 }{ \sqrt { d^{ 2 }_1+(ct)^{ 2 } }  }$

$x_{ d }=\int _{ 0 }^{ t=D/c }{ \cfrac { F_{ h } }{ m_{ p } }  } dt=\int _{ 0 }^{ t=D/c }{ \cfrac { q_{ p }q }{ 4\pi m_{ p }\varepsilon _{ o } }d_1 (d^{ 2 }_1+c^{ 2 }t^{ 2 })^{ -3/2 } } dt=\cfrac { q_{ p }q }{ 4\pi m_{ p }\varepsilon _{ o } } \left\{ \cfrac { t }{ d_1\sqrt { d^{ 2 }_1+c^{ 2 }t^{ 2 } }  }  \right\} ^{ D/c }_{ 0 }$

$x_{ d2}=\cfrac { q_{ p }q }{ 4\pi  m_{ p }\varepsilon _{ o }c } \left\{ \cfrac { D }{ d_1\sqrt { d^{ 2 }_1+D^{ 2 } }  }  \right\}$

$d_1=x_{d1}+d$

$x_{ d2}=\cfrac { q_{ p }q }{ 4\pi  m_{ p }\varepsilon _{ o }c } \left\{ \cfrac { D }{ (x_{d1}+d)\sqrt { (x_{d1}+d)^{ 2 }+D^{ 2 } }  }  \right\}$

If we estimate the optical path as the hypotenuse of a triangle  (It is usually wrong to estimated optical path to find path difference, because the actual path difference is already small.), the optical path difference between a straight photon and a deflected photon is,

$OP=\sqrt { (D_{ o }+D)^{ 2 }+(x_{ d1 }+x_{ d2 })^{ 2 } } -(D_{ o }+D)$

For destructive interference,

$\sqrt { (D_{ o }+D)^{ 2 }+(x_{ d1 }+x_{ d2 })^{ 2 } } -(D_{ o }+D)=\cfrac { (2n+1)\lambda  }{ 2 }$

where  $n=0, 1, 2, 3...$

$\cfrac{\sqrt { (D_{ o }+D)^{ 2 }+(x_{ d1 }+x_{ d2 })^{ 2 } }}{D_{ o }+D} -1=\cfrac { (2n+1)\lambda  }{ 2(D_{ o }+D) }$

$\cfrac{\sqrt { (D_{ o }+D)^{ 2 }+(x_{ d1 }+x_{ d2 })^{ 2 } }}{D_{ o }+D} =\cfrac {(2n+1)\lambda  }{ 2(D_{ o }+D) } +1$

$\cfrac{1}{cos(\theta)}=\cfrac { (2n+1)\lambda  }{ 2(D_{ o }+D) } +1$

$cos(\theta )=\cfrac { 2(D_{ o }+D) }{ (2n+1)\lambda +2(D_{ o }+D) }$

Since,

$\cfrac { 2(D_{ o }+D) }{ (2n+1)\lambda +2(D_{ o }+D) }<1$

we know that  $\theta$  has solutions.

when  $n$ is small,

$\cfrac { 2(D_{ o }+D) }{ (2n+1)\lambda +2(D_{ o }+D) }\approx 1$,    $\theta=0$


For constructive interference,

$\sqrt { (D_{ o }+D)^{ 2 }+(x_{ d1 }+x_{ d2 })^{ 2 } } -(D_{ o }+D)=n\lambda$

$n=0, 1, 2, 3,...$

$\cfrac{1}{cos(\theta)}=\cfrac{n\lambda}{D_{ o }+D}+1$

$cos(\theta)=\cfrac{D_{ o }+D}{n\lambda+D_{ o }+D}$

when $n$  is small,

$\cfrac{D_{ o }+D}{n\lambda+D_{ o }+D}\approx 1$   $\theta=0$

Both constructive and destructive interference mix and there is no pattern!  There is no observable interference pattern due to the interaction between the straight photons and deflected photons.

Lost is Lost...

Photon travels in a helical path.  Its phase is understood to be its relative position along its circular path.  For every $\lambda$ distance along its direction of travel, the photon completes one full revolution, a phase of  $2\pi$.  The following diagram shows how destructive and constructive interference might occur through collisions.

Over a longitudinal distance of  $\cfrac{\lambda}{2}$, the photon along it circular path reverses direction completely.  In this direction, it can collide with another photon at zero phase,

$m_pc-m_pc=0$

Momentum is zero after the collision and the kinetic energy is completely lost.

This is consistent with the conventional understanding of destructive interference.  In this case, a $\pi$ phase difference or a optical path difference of  half a wavelength can result in such destructive collisions.

In this model however,  there is no gain in energy during constructive interference.  It is unlikely to have collision, as both photons have the same velocity.

The lost energy during destructive interference does not re-appear during constructive interference.

Primero Uno, Number One

From the proportion mass table,

$\begin{matrix} \begin{matrix} m_p & 0 \\ 2m_p & 2^{n-2} \\ 4m_p & 2^{n-4}\\8m_p &2^{n-6} \\..&..\end{matrix} &  \\  &  \end{matrix}$
we see that the general particle,  $m_{p\,i}$

$\begin{matrix} 2^{i}m_p &2^{n-2i}  \end{matrix}$

when

$i=\cfrac{n}{2}$

we have

$m_{p\,n/2}=(\sqrt{2})^nm_p$

The single heaviest particle in the universe!  The top dog,  The Primero Onu, Number One.

Banana + Mochi = Banana Mochi

From the previous post "If The Universe Is A Mochi...", we would expect the universe to be roughly proportioned into,

Given,

$m_p=\cfrac{m}{2^n}$

$\begin{matrix} \begin{matrix} m_p & 0 \\ 2m_p & 2^{n-2} \\ 4m_p & 2^{n-4}\\8m_p &2^{n-6} \\..&..\\..&..\end{matrix} &  \\  &  \end{matrix}$
The most abundant elements is of the type  $2m_p$.

If we sum all the masses,

$0.m_p+2^{n-2}.2m_p+2^{n-4}.4m_p+2^{n-6}.8m_p+..$

And taking the limit  $n\rightarrow\infty$,

$\lim_{n\rightarrow\infty}\left\{2^{n-1}m_p(1+\cfrac{1}{2}+\cfrac{1}{4}+...)\right\}=\lim_{n\rightarrow\infty}\left\{2^{n}\cfrac{m}{2^n}\right\}=m$

which is the mass the big bang started with.

It is most interesting that this model suggests a binary count of masses in the universe.  That all masses are related to a constant ($m_p$)  by some power of 2,

$m_n=2^nm_p$,

and  $m_p$  itself being rare.

If The Universe Is A Mochi...

If the Big Bang is really BIG, we would expect the system to be driven forward towards greater entropy that,

$m_p=\lim_{n\rightarrow\infty}\left\{{\cfrac{m}{2^n}}\right\}$

each of velocity,

$v=\sqrt{2}c$

as we have seen that under both conservation laws of energy and momentum the split masses do not lose velocity and have equal velocity.  The process does not cost energy and is expected to go on util all particles are of mass  $m_p$.

What happens after the big bang has gained maximum entropy?  It begins to reverse itself.  The energy released are slowly being nullified.  (Both forward and reverse process are simultaneous.)  From the post  "If The Universe Is A Banana...",  in equation  (1) we find that for each particle  $m_p$  there is another of the reverse velocity.  Pairing them up for collisions we have,

$m_pv-m_pv=0=2m_pu$

$u=0$

and

$m_pv+m_pv=2m_pu$

$u=v$

The direct collisions head-on, destroys velocity and incur a kinetic energy loss.  The resulting mass coalesce and has twice the initial mass.  We define the energy loss as,

$\Delta L=\cfrac{1}{2}m_pv^2+\cfrac{1}{2}m_pv^2=m_pv^2$

For the side collisions where both masses travel in parallel and coalesce , there is no energy cost.

$\Delta L_s=0$

Both type of collision are equally likely and we expect half of the moving  $m_p$  to be involved in each type of collisions.

Hypothetically, the  $2^{n-1}$  pair of particles will give rise to  $2^{n-2}$  collisions of each type with a total energy loss of,

$Loss_1 = 2^{n-2}\Delta L$

and  produce $2^{n-1}$  particles of mass  $2*m_p$,  ($2^{n-2}$  pairs).   Half of this with non zero velocity  ($2^{n-3}$  pairs) in turn can similarly collide,

$2m_pv-2m_pv=0=4m_pu$

$u=0$

and

$2m_pv+2m_pv=4m_pu$

$u=v$

The associated loss per collision is given by,

$\Delta L_2=\cfrac{1}{2}2m_pv^2+\cfrac{1}{2}2m_pv^2=2m_pv^2=2\Delta L$

Since only half of the pairs of particles of mass $2m_p$  actually incur energy cost ( ($2^{n-4}$  pairs), we have the total loss as,

$Loss_2 = 2^{n-4}2\Delta L= 2^{n-3}\Delta L$

A total of $2^{n-3}$  masses,  $4m_p$ are produced from the previous collisions, of which half  ($2^{n-4}$) have non zero velocity, of which there are  $2^{n-5}$  pairs  colliding to produce $8m_p$,  and half of this,  $2^{n-6}$  collide at a cost of

$\Delta L_4=\cfrac{1}{2}.4m_pv^2+\cfrac{1}{2}.4m_pv^2=4\Delta L$

The total loss as a result of this type of collision is,

$Loss_4 = 2^{n-6}4\Delta L= 2^{n-4}\Delta L$

The total process loss is thus given by,

$LP=Loss_1+Loss_2+Loss_4...$

$LP=2^{n-2}\Delta L+2^{n-3}\Delta L+ 2^{n-4}\Delta L...$

$LP=2^{n-1}\sum^n_1{2^{-i}\Delta L}$

$LP=2^{n-1}m_p2c^2\sum^n_1{\cfrac{1}{2^i}}=2^{n}m_pc^2\sum^n_1{\cfrac{1}{2^i}}$

Since,  $m_p=\cfrac{m}{2^n}$,

$LP=mc^2\sum^n_1{\cfrac{1}{2^i}}$

Taking the limit  $n\rightarrow\infty$,

$LP=\lim_{n\rightarrow\infty}{mc^2\sum^n_1{\cfrac{1}{2^i}}}=mc^2$

which is the amount of energy we started with.

So, the big bang as a whole considering both forward entropy gain (particles split) and reverse entropy loss (particle coalescence) is stable.  The net energy sum is zero.

Feeling mochi safer already.

If The Universe Is A Banana...

A according to the big bang theory, the universe is a lump of very dense mass/energy,  $m$  that exploded.  Without lost of generality let say it first split into two, then into four, and again til infinitnum.  What is its final mass and velocity?

Consider conservation of energy and momentum

$mc^2=\cfrac{1}{2}mc^2+\cfrac{1}{2}mc^2$

$mc^2=\cfrac{1}{2}(\cfrac{m}{2})(\sqrt{2}c)^2+\cfrac{1}{2}(\cfrac{m}{2})(\sqrt{2}c)^2$

after  $n$  split, each particle will have energy,

$E_p=\left\{\cfrac{1}{2}\right\}^nmc^2$   ---(1)

and its mass is,

$m_p=\cfrac{m}{2^n}$

And the total number of particles after  $n$  split is,

$2^n$

Initially it is just a lump of mass with zero velocity.  Consider conservation of momentum after the first split, resulting in 2 masses $m/2$ each,

$0=\cfrac{m}{2}\sqrt{2}c+\cfrac{m}{2}\sqrt{2}c$  --- (1)

Subsequently, each particle starts with a momentum of

$P_p=\cfrac{m}{2}\sqrt{2}c$

the second split results in

$\cfrac{m}{2}\sqrt{2}c=\cfrac{m}{4}\sqrt{2}c+\cfrac{m}{4}\sqrt{2}c=\cfrac{m}{2}\cfrac{1}{2}\sqrt{2}c+\cfrac{m}{2}\cfrac{1}{2}\sqrt{2}c$

after  $n$ split, each particle will have momentum,

$P_p=\cfrac{1}{2}\left\{\cfrac{1}{2}\right\}^{n-1}m\sqrt{2}c=\left\{\cfrac{1}{2}\right\}^{n}m\sqrt{2}c$

Consider the total energy of the particle,

$E_p=PE_p+KE_p$

But since we started with  $PE_p=0$,

$E_p=KE_p=\cfrac{1}{2}mv^2=\cfrac{1}{2}mv.v=\cfrac{1}{2}P_p.v$

$\cfrac{1}{2}P_p.v=E_p=\left\{\cfrac{1}{2}\right\}^nmc^2=\cfrac{1}{2}.\cfrac{1}{2}\left\{\cfrac{1}{2}\right\}^{n-1}m\sqrt{2}c.v$

$v=\sqrt{2}c$

Up to here, if  $E=mc^2$  is our kinetic energy in time that denotes our existence, where  $c$  is our time speed along the time axis,  then it is possible to reach speed in space of up to  $\sqrt{2}c$.  And since the derivation up to this point, has not included any losses (both conservation laws apply), this is the max speed by which we can attained in space (a higher value of  $v$ implies $E>mc^2$, a lower value would imply loss, for which there is none). So, the maximum speed in free space is  $\sqrt{2}c$  and not  $c$.

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Here a similar graph where the absolute values of both gradients above the kink is less than 1.

As the derivative move toward -1, the band gap increases (because ln(|x|) has a zero at x=1 with a positive gradient).  In this case the higher temperature curve has a higher band gap.  As temperature decreases the band gap decreases.  The graph shows increasing temperature widen the band gap and move the emitted spectrum towards the ultra-violet band.

If quanta are created with $\cfrac{d\,T}{d\,t}$,  a strong variation in $T$,  $E_{BG}$ will vary through a wide bandwidth and is likely to be white light.

For Show

In the following graph, 1/x*(1-(1-1/x)^(1/2)) in blue is plotted together with 1/(2*x)*(1-(1-1/(2*x))^(1/2)) in black where x has been replaced with 2x.  This means, the black curve represents higher temperature; each value of x corresponds to 2x for the black curve.  Also plotted in the same corresponding color are the derivatives of the curves.  (The black curve should be above the blue curve if not manipulated so.)

The drop in band gap is illustrated on the derivative curves.  The band gap is from the top of the discontinuity all the way down to where its value approaches infinity.  It is noticed that the top of the discontinuity in the higher temperature curve dropped resulting in a drop in  $E_{BG}$.  This corresponds to a steeper gradient above the kink; in value, a lower value, more negative.

These curves are illustrative only.

Light Quantum Reduces Conductivity, Reflection=Re-Emission, Selective Reflection

How do quanta behave interacting with electrons at the band gap?

If the electron at the band gap is able to absorb the quantum, it will jump from the conduction band to the valence band, the result is a decrease in conductivity.  The same electron will re-emit the quantum and the material will appear as if reflecting the light quantum.

 If the electron is not able to absorb the quantum because its energy state simply does not add up, then it will not be able to re-emit the quantum and so the material will appear as if it does not reflect the light quantum.

Red light on green colored object has no reflection and the object appears black.

What happen to the quantum not absorbed?  It breaks up as it hit the nucleus, other electrons, etc. and dissipates.

So strictly speaking, the following account is wrong.

"For example, a red shirt looks red because the dye molecules in the fabric have absorbed the wavelengths of light from the violet/blue end of the spectrum. Red light is the only light that is reflected from the shirt. If only blue light is shone onto a red shirt, the shirt would appear black, because the blue would be absorbed and there would be no red light to be reflected."

Reflection is re-emission after absorption; the electron absorb the light quantum, transit down the band gap and when it returns from lower orbit it emit that quantum, and we see that as reflection.  Other colored quanta are not absorbed and so not re-emitted (reflected).  They simply dissipate in the lattice of the material.  In order to be reflected the quantum must first be absorbed.

Have a nice day.  Any one has a copper colored light?

Quantum After Quantum Not Wave

If we assume that collision is the only factor effecting the movement of space particles, then from the previous post "Shape Of Things To Come",  the particle does not lose its momentum, otherwise it would disappear within ten collisions.  The particle collide with one other particle each time and impart its whole velocity onto this particle,  itself come to rest immediately.  There is no change in speed direction; the quantum travels in a straight line.  (And so light travels in a straight line.)

So, the quantum as a whole travels without attenuation, as its particles transfer momentum without loss from collision to collision.

The next quantum is at a period  $T_e$ later,  as the electron perform SHM about its mean orbital radius  $r_{eo}$.   $T_e$  is the period of the electrons oscillating about  $r_{eo}$.  This period varies from material to material not necessarily at resonance.  A coil brought slowly to resonance by increasing its voltage frequency from zero, emit the same quantum throughout; it displays the same spectrum.  (Remember that the band gap is oscillating at twice the frequency of neighboring oscillating valence electrons, atom to atom.)

This suggests that we detect the "colors" of a quantum not by its frequency  $\cfrac{1}{T_e}$  but by the amount of energy it carries,  ie.  its $KE_s$.  However, each quantum is associated with photons over a range of frequencies as a result of the way photons interact with electrons that emit the quantum on transition across a band gap.  $\cfrac{1}{T_e}$  effects the intensity of the light we perceived,  obviously $\cfrac{1}{T_e}KE_s$ is power, Js^-1.

Disappointingly no shape and no wave.

Shape Of Things To Come

Let's take a look at the cascade of collisions again,

We can identify a series of forward velocities after  $n$  number of collisions,

$v\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n$,  

$v\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^2\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n$,  

$v\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^4\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n$...

The total momentum of this forward particles is,

$P_{ps}=m_sv\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n lim_{m\rightarrow\infty}\left\{ 1+\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^2+\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^4+...\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^{2m}+... \right\}$

$=m_sv \left\{\cfrac { 2}{ N_{ n }(\rho )+1 }\right\}^n lim_{m\rightarrow\infty} \sum^m_0{\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^{2m}}$

$P_{ps}=m_sv \left\{\cfrac { 2}{ N_{ n }(\rho )+1 }\right\}^n\cfrac{1}{1-\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^2}$

$=m_{ s }v\left\{ \cfrac { 2 }{ N_{ n }(\rho )+1 }  \right\} ^{ n }\cfrac { (N_{ n }(\rho )+1)^{ 2 } }{ (N_{ n }(\rho )+1)^{ 2 }-(N_{ n }(\rho )-1)^{ 2 } }$

$=m_{ s }v\cfrac { 2^{ n } }{ (N_{ n }(\rho )+1)^{ n-2 } } \cfrac { 1 }{ (N_{ n }(\rho )+1)^{ 2 }-(N_{ n }(\rho )-1)^{ 2 } }$

$=m_{ s }v\cfrac { 1 }{ N_{ n }(\rho ) } \left\{ \cfrac { 2 }{ (N_{ n }(\rho )+1) }  \right\} ^{ n-2 }$

If  $N_{ n }(\rho )>1$  the forward momentum dies down eventually as  $n\rightarrow\infty$,  after many collisions.

And a series of backward velocities,

$v\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n$,  

$v\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^3\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n$,  

$v\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^5\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n$...

The total momentum of this backward particles is,

$P_{ps\,r}=m_sv\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\} lim_{m\rightarrow\infty}\left\{ 1+\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^2+...\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^{2m}+... \right\}$

$=m_sv\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}  lim_{m\rightarrow\infty} \sum^m_0{\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^{2m}}$

$=m_{ s }v\left\{ \cfrac { 2 }{ N_{ n }(\rho )+1 }  \right\} ^{ n }\cfrac { N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 } \cfrac { (N_{ n }(\rho )+1)^{ 2 } }{ (N_{ n }(\rho )+1)^{ 2 }-(N_{ n }(\rho )-1)^{ 2 } }$

$=m_{ s }v\left\{ \cfrac { 2 }{ N_{ n }(\rho )+1 }  \right\} ^{ n-1 }\cfrac { N_{ n }(\rho )-1 }{ 2N_{ n }(\rho ) }$

$=m_{ s }v\left\{ \cfrac { 2 }{ N_{ n }(\rho )+1 }  \right\} ^{ n-1 }.\cfrac { 1 }{ 2 }\left\{ 1 -\cfrac { 1 }{ N_{ n }(\rho ) }  \right\}$

If  $N_{ n }(\rho )>1$ then the backward momentum also attenuate to zero.  When  $N_{ n }(\rho )=1$  there is no backward momentum, the particle has zero velocity.

From the post "Collisions And More Collisions" the last graph shows that when  $N_{ n }(\rho )=2$ velocity the attenuate very quickly within 10 collisions.  This would suggest that under normal circumstances  $N_{ n }(\rho )=1$.  That the forward momentum is carried from particle to particle without loss.

Collisions And More Collisions

Consider the particle of space,  $m_{ps}$ with velocity  $v$ in collision with its neighbors.  This collision is elastic and we have,

conservation of momentum,

$m_{ ps }v=N_{ n }(\rho )m_{ ps }v_{ n }+m_{ ps }v_{ s }$

where $N_n(\rho)$ is the number of neighbors the particle collides with,  a number that depends on density,  $\rho$.

$v=N_{ n }(\rho )v_{ n }+v_{ s }$    ---- (1)

and conservation of kinetic energy,

$\frac { 1 }{ 2 } m_{ ps }v^{ 2 }=N_{ n }(\rho )\frac { 1 }{ 2 } m_{ ps }v^{ 2 }_{ n }+\frac { 1 }{ 2 } m_{ ps }v^{ 2 }_{ s }$

$v^{ 2 }=N_{ n }(\rho )v^{ 2 }_{ n }+v^{ 2 }_{ s }$    --- (2)

Squaring equation  (1),

$v^{ 2 }=(N_{ n }(\rho )v_{ n }+v_{ s })^{ 2 }=N^{ 2 }_{ n }(\rho )v^{ 2 }_{ n }+v^{ 2 }_{ s }+2N_{ n }(\rho )v_{ n }.v_{ s }$

minus equation (2)

$0=\left\{N^{ 2 }_{ n }(\rho )-N_{ n }(\rho )\right\}v^{ 2 }_{ n }+2N_{ n }(\rho )v_{ n }.v_{ s }$

$0=v_{ n }N_{ n }(\rho )\left\{ (N_{ n }(\rho )-1)v_{ n }+2v_{ s } \right\}$

Since  $v_n$  and  $N_{ n }(\rho )$  are not zero,

$(N_{ n }(\rho )-1)v_{ n }+2v_{ s }=0$

$-v_{ s }=\cfrac { 1 }{ 2 } \left\{ N_{ n }(\rho )-1 \right\} v_{ n }$

From (1),

$v=N_{ n }(\rho )v_{ n }-\cfrac { 1 }{ 2 } \left\{ N_{ n }(\rho )-1 \right\} v_{ n }$

$v=\cfrac { 1 }{ 2 } \left\{ N_{ n }(\rho )+1 \right\} v_{ n }$

$v_{ n }=\cfrac { 2 }{ N_{ n }(\rho )+1 } v$

and

$v_{ s }=v-\cfrac { 2N_{ n }(\rho ) }{ N_{ n }(\rho )+1 } v=\left\{ 1-\cfrac { 2N_{ n }(\rho ) }{ N_{ n }(\rho )+1 }  \right\} v$

$v_{ s }=-\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }  v$

when  $N_{ n }(\rho )=1$

$v_{ n }=v$    and     $v_{ s }=0$

The moving particle stopped and the once stationary particle moves forward with velocity  $v$.

when  $N_{ n }(\rho )=2$

$v_{ n 1}=\cfrac{2}{3}v$,    $v_{ n 1}=\cfrac{2}{3}v$    and     $v_{ s }=-\cfrac{1}{3}v$

when  $N_{ n }(\rho )=3$

$v_{ n 1}=\cfrac{1}{2}v$,    $v_{ n 2}=\cfrac{1}{2}v$,    $v_{ n 3}=\cfrac{1}{2}v$     and    $v_{ s }=-\cfrac{1}{2}v$

The following diagram shows the cascade of collisions as time progresses,

The envelop of this cascade will always lead the shape of this velocity profile.  Particle collision within the envelop will not gain enough velocity to surpass the envelop.  As such we have a velocity curve as show below for the case of  $N_n(\rho)=2$

The peak of this curve is reduced by subsequent collisions, each time by the collision factor,

$C_c=\cfrac { 2 }{ N_{ n }(\rho )+1 }$

For particles in the reverse direction, they are first reduced by the reverse reduction factor,

$R_c=-\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }$

and then by  $C_c$  for each subsequent collision they encounter.

This velocity factor vs collision count curve is not the shape profile of the packet of energy.  For that we need to integrate over time.

Quantum and A Packet Of Space

Since the underlying principle that results in the kink in the  $r_e$  vs $T$  profile, is space density changing with temperature  (post "Science Fantasy, My Very Own...").  And it is postulated that space density decreases monotonously with increasing temperature,

$\cfrac{d\,r_e}{d\,T}\cfrac{d\,T}{d\,(d_s)}=\cfrac{d\,r_e}{d\,(d_s)}$

A discontinuity in $\cfrac{d\,r_e}{d\,T}$  will also result in a discontinuity in the rate of change of space density along the  $r_e$  direction.  $d_s$  is still continuous with  $r_e$  but a similar kink occurs in the graph  $d_s$  vs  $r_e$.

The following illustrative plot shows the discontinuity in red,  (the actual function plotted is 1/x*(1-(1-1/x)^(1/2)).)

If drag force is directly proportional to density,  $d_s$  and velocity squared,  $v^2$

$F=A\,d_s v^2$  where  $A$  is a constant of proportionality.

The work done against this force (by an electron) is,

$\int{F}d\,r_e=\int{A\,d_sv^2}d\,r_e$

$=A .\int { (d_{ s }) } d\, r_{ e }.v^{ 2 }-\int { A.\int { (d_{ s }) } d\, r_{ e }.2v }\, d\, r_{ e }$

If we were to define,

$m_{sA}=\int { (d_{ s }) } d\, r_{ e }$

as mass per unit area of space and,

$A_sm_s=Am_{sA}=A .\int { (d_{ s }) } d\, r_{ e}$  

where  $m_s$  is the mass of a packet of space, then the term,

$A .\int { (d_{ s }) } d\, r_{ e }.v^{ 2 }=A_sm_sv^{ 2 }$

is just a multiple of the of the $KE_s$ of a packet of space with mass  $m_s$  at velocity  $v$.  So,  the work done against drag, in part, increases the velocity of this packet of space,

Work done against drag  $F$,   $\int{F}d\,r_e=A_{ s }\, m_{ s }v^{ 2 }-2A_{ s }\int { \, m_{ s }.v } \, d\, r_{ e }$

At the kink point, integrating  from  $re1$  to  $re2=re1+\varepsilon$  where  $\varepsilon$  is small.

 $\int^{re2}_{re1}{F}\,d\,r_e=\left\{A_{ s }\, m_{ s }v^{ 2 }\right\}^{re1+\varepsilon}_{re1}-2A_{ s }\left \{ \, m_{ s }.v  \right\}^{re1+\varepsilon}_{re1}.\varepsilon$

Work done in passing over the kink is,

 $\int{F}\,d\,r_e|_{kink}=lim_{\varepsilon\rightarrow0}\left\{\left\{A_{ s }\, m_{ s }v^{ 2 }\right\}^{re1+\varepsilon}_{re1}-2A_{ s }\left \{ \, m_{ s }.v  \right\}.\varepsilon\right\}$

 $\int{F}\,d\,r_e|_{kink}=A_{ s }m_{ s }v^{ 2 }_{re2}-A_{ s }m_{ s }v^{ 2 }_{re1}=2A_s\left\{\cfrac{1}{2}{m_{ s }v^{ 2 }_{re2}-\cfrac{1}{2}m_{ s }v^{ 2 }_{re1}}\right\} =2A_s\Delta KE_s$

This discontinuity in velocity across the kink point is consistent with the treatment of electrons across the kink point.  For the case of electrons,  $r_e$  is a constant at the kink point, so  $PE_e$  is constant and so the change in total energy is attributed wholly to a change in  $KE_e$.  ie velocity is discontinuous across the kink point.

This means, the work done against the drag force at the kink point, can be seem as resulting in an increase in kinetic energy of a packet of space with mass  $m_s$.

Furthermore, this work done against the drag force is equal to the change in total energy of the electron on passing the kink point, because the $r_e$  vs  $T$  curve on which the kink reside was derived by considering the effect of temperature on drag at high velocity, in the first place. Therefore the  $KE_s$  gained by the packet of space at the kink point is equal to the band gap.

$E_{BG}=2A_s\Delta KE_s$

Metaphorically,  the electron collided with a packet of space with mass  $m_s$ at the kink point.  The packet of space gained kinetic energy,  $\Delta KE_s$  as given by the expression above, from the collision.

If we set  $2A_s=1$,  then the quantum is just a packet of space with kinetic energy.  This awkwardness result in part from considering electrons as point particles without the second and third dimension, but density as mass per unit volume in 3D.  And, in part from the undetermined constant of proportionality in the expression for drag force.

This packet of space has a boundary delimited by non zero velocities.  If space in made up of particles, then the extend over which the particles have non zero velocities constitute the body of this packet.

The unknow factor  $2A_s$  actually allows for multiple packets of space/quanta to be created at the kink from one transition of an electron across the band gap.   All quanta have the same velocity but their masses that was derived from an integral of space density, can be broken down into a sum of multiple terms.  This means the mass of individual quantum may not be equal, but their total sum is constrained by the integral of space density over  $r_e$.

$\int { (d_{ s }) } d\, r_{ e }=m_{sA1}+m_{sA2}+...$

And Alice broke the looking glass.  This is counter intuitive, because we would expect a collisions involving different masses to have different velocities.  In this case, part of the kinetic energy of the electron  $\Delta KE_s$,  (the work done against drag), is divided directly in proportion to their masses among the quanta created.  They all have the same velocity.

Does these packets of energy travel as waves?

Increase Photon Frequency

If photons are dipoles, here is a scheme to squeeze the radius of photons to smaller values, such that their frequencies are generally increased.

The expression for the total energy of a photon is  $m_pc^2$,  a sum of its linear KE and its KE in circular motion, BOTH at light speed  $c$.  It is independent of its radius  $r$.  The scheme above suggest that photon frequency can be increased by applying a current through a tapered coil.  If photons are dipoles, we can also estimate the relative Left Handedness or Right Handedness of them as  they occur naturally.

Have a nice day.

Why Photon Has No Mass?

If photons and light quanta are separate issue, it is possible that photons are particles (dipoles) at high speed from the Sun.

Daylight is the result of photons interacting with orbiting electrons.

Light from a electric torch is the result of directly exciting the electrons, the torch does not produce photons.

Light from combustion,  drives the electrons into oscillation along the radial direction through  $\cfrac{d^2T}{d\,t^2}$,  it may not produce photons.

Then there is photoelectric effect, it has to be from a light source that produces photons.  LEDs will not produce photoelectric effects.  Otherwise that will suggest electrons interacting directly with quanta of energy.   Which bring us to the question of what is the nature of such packets of energy.  Do they have a shape, a boundary, how fast do they travel, do they dissipate if left alone, how do they dissipate...

We see that from the post "Drag and A Sense of Lightness",  an electron at high speed in circular motion seems to lose mass given by the expression,

$m_e-Ar_e=m_{el}$

where $m_{el}$  is a newly defined reduced mass of a electron,  $A$ is the drag factor and  $r_e$  the radius of circular motion.

The expression comes from equating centripetal force with electrostatic force and drag,  The drag force provides for part of the centripetal.  The result is an apparent loss in mass, as the electron perform circular motion with radius  $r_e$ around the nucleus.

In the case of photon, the centripetal force is provided for fully by the drag as its speed approaches light speed.  This is the limiting case where the reduced mass is zero.  Consider the centripetal force on a photon in a helical path,  (Photons also have a transverse velocity,  $c^2$  that is not consider here.)

$\cfrac{m_pc^2}{r_p}=Ac^2$

as such

$m_p-Ar_p=0=m_{pl}$

Photon also has a reduced mass,  $m_{pl}$ when it is in circular motion down a helical path,  this reduced mass is zero.

At this point, photon can be anything that's a dipole.  $H$, hydrogen atom is a good guess.

White Light and The Photon Bus

In the case of an element producing white light, it is possible that the bandwidth of the quantum as suggested in end of the post "Mood Stone", is itself wide enough to contain the whole spectrum.  This bandwidth is the result of the electron and another valence electron from its neighboring atom both performing SHM about its radius  $r_e$, as such pushing and stretching their  $r_e$  vs  $T$  profile.  The band gap is changed continuously and as a result the quantum spread over a range of frequency.  Calcium is such an element.

In this case, green light will still reflect green,  as the amount of energy transferred to the electron depended on the photon, electrons that fail to reach the kink with enough energy will not emit any quantum.  But the reflection will contain a fair amount of white light, that result from the band gap changing dynamically.  A prism in the path of green light reflecting off a white calcium surface will show this.

Moreover, light dispersion occurs in all material and in the reflection there will be lower frequency components (post "Light Dispersion?)'.  For green incident light, yellow can be seen.  A green light incident-ed on a white surface obliquely, shows yellow on the far edge of the incident spot.

Still, there is this problem of photons themselves not being the quanta.  If photon energy can be parceled and dispensed, can photon absorb a quantum directly?  And transport the quantum along its way?

A photon bus!  A photon is self-propelling; to light speed or terminal velocity of free space.

Beam To Another Location

One limited way to teleport yourself, is to move along your trajectory in time as earth travels through space and spins.  More accurately, your are at a time in the past or future at your present location when earth is at a slightly different place.  So as you materialize sometime in the past or future, you are at a different place on earth.

Where?  That will take a whole lot of calculations.  Earth's spin, can carry you to anywhere along the latitude adjusted for altitude to account for earth trajectory around the Sun.

And the space time continuum continues.

Photons Don't Die

Photons are not light but they carry light, in a round about way.   Only by interacting with orbiting valence electrons are light quanta brought into existence.  Photon is a particle that carries energy and since it is self-propelling (post "Ladies and Gentlemen, We Have Light Speed!"), it regains its lost energy , and so, itself does not change in the energy processes that it is involved in. It never dies.

What is a photon???

Nature Of Energy, Mistaken Identity, Photon≠Quantum

There is a problem, not that this crazy idea seem to contradict the fact that green light reflects off a white surface green and that a red light reflects off the same white surface red and so color depends on illumination instead.  This contradiction is easy resolved by going more crazy,  that photon and electron interaction goes beyond the valence shell and interact within other electrons as well.  Each  different $r_e$  will have its own different kink point and a different associated band gap.  Thus emitting a different quantum and so a different color.  When all primary colors (RGB) are present, the element/material is white.

In the case of a white surface, it is likely that the different atoms of the molecules that constitute the material have different colored quantum all interacting with the incident photons at the same time.  The emitted quanta are of all primary colors of light, RGB; combined and we have white.  When a green light alone is shone on the surface, only the electrons at the appropriate  $r_e$ and the associated  $r_e$  vs  $T$  profile and band gap are affected.  The quanta emitted will be green.

The real problem is,  these emitted quanta are pure energy, whereas photons were modeled as spinning dipoles.

Are energy quanta dipoles?  The glaring fact is, photon is not the quantum.  The packet of energy emitted, is the result of an electron transition across the band gap because of a passing photon.  It is not the photon itself.

Mood Stone

From the previously post "Color of Material",  The color present by a material is determined by the discontinuity in gradient at the kink point in the  $r_e$  vs  $T$ profile.  A material that changes color noticeably with temperature means that the discontinuity in gradient,  ie. the band gap, changes widely with changing temperature.

In the case of a mood stone,  a few degrees change in temperature from ambient to body temperature shift the emitted frequencies from dark blue to red.  A decrease in emitted energy.

As temperature,  $T$  is increased,  temperature of the nucleus increases, and the temperature profile extending radially outwards shift upwards to increasing  $T$  and reaches further out to increasing  $r_e$.

The electron's  $r_e$  vs  $T$  profile stretches upwards and outwards correspondingly.

Since,  $T$  increases more than the small value of  $r_e$,  there is an decrease in the gradient,  $\cfrac{d\,r_e}{d\,T}$  atop of the kink point.  The gradient below the kink point is due to a square-root term and does not change. The result is a decrease in band gap and the emitted quantum has a lower energy corresponding to a lower frequency.

$grad T_L=\cfrac { d\, r_e }{ d T }|_{TL}$,    $grad T_H=\cfrac { d\, r_e }{ d T }|_{TH}$

The gradient atop of the kink is steeper when the temperature is higher.

$grad T_L>grad T_H\implies$

$0>ln(|\cfrac { d\, r_e }{ d T }|_{TL}|)>ln(|\cfrac { d\, r_e }{ d T }|_{TH}|)$

since,  $|\cfrac { d\, r_e }{ d T }|_{TL}|, |\cfrac { d\, r_e }{ d T }|_{TH}|<1$

$\left|ln(|\cfrac { d\, r_e }{ d T }|_{TL}|)\right|<\left|ln(\cfrac { d\, r_e }{ d T }|_{TH}|)\right|$

and From the post "Pag.Pag...Pag....Pag Dnab Ygrene", the band gap factor,

$E_{BG}=KE_{re}|_{r{e1}}\left\{{ln(\cfrac{ d\, r_e } { d T }|_{r{e1}})\over ln(\cfrac { d\, r_e }{ d T }|_{r{e2}})}-1\right\}$

where  $re1$  and  $re2$  are points just before and after the kink point.  We have,

$\left|{ln(|\cfrac{ d\, r_e } { d T }|_{kink}|)\over ln(|\cfrac { d\, r_e }{ d T }|_{TL}|)}-1\right|=\left|{ln(|\cfrac{ d\, r_e } { d T }|_{kink}|)\over| ln(|\cfrac { d\, r_e }{ d T }|_{TL}|)|}+1\right|$

and so,

$\left|{ln(|\cfrac{ d\, r_e } { d T }|_{kink}|)\over |ln(|\cfrac { d\, r_e }{ d T }|_{TL}|)|}+1\right|>\left|{ln(|\cfrac{ d\, r_e } { d T }|_{kink}|)\over |ln(|\cfrac { d\, r_e }{ d T }|_{TH}|)|}+1\right|$

$KE_{kink}$ also increases with temperature, but overall the band gap decreases.

The material changes to red color.

It is important that the electron move freely outwards,  corresponding to  $r_e$  stretching upwards.  If the valence electrons from neighboring atoms pushes against the profile as in the case of a close lattice structure, the change in gradient on top of the kink will be limited.

Moreover, given the fact that valence electrons are in SHM along the radial line, the band gap of the electron in the neighborhood will vary as the top of  the $r_e$  vs  $T$  profile is pushed by moving valence electron from adjacent atom(s).  The band gap will vary in a SHM matter and the result is the spread of the quantum over a range of energies.  So, the emitted energies spreads over a spectrum and is not a single spectra line.

The bandwidth of this spectrum, is twice the SHM frequency, as both electrons are moving.

Color of Material

From the previous post "Light Dispersion?", if a passing photon pushes an orbiting electron further into the nucleus but without subsequently ejecting the electron, then the electron is set into oscillation about its mean oribt  $r_{eo}$.

This electron on passing the kink point on the $r_e$  vs  $T$  profile  will then emit a packet of energy, a fixed quantum, until its oscillation is damped and it can no longer reach the kink point.  The size of this quantum of energy is fixed irrespective of the photon energy as long as it is able to provide enough energy that the electron passes beyond the kink on the electron's return.

$\Delta E_p(f_{th})>E_s(r_{kink},r_{eo})=PE_e(r_{kink})-PE_e(r_{eo})$

where  $\Delta E_p(f_{th})$  is the energy imparted onto the electron by the photon at  frequency  $f_{th}$ as they interact.

For oscillation without immediate ionization, this energy will have to be less than the ionization energy.  And  photon frequency is related to its radius by,

$c=2\pi.r_{th} f_{th}$

where  $r_{th}$  is the radius of the photon circular path at  frequency  $f_{th}$.

But if  the radius of the photon's helical path is too small, the photon may not be able to push the orbit electron further beyond the kink point.  So, we would expect the interaction between the photon and electron to be over a bandwidth of frequencies, for an emission of a quantum  (whether the atom is subsequently ionized or not).

$\Delta E_p(f_{lm})>E_s(r_{kink},r_{eo})=PE_e(r_{kink})-PE_e(r_{eo})$

Bandwidth,

$BW=f_{th}-f_{lm}=\cfrac{c}{2\pi}\left\{\cfrac{1}{r_{th}}-\cfrac{1}{r_{lm}}\right\}$

This energy quantum emitted by the material is perceived as its color!  Noted that oscillation is not required for emission of the quantum but repeated emissions requires the electron to oscillate about  $r_{eo}$,  its mean orbital radius about the nucleus, passing the kink point each time.

Does a photon then has color?  Color of light will be a misnomer.

Light Dispersion?

From the post "Like Wave, Like Particle, Not Attracted to Electrons", the energy required to move from  $r_{eo}$  to its final position  $r_{ef}$,

${E}_{s}={ m }_{ e }c^{ 2 } \{ln{( \cfrac{{r}_{eo}}{{r }_{ e f}})}+C({ r }_{ ef }-{ r }_{ e o})\}\ge PE_e$    ----(*)

The energy required to move from  $r\rightarrow\infty$ to  $r_{eo}$,

$PE_e(r_{eo})={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty  }{ \{ln{( \cfrac{{r }}{{r}_{eo}})}+C({ r }_{ eo}-{ r })\}}$

The energy required to move from  $r\rightarrow\infty$ to  $r_{ef}$,

$PE_e(r_{ef})={ m }_{ e }c^{ 2 }\lim _{ r\rightarrow \infty  }{ \{ln{( \cfrac{{r }}{{r}_{ef}})}+C({ r }_{ ef}-{ r })\}}$

And so,

${E}_{s}=\lim _{ r\rightarrow \infty  }\left\{PE_e(r_{ef})-PE_e(r_{eo})\right\}$

${E}_{s}={m_ec^2}\left\{ln{( \cfrac{{r_{eo} }}{{r}_{ef}}) +C(r_{ef}-r_{eo})}\right\}$

The electron is in circular motion about the nucleus, at  $r_{eo}$ it experiences the centripetal force,

$F_c=\cfrac{m_ec^2}{r_{eo}}=\cfrac{\partial\,{E}_{s}}{\partial\,r_e}={m_ec^2}\left\{ \cfrac{1} {{r}_{ef}{r_{eo} }}-C\right\}$

$C=\cfrac{1} {{r}_{ef}{r_{eo} }}-\cfrac{1}{r_{eo}}=\cfrac { 1-{ r }_{ ef } }{ { r }_{ ef }{ r_{ eo } } }$

Therefore,

${E}_{s}={m_ec^2}\left\{ln{( \cfrac{{r_{eo} }}{{r}_{ef}}) +\cfrac { 1-{ r }_{ ef } }{ { r }_{ ef }{ r_{ eo } } }(r_{ef}-r_{eo})}\right\}$

For the passing photon, if we consider only the direction along a radial line from  $r_{eo}$  to  $r_{ef}$,  since its longitudinal velocity does not effect net changes in energy in this direction.  (Actually if the interaction between the photon and electron is electrostatic than, the photon slows as it approaches the orbiting electron but speeds up again as it leave the vicinity of the electron; energy is conserved along its direction of travel).

The repulsion between the electron and the photon reduces the centripetal force on the photon.  It is expected that the photon path radius increases, .

$r_{po}\rightarrow r_{pf}$

together with a decrease in circular velocity and so, a drop in $KE_p$.  It is this drop in  $KE_p$  that accounts for the increase in $PE_e$  of the electron.  (The electron orbiting at light speed around the nucleus.)  The centripetal force, $F$  in account for photon/electron repulsion is,

$F=F_c-F_r=\cfrac{m_pc^2_{pf}}{r_{po}}-F_r=\cfrac{m_pv^2_{pf}}{r_{pf}}$

where $F_c$  is due to drag at light speed,  $F_r$  is the repulsion force,  $v_{pf}$  is the velocity of the photon at the furthest point in its orbit right over the atom.  The relative position of the photon and electron is unknown at this point, but  $F_{r\, max}$  is,

$F_{r\,max}=\cfrac{q_pe}{4\pi\varepsilon_o (r_{po}-r_{eo})^2}$

$\Delta E_p=\Delta KE_{p}=\cfrac{1}{2}m_p\left\{v^2_{pf}-c^2 \right\}=-{E}_{s}(r_{eo},\,r_{ef})$ --- (*)

From this we have,

$\cfrac { m_{ p }c^{ 2 }_{ pf } }{ r_{ po } } -\cfrac { q_{ p }e }{ 4\pi \varepsilon _{ o }(r_{ po }-r_{ eo })^{ 2 } } =\cfrac { m_{ p }v^{ 2 }_{ pf } }{ r_{ pf } }$

$m_{ p }v^{ 2 }_{ pf }={ r_{ pf } }\left\{ \cfrac { m_{ p }c^{ 2 }_{ pf } }{ r_{ po } } -\cfrac { q_{ p }e }{ 4\pi \varepsilon _{ o }(r_{ po }-r_{ eo })^{ 2 } }  \right\}$

substitue into  (*)

$\cfrac { 1 }{ 2 }{ r_{ pf } }\left\{ \cfrac { m_{ p }c^{ 2 }_{ pf } }{ r_{ po } } -\cfrac { q_{ p }e }{ 4\pi \varepsilon _{ o }(r_{ po }-r_{ eo })^{ 2 } }  \right\} -\cfrac { 1 }{ 2 } m_{ p }c^{ 2 }=-{E}_{s}(r_{eo},\,r_{ef})$

${ r_{ pf } }=A \left\{ \cfrac { 1 }{ 2 } m_{ p }c^{ 2 }-{ E }_{ s }(r_{ eo },\, r_{ ef }) \right\}$

where  $A$  is,

$A=\cfrac { 2 }{ \left\{ \cfrac { m_{ p }c^{ 2 }_{ pf } }{ r_{ po } } -\cfrac { q_{ p }e }{ 4\pi \varepsilon _{ o }(r_{ po }-r_{ eo })^{ 2 } }  \right\}  }$

This is the maximum value of  $r_{pf}$, simply because we set  $F_r$  to be maximum.  But since,

$c=2\pi .r .f$

there is a corresponding maximum decrease in frequency.

$f_{pf}=\cfrac{c}{2\pi.r_{pf}}$

In fact, since interaction between the photon and electron at different distances are possible,  there is a spread of frequency from  $f_{pf}$  to  $f_{po}$.  That photons emerging from the material has now a spread of energy, and the frequency range is,

$f_{po}≥f≥f_{pf}$

where  $f_{op}$  is the frequency of the incident photon passing through without interacting with any electrons.  If subsequently, these photons with a spread of frequencies are to eject more packets of energy,  the resulting quanta will show a spread of colors as each photon interacts with different electron at different  $r_e$,  $r_e$  vs  $T$  profile and kink point.

This might explain the spread of colors we see when white light enters a prism.  The resulting spectrum is continuous.  However, this model indicates that monochromatic light will also spread continuously down the frequency spectrum.  More importantly,  in this model, photons are self propelling dipoles that will regain the lost  $KE_p$  and returns to a circular velocity of light speed by itself.  So if we recombine the spread of light,  we will see white light further down its path with the same intensity!

Where do photons go to die?  Where's the photon graveyard?

Eye Beta, Peta

Obviously I am trying to be funny,

Does it work?

Listening To the Future, Future Radio

If we live in holographic world moving along the time line,

and since,  both  $e^{+iwt}$  and  $e^{-iwt}$  are admissible solutions to radio wave,  a radio wave travels back in time,  $e^{+iwt}$   in addition to one that travels forward,  $e^{-iwt}$  in time.  We may listen to the future by multiplying a suitable phase to the received signal at the antenna.

Have a nice day.

Missing Post About Photoelectric Ionization, A Shinny Issue

Here is it again,

The interaction between electron in orbit and a photon (a proposed dipole) is electrostatic.  The photon is in a helical path, on approach towards the atom, the orbital radius of the electron is squeezed to a lower value.  The energy is stored as developed in  "Like Wave, Like Particle, Not Attracted to Electrons", because of orbital speed at light speed cannot be increased further,

Consider the post "Pag.Pag...Pag....Pag Dnab Ygrene",  a packet of energy is emitted when the photon transit to a higher orbit pass a kink point in the  $r_e$  vs  $T$  profile, that means the light reflected off the surface of the metal is brighter than the incident light.  This is the common glare we experience from shiny surface.  In fact this is the reason way some surfaces are so shinny;  there are more photons on reflection than incident.

Next up, how much energy the photoelectric photon lose on passing?

Levitating Stones

From the post "Wait A Magnetic Moment" and the correction made in "Erratum, 2D Flat and Flatulent",

The magnetic moment of an electron in orbit,

$S_e=\sqrt{2}qc r_{es}$

If  $r_e$ has a time component,

$r_e=r_{eo}e^{i\omega t}$

then

$S_e=\sqrt{2}qc.r_{eo}e^{i\omega t}$

and  $S_e$  will resonate as  $r_e$ at,

$\omega^2_o= \cfrac { d\, r_{ e } }{ d\, T } \cfrac{\partial}{\partial\,r_e}\left(\cfrac { d^{ 2 }T }{ dt^{ 2 } }\right)$

From the post "Gravity Wave and Schumann Resonance",  Earth has a gravity wave at 7.489 Hz and Schumann resonances at 7.83 Hz.

If, assuming no damping,

$\cfrac { d\, r_{ e } }{ d\, T } \cfrac{\partial}{\partial\,r_e}\left(\cfrac { d^{ 2 }T }{ dt^{ 2 } }\right)=7.83^2$

is applied to a magnetic material where the electron moment is aligned into magnetic domains, we might just have a levitating magnet or lodestone.

$\cfrac { d^{ 2 }T }{ dt^{ 2 } }$  is probably provided by a heat source or indirectly by a electric charge.

$\cfrac { d\, r_{ e } }{ d\, T }$  is characteristic of the material dependent on temperature.

At the right combination of  a heat source and temperature, the material will levitate.

If a electrical voltage is to provide for  $\cfrac { d^{ 2 }T }{ dt^{ 2 } }$ then from the post "Rectified Waveform To The Rescue",

$\left|\cfrac { d^{ 2 }T }{ dt^{ 2 } }\right|=\omega_dT_o$

where  $T_o=V_oI_ocos(\theta)$,  $cos(\theta)$ is the power factor,  and  $V(t)=V_oe^{i\omega_d t}$.

It is unknown how the material will behave, however we can replace,

$\cfrac{\partial}{\partial\,r_e}\left(\cfrac { d^{ 2 }T }{ dt^{ 2 } }\right)$

with

$\omega_dT_o M_{t2}=\cfrac{\partial}{\partial\,r_e}\left(\cfrac { d^{ 2 }T }{ dt^{ 2 } }\right)$;

that is factor  $\omega_dT_o$ can be distilled.  $M_{t2}$ can be considered a material characteristic.

We have

$\left|\cfrac { d\, r_{ e } }{ d\, T }\right|\omega_dT_oM_{t2} =7.83^2$  or

$\omega_d=7.83^2.\cfrac{1}{\left|\cfrac { d\, r_{ e } }{ d\, T }\right|M_{t2}V_oI_ocos(\theta)}$

instead.

At a suitable cold temperature, adjust  $\omega_d$ as per above expression until the magnet/lodestone  levitate.  Make sure that the material is correctly orientated, that the magnetic domains, North pole, is downward.

Merry Go Round

Maybe the biggest mistake we can make is to think time is cyclic.

And round and round we go.  $i\beta$  is of course my personal time machine.

To Be Complex Is Phase

From the post "Rectified Waveform To The Rescue",  when the waveform is rectified,

$\omega_a=4\omega_d$.

${\cfrac { d^2 T }{ d\, t^2 }}=-\omega_d T_o$

$\xi^2=\cfrac{1}{2}(1-\cfrac{\omega_a}{\cfrac{1}{4}T_o})$

$\xi$  can be reduced to zero and beyond.

For

$\omega_a>\cfrac{1}{4}T_o$ for a rectified waveform driving force,

$\xi$  is complex.  What happens when  $\xi$  is complex?

Consider the energy loss in a damped system,

$F_{loss}=-2p.\cfrac{d\,x}{d\,t}$

$E_{loss}=\int{-F_{loss}}dx=\int{2p.\cfrac{d\,x}{d\,t}}dx=\int{2\xi\omega_o\cfrac{d\,x}{d\,t}}\cfrac{d\,x}{d\,t}dt$

$\cfrac{d\,E_{loss}}{d\,t}=2\xi\omega_o\left\{\cfrac{d\,x}{d\,t}\right\}^2$

Since  $\cfrac{d\,x}{d\,t}$ has a time component,  a complex  $i\xi$  would mean a  $\pi/2$  phase delay in  $\cfrac{d\,E_{loss}}{d\,t}$.

$2i\xi\omega_o\left\{\cfrac{d\,x}{d\,t}\right\}^2=2\xi\omega_o\left\{\cfrac{d\,x}{d\,t}\right\}^2e^{i\pi/2}=2\xi\omega_o\left|\cfrac{d\,x}{d\,t}\right|^2e^{i2wt}e^{i\pi/2}=2\xi\omega_o\left|\cfrac{d\,x}{d\,t}\right|^2e^{i(2wt+\pi/2)}$

Complex, Simply Yours

For the post "Rectified Waveform To The Rescue",

The response using a full normal wave is just the superposition of a rectified wave and an inverted rectified wave, but frequency halved.  The results does not contradict.

But the expression,

$\cfrac{d^2T}{d\,t^2}=i\omega_dV_oI_oe^{i2\omega_d t}cos(\theta)$

is a complex valued frequency  $i\omega_d$,  as needed in the post "Complex Frequency And SuperConductivity".  Which is dealt with,

$\cfrac{d^2T}{d\,t^2}=e^{i\pi/2}\omega_dV_oI_oe^{i2\omega_d t}cos(\theta)$

$\cfrac{d^2T}{d\,t^2}=\omega_dV_oI_oe^{i(2\omega_d t+\pi/2)}cos(\theta)$

A phase delay in time.

That is to say, a complex number, $e^{iA}$ multiplied to a time function  $e^{iwt}$,  travels back or forth in time by a phase  $A$.

Rectified Waveform To The Rescue

Once again, at resonance, for

$p=\xi\omega_o$

$\omega^2_n=\omega^2_o-2p^2=\omega^2_a$

$\omega^2_n=\omega^2_o-2\xi^2\omega^2_o=\omega^2_a$

$\omega^2_o(1-2\xi^2)=\omega^2_a$

From the post "What Am I Doing? Pressure Lamp"

$\omega^2_o=  {-\cfrac { d\, r_{ e } }{ d\, T }}_{kink} \cfrac{d^2}{d\,t^2} \left\{\cfrac{d\,T }{ d\,r_e}_{kink}  \right\} = {-\cfrac { d\, r_{ e } }{ d\, T }}_{kink}\cfrac{d\,T }{ d\,r_e}_{kink} \cfrac{d^2T}{d\,t^2}=-\cfrac{d^2T}{d\,t^2}$


when $T=time(t)T(r_e)$ is separable in time and $r_e$,

$\cfrac{d^2}{d\,t^2} \left\{\cfrac{d\,T }{ d\,r_e} \right\} =\cfrac{d\,T(r_e) }{ d\,r_e}_{kink} \cfrac{d^2time(t)}{d\,t^2}=\cfrac{d\,T }{ d\,r_e}_{kink}\cfrac{d^2time(t)}{d\,t^2}$

Therefore,

${ {-\cfrac { d\, r_{ e } }{ d\, T }}_{kink} \cfrac{d^2}{d\,t^2} \left\{\cfrac{d\,T }{ d\,r_e}_{kink}  \right\}(1-2\xi^2)=\omega^2_a}$ --- (*)

where ${\cfrac { d\, r_{ e } }{ d\, T }}_{kink}$ is negative.

Since the driving force is electrical, it is possible to apply rectified waveform instead of a pure sinusoidal.

In the both cases the applied frequencies doubles,  $\omega_a\rightarrow2\omega_a$,  For a normal rectified waveform,

$\cfrac{d^2T}{d\,t^2}$ is always negative

$\omega_a=2\omega$

and

$\cfrac{d^2}{d\,t^2} \left\{\cfrac{d\,T }{ d\,r_e}_{kink}\right\}=-T_o\omega^2\cfrac{d\,T }{ d\,r_e}_{kink}=-\cfrac{1}{4}T_o\omega^2_a\cfrac{d\,T }{ d\,r_e}_{kink}$

So,  expression (*) becomes,

$-\cfrac { 1 }{ 4 } T_o\omega^2_a(1-2\xi^2)=\omega^2_a$

$-\cfrac { 1 }{ 4 } T_o (1-2\xi^2)=1$

$\xi ^{ 2 }=\cfrac{1}{2} (1+\cfrac { 4 }{T_{ o } } )$

it is possible that  $\xi=1$.

When the inverted rectified waveform is applied,

$\cfrac{d^2T}{d\,t^2}$ is always positive

$\omega_a=2\omega$

and

$\cfrac{d^2}{d\,t^2} \left\{\cfrac{d\,T }{ d\,r_e}_{kink}\right\}=T_o\omega^2\cfrac{d\,T }{ d\,r_e}_{kink}=\cfrac{1}{4}T_o\omega^2_a\cfrac{d\,T }{ d\,r_e}_{kink}$

$\cfrac { 1 }{ 4 } T_o\omega^2_a(1-2\xi^2)=\omega^2_a$

$\cfrac { 1 }{ 4 }T_o (1-2\xi^2)=1$

$\xi ^{ 2 }=\cfrac{1}{2} (1-\cfrac { 4 }{T_{ o } } )$

It is not possible that,  $\xi=1$, since $T_o\gt 1$

To achieve $\xi=1$ would depend on the material and may not be achievable.  If it is achievable then the material can be made very conductive with a rectified voltage at resonance frequency applied to it at low temperature.

This is because, at critical damping, the electrons are pushed once over the kink.  A short pulse of this rectified waveform at the resonance frequency will increase the number of electrons in the conduction band and make the material more conductive.  Imagine an antenna pulse with this waveform before transmitting or receiving.

If not at critical damping, the reverse voltage on the negative cycle of the applied waveform will pull some of the electron back into the valence band, reducing conductivity.

On the practical side, simply switch the waveform to the inverted rectified wave and reduce  $\omega_a$  to half of resonance frequency at a suitable material temperature.  The glow should disappear immediately as the system is now critically damped.

When we consider in terms of current,  $I$   and voltages,  $V$,  of the oscillating electrical voltage system used to drive the system, things are even simplier.  Consider,

$\cfrac{d\,T}{d\,t}=VIcos(\theta)=\cfrac{1}{2}V_oI_oe^{i2\omega_d t}cos(\theta)$

$\cfrac{d^2T}{d\,t^2}=i\omega_dV_oI_oe^{i2\omega_d t}cos(\theta)$

where  $cos(\theta)$  is the power factor, and the applied voltage is a sinusoidal

$V=V_oe^{i\omega_dt}$

The system is actually driven at twice the dial frequency,  $\omega_d$.  ie

$\omega_a=2\omega_d$

We let,

$T_o=V_oI_ocos(\theta)$    then

$\omega^2_o=   {-\cfrac { d\, r_{ e } }{ d\, T }}_{kink} \cfrac{d^2}{d\,t^2} \left\{\cfrac{d\,T }{ d\,r_e}_{kink}  \right\}=-i \omega_dT_o=\omega_dT_oe^{-i\pi/2}$

A negative sign was introduced previously because the gradient was negative in order to oscillate, in this case, the negative sign appears as a phase lag.

$\omega^2_o= \cfrac{1}{2}\omega_aT_oe^{-i\pi/2}$

At resonance,

$\omega^2_o(1-2\xi^2)=\omega^2_a$

$\cfrac{1}{2}\omega_aT_oe^{-i\pi/2}(1-2\xi^2)=\omega^2_a$

$\xi^2=\cfrac{1}{2}(1-\cfrac{\omega_a}{\cfrac{1}{2}T_oe^{-i\pi/2}})$

$\xi^2=\cfrac{1}{2}(1+\cfrac{\omega_a}{\cfrac{1}{2}T_o})$

when,

$\omega_a=\cfrac{1}{2}T_o$,    $\xi=1$

And so, the electrons moved to the conduction band and jammed.

The advantage is  $w_a$ can be low but applied over a longer time to ensure that all electrons has moved to the conduction band.

When,

$\omega_a<\cfrac{1}{2}T_o$,

$\xi<1$

and the system will glow in resonance.  This suggest that when a piece of wire is heated with a current, the wire will stop glowing when the frequency of the applied voltage is increased beyond the point,  $\cfrac{1}{2}T_o$.  All power input is then just heat, no light.

When the waveform is rectified,  $\omega_a=4\omega_d$.

Since,

$\omega^2_o=-\cfrac{d^2T}{d\,t^2}$

$\omega^2_o(1-2\xi^2)=\omega^2_a=-\cfrac{d^2T}{d\,t^2}(1-2\xi^2)=\omega^2_a$

When ${\cfrac { d^2 T }{ d\, t^2 }}$ is always negative,

  $\xi^2=\cfrac{1}{2}(1-\cfrac{\omega_a}{\cfrac{1}{4}T_o})$

$\xi$  can be reduced to zero when

$\omega_a=\cfrac{1}{4}T_o$ for a rectified waveform.

When ${\cfrac { d^2 T }{ d\, t^2 }}$ is always positive for the inverted rectified waveform,

$\xi^2=\cfrac{1}{2}(1+\cfrac{\omega_a}{\cfrac{1}{4}T_o})$

it will serve the purpose of pushing electrons into the conduction band.

$\xi=1$,    when

$\omega_a=\cfrac{1}{4}T_o$ for a inverted rectified waveform.

The factor  $e^{i2\omega_d t}$  is not involved in the algebra because we are concern only with the extrema value,  a point on the waveform and how often this point occurs, ie. its frequency.

Unfortunately, all these are hypotheses subjected to human failings and dumb mathematical mistakes.

Complex Frequency And SuperConductivity

Consider this, from the post "Where Damping Is Light",

$\omega ^{ 2 }_{ a }=\cfrac { 2\xi ^{ 2 }\omega _{ o }^{ 2 } }{ T_{ o }-1 }$

$\cfrac { \omega ^{ 2 }_{ a } }{ \omega _{ o }^{ 2 } } =\cfrac { 2\xi ^{ 2 } }{ T_{ o }-1 }$

Since,  $\omega ^{ 2 }_{ o }=T_{ o }\omega ^{ 2 }_{ a }$,    $T_{ o }=\cfrac { \omega ^{ 2 }_{ o } }{ \omega ^{ 2 }_{ a } }$

$\cfrac { \omega ^{ 2 }_{ a } }{ \omega _{ o }^{ 2 } } =\cfrac { 2\xi ^{ 2 } }{ \cfrac { \omega ^{ 2 }_{ o } }{ \omega ^{ 2 }_{ a } } -1 }$

$1-\cfrac { \omega ^{ 2 }_{ a } }{ \omega _{ o }^{ 2 } } =2\xi ^{ 2 }$

$\xi = \cfrac { 1 }{\sqrt{ 2} }\sqrt { (1-\cfrac { \omega ^{ 2 }_{ a } }{ \omega _{ o }^{ 2 } } ) }$

If is it possible to adjust the damping ratio by changing the ratio of the frequency of the driving function and the system nature frequency  (ie.  incur loss by purposely not driving the system at resonance),

 $\xi =1$    when

$\omega ^{ 2 }_{ a } =-\omega _{ o }^{ 2 }$

when either  $\omega_o$  or  $\omega_a$  is complex.

When  $\xi =1$, the electron constantly get pushed up the kink into the conduction band without returning.  The material then become very conductive.

But is complex frequency possible?

Conduction Band, Valence Band Jamming Together

From the previously post "Where Damping Is Light",

$\xi ^{ 2 }=\cfrac { 1 }{ 2 } (1-\cfrac { 1 }{T_{ o }}  )$

and

$T_{ o } >1$

the system is always underdamped.

According to conventional description, it is consistent that,

the region above the kink is the conduction band.  The region below the kink is the valence band.  Normally,  room temperature is below the kink, most electrons are in the valence band.  Electrons get trapped in the conduction band because a quantum of energy is needed to fall back into the valence band (ref:  "Post Pag.Pag...Pag....Pag Dnab Ygrene").  Such electron contributes to the conductivity of the material.  The same quantum of energy also represents the energy band gap between the conduction and valence band.