Lingpoche, 凌波车

Consider this, every turn of time $t_g$ around the circular path, $t_g$ advances by $t_{\lambda\,c}$ along the axis of rotation.

$2\pi r=kt_{\lambda}$

$k$ represents the resistance along the time axis, for every $2\pi$ around the circular path, it advances by one unit along the axis.  If $k$ is measured in unit per radian, it is a constant.

In general, for a radial path of $r_v$,

$2\pi r_v=k_vt_{\lambda\,c}$

$k_v=2\pi\cfrac{ r_v}{t_{\lambda\,c}}$

$v_t=\cfrac{c}{k_v}=\cfrac{ct_{\lambda\,c}}{2\pi r_v}$

where

$\cfrac{\partial\,t_{g\,t}}{\partial\,t}=v_t$

But when $v_c$ is normal time speed.

$T_c=\cfrac{2\pi r_{c }}{c}=\cfrac{t_{\lambda\,c}}{v_c}$ --- (1)

$ct_{\lambda\,c}=2\pi r_{c }v_c$

Therefore,

$v_{ t }=\cfrac { r_{ c }v_{ c } }{ r_{ v } }$

When $r_v$ is varied in general, $f_v$, the frequency varies, $k_v$ meausred in unit per perimeter changes such that for every cycle around the circular path, time advances by $t_{\lambda\,c}$ still.  $k$ when measured in unit per radian (per $2\pi$) however, remains a constant.

Alternatively,

$T_v=\cfrac{2\pi r_{v }}{c}=\cfrac{t_{\lambda\,c}}{v_t}$ --- (2)

We formulate (1) divided by (2),

$\cfrac{T_c}{T_v}=\cfrac{r_{c }}{r_{v }}=\cfrac{v_t}{v_c}$

Since $r_c$ is small it seems that we can only go backwards in time with this.

$r_v\gt r_c$  

$v_t\lt v_c$

$T_c\lt T_v$

For every $T_v$, many $T_c$ would have passed, $v_c$ rushes forward and we travel back in time.

Maybe $v_t$ can be reduced further by counter rotating, in which case,

$T_v=\cfrac{2\pi r_{v }}{v}=\cfrac{t_{\lambda\,c}}{v_t}$

And,

$\cfrac{T_c}{T_v}=\cfrac{r_{c }}{r_{v }}\cfrac{v}{c}=\cfrac{v_t}{v_c}$

When,

${r_v}\gt{r_c}$

${r_v}c\gt\gt{r_c}v$

such that,

$v_t\lt\lt v_c$

$T_c\lt\lt T_v$

If it is possible for,

${r_v}\lt{r_c}$

then,

$v_t\gt v_c$

Normal time recedes, as we travel forward in time.

Time travels in a helix at constant speed.  Time advances after every cycle. $k$ measured in unit per radian is a constant. When the radius of the time circle is increased, at constant speed around the circular path, time advances is slowed.  Normal time is in constant flux, as such normal time goes forward as we travel back in time.  If we rotate the sonic cone anti-clockwise, $v_t$ slows down further.

What exactly is "lingpo"?  A sonic cone at 7.489 Hz.  Two of such cones to form an enclosure.

And off you go, in time.

Time Speed Again, Two Pie Or Not Two Pie

Consider again,

Let,

$r.\Delta \theta = k.\Delta t_c$

$\omega _{ c }=\cfrac { \partial \, \theta  }{ \partial \, t_{ c } } =\cfrac{k}{r}$

$\omega =\omega _{ c }\cfrac { \partial \, t_{ c } }{ \partial \, t } =\cfrac { \partial \, \theta  }{ \partial \, t_{ c } } \cfrac { \partial \, t_{ c } }{ \partial \, t } =\cfrac{k}{r} \cfrac { \partial \, t_{ c } }{ \partial \, t }$

$\left( \omega -\cfrac { k }{ r }  \right) \cfrac { \partial \, t_{ c } }{ \partial \, t } =0$

As, $r\rightarrow\infty$,   $k\ne0$

$\omega\cfrac { \partial \, t_{ c } }{ \partial \, t } =0$

since,

$\omega\ne0$

$\cfrac { \partial \, t_{ c } }{ \partial \, t } =0$

This implies that as time stretches out, time speed is zero.

when $r\ne\infty$,

$2\pi r=kt_{\lambda}$

$2\pi rf=kt_{\lambda}f=k\cfrac { \partial \, t_{ c } }{ \partial \, t }$

$\cfrac { \partial \, t_{ c } }{ \partial \, t }=\cfrac { 2\pi rf }{ k }=\cfrac { r\omega }{ k }=\cfrac{c}{k}$

What about,

$t_{ c }=2\pi r$

$\cfrac{t_c}{t}= \cfrac { 2\pi r }{ t } =c$

$\cfrac { \partial \, t_{ c } }{ \partial \, t } =2\pi \cfrac { \partial \, r }{ \partial \, t }=c$

This is wrong because $r$ for a given time speed is fixed.  Time $t_c$, will always coil around $x$ at a fixed $r$, gvien $\cfrac{\partial\,t_c}{\partial\,t}$,

$\cfrac{\partial\,r}{\partial\,t}=0$

$r$ does not increase and the time coil does not expand as time progresses.

Time is in a helix with a phase associated with its position on its circular path along its axis of travel.  If time returns to its original phase at a distance $t_{\lambda}$ ahead, after time period $T$,

$2\pi r=kt_{\lambda}$

$\cfrac { 2\pi r }{c}=T=\cfrac{k}{c}t_{\lambda}$

$t_{\lambda}=\cfrac{c}{k}T$

so, time linear speed $v_t$ is,

$v_t=\cfrac{c}{k}$

Time speed $v_t$, is not light speed, $c$.

Strained And Witchcraft

Consider the relation between time and space,

$\cfrac{2\pi.t_c}{t}=c$

$2\pi.t_c=ct$

$2\pi\cfrac{\partial\,t_c}{\partial\,t}=c$

$\cfrac{\partial\,t_c}{\partial\,t}=\cfrac{c}{2\pi}$

In a similar way and considering the direction of these dimensions,

$\cfrac{\partial\,t_T}{\partial\,t}=-i\cfrac{c}{2\pi}$

$\cfrac{\partial\,t_g}{\partial\,t}=i\cfrac{c}{2\pi}$

And

$\cfrac { \partial \, x }{ \partial \, t } =\cfrac { \partial \, x }{ \partial \, t_{ c } } \cfrac { \partial \, t_{ c } }{ \partial \, t } +\cfrac { \partial \, x }{ \partial \, t_{ g } } \cfrac { \partial \, t_{ g } }{ \partial \, t } +\cfrac { \partial \, x }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t }$ --- (*)

$\cfrac { \partial \, x }{ \partial \, t } =\cfrac{c}{2\pi}\left\{\cfrac { \partial \, x }{ \partial \, t_{ c } }+i\cfrac { \partial \, x }{ \partial \, t_{ g } }-i\cfrac { \partial \, x }{ \partial \, t_{ T } }\right\}$

If all time dimensions are equivalent,

$\cfrac { \partial \, x }{ \partial \, t_{ c } }=\cfrac { \partial \, x }{ \partial \, t_{ g } } =\cfrac { \partial \, x }{ \partial \, t_{ T } }$

$\cfrac { \partial \, x }{ \partial \, t } =\cfrac{c}{2\pi}\cfrac { \partial \, x }{ \partial \, t_{ c } }$

We also have

$\left(\cfrac { \partial \, x }{ \partial \, t } \right)^2=\left(\cfrac { \partial \, x }{ \partial \, t_c } \right)^2+\left(\cfrac { \partial \, x }{ \partial \, t_g } \right)^2+\left(\cfrac { \partial \, x }{ \partial \, t_T } \right)^2$

 $\cfrac { \partial \, x }{ \partial \, t } =\sqrt{3}\cfrac { \partial \, x }{ \partial \, t_{ c } }$

This is possible only if,

 $\cfrac { \partial \, x }{ \partial \, t } =\cfrac { \partial \, x }{ \partial \, t_c }= 0$

At this point we can swap $x$ for $t_c$, since they are both dummy variables and conclude by symmetry that when $\cfrac{\partial\,x}{\partial\,t}=\cfrac{c}{2\pi}$,  $\cfrac { \partial \,t_c}{ \partial \, t } =0$

However, for the fun of it,

When $x$ wraps around $t_c$,

$\cfrac{2\pi.x}{t}=c$

$2\pi.x=ct$

$2\pi\cfrac { \partial \, x }{ \partial \, t } =c$

and,

 $2\pi\cfrac { \partial \, x }{ \partial \, t_c } =c\cfrac{\partial\,t}{\partial\,t_c}$

So,

$\cfrac { \partial \, x }{ \partial \, t } =\cfrac { \partial \, x }{ \partial \, t_c } \cfrac{\partial\,t_c}{\partial\,t}$ --- (***)

Since,

$\cfrac { \partial \, x }{ \partial \, t_{ c } } =\cfrac { \partial \, x }{ \partial \, t_{ g } } \cfrac { \partial \, t_{ g } }{ \partial \, t_{ c } } +\cfrac { \partial \, x }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t_{ c } }$

As changes in $x$ and $t_c$ does not effect $t_g$ and $t_T$, and as far as $x$ is concern $t_g$ and $t_T$ are equivalent, !?!

$\cfrac { \partial \, x }{ \partial \, t_{ c } } =\cfrac{c}{2\pi}\cfrac{\partial\,t}{\partial\,t_c}=2\cfrac { \partial \, x }{ \partial \, t_{ g } } \cfrac { \partial \, t_{ g } }{ \partial \, t_{ c } } =2\cfrac { \partial \, x }{ \partial \, t_{T } } \cfrac { \partial \, t_{ T } }{ \partial \, t_{ c } }$

ie.

$\cfrac { \partial \, x }{ \partial \, t_{ c } }\cfrac{\partial\,t_c}{\partial\,t} =\cfrac{c}{2\pi}$ --- (**)

and

$\cfrac { \partial \, x }{ \partial \, t_{ c } }  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } }=2\cfrac { \partial \, x }{ \partial \, t_{ g } }$

From which we formulate,

$\cfrac { \partial \, x }{ \partial \, t_{ c } }  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t } =2\cfrac { \partial \, x }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }$ --- (1)

Similarly,

$\cfrac { \partial \, x }{ \partial \, t_{ c } }  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } }=2\cfrac { \partial \, x }{ \partial \, t_{ T } }$

$\cfrac { \partial \, x }{ \partial \, t_{ c } }  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } }\cfrac { \partial \, t_{ T } }{ \partial \, t } =2\cfrac { \partial \, x }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t }$ --- (2)

Therefore (1)+(2),

$\cfrac { \partial \, x }{ \partial \, t_{ c } } \left\{ \cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } }\cfrac { \partial \, t_{ T } }{ \partial \, t }\right\}=2\left\{\cfrac { \partial \, x }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+\cfrac { \partial \, x }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t }\right\}$

Substitute (**) into the above,

 $\cfrac {c }{ 2\pi } \left\{ \cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } }\cfrac { \partial \, t_{ T } }{ \partial \, t }\right\}=2\cfrac{\partial\,t_c}{\partial\,t}\left\{\cfrac { \partial \, x }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+\cfrac { \partial \, x }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t }\right\}$

But,

 $\cfrac{\partial\,x}{\partial\,t}=\cfrac { \partial \, x }{ \partial \, t_{ c } }\cfrac { \partial \, t_{ c } }{ \partial \, t }+\cfrac { \partial \, x }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+\cfrac { \partial \, x }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t }$

So,

  $\cfrac {c }{ 2\pi } \left\{ \cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } }\cfrac { \partial \, t_{ T } }{ \partial \, t }\right\}=2\cfrac{\partial\,t_c}{\partial\,t}\left\{\cfrac{\partial\,x}{\partial\,t}-\cfrac { \partial \, x }{ \partial \, t_{ c } }\cfrac { \partial \, t_{ c } }{ \partial \, t }\right\}$

But from (***),

$\cfrac { \partial \, x }{ \partial \, t } =\cfrac { \partial \, x }{ \partial \, t_c } \cfrac{\partial\,t_c}{\partial\,t}$

So,

$\cfrac {c }{ 2\pi } \left\{ \cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } }\cfrac { \partial \, t_{ g } }{ \partial \, t }+  \cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } }\cfrac { \partial \, t_{ T } }{ \partial \, t }\right\}=0$

But,

$\cfrac { \partial \, t_{ c } }{ \partial \, t } =\cfrac { \partial \, t_{ c } }{ \partial \, t_{ g } } \cfrac { \partial \, t_{ g } }{ \partial \, t } +\cfrac { \partial \, t_{ c } }{ \partial \, t_{ T } } \cfrac { \partial \, t_{ T } }{ \partial \, t }$

Thus,

$\cfrac {c }{ 2\pi } \cfrac { \partial \, t_{ c } }{ \partial \, t } =0$

$\cfrac { \partial \, t_{ c } }{ \partial \, t } =0$

Blasphemy, I know.  The point is,

when $\cfrac { \partial \, x }{ \partial \, t } =\cfrac{c}{2\pi}$, $\cfrac { \partial \, t_{ c } }{ \partial \, t } =0$ and when $\cfrac { \partial \, t_c }{ \partial \, t } =\cfrac{c}{2\pi}$, $\cfrac { \partial \, x }{ \partial \, t } =0$ are consistent.

The factor of $\cfrac{1}{2\pi}$ appearing before $c$ is the result of circular motion around the orthogonal axis.  A particle traveling in circular motion in space about the time axis is not traveling in time, so time speed equals zero.  Similarly, the same particle traveling in circular motion in time about the space axis has zero space speed.

This relationship can be rewritten as,

$v^2+v^2_t=\left(\cfrac{c}{2\pi}\right)^2$ --- (+)

where $v$ is the particle velocity in space, $v_t$ is the particle velocity in time and $c$ a constant.  Pythagoras' Theorem applies because $v$ is perpendicular to $v_t$.  Obviously expression (+) satisfies the boundary value conditions.

More importantly, the time speed limit occurs at $\cfrac{c}{2\pi}$ not at $c$, because the particle goes into circular motion.

More Portals

If two portals point directly at each other, then their forces should cancel BOTH outside and between the portal pair.

But Maxwell third and fourth equations suggest that we can just have the forces cancel outside the two face of the portals by letting one portal be passive.  The portal that we are traveling from is the active portal.

The active portal induces a corresponding force pair ($\oint{E_{\psi}}\,d\,\lambda$) in the passive portal that in turn generate a force that cancels with the force from the active portal, beyond the passive portal as expected.

Furthermore,

this is also a possible portal configuration.  Two of such double pillars facing each other, with high $g_T$ between one of the double pillars can generate a conduit between the portals.

Sound Portal

My very own mandala.

How far did the Tibetan go with sound technology at 7.489 Hz?  Apparently, very far.  But can manipulating gravitational wave open up a portal?

And to where?  From the post "Where's The Charge?",  if

$f=n\cfrac{c}{2\pi a_\psi}$,    $n=1$

So $f$ confine to a smaller $a_\psi$ will be trapped at a lower $h_n$.

$hf=2\pi a_{\psi}mcf=2\pi a_{\psi}mc7.489=mc^2+x.F$

$x.F\lt0$  (because $a_{\psi}$ that produced 7.489 Hz is the size of Earth.)

where $a_{\psi}$ is the radius of the confine into which sound waves of 7.489 Hz are directed as the diagram above shows.

Energy is being released.  Photons will be emitted.  $E_{\Delta h}\lt 0$  and $\cfrac{\partial\,E_{\Delta h}}{\partial\, t}\lt0$, from previously (Post "Cold Jump"),

$a_{ \psi \, n }=-\cfrac { c^{ 2 }E_{ \Delta h } }{ 2\pi  } \left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }$

 the direction of travel is opposite to the right hand screw rule.  We point the mandala, front face, in the direction of the corresponding portal, face front and jump into the portal and travel backwards.

Time is always clockwise in the direction of $x$, even if time is slowed.

Note:  Not the mandala itself but the construct indicated by it.

Spin Cool

Maybe $\psi$ can be depressed by spinning the two particles.

Centrifugal forces on the two particles will space them further apart and cause a depression in $\psi$ in the space between them.

Jump in.

But what is this particle pair?  These two particles are in orbit around each other, presenting a positive force density outwards.

Cold Jump

From the post "Time Travel Made Easy",

$\Delta_t=\cfrac{\partial\,t_{gf}}{\partial\,t}-\cfrac{\partial\,t_{gi}}{\partial\,t}=\cfrac{1}{mc^2}(\cfrac{\partial\,t_{gi}}{\partial\,t}E_{\Delta h}+t_{gi}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})$

From the post "Teleportation",

$t_{ gi }=\cfrac { 2\pi a_{ \psi \, n } }{ c }$

and

$\cfrac{\partial\,t_{gi}}{\partial\,t}=c$

We have,

$\cfrac { \partial \, t_{ gf } }{ \partial \, t } -c=\cfrac { 1 }{ mc^{ 2 } } (cE_{ \Delta h }+\cfrac { 2\pi a_{ \psi \, n } }{ c } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } )$

$\cfrac { 2\pi a_{ \psi \, n } }{ c } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } =mc^{ 2 }\left( \cfrac { \partial \, t_{ gf } }{ \partial \, t } -c \right) -cE_{ \Delta h }$

$a_{ \psi \, n }=\cfrac { mc^{ 3 } }{ 2\pi  } \left( \cfrac { \partial \, t_{ gf } }{ \partial \, t } -c \right) \left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }-\cfrac { c^{ 2 }E_{ \Delta h } }{ 2\pi  } \left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }$ --- (*)

The plot below shows the $\psi$ profile of two particles in pair, they are attracted to each other for $x\lt a_{\psi}$.  If $a_{\psi}$ is increased, $\psi$ in the space between the two particles is close to zero and negative.

When $\psi$ is depressed,

The plateau value of $\psi_{+n}+\psi_{-n}$ decreases further into the negative region.

From equation (*), we see that even without a time speed differential,

$\cfrac { \partial \, t_{ gf } }{ \partial \, t } -c=0$

$a_{ \psi \, n }=-\cfrac { c^{ 2 }E_{ \Delta h } }{ 2\pi  } \left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }$

And since $\psi\lt0$, any transitions from normal energy state, without the need to use high energy photons to first elevate the energy state, is negative,

 $E_{ \Delta h }\lt 0$,

but,

$\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }\lt0$

so the direction of travel is opposite to the right hand screw rule,

where $t_g$ around the spiral is slowed, ie $\cfrac{\partial\,t_{gf}}{\partial\,t}\lt0$.  Such a scheme avoids the use of high energy photons to first slowly increase the particle's energy state and then allow the energy state to fall in a cascade.

Cool, very cool.  How then to depress $\psi$?

All Of Max

We can of course superimpose the results from "Two Wrongs Make Both Wrong, None Right", and obtain,

$\cfrac { d\,  }{ d\, t } \oint { E_{\psi}  } \, d\, A_{ l }+\cfrac { 1 }{\varepsilon _{ new }}\oint{ J_{\psi}}\,d\,A_l=\oint _{ 2\pi a_{ \psi  } }{ F_{ \psi  } } \, d\, \lambda$

Replacing $\varepsilon _{ new }$ with $\varepsilon _{ o}$ and multiplying by $\mu_{o}\varepsilon_{o}$,

$\mu_{o}\varepsilon_{o}\cfrac { d\,  }{ d\, t } \oint { E_{\psi}  } \, d\, A_{ l }+\mu_{o}\oint{ J_{\psi}}\,d\,A_l=\mu_{o}\varepsilon_{o}\oint _{ 2\pi a_{ \psi  } }{ F_{ \psi  } } \, d\, \lambda$

$\mu_{o}\varepsilon_{o}\cfrac { d\,  }{ d\, t } \oint { E_{\psi}  } \, d\, A_{ l }+\mu_{o}\oint{ J_{\psi}}\,d\,A_l=\cfrac{1}{c^2}\oint _{ 2\pi a_{ \psi  } }{ F_{ \psi  } } \, d\, \lambda$

which is the fourth Maxwell's Equations for a particle.

Since $a_{\psi}$ is the radius of a circle, when we rotate $A_l$ by $2\pi$ along any axis of symmetry through $A_l$, we see that $E_{\psi}$ cancels with $-E_{\psi}$ when $A_l$ has been rotated by $\pi$. Similarly, $J_{\psi}$ cancels with $-J_{\psi}$ at the rotation of $\pi$.  When so rotated, the integral $\oint_{2\pi a_\psi}{F_{\psi}}\,d\,\lambda$ describes a closed surface area containing the volume defined by $A_l$ after $2\pi$ rotation.  As such,
$\require{cancel}$
$\cancelto{0}{\mu_{o}\varepsilon_{o}\cfrac { d\,  }{ d\, t } \oint { E_{\psi}  } \, d\,V}+\cancelto{0}{\mu_{o}\oint{ J_{\psi}}\,d\,V}=\cfrac{1}{c^2}\oint _{ 4\pi a^2_{ \psi  } }{ F_{ \psi  } } \, d\, A_s=0$

This is the second Maxwell's equation, Gauss's law for magnetism.

Consider the fourth Maxwell equation again, if we set $J_\psi=0$ and integrate along $a_{\psi}$, the radius of the circle defined by the perimeter $A_l$,
$\require{cancel}$
$\mu _{ o }\varepsilon _{ o }\int { \cfrac { d\,  }{ d\, t } \oint { E_{ \psi  } } \, d\, A_{ l } } dx+\mu _{ o }\int { \oint {\cancelto{0}{ J_{ \psi  }} } \, d\, A_{ l } } dx=\cfrac { 1 }{ c^{ 2 } } \int { \oint _{ 2\pi a_{ \psi  } }{ F_{ \psi  } } \, d\, \lambda  } dx$

The loop $2\pi a_\psi$ becomes an area $A_l=\pi a^2_{\psi}$, therefore,

$\mu _{ o }\varepsilon _{ o }\int { \cfrac { d\,  }{ d\, t } \oint { E_{ \psi  } } \, d\, A_{ l } } dx=\cfrac { 1 }{ c^{ 2 } }  { \oint _{ \pi a^2_{ \psi  } }{ F_{ \psi  } } \, d\, A_l  }$

It is totally arbitrary that we move at speed $c$, and since $x$ is perpendicular to $\lambda$ and $A_l$.

Using $A_l$ are the reference direction

$\cfrac { dx }{ dt } =ic$

$\mu _{ o }\varepsilon _{ o }ic\int { \cfrac { d\,  }{ d\, t } \oint { E_{ \psi  } } \, d\, A_{ l } } dt=\cfrac { 1 }{ c^{ 2 } } \oint _{\pi a^2_{\psi}}{ F_{ \psi  } } dA_{ l }$

$c=\cfrac { 1 }{ \sqrt { \mu _{ o }\varepsilon _{ o } }  }$

$ic\oint { E_{ \psi  } } \, d\, A_{ l }= \oint_{\pi a^2_{\psi}} { F_{ \psi  } } dA_{ l }$

In a similar way, we consider $A_l$ as the integral of a loop $2\pi a_{\psi}$ outwards along $dx$.  This loop is obtained by integrating along $d\lambda$ for $2\pi a_{\psi}$.

$ic\int\oint _{2\pi a_\psi}{ E_{ \psi  } } \, d\, \lambda dx= \oint _{\pi a^2_{\psi}} { F_{ \psi  } } dA_{ l }$

Similarly, $\cfrac { dx }{ dt } =ic$

$(ic)^2\int\oint_{2\pi a_\psi} { E_{ \psi  } } \, d\, \lambda d\,t= \oint _{\pi a^2_{\psi}} { F_{ \psi  } } dA_{ l }$

$\oint_{2\pi a_\psi} { E_{ \psi  } } \, d\, \lambda =-\cfrac{1}{c^2} \cfrac{d}{d\,t}\left\{\oint  _{\pi a^2_{\psi}}{ F_{ \psi  } } dA_{ l }\right\}$

This is the third Maxwell's Equation.  Consider again the fourth Maxwell equation, when we set $F_\psi=0$

$\mu _{ o }\varepsilon _{ o }\cfrac { d\,  }{ d\, t } \oint_{\pi a^2_\psi} { E_{ \psi  }}  \, d\, A_{ l }+\mu _{ o }\oint _{\pi a^2_\psi}{ J_{ \psi  } } \, d\, A_{ l }=\cfrac { 1 }{ c^{ 2 } } \oint _{ 2\pi a_{ \psi  } }{ \cancelto{0}{F_{ \psi  } }} \, d\, \lambda$

If we consider $A_l$ to be from the integration of $2\pi a_{\psi}$ along the radius $x$, using $A_l$ as reference,

$i\lambda=A_l$  and $\cfrac{dx}{d\,t}=ic$

$\varepsilon _{ o }\cfrac { d\,  }{ d\, t }\int{ \oint _{ 2\pi a_{ \psi  } } { E_{ \psi  }}}  \, d\, i\lambda d\,x+\oint _{\pi a^2_\psi}{ J_{ \psi  } } \, d\, A_{ l }=0$

$\varepsilon _{ o }(i)^2c\cfrac { d\,  }{ d\, t }\int{ \oint  _{ 2\pi a_{ \psi  } }{ E_{ \psi  }}}  \, d\, \lambda d\,t+\oint _{\pi a^2_\psi}{ J_{ \psi  } } \, d\, A_{ l }=0$

$\varepsilon _{ o }c\oint _{ 2\pi a_{ \psi  } } { E_{ \psi  }}  \, d\, \lambda=\oint _{\pi a^2_\psi}{ J_{ \psi  } } \, d\, A_{ l }$

$J_{ \psi  }=\cfrac { 1 }{ 2\pi a^{ 2 }_{ \psi  } } \cfrac { d\, q_{ \psi  } }{ d\, t }$

Substitute in $J_{\psi}$ and integrate over time, t.

$\varepsilon _{ o }c\int { \oint _{ 2\pi a_{ \psi  } } { E_{ \psi  } }  } \, d\, \lambda d\, t=\oint _{ \pi a^{ 2 }_{ \psi  } }{ \int { \cfrac { 1 }{ 2\pi a^{ 2 }_{ \psi  } } \cfrac { d\, q_{ \psi  } }{ d\, t }  }  } d\, t\, d\, A_{ l }$

$E_\psi$ is constant,

$\varepsilon _{ o }ct\oint  _{ 2\pi a_{ \psi  } }{ E_{ \psi  } } d\, \lambda =\oint _{ \pi a^{ 2 }_{ \psi  } }{ \cfrac { q }{ 2\pi a^{ 2 }_{ \psi  } }  } \, d\, A_{ l }$

$\varepsilon _{ o }\oint  _{ 2\pi a_{ \psi  } }{ E_{ \psi  } } d\, \lambda =\oint _{ \pi a^{ 2 }_{ \psi  } }{ \cfrac { q }{ 2\pi a^{ 2 }_{ \psi  }ct }  } \, d\, A_{ l }$

where $vol=2\pi a^{ 2 }_{ \psi  }ct$ is the volume transcribed by $J$ with a base of $2\pi a^{ 2 }_{ \psi  }$  over time $t$.

With $t$ small,

$\rho=\cfrac { q_\psi }{ 2\pi a^{ 2 }_{ \psi  }ct }$

where $\rho$ is the volume charge density.  So,

$\varepsilon _{ o }\oint_{2\pi a_{\psi}} { E_{ \psi  } } d\, \lambda =\oint _{ \pi a^{ 2 }_{ \psi  } }{ \rho  } \, d\, A_{ l }$

We then rotate $2\pi a_{\psi}$ into a surface containing the volume define by $A_l$ as $A_l$ rotates correspondingly,

$\oint_{4\pi a^2_{\psi}} { E_{ \psi  } } d\, A_l=\cfrac{1}{\varepsilon _{ o }}\oint { \rho  } \, d\,V$

This is the first Maxwell's Equation, Gauss's Law.

Thanks to the fact that $a_{\psi}$ describes circles and spheres, we have all four Maxwell's Equations for particles.  $E_\psi$ is the field around the particle, $F_\psi$ is a field analogous to the $B$ field of charges, produce by the particle in motion.

Two Wrongs Make Both Wrong, None Right

From the post "Maxwell, Planck And Particles",

 $\cfrac { 1 }{ \varepsilon _{ o }}\oint{ \cfrac { 1 }{ 4\pi a^2_{\psi} }}\cfrac{d\,q_{\psi}}{d\,t}\,d\,A_l= \oint_{2\pi a_{\psi}}{\cfrac{d\,(2 \pi mc)}{d\,t}}\,d\,\lambda= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

If, we formulate instead,

$\oint { \cfrac { d\,  }{ d\, t } \left\{ \cfrac { q_{ \psi } }{ 4\pi \varepsilon _{ o }a^{ 2 }_{ \psi  } }  \right\}  } \, d\, A_{ l }=\oint _{ 2\pi a_{ \psi  } }{ \cfrac { d\, (2\pi mc) }{ d\, t }  } \, d\, \lambda \, =\oint _{ 2\pi a_{ \psi  } }{ F_{ \psi  } } \, d\, \lambda$

$E_{\psi}=\cfrac { q_{ \psi} }{ 4\pi \varepsilon _{ o }a^{ 2 }_{ \psi  } }$

$\cfrac { d\,  }{ d\, t } \oint { E_{\psi}  } \, d\, A_{ l }=\oint _{ 2\pi a_{ \psi  } }{ F_{ \psi  } } \, d\, \lambda$

This expression is without any appendages because if the $2\pi$ factor missing from the circular momentum term is the only reason

$\varepsilon _{ o }\ne\mu_o$,

then,

$\varepsilon _{ o }=\mu_o=\cfrac{1}{c}$  

after we made the correction, which would set,

$Z_o=1=\sqrt{\cfrac{\mu_o}{\varepsilon_o}}$

that space present equal resistance to $B$ and $E$ field.  With such a change to $\varepsilon_o$, the numerical value of $q$ changes also.

Let, consider again,

 $\cfrac { 1 }{ \varepsilon _{ o }}\oint{ \cfrac { 1 }{ 4\pi a^2_{\psi} }}\cfrac{d\,q_{\psi}}{d\,t}\,d\,A_l= \oint_{2\pi a_{\psi}}{\cfrac{d\,(2 \pi mc)}{d\,t}}\,d\,\lambda$

$\cfrac { 1 }{ 4\pi\varepsilon _{ o }}\oint{ \cfrac { 1 }{ 2\pi a^2_{\psi} }}\cfrac{d\,q_{\psi}}{d\,t}\,d\,A_l= \oint_{a_{\psi}}{\cfrac{d\,( mc)}{d\,t}}\,d\,\lambda= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

without the correction to circular momentum.

Since,

$J_{\psi}=\cfrac { 1 }{ 2\pi a^2_{\psi} }\cfrac{d\,q_{\psi}}{d\,t}$

 $\cfrac { 1 }{ 4\pi\varepsilon _{ o }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

And then we make the adjustment to $\varepsilon _{ o }$, $4\pi\varepsilon _{ o }\rightarrow\varepsilon _{ new }$  to correct for the mssing $2\pi$ factor for circular momentum.

 $\cfrac { 1 }{\varepsilon _{ new }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

 Since, $\mu_{new}=\varepsilon_{new}=\cfrac{1}{c}$

 $\cfrac { 1 }{\mu_{ new }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

Why not simply make the adjustment $mc\rightarrow2\pi mc$, and obtain,

$\cfrac { 1 }{ 2\varepsilon _{ o }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

as in the post "Maxwell And Particles"???

This expression is wrong because it is still using the old definitions of $\mu_o$ and $\varepsilon_o$.

Maxwell, Planck And Particles

From the post "Just Lots of Colors, Retro Disco" Jun2014, where the photon is modeled as a dipole, and the posts "de Broglie Per Unit Volume" and "de Broglie Per Person",

$h=\cfrac { 1 }{ 8\pi \varepsilon _{ o } } q=2\pi a_{ \psi \,  }mc$

where $q$ is the dipole charge.

$h=\cfrac { 1 }{ 2 \pi\varepsilon _{ o }} \cfrac { q }{ 4\pi a^2_{\psi} }\pi a^2_{\psi}= \oint_{2\pi a_{\psi}}{mc}\,d\,\lambda$

where the closed looped integral is around $2\pi a_{\psi}$.

$h=\cfrac { 1 }{ 2 \pi\varepsilon _{ o }}\oint{ \cfrac { q }{ 4\pi a^2_{\psi} }}\,d\,A_l= \oint_{2\pi a_{\psi}}{mc}\,d\,\lambda$

where the area integral is over the area, $A_{a_{\psi}}=\pi a^2_{\psi}$ surrounded by $2\pi a_{\psi}$.

Differentiating with respect to time,

$\cfrac{d\,h}{d\,t}=\cfrac { 1 }{ 2 \pi\varepsilon _{ o }}\oint{ \cfrac { 1 }{ 4\pi a^2_{\psi} }}\cfrac{d\,q}{d\,t}\,d\,A_l= \oint_{2\pi a_{\psi}}{\cfrac{d\,mc}{d\,t}}\,d\,\lambda$ --- (*)

Compare this with Ampere's Circuital Law, (The fourth Maxwell's Equation with time varying component set to zero),

$\mu_o\oint{J}\,d\,A_l=\oint_{2\pi l}{B} \,dl$

In the first place, the factor of $\cfrac{1}{2\pi}$ in expression (*) suggests that for a body in circular motion at speed $c$, its momentum is not,

$p=mc$   but   $p=2\pi mc$

Consider the centripetal force, $F$ that accounts for circular motion,

$p=\int { F } dt=\int { \cfrac { mv^{ 2 } }{ r }  } dt$

$\because \cfrac{d\,p}{d\,t}=F$

Since,

$\theta =\omega t$   and $v=rw$

$p=\int { \cfrac { m\left( r\omega  \right) ^{ 2 } }{ r }  } \cfrac { 1 }{ \omega  } d\theta =\int { mr\omega  } d\theta$

Over one period $\theta=0\rightarrow\theta=2\pi$,

$p=2\pi mv$

And we got it all wrong.  This is important!  We have instead,

$\cfrac{d\,h}{d\,t}=\cfrac { 1 }{ \varepsilon _{ o }}\oint{ \cfrac { 1 }{ 4\pi a^2_{\psi} }}\cfrac{d\,q}{d\,t}\,d\,A_l= \oint_{2\pi a_{\psi}}{\cfrac{d\,(2 \pi mc)}{d\,t}}\,d\,\lambda$

Secondly, for a photon $m=0$, but a change in its $\psi$ is momentum (from the posts "Bouncy Balls, Sticky Balls, Transfer Of Momentum" and "de Broglie Per Unit Volume"), so

$\cfrac{d\,(2\pi mc)}{d\,t}=F_\psi$

is a force that acts on $\psi$.  Depending on the nature of the photon it may be $\psi$ along $t_T$,$t_g$ or $t_c$ and one other space dimension, for which the photon is defined.  $B$ is perpendicular to $J$ in Maxwell's equation.  The direction of

$\cfrac{d\,q}{d\,t}$ is also perpendicular to $F_\psi$

The nature of $F_\psi$ depends on the nature of $q$.  $q$ is thus qualified as, $q_{t_T}$, $q_{tg}$ and $q_{tc}$ depending on the dimensions between which the photon is oscillating.  In all cases, these include a space dimension, the other being a time dimension, $t_T$,$t_g$ or $t_c$.  So a photon in motion exert a force around it, this force, $F_\psi$ is a force in space (ie.  $F=ma$) and also along one of the time axes $t_T$,$t_g$ or $t_c$.  Previously, in Maxwell's Equation, the effect of the force in the time dimension in which the charge is oscillating has been ignored.  So expression (*) in the case of a photon, ph($t_c$,$t_g$) is,

 $\cfrac { 1 }{ 2\varepsilon _{ o }}\oint{ \cfrac { 1 }{ 2\pi a^2_{\psi} }}\cfrac{d\,q_{t_g}}{d\,t}\,d\,A_l= \oint_{2\pi a_{\psi}}{\cfrac{d\,(2 \pi mc)}{d\,t}}\,d\,\lambda= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

$F_{\psi}$ then acts on $t_g$ and a space dimension.

This equation for particles in motion through an area $A_l$ exert a force field, effecting similar particles, around a loop $2\pi a_{\psi}$ is just a restatement of Ampere's Law but extended to all particles.  It may be possible that all of Maxwell's equations can be similarly extended.

Let,

$J_{\psi}=\cfrac { 1 }{ 2\pi a^2_{\psi} }\cfrac{d\,q_{t_T}}{d\,t}$

 $\cfrac { 1 }{ 2\varepsilon _{ o }}\oint{ J_{\psi}}\,d\,A_l= \oint_{2\pi a_{\psi}}{F_\psi}\,d\,\lambda$

This however is using the old definition of $\mu_o$ and $\varepsilon_o$.

More importantly, the correct momentum in circular motion,

$p=2\pi mc=m2\pi c$

makes $\mu_o$ and $\varepsilon_o$ equivalent but orthogonal.

Consider,

$\varepsilon_{new}=i\mu_{new}$

$\sigma ^2=(\cfrac{1}{\sqrt{2}}\varepsilon_{new})^2+(\cfrac{1}{\sqrt{2}}\mu_{new})^2$ --- (**)

$\sigma ^2=\varepsilon^2_{new}=\varepsilon_{new}\mu_{new}$

When $c$ is not in circular motion,

$c=\cfrac{1}{\sqrt{\mu_{new}\varepsilon_{new}}}=\cfrac{1}{\sigma}$

as such $\sigma=\varepsilon_{new}=\cfrac{1}{c}$ is the resistance to $c$.

When $c$ is in circular motion, where $\sigma$ has been resolved into two components as in (**).  Along one of this component,

$\varepsilon_o=\cfrac{1}{\sqrt{2}}\varepsilon_{new}$

$\mu_o$ wrongfully incoperated a factor of $\cfrac{1}{2\pi}$ when momentum was taken to be $mv$ and not $2\pi mv$.  So,

$2\pi \mu_o=\cfrac{1}{\sqrt{2}}\mu_{new}$

And so,

$\mu_o\varepsilon_o=\cfrac{1}{4\pi}\mu_{new}\varepsilon_{new}$

This means in the expression for $c$,

$c=\cfrac{1}{\sqrt{\mu_{new}\varepsilon_{new}}}=\cfrac{1}{\sqrt{4\pi\mu_o\varepsilon_o}}$

$\mu_o\varepsilon_o$ has to be adjusted by a factor of $4\pi$ because the $2\pi$ factor needed when momentum is circular was not accounted for. ($2\varepsilon_o\rightarrow\varepsilon_o$ and $2\pi\mu_o\rightarrow\mu_o$)

$\mu_o$ is defined as $4\pi \times10^{-7}$ NA^-2.  This is wrong for failing to account for the fact that in circular motion, the momentum is $2\pi mv$ and not $mv$.

It is correct and consistent to set

$\mu_o=\varepsilon_o=\cfrac{1}{c}$

at the same time, all momentum $p$ in circular motion is replaced with $2\pi p$, ie.

$p_{cir}=2\pi p$

where $p_{cir}$ is the momentum in circular motion. Equivalently, all velocity $v$ in circular motion is replaced with $2\pi v$.

$v_{cir}=2\pi v$

Equivalently,

$p_{cir}=2\pi p=2\pi mv=m2\pi v=mv_{cir}$

Correcta, Errata...

Note: This derivation for Ampere's Law applicable to all particles is not rigorous, but indicative none the less.

Squeezing Photons

From the post "Just Lots of Colors, Retro Disco" on Jun 2014, photon modeled as dipole that can resonate at

$f_{reso}=15.088 kHz$

From the post "Increase Photon Frequency" (Sep 2014), we see that photons can be squeeze to high frequency with the use of a tapered coil.  The coil will be more effective operating at a frequency of 15.088 kHz.

This allows photons of higher $a_\psi$ to be squeezed to lower $a_\psi$.

This might allow photons needed to energize for teleportation be produced without reaching for ultra ultra high frequencies (10⁴²Hz).

Squeeze and telepop.

More ScFi, Resonator And Wrap Speed

Consider the expression for $E_{ \Delta h }$,

$E_{ \Delta h }=\int { F_{ \Delta hi } } da_{ \psi \, i }=mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } (a_{ \psi \, f }-a_{ \psi \, i })$

Differentiating with respect to $a_{ \psi \, i }$,

$\cfrac { \partial \, E_{ \Delta h } }{ \partial \, a_{ \psi \, i } } =F_{ \Delta hi }=-mc^{ 2 }\cfrac { a_{ \psi \, f } }{ a^{ 2 }_{ \psi \, i } }$

And consider,

$E_{ \Delta h }=\int { F_{ \Delta hf } } da_{ \psi f }=mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } (a_{ \psi \, f }-a_{ \psi \, i })$

Differentiating with respect to $a_{ \psi \, f }$,

$\cfrac { \partial \, E_{ \Delta h } }{ \partial \, a_{ \psi \, f } } =F_{ \Delta hf }=mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } }$

So, when both $a_{ \psi \, i }$ and $a_{ \psi \, f }$ are changing,

$\cfrac { \partial \, E_{ \Delta h } }{ \partial \, a_{ \psi  } } =\cfrac { \partial \, E_{ \Delta h } }{ \partial \, a_{ \psi \, f } } +\cfrac { \partial \, E_{ \Delta h } }{ \partial \, a_{ \psi \, i } } =F_{ \Delta h\, T }=mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } \left( 1-\cfrac { a_{ \psi \, f } }{ a_{ \psi \, i } }  \right)$

And we consider the second derivative,

$\cfrac { \partial ^{ 2 }E_{ \Delta h } }{ \partial \, a^{ 2 }_{ \psi  } } = \cfrac { \partial \,  }{ \partial \, a_{ \psi  } }\left\{\cfrac { \partial \, E_{ \Delta h } }{ \partial \, a_{ \psi \, f } } +\cfrac { \partial \, E_{ \Delta h } }{ \partial \, a_{ \psi \, i } }\right\}=F^{ ' }_{ \Delta h\,T}$

$F^{ ' }_{ \Delta h\,T}=-mc^{ 2 }\cfrac { 1 }{ a^{ 2 }_{ \psi \, i } } +2mc^{ 2 }\cfrac { a_{ \psi \, f } }{ a^{ 3 }_{ \psi \, i } } -mc^{ 2 }\cfrac { 1 }{ a^{ 2 }_{ \psi \, i } }$

$F^{ ' }_{ \Delta h\,T}=2mc^{ 2 }\cfrac { 1 }{ a^{ 2 }_{ \psi \, i } } \left( \cfrac { a_{ \psi \, f } }{ a_{ \psi \, i } } -1 \right)  =-k$

And an approximation,

$F_{ \Delta h\, T }=F^{ ' }_{ \Delta h }.\Delta a_{\psi}$

If such a system is prompted to oscillate,

$\omega _{ n }=\sqrt { \cfrac { k }{ m }  } =\sqrt { 2 } c\sqrt { \cfrac { a_{ \psi \, f } }{ a^{ 3 }_{ \psi \, i } }  }$

where $\omega _{ n }$ is the natural frequency of the system.  Naturally, such a system is damped as it emits photons, the loss due to damping is,

$P_{loss}=- \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }=-m\nu\left(\cfrac{d\,a_{\psi\,i}}{d\,t}\right)^2$

where $\nu$ (rads^-1) is the damping factor using the model $f(x,\dot{x})=-kx-m\nu\dot{x}$ and the rate at which $f(x,\dot{x})$ does work is,

$P=f.\dot{x}=-kx\dot{x}-m\nu\dot{x}^2=P_{osc}+P_{loss}$

And consider, $a_{\psi\,i}$ making a round trip to $a_{\psi\,f}$ and back in time $T_d=2\pi\cfrac{1}{\omega_d}$,

$\cfrac{d\,a_{\psi\,i}}{d\,t}=\cfrac{2(a_{\psi\,i}-a_{\psi\,f})}{T_d}=\cfrac{1}{\pi}(a_{\psi\,i}-a_{\psi\,f})\omega_d$

where $\omega_d$ is the damped resonance frequency.  We have,

$\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }=m\nu\left(\cfrac{d\,a_{\psi\,i}}{d\,t}\right)^2=m\nu\cfrac{1}{\pi^2}(a_{\psi\,i}-a_{\psi\,f})^2\omega^2_d$

As,

$\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } =\cfrac { \partial \, E_{ \Delta h } }{ \partial \, a_{ \psi \, f } }\cfrac{d\,a_{\psi\,f}}{d\,t} +\cfrac { \partial \, E_{ \Delta h } }{ \partial \, a_{ \psi \, i } }\cfrac{d\,a_{\psi\,i}}{d\,t}$

Assuming $\cfrac{d\,a_{\psi\,f}}{d\,t}=\cfrac{d\,a_{\psi\,i}}{d\,t}$, that the external driving force affects both equally,

$mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } \left( 1-\cfrac { a_{ \psi \, f } }{ a_{ \psi \, i } }  \right)=m\nu\cfrac{1}{\pi}(a_{\psi\,i}-a_{\psi\,f})\omega_d$

Since, $a_\psi$ at $a_{ \psi \, f }$ loses almost all its energy, the amplitude of the wave at $a_{ \psi \, f }$, $A_f$ is small,

$\nu\approx 1$ rads^-1

then,

$w_d=\pi c^2\cfrac{1}{a^2_{ \psi \, i } }$

which is a very high frequency, (10³⁴Hz).

At this frequency, the particle will be "resonating" with high output of photons emitted as a result of the energy state transition $a_{ \psi \, i } \rightarrow a_{ \psi \, f }$.

If this high frequency, $\omega_d$ can be achieved, we have a photon resonator that might drive that particular particle to wrap speed.

Note:  What? Wrong to replace $\dot{x}$ with average speed?

645 Hard Science Fiction

645 pages of hard science... fiction, already.  Ah, Ah, Ah.  Maybe we should attempt at science not already in science fiction.

Until Next time.

The Lights Is On And Off You Go

In the post "Charge Photons Creation", it was suggested that charges on relaxation from an excited state emit photons of the nature, ph($t_c$,$t_g$) and ph($t_c$,$t_T$) where $t_c$ is the time axis on with the photons exist and $t_g$, or $t_T$ is the time axis on which they oscillates.

If the process is reversible, then a photon of the correct nature will impart energy onto charges of the corresponding nature ch($t_g$) or ch($t_T$).

In the post "Amplitude, $A_n$", it was suggested that $\psi$ at $a_{\psi\,n}$, makes a state transition to the next level, $a_{\psi\,n+1}$ when its amplitude has gain enough magnitude.

It might be possible to bathe a particle in photons of the correct nature such that its $a_{\psi\,n}$ increases slowly, and then induce a sudden collapse such that we have,

$\cfrac{\partial\,E_{\Delta h}}{\partial\,t}\lt0$

and very large.

As long as the photons shines on the particle,

$\cfrac{\partial\,E_p}{\partial\,t}=\cfrac{\partial\,E_{\Delta h}}{\partial\,t}\gt0$

where $E_p$ is the photon energy,

this itself maybe enough to stop the relaxation of energy states, as $a_{\psi\,n}$ is prompted to the next higher level continuously.  Conversely if the photons are suddenly switched off, the absence of a driving force that increases $a_{\psi\,n}$ continuously maybe enough to trigger a collapse of energy states, as the particle is now at a very high $a_{\psi\,n}$.  Otherwise, an increase in $\psi$ from a magnetic/electric field around the particle as the photons are switched off, might prompt a continuous energy state relaxation.  Such continuous relaxations are accompanied by emissions of photons at light speed, and are expected to produce a large negative $\cfrac{\partial\,E_{\Delta h}}{\partial\,t}$.

Teleportation/Warp speed could be as simple as turning a switch on and off.  The presence of an external field that would effect $E_{\Delta h}$ of the particle is important because its rate of change with time must follow the profile,

$\cfrac{\partial\,E_{\Delta h}}{\partial\,t}=\cfrac{A}{t}$    $A\lt0$

over the cascade of transitions to lower $a_{\psi\,n}$.

How then to generate massive amount of photons of high energy, of a particular nature?

Imagine such a massive photon generator at the center of a spaceship.  When the generator is on, photons from it energize the whole craft to high $a_{\psi\,n}$.  When the generator is suddenly switched off, all the particles that constitute the craft relax to a lower energy state under a force field, $B$ or $E$, that is turned on as the generator is switched off.  The craft is teleported.

But to which direction?  In the case of a portal, the nature of $t_c$ around $x$ dictates that $t_c$ slows down in a circular plane perpendicular to $x$.  Photons are inject from a circular perimeter clockwise, toward the center to slow time, $t_c$ within the circle.  The direction of travel is then into the circle, perpendicular to the plane of the circle.

In the case of a photon generator, the photons need be radiated outwards in a clockwise circular manner from stern to bow, then the spaceship will be driven forward.

The geometry of the photons, around the photon generator at the center, dictates the direction of travel.

Jump!

Undefined At What? High Five!

Fortunately, we are able to formulate,

$\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }=\cfrac{A}{t+t^{-}}$

such that $t^{-}\rightarrow 0$ as $t\rightarrow 0^{-}$ and we have the following illustrative plot.

where,

$\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }=\cfrac{A}{t}$,  and $A\lt0$

In practice, $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \nrightarrow -\infty$ on the throw of the switch, but it can be a high value that,

${a_{ \psi \, n }}_o=\cfrac { 1 }{ 2\pi  } mc^{ 4 }\left( \left|\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right| \right) ^{ -1 }$

is sufficiently large.

The question remains, how to induce $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }$ in a particle.

Stargate

If we stare down one of the space dimension with one time dimension curled around it,

where $r$ is very small.  $t_g$ is at light speed, $v=c$.  If it is possible to slow down time,

$\cfrac{\partial\,t_{gf}}{\partial\,t}\lt c$   (in ticks per second)

such that,

$\cfrac{\partial\,t_{gf}}{\partial\,t}=r\omega$

$r=\cfrac{1}{\omega}\cfrac{\partial\,t_{gf}}{\partial\,t}\lt\cfrac{c}{\omega}$

where both $\omega$ and $r$ are physically manageable number, consider the equation from the post "Time Travel Made Easy" again,

$\cfrac{\partial\,t_{gf}}{\partial\,t}=\cfrac{1}{mc^2}(\cfrac{\partial\,t_{gi}}{\partial\,t}E_{\Delta h}+t_{gi}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})+\cfrac{\partial\,t_{gi}}{\partial\,t}$ ---(*)

$\cfrac{\partial\,t_{gi}}{\partial\,t}=c$

$\cfrac { \partial \, t_{ gf } }{ \partial \, t } =\cfrac { 1 }{ mc^{ 2 } } (cE_{ \Delta h }+t_{ gi }\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } )+c$

$t_{ gi }\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } =mc^{ 2 }\cfrac { \partial \, t_{ gf } }{ \partial \, t } -cE_{ \Delta h }-mc^{ 3 }$

$t_{ gi }=\cfrac { 2\pi a_{ \psi \, n } }{ c }$

$a_{ \psi \, n }\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } =\cfrac { 1 }{ 2\pi  } \left\{ mc^{ 3 }\cfrac { \partial \, t_{ gf } }{ \partial \, t } -c^{ 2 }E_{ \Delta h }-mc^{ 4 } \right\}$

$\begin{equation*}a_{ \psi \, n }=\cfrac { 1 }{ 2\pi  } \left\{ -mc^{ 4 }\left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }+mc^{ 3 }\cfrac { \partial \, t_{ gf } }{ \partial \, t } \left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }-c^{ 2 }E_{ \Delta h }\left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 } \right\} \end{equation*}$

Since,

$\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \lt0$

$\begin{equation*}a_{ \psi \, n }=\cfrac { 1 }{ 2\pi  } \left\{ mc^{ 4 }\left( \left| \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right|  \right) ^{ -1 }-mc^{ 3 }\cfrac { \partial \, t_{ gf } }{ \partial \, t } \left( \left| \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right|  \right) ^{ -1 }+c^{ 2 }E_{ \Delta h }\left( \left| \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right|  \right) ^{ -1 } \right\} \end{equation*}$

replacing ${a_{ \psi \, n }}_o=\cfrac { 1 }{ 2\pi  } mc^{ 4 }\left( \left|\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right| \right) ^{ -1 }$

$a_{ \psi \, n }={ a_{ \psi \, no } } \left\{ 1-\cfrac { 1 }{ c } \cfrac { \partial \, t_{ gf } }{ \partial \, t } +\cfrac { 1 }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } t  \right\}$

From (*)

$\cfrac { \partial \, t_{ gf } }{ \partial \, t } =\cfrac { 1 }{ mc^{ 2 } } (\cfrac { \partial \, t_{ gi } }{ \partial \, t } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } t+\cfrac { \partial \, t_{ gi } }{ \partial \, t } t\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } )+\cfrac { \partial \, t_{ gi } }{ \partial \, t }$

$\cfrac { \partial \, t_{ gf } }{ \partial \, t } =\left\{ \cfrac { 2t }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } +1 \right\} \cfrac { \partial \, t_{ gi } }{ \partial \, t }$ --- (**)

So,

$a_{ \psi \, n}={ a_{ \psi \, no } } \left\{ 1-\left\{ \cfrac { 2t }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } +1 \right\}+\cfrac { 1 }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } t  \right\}$

$a_{ \psi \, n }={ a_{ \psi \, no } } \left\{-\cfrac { t }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }   \right\}$

$a_{ \psi \, n }=\cfrac{1}{2\pi}c^2.t$

The particle travels at a forward velocity of,

$v_{tele}=\cfrac{1}{2\pi}c^2$

Is this speed possible?  Yes, time speed has been slowed down, that implies greater than light speed space travel.  However we are NOT traveling through space this way.

At time $t=0^{-}$, just before $t=0$,
$\require{cancel}$
$a_{ \psi \, n, \,t=0}={ a_{ \psi \, no } } \left\{ 1-\cfrac { 1 }{ c } \cfrac { \partial \, t_{ gf } }{ \partial \, t } +\cancelto{0}{\cfrac { 1 }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } t } \right\}$

$a_{ \psi \, n,\,t=0 }={ a_{ \psi \, no } } \left\{ 1-\cfrac { 1 }{ c } \cfrac { \partial \, t_{ gf } }{ \partial \, t }  \right\}$

in this case, $\cfrac { \partial \, t_{ gf } }{ \partial \, t }\ne0$, and the teleported distance is $a_{ \psi \, n\,t=0 }$.  As with the case considered in the post "Teleportation", with

$\cfrac { \partial \, t_{ gf } }{ \partial \, t }$=constant

 the particle return velocity is,

$a_{ \psi \, no } \cfrac { 1 }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } t=-\cfrac{1}{2\pi}c^2.t=v_{tele}.t$

$v_{tele}=-\cfrac{1}{2\pi}c^2$

At $t\gt0$ however,

$v_{tele}=\cfrac{1}{2\pi}c^2$

It seems paradoxical that the particle can be at different locations and have different velocities just before and after $t=0$.  The trick here is,

$\cfrac { \partial \, t_{ gf } }{ \partial \, t } =\left\{ \cfrac { 2t }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } +1 \right\} \cfrac { \partial \, t_{ gi } }{ \partial \, t }$ = constant

The expression $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }.t$ is held constant such that 

$\cfrac { \partial \, t_{ gf } }{ \partial \, t } =\left\{ \cfrac { 2t }{ mc^{ 2 } } \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } +1 \right\} \cfrac { \partial \, t_{ gi } }{ \partial \, t }$

from (**) is a constant.  This means

 $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }=\cfrac{A}{t}$

where $A$ is a constant.

This way, only the situation for $t=0^{-}$ applies and as will be discussed below $t\ngtr0$.

The problem is, if $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }$ is induced in a particle, how to turn it off when it is at a distance $a_{ \psi \, n,\,t=0 }$ away?

Simple, we beam the particle through another portal that induces $-\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }$.  After passing through the second portal, $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }=0$.  Equation (*) collapses to,

$\cfrac{\partial\,t_{gf}}{\partial\,t}=\cfrac{\partial\,t_{gi}}{\partial\,t}$

as at $t=0$,  $E_{\Delta h}=0$.

And the particle does not return from $x=a_{ \psi \, n,\,t=0 }$,  $t\ngtr0$.

Since time, $t_g$ is curled around $x$ where $it_g=x$ as required in the post "Time Travel Made Easy".  $x$ is perpendicular to the circular plane containing $t_g$.  The portal is directional.  This sort of teleportation unfortunately is restricted to line of sight.  Unless we can draw a straight line between the two points, teleportation is not possible.

And the billion dollar question is, how to induce $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }=\cfrac{A}{t}$ in a particle?

Teleportation

Why did the Philadelphia Experiment traveled into the future time?  From the post "Rotating Al Again", if they have just rotated the ($t_c$-$t_T$) plane to the ($t_g$-$t_T$) plane by rotating the projection of $t_{now}$ on the ($t_c$-$t_T$) plane by 60^o onto the ($t_g$-$t_T$) plane, and the space dimensions are coiled up along the time axes such that any space dimension is perpendicular the time dimension and at the same time orthogonal to all other dimensions, space and time cannot cross nor superimpose no matter what rotation or transformations.

In our 3 space dimension world time flows freely, continuously.  If we are able to rotate/transform a space dimension on to a time dimension then space along that dimension will flow freely.  The time dimension that has been swapped with a space dimension will stand still unless kinetic energy is applied.

From the post "Eternal Embrace",  time and space loop around each other.  As we shorten time, space elongates.  It might be possible that using the equation from the post "Time Travel Made Easy",

$\cfrac{\partial\,t_{gf}}{\partial\,t}=\cfrac{1}{mc^2}(\cfrac{\partial\,t_{gi}}{\partial\,t}E_{\Delta h}+t_{gi}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})+\cfrac{\partial\,t_{gi}}{\partial\,t}$

and setting

$\cfrac{\partial\,t_{gf}}{\partial\,t}=0$

$\cfrac{1}{mc^2}(\cfrac{\partial\,t_{gi}}{\partial\,t}E_{\Delta h}+t_{gi}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})+\cfrac{\partial\,t_{gi}}{\partial\,t}=0$

$t_{ gi }\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } =-\left( E_{ \Delta h }+mc^{ 2 } \right) \cfrac { \partial \, t_{ gi } }{ \partial \, t }$

but,

$\cfrac { \partial \, t_{ gi } }{ \partial \, t } =c$

So,

$t_{ gi }\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } =-c\left( E_{ \Delta h }+mc^{ 2 } \right)$

since,

$t_{ gi }\gt0$,   $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \lt0$

Furthermore,

$t_{ gi }=\cfrac { 2\pi a_{ \psi \, n } }{ c } =-c\left( E_{ \Delta h }+mc^{ 2 } \right) \left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }$

$a_{ \psi \, n }=-\cfrac { 1 }{ 2\pi  } c^{ 2 }\left( E_{ \Delta h }+mc^{ 2 } \right) \left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }$

Initially when, $E_{ \Delta h }=0$

$a_{ \psi \, n }=-\cfrac { 1 }{ 2\pi  } mc^{ 4 }\left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }$

As $\left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right)$ is negative,

${a_{ \psi \, n }}_o=\cfrac { 1 }{ 2\pi  } mc^{ 4 }\left( \left|\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right| \right) ^{ -1 }$ --- (*)

So, depending on the value of $\left( \left|\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \right| \right) ^{ -1 }$,  the particle is transported to a distance defined by equation(*).  Afterwards,

$E_{ \Delta h }=\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } t$

$a_{ \psi \, n }=-\cfrac { 1 }{ 2\pi  } c^{ 2 }\left\{ t+mc^{ 2 } \left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 } \right\}$

$a_{ \psi \, n }=-\cfrac { 1 }{ 2\pi  }mc^{ 4 }\left( \cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }  \right) ^{ -1 }-\cfrac { 1 }{ 2\pi  } c^{ 2 }t$

$a_{ \psi \, n }={a_{ \psi \, n }}_o-\cfrac { 1 }{ 2\pi  } c^{ 2 }t$

After the initial jump, if the particle is still subjected to $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t } \lt0$, as time passes, the particle travels back to its origin with a velocity $v_{tele}$,

 $v_{tele}=-\cfrac { 1 }{ 2\pi  } c^{ 2 }$

If $\cfrac { \partial \, E_{ \Delta h } }{ \partial \, t }$ is set to zero immediately at ${a_{ \psi \, n }}_o$ then the particle remains at ${a_{ \psi \, n }}_o$.

Teleportation!  Warp Speed!  When the particle is allowed to return, on a smaller scale, Brownian motion.

Invisible, Right Here!

What?  Expecting invisibility in the previous post "Now You See Me, Now You Don't"?  Here is the Time Flow Differential Equation again, but this time along $t_c$,

$\Delta_t=\cfrac{1}{mc^2}(\cfrac{\partial\,t_{ci}}{\partial\,t}E_{\Delta h}+t_{ci}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})$

What if $t_c$ is sufficiently slowed?  If optics are charge phenomenon, due to photons that exist along the $t_c$ time dimension, then slowing $t_c$ slow down the photons.  Our eyes and equipments may not be able to detect such slowed photons, and we effectively turn invisible.

Remember, you are in my dream.

What would changing $\cfrac{\partial\,t_T}{\partial\,t}$ do?

$\Delta_t=\cfrac{1}{mc^2}(\cfrac{\partial\,t_{Ti}}{\partial\,t}E_{\Delta h}+t_{Ti}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})$

Maybe it will stop aging.

Wouldn't $t_g$ effect gravity and gravity only,

$\Delta_t=\cfrac{1}{mc^2}(\cfrac{\partial\,t_{gi}}{\partial\,t}E_{\Delta h}+t_{gi}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})$

Floating around invisible, totally ignored and aging slowly.  I'm a see through plastic banana balloon.

Now You See Me, Now You Don't

From the previous post "Time Travel Made Easy",

$\Delta_t=\cfrac{1}{mc^2}(\cfrac{\partial\,t_{gi}}{\partial\,t}E_{\Delta h}+t_{gi}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})$

$E_{\Delta h}$ occurs naturally as state transitions of a particle, sometime with the emission of a photon.  Time travel is thus a natural process and happens all the time.

Such phasing in and out of time might be the root of Brownian motions.  The key is the frequency of the random motion and the frequency of energy state transitions.  If these phenomena are related, these frequencies are equal if not of the same nature, assuming that all $E_{\Delta h}$ results in observable discrepancies in time and position.

Note:  If there is a flash of light on time travel, photons are emitted and $\cfrac{\partial\,E_{\Delta h}}{\partial\,t}\lt0$, the time travel direction is backward in time.  If there is a flash of light on arrival from time travel, $\cfrac{\partial\,E_{\Delta h}}{\partial\,t}$ is still negative, the time traveler is from the future.

$E_{\Delta h}$ is the change in energy experienced by the particle, an external agent effecting such energy changes on the particle will have to do the opposite; absorb energy, the external agent itself gaining energy for the particle to have a negative $\cfrac{\partial\,E_{\Delta h}}{\partial\,t}$.

Time Travel Made Easy

From the post "What Does It Mean to Be Excited",

$E_{\Delta h}=mc^2(\cfrac{a_{\psi\,f}}{a_{\psi\,i}}-1)$

and from the previous post "Charge Photons Creation",

$iT=\cfrac{2\pi a_{\psi\,n}}{c}=t_g$

$\cfrac{a_{\psi\,f}}{a_{\psi\,i}}=\cfrac{t_{gf}}{t_{gi}}$

We have,

$E_{\Delta h}=mc^2(\cfrac{t_{gf}}{t_{gi}}-1)$

$t_{gf}=t_{gi}(\cfrac{E_{\Delta h}}{mc^2}+1)$

But what is $t_{gf}$ and $t_{gi}$?  However,

$\cfrac{\partial\,t_{gf}}{\partial\,t}=\cfrac{1}{mc^2}(\cfrac{\partial\,t_{gi}}{\partial\,t}E_{\Delta h}+t_{gi}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})+\cfrac{\partial\,t_{gi}}{\partial\,t}$

This suggests that the passage of time can be affected by an absolute change in energy, $E_{\Delta h}$ or an high rate of change in  $E_{\Delta h}$,  $\cfrac{\partial\,E_{\Delta h}}{\partial\,t}$.  In both cases, we are restrained by $mc^2$.

$\Delta_t=\cfrac{\partial\,t_{gf}}{\partial\,t}-\cfrac{\partial\,t_{gi}}{\partial\,t}=\cfrac{1}{mc^2}(\cfrac{\partial\,t_{gi}}{\partial\,t}E_{\Delta h}+t_{gi}\cfrac{\partial\,E_{\Delta h}}{\partial\,t})$

It is possible to increase or decrease time flow by changing the sign of $\cfrac{\partial\,E_{\Delta h}}{\partial\,t}$.  When $\Delta_t\gt0$, the rest of the world flows backward and we travel forward in time; when $\Delta_t\lt0$, the rest of the world flows forward and we travel back in time.

Changing $\cfrac{\partial\,E_{\Delta h}}{\partial\,t}$ is the same as manipulating $\psi$ as discussed in the post "Time Travel By Manipulating ψ".  The relationship presented here is much more clear.

Charge Photons Creation

From the post "Why Should Amplitude Remain Constant?", it was argued that a change in parameter $n_i\rightarrow n_f$ changes $\lambda_n$ and $a_{\psi\,n}$ but does not change $A_n$.  $\lambda_n$ is orthogonal to $a_{\psi\,n}$ and both are orthogonal to $A_n$.

Since,

$c=f_n\lambda_n=\cfrac{\lambda_n}{T}$

$T=\cfrac{\lambda_n}{c}$

where $T$ is along the $t_c$ time dimension.

$2\pi a_{\psi\,n}=i\lambda_n$

$\lambda_n=-i2\pi a_{\psi\,n}$

So,

$T=\cfrac{-i2\pi a_{\psi\,n}}{c}$

$iT=\cfrac{2\pi a_{\psi\,n}}{c}$

$iT$ is a time dimension perpendicular to $T$.  It is either $t_T$ or $t_g$.  $iT$ is not any of the two space dimensions as $a_{\psi\,n}$ does not affect $A_n$.

$iT$ is separately,

$iT=t_T$

 $iT=t_g$

for the two type of charges.

A change in $a_{\psi\,n}$, therefore changes $t_T$ or $t_g$, and a change in $\lambda_n$ changes $t_c$.

As long as $a_{\psi\,n}$ and $\lambda_n$ are valid solutions to the wave equation,  the particle is still a wave in 2 space dimensions and one time dimension $t_g$ or $t_T$ after the changes (ie the particle is not destroyed).  That means $v=c$ along both $t_c$ and, $t_T$ or $t_g$.

If a change forces $v\gt c$ along $t_g$ or $t_T$, the energy is released along that dimension instead.  $v$ remains constant at $c$.  If a changes results in $v\lt c$, energy is absorbed along the same time dimension, such that $v$ increases to $c$ again.  In both changes, other orthogonal dimensions are not affected.  The particle has no access to other dimensions in which it is not oscillating, and not existing (The charge exist along $t_c$).  So the charge has access to $t_c$, two space dimensions and $t_T$ or $t_g$.

A slow down in time $t_c$ is associated with greater then light speed, $v_s\gt c$ in space.  When time returns to normal, $v_s=c$, the particle velocity is at light speed.  (Time speed changes after the particle has reach light speed in space).  As such a change along $t_c$, the time dimension in which the particle exist, is coupled to the space dimension, ($s$) along which it travels ($c^2=v^2_s+v^2_{tc}$).  As the particle returns to light speed in $t_c$, it is in light speed in $s$.

Taken altogether,  changes in $a_{\psi\,n}$ and $\lambda_n$ of a particle creates another wave in the same set of dimensions as the particle, as long as the final $a_{\psi\,f}$ and $\lambda_f$ are still valid solutions to the wave equation.  If the particle is destroyed, there is then no light speed constrain on the dimensions involved and no emission/radiation of energy would be required.  That the original particle is not destroy is itself a constrain leading to the creation of the new particle.

The diagram above shows that changes in $a_{\psi\,n}$ and $\lambda_n$ of charges lead to the creations of two types of photons oscillating along $t_g$ or $t_T$, existing along $t_c$.  These are charge photons associated with each type of charge.

Makan Time!  If you are Swedish it means something else.

Geometric Mean And Trickery

Consider a transition from $n_i\rightarrow n_f$,

$E_{ \Delta h 1}=mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } (a_{ \psi \, f }-a_{ \psi \, i })$

Consider another transition from $a_{ \psi \, f }+a_{ \psi \, i } \rightarrow a_{ \psi \, f }$,  (This is an extrapolation, it is not proven that $a_{ \psi \, f }+a_{ \psi \, i }$ is a solution.)

$E_{ \Delta h 2}=\cfrac { mc^{ 2 } }{ a_{ \psi \, f }+a_{ \psi \, i } } (a_{ \psi \, f }-(a_{ \psi \, f }+a_{ \psi \, i } ))=\cfrac { mc^{ 2 } }{ a_{ \psi \, f }+a_{ \psi \, i } } (-a_{ \psi \, i })$

such that, $a_{\psi\,i}$ is in between the paths of $E_{ \Delta h 1}$ and $E_{ \Delta h 2}$.

If we take the geometric mean of the two,

$\bar{E}_{\Delta h}=\sqrt{E_{ \Delta h 1}.E_{ \Delta h 2}}=mc^2\sqrt{\cfrac{a_{ \psi \, i }-a_{ \psi \, f }}{a_{ \psi \, f }+a_{ \psi \, i } }}$

and let

$E_o=-\cfrac{\bar{E}_{\Delta h}}{a_{\psi\,i}}=-mc^2\cfrac{1}{a_{\psi\,i}}\sqrt{\cfrac{a_{ \psi \, i }-a_{ \psi \, f }}{a_{ \psi \, f }+a_{ \psi \, i } }}$  Jm^-1

We have an expression for $E_o$.  A negative sign is necessary because a change from a higher value of $a_\psi$ to a lower value of $a_\psi$ (ie. a positive value of $(a_{ \psi \, i }-a_{ \psi \, f })$ ) results in a loss of energy.

Note:  If,

$E_{ \Delta h\,f}=2\pi a_{\psi\,f}mc.f$

$E_{ \Delta h\,i}=2\pi a_{\psi\,i}mc.f$

And,

$E_{ \Delta h\,i}+E_{ \Delta h\,f}=2\pi a_{\psi\,i}mc.f+2\pi a_{\psi\,f}mc.f$

$2\pi (a_{\psi\,i}+a_{\psi\,f})mc.f=E_{ \Delta h\,i+f}$

And,

$E_{ \Delta h\,i+f}=\cfrac { mc^{ 2 } }{ a_{ \psi \, f }+a_{ \psi \, i } } (a_{ \psi \, f }-(a_{ \psi \, f }+a_{ \psi \, i } ))$ --- (*)

This does not proof that $a_{ \psi \, f }+a_{ \psi \, i }$ is a solution to $x$ for $\psi$ but it shows that (*) is a valid expression.

Simply $\hbar$

Consider,

$E_{\Delta h}=mc^2(\cfrac{a_{\psi\,f}}{a_{\psi\,i}}-1)$

In the case of a photon, $a_\psi$ is the radius of its helical path.  This shows that a change in radius is accompanied by an exchange of energy.

and

$2\pi a_{\psi\,n}= \lambda_{n}$

So,

$E_{ \Delta h }=mc^{ 2 }(\cfrac { \lambda _{ f } }{ \lambda _{ i } } -1)$

$E_{ \Delta h }=mc^{ 2 }(\cfrac { f_{ i } }{ f_{ f } } -1)$

Which suggests a photon can change frequency, given an appropriate energy input, $E_{ \Delta h }$.  If $E_{ \Delta h }$ resulted in an emission of a photon,

$h_ff_p=-E_{ \Delta h }=mc^{ 2 }(1-\cfrac { f_{ i } }{ f_{ f } })$

$\lambda_fmcf_p=\cfrac{1}{f_f}mc^2f_p=mc^{ 2 }(1-\cfrac { f_{ i } }{ f_{ f } })$

$f_p=f_f(1-\cfrac { f_{ i } }{ f_{ f } })=f_f-f_i$

Surprise! Surprise!  This suggests that an emission of a photon occurs only if the lower energy states have been folded up, where a lower energy state actually have higher frequency.  This is illustrated on the plot below,

Frequency Intersections

The curve marked $n=1$ in red, is folded up and has a higher intersection for frequency with the line $y=x$.  Transition from all other states with lower frequencies will result in an emission of a photon. Transition from states $n=10$ however, will not result in a photon emission because the associated frequency is higher than that for state $n=1$.

And,

$\cfrac{1}{\lambda_p}=\cfrac{1}{\lambda_f}-\cfrac{1}{\lambda_i}$

for photon emission.

Since,

$h_if_i=mc^2$

$h_ff_f=mc^2$

$E_{ \Delta h }=mc^{ 2 }(\cfrac { h_{ f } }{ h_{ i } } -1)$

which relates the change in energy as a result of a change in $h_n$ directly.

We still don't have justifications for $E_o$

Touch And Go

From the previous post "Amplitude $A_n$", it is also possible that a state transition, $n_f\rightarrow n_i$, occurs when,

$r=a_{\psi\,i}$

then,

$a^2_{\psi\,i}=a^2_{\psi\,f}+A^2_f$

$A_f=\sqrt{a^2_{\psi\,i}-a^2_{\psi\,f}}$

In which case we still have,

$E_{ \Delta n }\propto A_f$,  because $\psi$ is already energy density.

$E_{ \Delta n }=E_{ o }\sqrt { a^{ 2 }_{ \psi \, i}-a^{ 2 }_{ \psi \, f} }$

And we define $E_o$ as,

$E_{ o }=-mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } \sqrt { \cfrac { (a_{ \psi \, i }-a_{ \psi \, f }) }{ (a_{ \psi \, i }+a_{ \psi \,f}) }  }$

In this case a state transition occurs when a lower $\psi$ wave touches the path of an outer wave.  This would occur before the perimeters of the paths are equal.

Amplitude, $A_n$

Obviously,

$|\psi_n|=A_n$

The following diagram shows two energy states, at $a_{\psi\,i}$ and $a_{\psi\,f}$.

As the particle gain energy, $A_f$ increases, and

$r^2=a^2_{\psi\,f} + A^2_f$ --- (*)
$r$ increases.

Could it be that when the perimeters are equal,

$2\pi \sqrt { \cfrac { ({ a^2_{\psi\,f} }+r^{ 2 }) }{ 2 }  }=2\pi a_{\psi\,i}$

a transit of energy states $n_f\rightarrow n_i$ occurs;  ie when

$r^{ 2 }=2a^2_{\psi\,i}-a^2_{\psi\,f}$

By substituting $r$ into (*)

$A_f=\sqrt{2}\sqrt{a^2_{\psi\,i}-a^2_{\psi\,f}}$

$A_i=0$

Just on transition, $A_i$ is small.  This can explain a lower $a_\psi$ to a higher $a_\psi$ transition, but what about the reverse?

The reverse occurs when $a_{\psi\,i}$ collapses back to $a_{\psi\,f}$ but the amplitude, $A_i\rightarrow A_f$  remains small.

The excess energy as a result of a reduced $A_f$ at $a_{\psi\,f}$ is emitted as a photon.  In the limit when $A_i\rightarrow 0$, this excess energy is just the rms value of $A_f$ (not $A^2_f$ as $\psi$ is already energy),

$E_{\Delta n}=\cfrac{1}{\sqrt{2}}A_f=\sqrt{a^2_{\psi\,i}-a^2_{\psi\,f}}$

$A_f$ is measured in meters (m), so more correctly,

$E_{\Delta n}=E_o\sqrt{a^2_{\psi\,i}-a^2_{\psi\,f}}$

where $E_o$ (Jm^-1) maps meters (m) to Joules (J) in the diagram above.

Is this the same result as before?

$E_{ \Delta n }=E_{ o }\sqrt { a^{ 2 }_{ \psi \, i }-a^{ 2 }_{ \psi \, f } }$

$E_{ \Delta n }=E_{ o }\sqrt { (a_{ \psi \, i }-a_{ \psi \, f })(a_{ \psi \, i}+a_{ \psi \, f }) }$

If $E_o$ is,

$E_{ o }=-mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } \sqrt { \cfrac { (a_{ \psi \, i}-a_{ \psi \, f }) }{ (a_{ \psi \, i }+a_{ \psi \, f }) }  }$

then,

$E_{ \Delta n }=-mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } \sqrt { \cfrac { (a_{ \psi \, i }-a_{ \psi \, f }) }{ (a_{ \psi \,i }+a_{ \psi \,f }) }  }\sqrt { (a_{ \psi \, i}-a_{ \psi \, f })(a_{ \psi \, i }+a_{ \psi \,f }) }$

$E_{ \Delta n }=mc^{ 2 }(\cfrac { a_{ \psi \, f } }{ a_{ \psi \, i } } -1)$

Since,

$a_{\psi\,n}=e^{ -\cfrac { Aa_{\psi\,n}+C }{ n^{ 2 } }}$

$E_{\Delta n}=mc^2\left\{e^{ -\cfrac { Aa_{\psi\,f}+C }{ n^{ 2 }_f }+\cfrac { Aa_{\psi\,i}+C }{ n^{ 2 }_i }  }-1\right\}$

which is the same expression for $E_{\Delta n}$ in the post "What Does It Mean to Be Excited".

But why is,

$E_{ o }=-mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } \sqrt { \cfrac { (a_{ \psi \, i }-a_{ \psi \, f }) }{ (a_{ \psi \,i}+a_{ \psi \, f }) }  }$  (Jm^-1)???

Unless $E_o$ is justified, this scenario is just probable.

Note:  The perimeter of a ellipse is used here,

$P=2\pi\sqrt{\cfrac{a^2+b^2}{2}}$

where $a$ and $b$ are the radii of the ellipse.

But...

There is a problem, from the post "Schrödinger's Equation", $\psi$ is actually energy density, in which case, the collapsing atom paradox is unresolved.  Unless the proposed radiated energy from an orbiting electron is actually $\psi$ that remains around the particle, and does not actually radiated away. $\psi$ is the energy field around the particle, the negative gradient of which is the force field.  In the case of an electron, this force field is the $E$ field.

The particles are still not orbiting about any nucleus, although from the post "Boundary Between Wave And Particle Interaction", they can be superimposed onto a common center, provided that they interact as particles and not affect each others' $\psi$.

It is still unanswered why the particle make energy transitions in the first place.  What happen to $\psi$ when a particle gain energy?

Why Should Amplitude Remain Constant?

The amplitude of the wave, $A_n$ is perpendicular to both $\lambda_n$ and $a_{\psi\,n}$.  If there are forces changing $\lambda_n$ and $a_{\psi\,n}$, these forces will have no effect on $A_n$.

So, because there is no force acting on $A_n$, $A_n$ remains unchanged and the system loses energy as the particle transit from a higher $a_\psi$ to a lower $a_\psi$.

$\lambda_n$ and $a_{\psi\,n}$ changes when we set $n_i \rightarrow n_f$, but this by itself does not change $A_n$.

What Does It Mean to Be Excited

From the post "de Broglie Per Person"

$2\pi a_{ \psi  }=n\lambda$

$h_n=2\pi a_{ \psi  }mc=mc.n\lambda$

with $n=1$.

$2\pi a_{ \psi  }=\lambda_1$

$h_1=2\pi a_{ \psi  }mc=mc.\lambda_{1}$

At $_na_\psi$,

$h_n=2\pi a_{\psi\,n}.mc=mc.\lambda_n$

When a particle is excited we have,

$h_{f}\rightarrow h_{i}$

and

$\lambda_{f}\rightarrow\lambda_{i}$

And when the particle returns from an excited state, $\lambda_{i}$ is applied to $h_{f}$ and we formulate,

$E_{i\rightarrow f}=h_{f}f_{i}=2\pi a_{ \psi\,_f  }.mc.f_{i}=\lambda_{f}mc.\cfrac{c}{\lambda_i}$

which is the energy of the particle at $h_f$.

Since,

$2\pi a_{\psi\,n}=2\pi e^{ -\cfrac { Aa_{\psi\,n}+C }{ n^{ 2 } }  }= \lambda_{n}$

$\cfrac{\lambda_{f}}{\lambda_{i}}= e^{ -\cfrac { Aa_{\psi\,f}+C }{ n^{ 2 }_f }+\cfrac { Aa_{\psi\,i}+C }{ n^{ 2 }_i }  }$  --- (*)

We have,

$E_{i\rightarrow f}=mc^2.e^{ -\cfrac { Aa_{\psi\,f}+C }{ n^{ 2 }_f }+\cfrac { Aa_{\psi\,i}+C }{ n^{ 2 }_i }  }=\psi_f+x.F$

The energy required at $n_f$ is $E_{i\rightarrow f}$, the energy in excess is $x.F$, the work function.

$E_{\Delta n}=E_{i\rightarrow f}-E_f=mc^2.e^{ -\cfrac { Aa_{\psi\,f}+C }{ n^{ 2 }_f }+\cfrac { Aa_{\psi\,i}+C }{ n^{ 2 }_i }  }-mc^2=x.F$

$E_{\Delta n}=mc^2\left\{e^{ -\cfrac { Aa_{\psi\,f}+C }{ n^{ 2 }_f }+\cfrac { Aa_{\psi\,i}+C }{ n^{ 2 }_i }  }-1\right\}=x.F$ --- (1)

where $E_{\Delta n}$ is the energy gained by the particle.  In this case the, the particle loses energy and $E_{\Delta n}$ is negative.

In this derivation, the Planck constant $h$, is formulated for each solution of $a_\psi$.  The resulting $h_i$ is applied at the appropriate level to obtain $\psi_i+x.F$, the work function.  $x.F$ is emitted as a photon.

Alternatively,

$\Delta E=E_{f}-E_{i}=mc^2-mc^2=0$

What happened?

When we apply $h_f$ to $\lambda_i$, we are suggesting that a wave at $\lambda_i$ is caught at energy level $h_f$.  $\lambda_i$ adjusts itself to $\lambda_f$ and there are energy changes in the process.  When $\lambda_i$ is due to a larger $a_{\psi\,n}$, its amplitude is also smaller  (cf. post "H Bar And No Bar"; this amplitude need to be quantified).  When $\lambda_i$ constrict to $\lambda_f$ at energy state $n=f$ it does so at smaller amplitude, it loses energy.

This lost in energy appears as the work function $x.F$ and is emitted as a photon.  The material cools.  The wave will regain its amplitude as heat is applied.  In the diagram above we considered only the fundamental frequency, when $2\pi a_{\psi\,n}=\lambda_n$.

In the post "Particle Spectrum", we took reference at $n=1$, where $h_o=h_1$.  The expressions for $E_{\Delta n}$ and $\lambda_p$ are valid when the final energy state is $n=1$.

Intuitively, $\psi$ has a longer path length at $n_i$, the difference in energy $E_{\Delta h}$ due to a change in $h_n$ as the wave move from $n_i$ to $n_f$ is,

$E_{\Delta h}=h_{f}f_i-h_{i}f_i=\cfrac{c}{\lambda_i}(h_{f}-h_{i})=\cfrac{c}{2\pi a_{\psi\,i}}(h_{f}-h_{i})$

$E_{\Delta h}=\cfrac{1}{a_{\psi\,i}}mc^2(a_{\psi\,f}-a_{\psi\,i})$

$E_{\Delta h}=mc^2(\cfrac{a_{\psi\,f}}{a_{\psi\,i}}-1)$

$E_{\Delta h}=mc^2\left(e^{ -\cfrac { Aa_{\psi\,f}+C }{ n^{ 2 }_f }+\cfrac { Aa_{\psi\,i}+C }{ n^{ 2 }_i }  }-1 \right)$

Which is the same expression as (1).  Both derivations are equivalent.

$\psi$ of a particle is excited to $n_i$ energy state, $a_\psi=a_{\psi\,i}$.  On its return to a lower $n_f$ energy state, $a_\psi=a_{\psi\,f}$ where $a_{\psi\,f}\lt a_{\psi\,i}$, it loses energy in the form of a photon.

This is consistent with the observation of spectra lines, but this particle is not orbiting around any nucleus.

To be excited is then $a_{\psi\,i}\rightarrow a_{\psi\,f}$ where $a_{\psi\,f}\gt a_{\psi\,i}$; to have a higher $a_\psi$.

Note: $E_{\Delta h}$ and $E_{\Delta n}$ are not the difference in energy as the particle move from $n_i$ to $n_f$ energy state.  But are energy changes as a result of $h_n$ changes between energy states.

In all cases, for,

$n_i\gt n_f$,    $f_i\gt f_f$

which is consistent with the post "Discreetly, Discrete λ And Discrete Frequency, f", unless the frequency profile is folded upwards.