An Interesting Force.

A plot of the force between electrons in parallel orbits given by,

$F_a=-\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { r-r_e }{ \left\{ (2a_{ e })^{ 2 }+(r-r_e)^{ 2 } \right\} ^{ 3/2 } } d\,r$

is provided below.

It is both repulsive (positive) and attractive (negative) depending on the value of $r_e$.  A most interesting force.

Ionization Redefined

If the force between an orbiting electron and the nucleus is repulsive what then is ionization energy, $E_i$?

${E}_{i}={m_ec^2}\left\{ln{( \cfrac{{r_{eo} }}{{r}_{ef}}) +\cfrac { 1-{ r }_{ ef } }{ { r }_{ ef }{ r_{ eo } } }(r_{ef}-r_{eo})}\right\}-\cfrac{q^2}{4\varepsilon_o}\cfrac{1}{r_{e}}+G\cfrac { m_{ a }m_{ e } }{ r_{ e } }+\int^{\infty}_{r_e}{\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { r-r_e }{ \left\{ (2a_{ e })^{ 2 }+(r-r_e)^{ 2 } \right\} ^{ 3/2 } }} d\,r$

The terms are,

${m_ec^2}\left\{ln{( \cfrac{{r_{po} }}{{r}_{pf}}) +\cfrac { 1-{ r }_{ pf } }{ { r }_{ pf }{ r_{ po } } }(r_{pf}-r_{po})}\right\}$  

work done against the centripetal force, while the circular velocity remains constant.

$\cfrac{q^2}{4\varepsilon_o}\cfrac{1}{r_{e}}$

work recovered from the repulsive force of the nucleus $B$ orbit.

$G\cfrac { m_{ a }m_{ e } }{ r_{ e } }$

work done against gravitational attraction.

and

$\int^{\infty}_{r_e}{\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { r-r_e }{ \left\{ (2a_{ e })^{ 2 }+(r-r_e)^{ 2 } \right\} ^{ 3/2 } }} d\,r$

$=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } }\cfrac{1}{\sqrt{4a^2_e}}=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } }\cfrac{1}{2a_e}$

work done against the attraction of neighboring electron $B$ orbits.

The work against centripetal force explodes immediately.  The centripetal force is provided by drag at terminal velocity.  Work done against the centripetal force is essentially work against drag.  The electron gains velocity initially as a result of electrostatic attraction.  Upon achieving light speed, drag curls its path into a helix. The electron perform circular motion in the plane perpendicular to its direction of travel.

To resolve this issue, we see that this work is zero for incremental change in $r_{ef}$, ie. when

$r_{ef}=r_{eo}+\Delta r$    as  $\Delta r\rightarrow0$

this work is zero.  In other words, move slowly lo.

And we are left with,

${E}_{i}=-\cfrac{q^2}{4\varepsilon_o}\cfrac{1}{r_{e}}+G\cfrac { m_{ a }m_{ e } }{ r_{ e } }+\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } }\cfrac{1}{2a_e}$

$E_i=G\cfrac { m_{ a }m_{ e } }{ r_{ e } } +\cfrac { q^{ 2 } }{ 4\varepsilon _{ o } } \left\{ \cfrac{n_e}{2a_e} -\cfrac { 1 }{ r_{ e } }  \right\}$

which boldly is the new expression for ionization energy.  $n_e$ is the number of electrons in the stack minus one,  $a_e$ is the radius of an electron and  $m_a$ is the mass of the nucleus.

Stacked Orbits

Such stacked orbits means that the electrons are held more firmly in orbit and we see a decrease in reactivity across the periodic table as electrons are added to each consecutive elements.  If not for the Lorentz's force, each additional electron should be more readily removed from the atom and thus makes the atom more reactive.  The existence of $B$ orbit and the resulting stacked orbits provide an explanation for the increase in stability across the periodic table.

The nucleus is not involved in this stability.

A Brand New Attractive Force

Consider two electrons in parallel orbits,

$F_{ L }sin(\theta )=F_{ a }$

The attractive force between the electrons is the Lorentz's force.  $F_a$ is the force in the plane of the orbit that keeps the orbital radius constant due to the attraction of the neighboring orbit.

$tan(\theta )=\cfrac {\Delta r }{ 2a_{ e } }$

$F_L=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o }(2a_{ e })^{ 2 } } cos^2(\theta )=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o }(2a_{ e })^{ 2 } } \cfrac { 4a^2_e }{  (2a_{ e })^{ 2 }+(\Delta r)^{ 2 } } =\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { 1 }{ (2a_{ e })^{ 2 }+(\Delta r)^{ 2 } }$

Therefore,

$F_{ a }=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o }(2a_{ e })^{ 2 } } \cfrac { (2a_{ e })^{ 2 } }{ (2a_{ e })^{ 2 }+(\Delta r)^{ 2 } } \cfrac {(\Delta r)}{ \sqrt { (2a_{ e })^{ 2 }+(\Delta r)^{ 2 } }  }$

$F_{ a }=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { (\Delta r)}{ \left\{ (2a_{ e })^{ 2 }+(\Delta r)^{ 2 } \right\} ^{ 3/2 } }$

$\Delta r= r-r_e$

$F_{ a }=\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { r-r_e }{ \left\{ (2a_{ e })^{ 2 }+(r-r_e)^{ 2 } \right\} ^{ 3/2 } }$

A brand new attractive force! And we have,

$\cfrac { m_{ e }c^{ 2 } }{ r_{ e } } =Ac^{ 2 }+G\cfrac { m_{ a }m_{ e } }{ r^{ 2 }_{ e } }+\cfrac { n_{ e }q^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { r-r_e }{ \left\{ (2a_{ e })^{ 2 }+(r-r_e)^{ 2 } \right\} ^{ 3/2 } }  -\cfrac { Zq^{ 2 } }{ 4\varepsilon _{ o }(r_{ e }-r_{ p })^{ 2 } } -\cfrac { T_{ n }T_{ e } }{ 4\pi \tau_{ o }r^{ 2 }_{ e } }$

a new expression for the forces in action on an orbiting electron.  Under normal circumstances,

$r=r_e$    and    $F_a=0$

But this force will definitely play a part in ionization.

$2a_e$ is twice the radius of an electron, and $n_e$ the number of electrons in orbit minus one.  We are assuming that the orbiting electrons are stacked closest possible in parallel orbits.

Wanted: Attractive Force. Electrostatic Need Not Apply

There is no other way.  The electron is attracted to the nucleus, it is accelerated to light speed and begins curl in the direction perpendicular to its forward motion.  It transcribe a helical path.  It is the drag force that provides the centripetal force that rotates the electron in the direction perpendicular to its direction of travel.  The magnetic repulsive force in the vicinity of the nucleus and thermal gravity pushes the electron way, and the new centripetal force is given by,

$\cfrac { m_{ e }c^{ 2 } }{ r_{ e } } =Ac^{ 2 }+G\cfrac { m_{ a }m_{ e } }{ r^{ 2 }_{ e } } -\cfrac { Zq^{ 2 } }{ 4\varepsilon _{ o }(r_{ e }-r_{ p })^{ 2 } } -\cfrac { T_{ n }T_{ e } }{ 4\pi \tau_{ o }r^{ 2 }_{ e } }$

where $Z$ is the atomic number, $m_a$ is the atomic mass, and $r_p$ is the proton spin radius.

The equation does not have the electrostatic term,

$+\cfrac { Zq^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ e } }$

as the charges are in constant motion.

The gravitational term,

$G_Z=G\cfrac { m_{ a }m_{ e } }{ r^{ 2 }_{ e } }$

plays a much more important role.

$Ac^{ 2 }r^{ 2 }_{ e }-m_{ e }c^{ 2 }r_{ e }+\left(Gm_{ a }m_{ e }-\cfrac { Zq^{ 2 } }{ 4\varepsilon _{ o } } \cfrac { r^{ 2 }_e }{ (r_{ e }-r_{ p })^{ 2 } } -\cfrac { T_{ n }T_{ e } }{ 4\pi \varepsilon _{ o } } \right)=0$ --- (*)

we see that the equation for $r_e$ is still essentially a quadratic equation.  For heavy nucleus elements, it is expected that the gravitation term is high enough to give a positive y-intercept to equation (*).  A positive y-intercept will allow the parabola to intercept the x-axis twice and give two positive solutions to $r_e$.  The lower solution corresponds to the an orbit in the valence band and the higher solution is in the conduction band.  An electron can make the transition from the valence band to the conduction band, both valid $r_e$ solutions to the force equation (*).  This behavior is expected of heavier elements but not all elements.

Increasing temperature can bring the y-intercept down, and eventually a possible zero valued solution to $r_e$.  This temperature would correspond to $T_c$, the temperature for the formation of plasma.

Decreasing temperature or increasing the nucleus mass, ie the $G_Z$ term, moves the curve upwards and moves the two roots closer until a double root is created.  At the double root, the kink point, the root and the minima all coincide.  The element is very conductive, very shinny as the electron readily crosses kink point and emit a packet of energy.

But it is commonly believed that $G_Z$ is small and does not play a part in this situation.  Unless there is another attractive force in the equation, I believe common opinion is wrong, gravity plays a much more important role and cannot be ignored.  Electrostatic attraction between the charges cannot account for a positive y-intercept even if it is admitted because it is already smaller than the magnetic repulsive force by a factor of $\pi$.

Let's look for another attractive force that hold the electron to its orbit of $r_e$.

Why Half The Intensity, Albert?

If the electron and the nucleus are held apart by a repulsive force then the work done as a photon passes is,

${E}_{s}={m_pc^2}\left\{ln{( \cfrac{{r_{po} }}{{r}_{pf}}) +\cfrac { 1-{ r }_{ pf } }{ { r }_{ pf }{ r_{ po } } }(r_{pf}-r_{po})}\right\}=-\cfrac{q^2}{4\varepsilon_o}\left\{\cfrac{1}{r_{ef}}-\cfrac{1}{r_{eo}}\right\}=\Delta PE_{re}$

where  $r_{pf}$ and $r_{pf}$ are the initial and final radius of the photon in its helical path, $r_{eo}$ and $r_{ef}$ are the initial and final radius of the electron orbit around the nucleus.  The expression on the left was derived in the post "Light Dispersion?" from considering the work done along the centripetal force in moving the photon to a different orbital radius.  The expression on the right is the potential energy change in a field of magnetic repulsive force $F_L=-i\pi qE$, also the Lorentz's force between the electron and the nucleus.  The negative sign indicates that a gain in energy in one equals the lost of energy in the other.

Up till now, the only postulation as to why a photon and an orbiting electron repel each other is thermal gravity between two hot bodies. If we examine the likely magnetic interaction between an orbiting electron and a passing photon,

We find that when the electron orbits are also in the same anti-clockwise sense looking onward to the direction of the approaching photon, the particles should be attracted to each other based on the their interacting magnetic fields alone.  The electrons are pulled further from their orbits and the photons pulled down to a smaller radii.  The electron on reaching high orbit is not going to to be ionized.  The photon is still above the electron, the charge cannot escape the hold of the atom.  Given the $\cfrac{1}{r_e}$ relationship for $PE_{re}$, the energy change from moving a distance $\Delta r_e$ is always higher downwards than upwards.  (When the photon has a smaller radius, the electrons are pulled down to smaller orbital radii. This can cause ionization.)

The orbits can of course be re-oriented to an clockwise sense, in which case the particles are like opposite current in parallel wires and be repelled from one another.  This was the scenario proposed previously where the particles repel each other resulting in diffraction, ionization, glare, etc. (In the case of a smaller photon radial path the electrons are pushed upwards to higher orbit. This will not cause ionization.)

Note: From the above expression by setting $r_{ef}\rightarrow\infty$ for ionization and $r_{po}\lt r_{re}$, it is not possible to prove conclusively for all cases that the atom cannot be ionized by pushing the electron to higher orbit.

Since, both orientations are equally likely and all orientations in between, any photon at and beyond the threshold frequency (at the appropriate interaction radius $r$) on approach to an atom has only half the chance of ejecting a electron.  As a whole, only half the photons would have a photoelectric effect.  This conclusion is the same as we consider the relative size of the orbital radius and helical path radius.

If the interaction is purely electrostatic, we would expect the photon to eject the electron in both orientations and that all photons at the threshold frequency would cause a photoelectric effect.

We have assumed that the electron orbits re-orientate themselves slowly, not at the light speed that the photon is approaching.  We have also ignored the $B$ field due to the spinning positive charge which is always perpendicular to the $B$ field due to the negative charge as they both travel down the helical path.

It is possible that the photon-electron interaction is limited to photon radius larger than the electron orbit, ie $r_{ph}\gt r_e$.  The following graph shows why pushing the electron down is likely to cause ionization.

Given a $\cfrac{1}{r_e}$ profile for $PE_{re}$, an equal distance upwards and downwards, the electron pushed downwards gain more energy.

$\Delta PE_{ \Delta r- }>\Delta PE_{ \Delta r+  }$

 If this energy gain is greater than ionization energy, the electron will break the hold of the nucleus.

$\Delta PE_{ \Delta r- }>\Delta PE_{ ion }\implies ionization$

Pushing the electron out is not going to cause ionization unless $\Delta r_e\rightarrow\infty$.

Happy 500!

An electron in $B$ orbit around the positive nucleus experiences a repulsive force given by,

$\cfrac{qq}{4\pi\varepsilon_or^2_e}\rightarrow\pi \cfrac{qq}{4\pi\varepsilon_or^2_e}=\cfrac{qq}{4\varepsilon_or^2_e}$

Two positive charges in parallel $B$ orbit experience an attractive force given by,

$F_a=-\cfrac{qq}{4\varepsilon_or^2_p}$

Similarly, two negative charges in parallel $B$ orbit experience an attractive force given by,

$F_a=-\cfrac{qq}{4\varepsilon_or^2_e}$

The charges are in orbit because of the initial speed it acquired due to electrostatic attraction between opposite charge.  They are held in orbit and kept from collapsing by a magnetic repulsive force.

What happen to two like charges then?  Do they coalesce?  When they are stopped in their $B$ orbits, they will immediately repel apart. An instant disintegration!

This might be the rationale behind a particle collider.

Passing It On...

From the post "Catastrophe! No Small Change",

$U_B=\cfrac{\mu_o}{32}\left( \cfrac { e }{ \pi a_{ e } }  \right) ^{ 2 }\left( sin(2\phi )sin(\phi /2)\cfrac { d\phi  }{ dt }  \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }$

where,

$h(\phi ,\theta )=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }$

What interest us here is the term,

$\left(sin(2\phi )sin(\phi /2)\cfrac { d\phi  }{ dt }\right)^2$

where $\phi$ is revolves along the orbit of the charge.  Imagine a cylinder defined by this motion at a height of  $y$ perpendicular to the plane of the orbit.  The wall of such a cylinder cuts the $B$ field established by the moving charge perpendicularly, at the highest point of the loop center at the circumference perpendicular to the tangent of the orbit.

Considering the $B$ field alone, the force line does no work as the motion is in a perpendicular direction, and the charge in constant $\cfrac{d\,\phi}{d\,t}$ around the orbit experience no change in kinetic energy.

But,

$\cfrac{d\,U_B}{d\,\phi}=A\cfrac{d}{d\,\phi}\left\{\left( sin(2\phi )sin(\phi /2)\cfrac { d\phi  }{ dt }  \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }\right\}$

where $A$ is the preceding constant and  $\theta_o$ is up to where the cylinder cuts the $B$ field at $y$.

Consider

$\Omega=\cfrac{d}{d\,\phi}\left\{\left( sin(2\phi )sin(\phi /2)\cfrac { d\phi  }{ dt }  \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }\right\}$

Consider the first term,

$2sin(2\phi )sin(\phi /2)\cfrac { d\phi  }{ dt }\left \{2cos(2\phi)sin(\phi /2)\cfrac { d\phi  }{ dt }+\cfrac{1}{2}sin(2\phi )cos(\phi /2)\cfrac { d\phi  }{ dt }\right\}$

$=\left\{ sin(4\phi )sin^{ 2 }(\phi /2)+\cfrac { 1 }{ 2 } sin^{ 2 }(2\phi )sin(\phi ) \right\} \left( \cfrac { d\phi  }{ dt }  \right) ^{ 2 }$

Consider the second term,

$2\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \cfrac{d}{d\,\phi}\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) }$

$=2\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \cfrac{d}{d\,\phi}\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } {  (cos(\phi )sin^{ 3 }(\theta )d\phi ) }$

From the post "Catastrophe! No Small Change",

$cos(\phi )sin^{ 3 }(\theta )d\phi =dcos(\theta )$,    so,

$=2\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) }.{ \cfrac { 1 }{ cos^{ 4 }(\theta_o ) } h(\phi ,\theta_o ) } {cos(\phi )sin^{ 3 }(\theta_o ) }$

Therefore,

$\Omega=\left( sin(2\phi )sin(\phi /2)\cfrac { d\phi  }{ dt }  \right)\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right). \\\left[\cfrac { d\phi  }{ dt } \left \{2cos(2\phi)sin(\phi /2)+\cfrac{1}{2}sin(2\phi )cos(\phi /2)\right\}\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right)+{ \cfrac { 2 }{ cos^{ 4 }(\theta_o ) } h(\phi ,\theta_o ) } {cos(\phi )sin^{ 3 }(\theta_o ) }\left( sin(2\phi )sin(\phi /2)\cfrac { d\phi  }{ dt }  \right)\right]$

And we have the change in energy with time as,

$\cfrac{d\,U_B}{d\,t}=\cfrac{d\,U_B}{d\,\phi}\cfrac { d\phi  }{ dt }=A\Omega\cfrac { d\phi  }{ dt }$

Clearly  $\Omega\ne0$ for all $\phi$.  This implies that the charge while in orbit constantly absorb and radiate energy.  A plot of,

 $O=\left(sin(2\phi )sin(\phi /2)\right)^2$  and $sin(\phi/2)$ below clearly shows the periodicity in $U_B$ of  $2\pi$.

And that the net change in $U_B$ over one period is zero as the graph returns to the same point (zero) at the end of the period.

Together with the previous post "Diffusion, Only If...",  we have a non zero expression for $A(x,\dot{x})$,

$\cfrac { \partial\,P }{ \partial\,t }=-\cfrac{1}{2}A(x,\dot{x})=\cfrac{d\,U_B}{d\,t}$

This is the mechanism by which energy is propagated throughout a material.  The orbiting charges are constantly radiating and absorbing energy, with no net change over a period of the orbiting motion.

Disclaimer: Amateur At Work

I need to check my Maths...Hold your horses!  Still checking pressure lamp.

Diffusion, Only If...

Given,

$L= kinetic\, energy-potential\, energy = T-V$

Consider,

$\cfrac { \partial \, L }{ \partial \, \dot { x }  } =\cfrac { \partial \, L }{ \partial \, \, x } \cfrac { \partial \, x }{ \partial \, \dot { x }  } +\cfrac { \partial \, L }{ \partial \, t } \cfrac { \partial \, t }{ \partial \, \dot { x }  }$

as $L=f(x,\dot{x},t)$

Take derivative wrt $t$,

$\cfrac { \partial  }{ \partial \, t } \{ \cfrac { \partial \, L }{ \partial \, \dot { x }  } \} =\cfrac { \partial  }{ \partial \, t } \{ \cfrac { \partial \, L }{ \partial \, x }  \cfrac { \partial \, x }{ \partial \, \dot { x }  }  +\cfrac { \partial \, L }{ \partial \, t } \cfrac { \partial \, t }{ \partial \, \dot { x }  } \}$

Consider the second term,

$\cfrac { \partial  }{ \partial \, t } \{ \cfrac { \partial \, L }{ \partial \, t } \cfrac { \partial \, t }{ \partial \, \dot { x }  } \} =\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } \cfrac { \partial \, t }{ \partial \, \dot { x }  } -\cfrac { \partial \, L }{ \partial \, t } \cfrac { 1 }{ \{ \cfrac { \partial \dot { x } \,  }{ \partial \, t } \} ^{ 2 } } \cfrac { \partial ^{ 2 }\dot { x } \,  }{ \partial \, t^{ 2 } }$

Consider the first term,

$\cfrac { \partial \,  }{ \partial \, t } \left\{ \cfrac { \partial \, L }{ \partial \, \, x } \cfrac { \partial \, x }{ \partial \, \dot { x }  }  \right\} =\cfrac { \partial \,  }{ \partial \, t } \left\{ \cfrac { \partial \, L }{ \partial \, \, x }  \right\} \cfrac { \partial \, x }{ \partial \, \dot { x }  } +\cfrac { \partial \, L }{ \partial \, \, x } \cfrac { \partial \,  }{ \partial \, t } \left\{ \cfrac { \partial \, x }{ \partial \, \dot { x }  }  \right\}$

$=\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial \, \, x } \cfrac { \partial \, x }{ \partial \, \dot { x }  } +\cfrac { \partial \, L }{ \partial \, \, x } \cfrac { \partial  }{ \partial \, \dot { x }  } \left\{ \cfrac { \partial \, x }{ \partial \, t }  \right\}$

$=\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial \, \, x } \cfrac { \partial \, x }{ \partial \, \dot { x }  } +\cfrac { \partial \, L }{ \partial \, \, x }$

For the Euler-Lagrange equation to be true,

$\cfrac { \partial  }{ \partial \, t } \{ \cfrac { \partial \, L }{ \partial \, \dot { x }  } \} =\cfrac { \partial \, L }{ \partial \, x }$

to be true, we have,

$\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial \, \, x } \cfrac { \partial \, x }{ \partial \, \dot { x }  } +\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } \cfrac { \partial \, t }{ \partial \, \dot { x }  } -\cfrac { \partial \, L }{ \partial \, t } \cfrac { 1 }{ \{ \cfrac { \partial \dot { x } \,  }{ \partial \, t } \} ^{ 2 } } \cfrac { \partial ^{ 2 }\dot { x } \,  }{ \partial \, t^{ 2 } }   =0$

But,

$\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial \, \, x } =\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial \dot { x }  } \cfrac { \partial \, \dot { x }  }{ \partial \, x } +\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial t } \cfrac { \partial \, { t } }{ \partial \, \dot { x }  } \cfrac { \partial \, \dot { x }  }{ \partial \, x }$

We have,

$\cfrac { \partial ^{ 2 }\, L }{ \partial \, t\partial \dot { x }  } +2\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } \cfrac { \partial \, t }{ \partial \, \dot { x }  } -\cfrac { \partial \, L }{ \partial \, t } \cfrac { 1 }{ \{ \cfrac { \partial \dot { x } \,  }{ \partial \, t } \} ^{ 2 } } \cfrac { \partial ^{ 2 }\dot { x } \,  }{ \partial \, t^{ 2 } }  =0$

$\cfrac { \partial \, L }{ \partial \, x } +2\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } \cfrac { \partial \, t }{ \partial \, \dot { x }  } -\cfrac { \partial \, L }{ \partial \, t } \cfrac { 1 }{ \{ \cfrac { \partial \dot { x } \,  }{ \partial \, t } \} ^{ 2 } } \cfrac { \partial ^{ 2 }\dot { x } \,  }{ \partial \, t^{ 2 } } =0$

$\cfrac { \partial \, L }{ \partial \, x } =\cfrac { 1 }{ \ddot { x }  } \left\{ \cfrac { \dddot { x }  }{ \ddot { x }  } \cfrac { \partial \, L }{ \partial \, t } -2\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } }  \right\}$

If  $\dddot { x }=0$, ie the forces in the fields do not change with time,

$\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } =-\cfrac{\ddot { x } }{2}\cfrac { \partial \, L }{ \partial \, x }$

That going through a $L$ gradient of $\cfrac { \partial \, L }{ \partial \, x }$ with acceleration $\ddot{x}$, the second rate of change in $L$ experienced by the body is given by the expression above.

Obviously when $\sum F=0$,  $\ddot{x}=0$  and  $\cfrac { \partial ^{ 2 }\, L }{ \partial \, t^{ 2 } } =0$ this implies,

$\cfrac { \partial \, L }{ \partial \, t } = A(x,\dot{x})$

$A$ is a function in $x$ and $\dot{x}$ not necessarily zero.  Consider the total energy of the system,

$E=L+2V$

$\cfrac { \partial\,E }{ dt } =\cfrac { \partial\,L }{ \partial\,t } +2\cfrac { \partial\,P }{\partial\,t }$

when $\ddot{x}=0$,  $\dot{x}$ is constant, $T$ is a constant,

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

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

the change in $L$ is solely due to the change in $P$

$\cfrac { \partial\,P }{ \partial\,t }=-\cfrac{1}{2}\cfrac { \partial\,L }{ \partial\,t }=-\cfrac{1}{2}A(x,\dot{x})=\cfrac{1}{4}\int{\ddot { x } \cfrac { \partial \, L }{ \partial \, x } }dt$ ---(**)

An expression in $\ddot{x}$ is different from the value of $\ddot{x}$.

$\cfrac { d^2P }{ dt^2 }=\cfrac{1}{4}\ddot { x } \cfrac { \partial \, L }{ \partial \, x }=0$

as we started with $\ddot{x}=0$

This suggests that when a body is in circular motion (for example electrons, protons in orbit), $\ddot{x}$ in the direction of $\dot{x}$ is zero, the centripetal force does no work and $\dot{x}$ is a constant, there can still be a change in potential energy with time given by expression (**).  When

$\cfrac { \partial\,P }{ \partial\,t }\gt0$    energy is absorbed,

$\cfrac { \partial\,P }{ \partial\,t }\lt0$     energy is radiated

Is this the explanation why energy tend to diffuse throughout a material?  Because the orbiting particles are constantly absorbing and radiating energy?  Only if,

$A(x,\dot{x})\ne0$.

In general this is possible when the system is periodic as in circular motion where over one period, the net gain in energy is zero.  Which brings us to $\phi$ from the post "Catastrophe! No Small Change"...until next time...

Taking My Time

Given any $x$,

$\cfrac { \partial x\,  }{ \partial  x } =1$

$\cfrac { \partial ^{ 2 }\, x }{ \partial \, x^{ 2 } } =0=\cfrac { \partial ^{ 2 }\, x }{ \partial \, x\partial t } \cfrac { \partial t\,  }{ \partial \, x } =\cfrac { \partial ^{ 2 }\, x }{ \partial t\partial t } \cfrac { \partial t\,  }{ \partial \, x } \cfrac { \partial t\,  }{ \partial \, x } =\cfrac { \partial ^{ 2 }\, x }{ \partial t^{ 2 } } \cfrac { \partial t\,  }{ \partial \, x } \cfrac { \partial t\,  }{ \partial \, x }$

So for all $x$,

$\cfrac { \partial ^{ 2 }\, x }{ \partial t^{ 2 } } \cfrac { \partial t\,  }{ \partial \, x } \cfrac { \partial t\,  }{ \partial \, x } =0$

ie,

$\cfrac{a}{v^2}=0$

In fact,

$\cfrac { \partial ^{ 2 }x\quad  }{ \partial \, x^{ 2 } } =0$

$=\cfrac { \partial \, \quad  }{ \partial \, x } \{ \cfrac { \partial x\,  }{ \partial \, x } \}$

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

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

$=\cfrac { \partial \, \quad  }{ \partial \, t } \{ \cfrac { \partial x\,  }{ \partial \, t } \} \cfrac { \partial t\,  }{ \partial \, x } +\cfrac { \partial \, \quad  }{ \partial \, t } \{ \cfrac { \partial t\,  }{ \partial \, x } \} \cfrac { \partial x\,  }{ \partial \, t }$

$= \cfrac { \partial ^{ 2 }x\, \quad  }{ \partial \, t^{ 2 } } \cfrac { \partial t\,  }{ \partial \, x } -\cfrac { 1 }{ \{ \cfrac { \partial x\,  }{ \partial \, t } \} ^{ 2 } } \cfrac { \partial \, \quad  }{ \partial \, t } \{ \cfrac { \partial x\,  }{ \partial \, t } \} \cfrac { \partial x\,  }{ \partial \, t }$

$=\cfrac { \partial ^{ 2 }x\, \quad  }{ \partial \, t^{ 2 } } \cfrac { \partial t\,  }{ \partial \, x } -\cfrac { 1 }{ \{ \cfrac { \partial x\,  }{ \partial \, t } \}  } \cfrac { \partial ^{ 2 }x\, \quad  }{ \partial \, t^{ 2 } }$

$=\cfrac { \partial ^{ 2 }x\, \quad  }{ \partial \, t^{ 2 } } \cfrac { \partial t\,  }{ \partial \, x } -\cfrac { \partial ^{ 2 }x\, \quad  }{ \partial \, t^{ 2 } } \cfrac { \partial t }{ \partial x }=0$

non sequitur.

Orbit Small

From the post "Electron Orbit B Field II",

$E=E_oe^{-\cfrac{x}{a_e}}$

$x$ is measured from the surface of the charge, where $a_e$ is the radius of the charge. So

$E_oe^{-\cfrac{0}{a_e}}=E_o$

$E_o$ is on the surface of the charge.

It is then possible that  $a_e\gt r_e$, as the measurement of $x$ does not include $a_e$.

Smooth Flow

From the post "Catastrophe No Small Change",

The graphform,

$en(\theta)=\cfrac{sin^3(\theta)}{cos^2(\theta)}e^{-\cfrac{4r_e}{a_ecos(\theta)}}$

defines the envelope of the radiated energy.

$\cfrac{d\,en(\theta)}{d\,\theta}=tan(x){ e }^{ -\cfrac { 4r_{ e }sec(x) }{ a_{ e } }  }\left( sin(x)+2tan(x)sec(x)-\cfrac { 4r_{ e } }{ { a }_{ e } } tan^{ 3 }(x) \right)$

A plot of this graph whose zeros are the maxima of the envelop, with $\cfrac{r_e}{a_e}$ varying from 1 to 2, is shown below.  There are three zeros, two points of inflection at 0 and π/2 and one maximum where the gradient of the derivative is negative.

For the maximum points,

$sin(x)+2tan(x)sec(x)-\cfrac { 4r_{ e } }{ { a }_{ e } } tan^{ 3 }(x)=0$

For the points of inflections,

$sin(x)=0$

Do not steal my money. No discontinuity.  We are dealing with radiated energy from varying magnetic field.  Previously when we encountered a discontinuity in the gradient of  $r_e$ vs $T$ we were dealing with electrostatic, centripetal force and thermal gravity, even gravity itself.

May be there is a similar bandgap, a magnetic bandgap that is not visible here.

No Need For Packets, All Eat Here

From the previous post "Catastrophe! No Small Change",

$\bar { U_{ B } } =\cfrac { \mu _{ o }\omega ^{ 2 }e^{ 2 }a_e }{ 8\pi ^{ 3 }r^{ 3 }_{ e } }  \cfrac { sin^{ 3 }(\theta _{ o }) }{ cos^{ 2 }(\theta _{ o }) } e^{ -\cfrac { 4r_{ e } }{ a_{ e }cos(\theta _{ 0 }) }  }$

The exponential term is very sensitive to the relative values of  $\cfrac{r_e}{a_e}$  and by reciprocity, $\cfrac{r_p}{a_p}$.  A illustrative plot of $U_B$ using varies ratio of $\cfrac{r_e}{a_e}$ is given below,  The actual graph plotted is (sin(x))^3*e^{-4*a/cos(x)}*1/(cos(x))^2 where  1<a<2, in steps of 0.05.

It seems that the charge's orbit is only slightly larger than its size in order to give the radiation profile obtained experimentally.

Catastrophe! No Small Change

If from the post "Electron Orbit B Field II",

$\oint{B}d\,r=2\pi rB_o=\mu _{ o }\varepsilon _{ o }\cfrac { \partial \, E_{ A } }{ \partial t }$

$2\pi rB_o=\mu _{ o }\varepsilon _{ o }2\pi E_{ o }cos(\phi /2)\cfrac { d\, x_{ \bot  } }{ d\, t } \cfrac { x^{ 2 }_{ \bot  } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }$

If we retain $r$ and label it $r_p$, the orbital radius of the nucleus then,

$r_pB_o=\mu _{ o }\varepsilon _{ o }E_{ o }cos(\phi /2)\cfrac { d\, x_{ \bot  } }{ d\, t } \cfrac { x^{ 2 }_{ \bot  } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }$

$r_pB_o=\mu _{ o }\varepsilon _{ o }(\cfrac{e}{4\pi\varepsilon_or_e^2})cos(\phi /2)\cfrac { d\, x_{ \bot  } }{ d\, t } \cfrac { x^{ 2 }_{ \bot  } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }$

Since,

$\cfrac { d\, x_{ \bot  } }{ d\, t } x^{ 2 }_{ \bot  } =r_{ e }cos(\phi )\cfrac { d\phi  }{ dt }(2r_{ e }sin(\phi /2))^2$

$\cfrac { d\, x_{ \bot  } }{ d\, t } x^{ 2 }_{ \bot  } =4r^{ 3 }_{ e }cos(\phi )\cfrac { d\phi  }{ dt } sin^{ 2 }(\phi /2)$

And so,

$r_{ p }B_{ o }=\\\mu _{ o }\varepsilon _{ o }(\cfrac { e }{ 4\pi \varepsilon _{ o }a_{ e }r_{ e }^{ 2 } } )cos(\phi /2)(4r^{ 3 }_{ e }cos(\phi )\cfrac { d\phi  }{ dt } sin^{ 2 }(\phi /2))\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) }$

$B_{ o }=\cfrac { \mu _{ o }e }{ 4\pi a_{ e } } \cfrac { r_{ e } }{ r_{ p } } sin(2\phi )sin(\phi /2)\cfrac { d\phi  }{ dt } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) }$

$B_o$ is generated by the electron in circular motion of radius $r_e$, so $B_o$ is proportional to $r_e$.  Such a field decreases with distance from the plane containing the orbit of  the electron, so $B_o$ is inversely proportional to $r_p$.  Consider the energy contained in such a field.

$U_B=\cfrac{1}{2}\cfrac{B^2_o}{\mu_o}$

When the nucleus has a positive charge or a net positive charge of one by reciprocity,

$r_e=r_p$  and so,  $\left(\cfrac{r_e}{r_p}\right)^2=1$

$U_B=\cfrac{1}{2\mu_o}\left( \cfrac { \mu _{ o }e }{ 4\pi a_{ e } }  \right) ^{ 2 }\left( sin(2\phi )sin(\phi /2)\cfrac { d\phi  }{ dt }  \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }$

$U_B=\cfrac{\mu_o}{32}\left( \cfrac { e }{ \pi a_{ e } }  \right) ^{ 2 }\left( sin(2\phi )sin(\phi /2)\cfrac { d\phi  }{ dt }  \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }$

When such a system is subjected to heat, both $r_e$ and $r_p$ will oscillate and so does the magnetic field.  The value $\theta_o$ will also varies as $r_p$ changes.  When $B_o$ collapses from a higher value to a lower value, energy is radiated.  This radiated energy varies with the value of $\theta_o$, given by the expression above.

The average energy over one period, (power) is given by,

$\bar{U_B}=\cfrac{1}{T}\int^{T}_{0}{U_B}dt=\cfrac{1}{T}.\int^{T}_{0}{\cfrac{\mu_o}{32}\left( \cfrac { e }{ \pi a_{ e } }  \right) ^{ 2 }\left( sin(2\phi )sin(\phi /2)\cfrac { d\phi  }{ dt }  \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }}dt$

$\bar{U_B}=\cfrac{1}{T}\cfrac{\mu_o}{32}\left( \cfrac { e }{ \pi a_{ e } }  \right) ^{ 2 }.\int^{\phi_T}_{0}{\left( sin(2\phi )sin(\phi /2)  \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }}\cfrac { d\phi  }{ dt } d\phi$

assuming that $\cfrac { d\phi  }{ dt }=\omega$ is a constant.

$\bar{U_B}=\cfrac{\omega}{T}\cfrac{\mu_o}{32}\left( \cfrac { e }{ \pi a_{ e } }  \right) ^{ 2 }\int^{\phi_T}_{0}{\left( sin(2\phi )sin(\phi /2)\right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 }}d\phi$

This is the integral of two functions involving  $\phi$ since,

$h(\phi,\theta)=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }$

Consider,

$\Psi=\int { \left( sin(2\phi )sin(\phi /2) \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 } } d\phi$

$\Psi=\int{ \left( 4sin(\phi /2)cos(\phi /2)cos(\phi )sin(\phi /2) \right) ^{ 2 }\left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 } } d\phi$

$\Psi=-\int { \left( 4sin^2(\phi /2)cos(\phi ) \right) ^{ 2 }\left(\cfrac { a_{ e } }{ r_{ e } }\right)^2.\\ \left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 3 }(\theta ) } \left( -h(\phi ,\theta )\cfrac { r_{ e } cos(\phi /2)}{ a_{ e }cos(\theta ) }  \right)  } { d\, (cos(\theta )) } \right) ^{ 2 } } d\phi$

Becasue,

$h(\phi ,\theta )=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }$

$\cfrac { dh(\phi ,\theta ) }{ d\phi  } =e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }\left(-\cfrac { 2r_{ e }}{ a_{ e }cos(\theta ) }\right) cos(\phi /2)\cfrac { 1 }{ 2 }$

$\cfrac { dh(\phi ,\theta ) }{ d\phi  }=-h(\phi ,\theta )\cfrac { r_{ e }cos(\phi /2) }{ a_{ e }cos(\theta ) }$

We have,

$\Psi=-\int { \left( 4sin^2(\phi /2)cos(\phi ) \right) ^{ 2 }\left(\cfrac { a_{ e } }{ r_{ e } }\right)^2 \left( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 3}(\theta ) } \left( \cfrac { dh(\phi ,\theta ) }{ d\phi  }  \right)  } { d\, (cos(\theta )) } \right) ^{ 2 } } d\phi$

$\Psi=-\int { 16sin^4(\phi /2)cos^2(\phi )\left(\cfrac { a_{ e } }{ r_{ e } }\right)^2\left( \cfrac { d }{ d\phi  } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 3}(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) } \right) ^{ 2 } } d\phi$

From,

$\cfrac { r_{ p } }{ x_{ \bot  } } =tan(\theta )$

$x_{ \bot  }=2r_{ e }sin(\phi /2)cos(\phi /2)=r_{ e }sin(\phi )$

$r_{ p }=r_{ e }sin(\phi )tan(\theta ),\quad but\quad r_{ p }=r_{ e }$

$cot(\theta )=sin(\phi )$

$-sin(\theta )d\theta =cos(\phi )sin^{ 3 }(\theta )d\phi =dcos(\theta )$ --- (*)

Despite of this relationship, $\phi$  and  $\theta$ are still independent variable.  The integration over $\theta$ is taken for each value of  $\phi$ as $\phi$ revolves around the orbit defined by $r_e$.

Substitute (*) into $\Psi$,

$\Psi=-16\left(\cfrac { a_{ e } }{ r_{ e } }\right)^2 \int { sin^4(\phi /2)cos^2(\phi )\left( \cfrac { d }{ d\phi  } \int _{ 0 }^{ \theta_o}{ \cfrac { 1 }{ cos^{ 3 }(\theta ) }  } h(\phi ,\theta ) { cos(\phi )sin^{ 3 }(\theta )d\phi  } \right) ^{ 2 } } d\phi$

$\Psi=-16\left(\cfrac { a_{ e } }{ r_{ e } }\right)^2\int { sin^4(\phi /2)cos^2(\phi )\left( \cfrac { 1 }{ cos^{ 2 }(\theta ) } h(\phi ,\theta ){ sin^{ 3 }(\theta ) } \right) ^{ 2 } } d\phi$

$\Psi=-16\left(\cfrac { a_{ e } }{ r_{ e } }\right)^2\cfrac { sin^{ 6 }(\theta ) }{ cos^{ 4 }(\theta ) } \int {sin^4(\phi /2)cos^2(\phi )\left( h(\phi ,\theta ) \right) ^{ 2 } } d\phi$

$\Psi= 4\cfrac { a^3_{ e } }{ r^3_{ e } } \cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) }. \int { sin^{ 3 }(\phi /2)cos(\phi )\left( h(\phi ,\theta ) \right) ^{ 2 } } \left( -\cfrac { 4r_{ e }sin(\phi /2) }{ a_{ e } } \cfrac { 1 }{ cos^{ 2 }(\theta ) }  \right) cos(\phi )sin^{ 3 }(\theta )d\phi$

Substitute $*$ into $\Psi$ again,

$\Psi= 4\cfrac { a^3_{ e } }{ r^3_{ e } } \cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) } \int { sin^{ 3 }(\phi /2)cos(\phi )\left( h(\phi ,\theta ) \right) ^{ 2 }\left( -\cfrac { 4r_{ e }sin(\phi /2) }{ a_{ e } } \cfrac { 1 }{ cos^{ 2 }(\theta ) }  \right)  } dcos(\theta )$

$\Psi= -4\cfrac { a^3_{ e } }{ r^3_{ e } } \cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) } \int{ sin^{ 3 }(\phi /2)cos(\phi )\left( h(\phi ,\theta ) \right) ^{ 2 }\left( -\cfrac { 4r_{ e }sin(\phi /2) }{ a_{ e } } \cfrac { sin(\theta ) }{ cos^{ 2 }(\theta ) }  \right)  } d\theta$

Because,

$h(\phi ,\theta )^2=e^{ -\cfrac { 4r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }$

$\cfrac { d\left( h(\phi ,\theta ) \right) ^{ 2 } }{ d\theta  } =h(\phi ,\theta )^2\left(-\cfrac { 4r_{ e }sin(\phi /2) }{ a_{ e }}\cfrac{sin(\theta) }{cos^2(\theta ) }\right)$

And so,

$\Psi=- 4\cfrac { a^3_{ e } }{ r^3_{ e } } \cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) } \int { sin^{ 3 }(\phi /2)cos(\phi )\cfrac { d\left( h(\phi ,\theta ) \right) ^{ 2 } }{ d\theta  }  } d\theta$

$\Psi=-4 \cfrac { a^3_{ e } }{ r^3_{ e } } \cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) } \cfrac { d }{ d\theta  } \int { sin^{ 3 }(\phi /2)cos(\phi ) } \left( h(\phi ,\theta ) \right) ^{ 2 }d\theta$

$\Psi= -4\cfrac { a^3_{ e } }{ r^3_{ e } } \cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) } sin^{ 3 }(\phi /2)cos(\phi )\left( h(\phi ,\theta ) \right) ^{ 2 }$

And the average energy over one period (power) in the $B$ field is,

$\bar{U_B}=-\cfrac{\omega}{T}\cfrac{\mu_o}{8}\left( \cfrac { e }{ \pi a_{ e } }  \right) ^{ 2 }\cfrac { a^3_{ e } }{ r^3_{ e } }\left\{\cfrac { sin^{ 3 }(\theta ) }{ cos^{ 2 }(\theta ) }\right\}^{\theta_o}_{0}2 \left\{sin^{ 3 }(\phi /2)cos(\phi )\left( h(\phi ,\theta_o ) \right) ^{ 2 }\right\}^{\pi}_{0}$

$\bar { U_{ B } } =-\cfrac { \omega ^{ 2 } }{ 2\pi  } \cfrac { \mu _{ o } }{ 8 } \left( \cfrac { e }{ \pi  }  \right) ^{ 2 }\cfrac { a_{ e } }{ r^{ 3}_{ e } } \cfrac { sin^{ 3 }(\theta _{ o }) }{ cos^{ 2 }(\theta _{ o }) } 2\left\{ -1.\left( h(\pi ,\theta _{ o }) \right) ^{ 2 } \right\}$

$\bar { U_{ B } } =\cfrac { \mu _{ o }\omega ^{ 2 }e^{ 2 }a_{ e } }{ 8\pi ^{ 3 }r^{ 3 }_{ e } }  \cfrac { sin^{ 3 }(\theta _{ o }) }{ cos^{ 2 }(\theta _{ o }) } \left( h(\pi ,\theta _{ o }) \right) ^{ 2 }$

$\bar { U_{ B } } =\cfrac { \mu _{ o }\omega ^{ 2 }e^{ 2 } a_{ e }}{ 8\pi ^{ 3 }r^{ 3 }_{ e } }  \cfrac { sin^{ 3 }(\theta _{ o }) }{ cos^{ 2 }(\theta _{ o }) } e^{ -\cfrac { 4r_{ e } }{ a_{ e }cos(\theta _{ 0 }) }  }$

A plot of (sin(x))^3*e^{-1.1/cos(x)}*1/(cos(x))^2  for  0<x<π/2 is given below,

There is no catastrophe.  When subjected to high perturbations,( high temperature etc), both orbits oscillate.  High $\theta_o$ corresponds to high orbit.  When the charge return from high values of $\theta_o$, higher amount of energy are released as the potential energy difference between those at high orbits and that at the centers of the oscillation are higher.  So, high $\theta_o$ corresponds to higher radiated energy.

If this interpretation is correct then there is no need for discrete amount of energy either.

Capillary Action And Camera!

Here's my take on capillary action,

The logic is naive; capillary action act against gravity, thermal gravity is anti gravity.  So, we set off to investigate temperature and capillary action.  The metal should be of high heat capacity such that the portion surrounding the insulating fluid does not cool to the temperature of the insulating fluid.

An insulating fluid is used to maintain the the temperature gradient between the metal and the fluid as the measurement progresses.

Increase $T_m$ and plot Temperature of Metal vs Height of Fluid; $T_m$ (K) vs $h$ (m) with  $d$, diameter of metal tube bore and $T_f$ temperature of fluid as parameters.  Viscosity plays a part as fluid in thermal contact with the metal wall experiences a net upward force due to thermal gravity and drags the fluid in the center of the bore upwards.

Maybe you can make a droplet of fluid flow upwards with a temperature gradient decreasing upwards on a hot surface with low drag.

Have a nice day.

And Mochi Sticks Too

All the preceding posts just to answer why the positive charges at the nucleus stick together.

They stick, because they are revolving in parallel $B$ orbits. The attractive force between two orbiting charges in parallel orbits is,

$F_L=\pi q|v|B=-i\pi qE$

where $q$ is the charge in one orbit, $v$ the velocity of the charge, $B$ the magnetic field established by the other orbiting charge and $E$ the electrostatic field at distance $r$ where $B$ is, due to the other moving charge, and $r$ is also the distance between the two orbits.  $-i$ rotates the force $F_L$ in the plane perpendicular to $B$ in the anticlockwise direction.

$F_L$ is the result of two interacting $B$ fields.  The maximum interaction occurs when the $B$ field lines are parallel.   The velocity $v$ of the charge placed in the established $B$ field was made perpendicular to the already established $B$ field lines such that the $B$ field set up by this moving charge is parallel to the first.

In the derivation of $F_L$ in the post "Lorentz's without q", a moving negative charge was used that established a $E$ field downwards.  A moving positive charge was then added to the situation to obtain $F_L$.

If $v$ in rotated, $v\rightarrow -iv$  to the same direction as the moving negative charge then $F_L$ is along  $-E$ as the expression shows.  This is consistent with the observed fact that parallel wires with opposite current repulse each other.

There is no relative velocity concerns here, as long as one charge is moving, it establishes a $B$ field that will interact with another moving charge.

Have a nice day.

Stable But Not Too Stable

Continuing from the post "Tie Me Not, Not Too Tight",

$\cfrac { \partial ^{ 2 }\, B_{ o } }{ \partial \, r_{ e }^{ 2 } } =2\cfrac { \mu _{ o }\varepsilon _{ o }\omega E_{ o } }{ a_{ e } } f(\phi )cot(\theta _{ o })\left\{ Ar^{ 2 }_{ e }-Br_{ e }+C \right\}$

where,

$C=\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }$

$B=\cfrac { 4sin(\phi /2) }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }$

$A=2\left( \cfrac { sin(\phi /2) }{ a_{ e } }  \right) ^{ 2 }\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }$

$A\gt0$  and  $C\gt0$  for  $0\lt\theta\lt\pi/2$ but $B$ depends on  $\phi/2$.

$B^{ 2 }-4AC=16\left( \cfrac { sin(\phi /2) }{ a_{ e } }  \right) ^{ 2 }\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }\\\left[ \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }-\cfrac { 1 }{ 2 } cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) } \right]$

From the graph of $Ar^{ 2 }_{ e }-Br_{ e }+C$ which determines the sign of the second derivative,

The second derivative is negative over a range of $r_e$, beyond this range it is positive.  Under normal circumstances at,

$\cfrac { \partial \, B_{ o } }{ \partial \, r_{ e } } =0$

a small increase in $r_e$ result in a negative change in $\cfrac { \partial \, B_{ o } }{ \partial \, r_{ e } }$ as

$\cfrac { \partial ^{ 2 }\, B_{ o } }{ \partial \, r_{ e }^{ 2 } }\lt0$.

Since we started with zero, a negative change means that the derivative itself is negative,

$\cfrac { \partial \, B_{ o } }{ \partial \, r_{ e } }\lt0$

and the system is stable.

However, if there is energy input that move $r_e$ beyond this range, where the second derivative becomes positive, an increase in $r_e$ will result in a positive $\cfrac { \partial \, B_{ o } }{ \partial \, r_{ e } }$.  $B_o$ increases and the repulsion between the spinning nucleus and the revolving electron increases causing $r_e$ to increase further.  The electron breaks away and ionization occurs.

Similarly, if $r_e$ is moved below the stable region, an decrease in $r_e$ results in a positive $\cfrac { \partial \, B_{ o } }{ \partial \, r_{ e } }$ that decreases $B$ correspondingly.  A reduction in repulsion between the spinning nucleus and the revolving electron decreases $r_e$ further and the electron collapses into the nucleus.  We have plasma.

So, $r_e$ has a range of stable values beyond which the particle breaks away at the increasing end or collapses at the decreasing end.

Energy input is required to move $r_e$.  This can be achieved by decreasing drag through increasing temperature, by pushing the orbiting electron inwards/outwards using photons at resonance, decreasing/increasing the $B$ field between the nucleus and electron by applying an opposing/parallel $B$ field, etc.

Orbit In Blue

The positive nucleus in orbit around the $B$ orbit generates a similar B field that interacts with the electron in a reciprocal way.  The electron begins to spin because any transverse motion from its orbit around the nucleus will cut this B field and result in a Lorentz's force in the transverse plane (a plane in the direction perpendicular to the blue field lines and the relative motion).  This force acts perpendicular to the transverse velocity component and causes the electron path to curve.  Given a strong B field, the electron spins in the transverse plane.

Which give definition to the confining sphere around the electron.  The electron has two degrees of freedom on the sphere, one due to its orbit around the nucleus and the other its motion on the reciprocal orbit, $B_r$.  The confining sphere may not be a sphere at all, but a confining volume.

Cart First Because The Cow Is Driving

If we were to make the nucleus stationary,

then the $B$ orbit will be seen as revolving in and out of the confining spheres the surface of which demarcate the presence of the nucleus and the electron.

Believe It Or Not

Here is another illustration for  $B$ orbit.

Movement is more relative motion between electron and the nucleus.  Although, the argument for a stationary nucleus because of heavier mass (inertia) is not valid because the forces involved are very high.

Tie Me Not, Not Too Tight

For $\theta_o$  such that $B_o$  is at the extrema, we look at

$g(\phi,\theta_o)= \cfrac { cos(\theta _{ o }) }{ sin(\theta _{ o }) } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }$

$g(\phi,\theta_o)=cot(\theta_o)\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }$

where,

$h(\phi,\theta)=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }$

Consider,

$\cfrac{\partial\,g}{\partial\,\theta_o}=\cfrac{\partial\,}{\partial\,\theta_o}\left\{cot(\theta_o)\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }\right\}$

$\cfrac { \partial \, g }{ \partial \, \theta _{ o } } =-csc^{ 2 }(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) }\\+cot(\theta _{ o })\cfrac { \partial \,  }{ \partial \, \theta _{ o } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } (-sin(\theta )){ d\, (\theta ) }$

$\cfrac { \partial \, g }{ \partial \, \theta _{ o } } =-csc^{ 2 }(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) }-\cfrac { cos(\theta_o) }{ sin(\theta_o) } \cfrac { 1 }{ cos^{ 4 }(\theta _{ o }) } h(\phi ,\theta _{ o })sin(\theta_o)$

$\cfrac { \partial \, g }{ \partial \, \theta _{ o } } =-csc^2(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) }-\cfrac { 1 }{ cos^{ 3 }(\theta _{ o }) } h(\phi ,\theta _{ o })$

Consider the extrema,

$\cfrac { \partial \, g }{ \partial \, \theta _{ o } } =0$

$\cfrac{1}{sin^2(\theta _{ o })}\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta ) } { d\, (cos(\theta )) }+\cfrac { 1 }{ cos^{ 3 }(\theta _{ o }) } h(\phi ,\theta _{ o })=0$

Since,

$h(\phi,\theta)\gt0$  and  $g(\phi,\theta_o)\gt0$ for $0\lt\theta_o\lt\pi/2$ and all $\phi$

A plot of  (1/sin(x))^2 and (1/cos(x))^3 shows that,

which means by itself  $g(\phi,\theta_o)$ has no extrema in the range  $0\lt\theta_o\lt\pi/2$.  Since, its first derivative of  $g(\phi,\theta_o)$ is negative  $B_o$  decreasse with  $(\theta_o)$.  The following diagram shows an orbiting electron driving a B field.

The electron approaches the nucleus because of electrostatic attract, before it collides with the nucleus, the electron reaches the terminal speed (light speed) and begins a helical path towards the nucleus.  Right over the nucleus, the relative motion of the revolving electron and the nucleus causes the nucleus to spin.  This creates a repulsive Lorentz's force that pushes the particles apart.  The electron and nucleus do not collide.  As a result of its spin, the nucleus interacts with the B field established by the revolving electron (initially part of its helical path) and revolve around a $B$ orbit.  In turn, the revolving nucleus generates a B field that causes the electron to spin and revolve around a second  $B_r$ orbit (Refer to post "Orbit In Blue").  $r_p$  do not effect  $r_e$ in reciprocal.  $r_p$ is the result of a centripetal force provided for by the B field established by the electron in orbit $r_e$.  The nucleus will be accelerated to terminal speed and reaches a minimum $r_p$ value such that the centripetal force balances the Lorentz's force (B field) and drag at terminal speed.  The radius $r_e$ affects $r_p$; $r_p$ does not affects $r_e$.

From the post "Electron Orbit B Field II",

$B_{ o }=\cfrac { \mu _{ o }\varepsilon _{ o }\omega r^{ 2 }_{ e }E_{ o } }{ a_{ e } } f(\phi )g(\phi ,\theta _{ o })$

We consider $B_o$ at its maximum,

$\cfrac{\partial\,B_o}{\partial\,r_e}=\cfrac{\partial\,}{\partial\,r_e}\left\{\cfrac { \mu _{ o }\varepsilon _{ o }\omega r^{ 2 }_{ e }E_{ o } }{ a_{ e } } f(\phi )g(\phi ,\theta _{ o })\right\}$

$\cfrac { \partial \, B_{ o } }{ \partial \, r_{ e } } =2\cfrac { \mu _{ o }\varepsilon _{ o }\omega r_{ e }E_{ o } }{ a_{ e } } f(\phi )g(\phi ,\theta _{ o })+\cfrac { \mu _{ o }\varepsilon _{ o }\omega r^{ 2 }_{ e }E_{ o } }{ a_{ e } } f(\phi )\cfrac { \partial \, g(\phi ,\theta _{ o })\,  }{ \partial \, r_{ e } }$

$\cfrac { \partial \, B_{ o } }{ \partial \, r_{ e } } =\cfrac { \mu _{ o }\varepsilon _{ o }\omega r_{ e }E_{ o } }{ a_{ e } } f(\phi )\left[ 2g(\phi ,\theta _{ o })+r_{ e }\cfrac { \partial \, g(\phi ,\theta _{ o })\,  }{ \partial \, r_{ e } }  \right]$

$\cfrac { \partial \, g(\phi ,\theta _{ o })\,  }{ \partial \, r_{ e } } =cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } \cfrac { \partial \, h(\phi ,\theta )\,  }{ \partial \, r_{ e } }  } { d\, (cos(\theta )) }$

$\cfrac { \partial \, h(\phi ,\theta )\,  }{ \partial \, r_{ e } } =e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }.-\cfrac { 2sin(\phi /2) }{ a_{ e }cos(\theta ) } =-h(\phi ,\theta )\, \cfrac { 2sin(\phi /2) }{ a_{ e }cos(\theta ) }$

Therefore,

$\cfrac { \partial \, g(\phi ,\theta _{ o })\,  }{ \partial \, r_{ e } } =-cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta )\, \cfrac { 2sin(\phi /2) }{ a_{ e }cos(\theta ) }  } { d\, (cos(\theta )) }$

$\cfrac { \partial \, g(\phi ,\theta _{ o })\,  }{ \partial \, r_{ e } } =-\cfrac { 2sin(\phi /2) }{ a_{ e } } cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }$

So,

$\cfrac { \partial \, B_{ o } }{ \partial \, r_{ e } } =\cfrac { \mu _{ o }\varepsilon _{ o }\omega r_{ e }E_{ o } }{ a_{ e } } f(\phi )\\\left[ 2g(\phi ,\theta _{ o })-r_{ e }\cfrac { 2sin(\phi /2) }{ a_{ e } } cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) } \right]$

For extrema,

$\cfrac { \partial \, B_{ o } }{ \partial \, r_{ e } } =0$

When

$\cfrac { a_{ e } }{ sin(\phi /2) } g(\phi ,\theta _{ o })=r_{ e }cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }$ --- (*)

$r_{ e }=\cfrac { a_{ e } }{ sin(\phi /2) } \cfrac { \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) } }{ \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) } }$

Consider the second derivative,

$\cfrac { \partial ^{ 2 }\, B_{ o } }{ \partial \, r_{ e }^{ 2 } } =2\cfrac { \mu _{ o }\varepsilon _{ o }\omega E_{ o } }{ a_{ e } } f(\phi )g(\phi ,\theta _{ o })+2\cfrac { \mu _{ o }\varepsilon _{ o }\omega r_{ e }E_{ o } }{ a_{ e } } f(\phi )\cfrac { \partial \, g(\phi ,\theta _{ o }) }{ \partial \, r_{ e } } +2\cfrac { \mu _{ o }\varepsilon _{ o }\omega r_{ e }E_{ o } }{ a_{ e } } f(\phi )\cfrac { \partial \, g(\phi ,\theta _{ o })\,  }{ \partial \, r_{ e } } +\cfrac { \mu _{ o }\varepsilon _{ o }\omega r^{ 2 }_{ e }E_{ o } }{ a_{ e } } f(\phi )\cfrac { \partial ^{ 2 }\, g(\phi ,\theta _{ o })\,  }{ \partial \, r_{ e }^{ 2 } }$

$\cfrac { \partial ^{ 2 }\, B_{ o } }{ \partial \, r_{ e }^{ 2 } } =\cfrac { 2\mu _{ o }\varepsilon _{ o }\omega E_{ o } }{ a_{ e } } f(\phi )\left[ g(\phi ,\theta _{ o })+2r_{ e }\cfrac { \partial \, g(\phi ,\theta _{ o })\,  }{ \partial \, r_{ e } } +\cfrac { r^{ 2 }_{ e } }{ 2 } \cfrac { \partial ^{ 2 }\, g(\phi ,\theta _{ o })\,  }{ \partial \, r_{ e }^{ 2 } } \right]$

$\cfrac { \partial ^{ 2 }\, g(\phi ,\theta _{ o })\,  }{ \partial \,r_{ e }^{ 2 } } =\left( \cfrac { 2sin(\phi /2) }{ a_{ e } }  \right) ^{ 2 }cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }$

So,

$\cfrac { \partial ^{ 2 }\, B_{ o } }{ \partial \, r_{ e }^{ 2 } } =\cfrac { 2\mu _{ o }\varepsilon _{ o }\omega E_{ o } }{ a_{ e } } f(\phi )\left[ g(\phi ,\theta _{ o })-r_{ e }\cfrac { 4sin(\phi /2) }{ a_{ e } } cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }+2r^{ 2 }_{ e }\left( \cfrac { sin(\phi /2) }{ a_{ e } }  \right) ^{ 2 }cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) } \right]$

$\cfrac { \partial ^{ 2 }\, B_{ o } }{ \partial \, r_{ e }^{ 2 } } =2\cfrac { \mu _{ o }\varepsilon _{ o }\omega E_{ o } }{ a_{ e } } f(\phi )cot(\theta _{ o })\left[ \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }\\-r_{ e }\cfrac { 4sin(\phi /2) }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }\\+2r^{ 2 }_{ e }\left( \cfrac { sin(\phi /2) }{ a_{ e } }  \right) ^{ 2 }\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) } \right]$

$\cfrac { \partial ^{ 2 }\, B_{ o } }{ \partial \, r_{ e }^{ 2 } } =2\cfrac { \mu _{ o }\varepsilon _{ o }\omega E_{ o } }{ a_{ e } } f(\phi )cot(\theta _{ o })\left\{ Ar^{ 2 }_{ e }-Br_{ e }+C \right\}$

where,

$C=\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }$

$B=\cfrac { 4sin(\phi /2) }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }$

$A=2\left( \cfrac { sin(\phi /2) }{ a_{ e } }  \right) ^{ 2 }\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }$

$A\gt0$  and  $C\gt0$  for  $0\lt\theta\lt\pi/2$ but $B$ depends on  $\phi/2$.

$B^{ 2 }-4AC=16\left( \cfrac { sin(\phi /2) }{ a_{ e } }  \right) ^{ 2 }\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }\\\left[ \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }-\cfrac { 1 }{ 2 } cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) } \right]$

If $B^{ 2 }-4AC\gt0$ then $\cfrac { \partial ^{ 2 }\, B_{ o } }{ \partial \, r_{ e }^{ 2 } }$ can be negative, since $A\gt0$ (part of the graph of $Ar^{ 2 }_{ e }-Br_{ e }+C$ is below the x-axis).  However this depends on the value of  $\theta_o$ and $cot(\theta_o)$.  $cot(\theta_o)$ is a monotonously decreasing function, that means for larger values of  $\theta_o$ or $r_p$,  $B_o$ is maximum.

Applying the condition for extrema, from  (*)

$\cfrac { a_{ e } }{ sin(\phi /2) } g(\phi ,\theta _{ o })=r_{ e }cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }$

$\cfrac { a_{ e } }{ r_{ e }sin(\phi /2) } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }=\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) }$

For a possible maxima in $B_o$, $B^{ 2 }-4AC\gt0$ implies,

$\cfrac { a_{ e } }{ r_{ e }sin(\phi /2) } -\cfrac { 1 }{ 2 } cot(\theta _{ o })\gt0$

$cot(\theta _{ o })\lt\cfrac { 2a_{ e } }{ r_{ e } } \cfrac { 1 }{ sin(\phi /2) }$

$\theta_o$  must be larger than,

$\theta_{ o }\gt cot^{ -1 }\left( \cfrac { 2a_{ e } }{ r_{ e } } \cfrac { 1 }{ sin(\phi /2) }  \right)$

where  $a_e$ is the size of the electron and  $r_e$ the electron orbit.  By sysmmetry, for the positive nucleus,

$\theta_{ oe }\gt cot^{ -1 }\left( \cfrac { 2a_{ p } }{ r_{ p } } \cfrac { 1 }{ sin(\phi /2) }  \right)$

where  $a_p$  is the size of the nucleus and  $r_p$  the radius of its orbit and $\theta_{oe}$ is the angle subtended by the electron's orbit and the moving nucleus as it revolves along it own orbit.

In order to have a maximum $B_o$ which corresponds to a maximum centripetal force, the respective $\theta$s have minimum values.  A smaller $\theta$ means a smaller radius that would require a higher centripetal force which $B_o$ at maximum can not provide for.

The conclusion is, the orbit does not collapse.

From the fist derivative of $B_o$ wrt $r_e$,

$\cfrac { \partial \, B_{ o } }{ \partial \, r_{ e } } =\cfrac { \mu _{ o }\varepsilon _{ o }\omega r_{ e }E_{ o } }{ a_{ e } } f(\phi )\\\left[ 2g(\phi ,\theta _{ o })-r_{ e }\cfrac { 2sin(\phi /2) }{ a_{ e } } cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\,  } { d\, (cos(\theta )) } \right]$

$cot(\theta _{ o })$ appears in both $g(\phi ,\theta _{ o })$ and the negative term.  An increasing  $r_e$ eventually results in a negative rate of change in $B_o$ that decreases $B_o$.  A decrease in $B_o$ reduces the repulsion between the spinning nucleus and the revolving electron. And $r_e$ decreases and the electron does not fly off.  Similarly, a small $r_e$ eventually results in a positive rate of change in $B_o$ that increases $B_o$.  The repulsive force between the spinning nucleus and the revolving electron increases.  The electron is pushed away and does not collide into the nucleus.

Once again the conclusion is, the orbit does not collapse.

B Orbits Here To Stay

The expression,

$\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  } } { d\, (cos(\theta )) }$

is an interesting one.  And has solution,

$z=cos(\theta)$

$\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ z^{ 4 } } e^{ -\cfrac { x }{ a_{ e }z }  } } { d\, z}=\cfrac { 1 }{ z^{ 4 } } \left\{ \cfrac { x }{ { a }_{ e } } { E }_{ i }(-\cfrac { x }{ { a }_{ e }z } )+xe^{ -\cfrac { x }{ { a }_{ e }z }  } \right\} +c$

where  $E_i(x)$  is the polyexponential function,

 OR

$\cfrac { a_{ e }(x^{ 2 }+2a_{ e }xz+2a_{ e }^{ 2 }z^{ 2 }) }{ x^{ 3 }z^{ 2 } } e^{ (-\cfrac { x }{ a_{ e }z } ) }$

And so,

$\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  } } { d\, (cos(\theta )) }=\left[\cfrac { a_{ e } }{ x^{ 3 }cos^{ 2 }(\theta ) } \left\{x^{ 2 }+2a_{ e }xcos(\theta )+2a_{ e }^{ 2 }cos^{ 2 }(\theta )\right\}e^{ (-\cfrac { x }{ a_{ e }cos(\theta ) } ) }\right]^{\theta_o}_{0}$

Which has a DC term,

$\cfrac { a_{ e } }{ x^{ 3 } } \left\{ x^{ 2 }+2a_{ e }x+2a_{ e }^{ 2 } \right\} e^{ (-\cfrac { x }{ a_{ e } } ) }$

This term suggests that the $B$ field is always present and the positively charged nucleuii are held permanently in the $B$ orbits.

1/x^4*e^(-a/(2*x)) for 0.5≤a≤5,  1/x^4*e^(-2.5/(2*x)),  and  1/x^5*e^(-2.5/(2*x)) are plotted below.

Since,  $x=cos(\theta)$, $x\le1$ the graph $\cfrac{1}{x^5}\gt\cfrac{1}{x^4}$.  And

$\cfrac{1}{x^4}e^{a/(2x)}$ increases with decreasing $a$ which in our original expression is $x$, the distance between the center of the loop and the orbiting negative charge.

Electron Orbit B Field II

This post is wrong.  Please refer to the post "Not To Be taken Too Seriously, Please" dated 14 May 2016. 

Continuing from the post "Electron Orbit B Field", we replace

$E=\cfrac{e}{4\pi\varepsilon_o d^2}$  with,

$E=E_oe^{-\cfrac{x}{a_{e}}}$,

where $a_e$  is the radius of the electron, and

$E_o=\cfrac{e}{4\pi\varepsilon_o a^2_e}$

This exponential form of  $E$  was developed analogously, using exponential gravity as guide.

Total  $E$  over flat circular x-section when the charge is at a distance  $x$  from the loop,

$E_A=\int_{0}^{r} {2\pi xtan(\theta). E_oe^{-\cfrac{x}{a_{e}cos(\theta)} } }d\,r$

$E_{ A }=2\pi E_{ o }\int _{ 0 }^{ r }{ xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } } d\, r$

$\cfrac { \partial \, E_{ A } }{ \partial t } =2\pi E_{ o }\cfrac { \partial \, }{ \partial t }\int _{ 0 }^{ r }{ xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } } d\, r$

Cosinder the term,

$\Pi=\cfrac { \partial \, }{ \partial t }\int _{ 0 }^{ r }{ xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } } d\, r$

$\Pi=\int _{ 0 }^{ r }{ \cfrac { d\, x }{ d\, t } tan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} }} \\+xtan(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} }\cfrac { -1 }{ a_{ e }cos(\theta ) } \cfrac { d\, x }{ d\, t } \\+xsec^{ 2 }(\theta )\cfrac { d\, \theta  }{ d\, t } e^{-\cfrac{x}{a_{e}cos(\theta)} }d\, r$

$\Pi=\int _{ 0 }^{ r }{ e^{-\cfrac{x}{a_{e}cos(\theta)} }\left\{ { \cfrac { d\, x }{ d\, t } tan(\theta ) }-xtan(\theta )\cfrac { 1 }{ a_{ e }cos(\theta ) } \cfrac { d\, x }{ d\, t } +xsec^{ 2 }(\theta )\cfrac { d\, \theta  }{ d\, t }  \right\}  } d\, r$

Since,

$\cfrac { r }{ x } =tan(\theta )$

$-\cfrac { r }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } =sec^{ 2 }(\theta )\cfrac { d\, \theta  }{ d\, t }$

$xsec^{ 2 }(\theta )\cfrac { d\, \theta  }{ d\, t }=-\cfrac { r }{ x } \cfrac { d\, x }{ d\, t }=-\cfrac { d\, x }{ d\, t }tan(\theta)$

Substitute into the expression for  $\Pi$,

$\Pi=\int _{ 0 }^{ r }{ e^{-\cfrac{x}{a_{e}cos(\theta)} }\left\{ { \cfrac { d\, x }{ d\, t } tan(\theta ) }-xtan(\theta )\cfrac { 1 }{ a_{ e }cos(\theta ) } \cfrac { d\, x }{ d\, t } -\cfrac { d\, x  }{ d\, t }tan(\theta )  \right\}  } d\, r$

$\Pi=-\cfrac { d\, x }{ d\, t } \cfrac { x }{ a_{ e } } \int _{ 0 }^{ r }{ tan(\theta )sec(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } } d\, r$

and

$\cfrac { r }{ x } =tan(\theta )$

$\cfrac { 1 }{ x } =sec^{ 2 }(\theta )\cfrac { d\, \theta  }{ dr }$

$dr=xsec^{ 2 }(\theta ){ d\, \theta  }$

$\Pi=-\cfrac { d\, x }{ d\, t } \cfrac { x }{ a_{ e } } \int _{ 0 }^{ r }{ tan(\theta )sec(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} } }xsec^{ 2 }(\theta ){ d\, \theta  }$

Substitute  $\Pi$  back into  $E_A$,

$\cfrac { \partial \, E_{ A } }{ \partial t } =-2\pi E_{ o }\cfrac { d\, x }{ d\, t } \cfrac { x }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ tan(\theta )sec(\theta )e^{-\cfrac{x}{a_{e}cos(\theta)} }} xsec^{ 2 }(\theta ){ d\, \theta  }$

$\cfrac { \partial \, E_{ A } }{ \partial t } =-2\pi E_{ o }\cfrac { d\, x }{ d\, t } \cfrac { x^{ 2 } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ sec^{ 4 (\theta )}e^{-\cfrac{x}{a_{e}cos(\theta)} } } sin(\theta ){ d\, \theta  }$

$\cfrac { \partial \, E_{ A } }{ \partial t } =2\pi E_{ o }\cfrac { d\, x }{ d\, t } \cfrac { x^{ 2 } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac{1}{cos^{ 4 }(\theta )}e^{-\cfrac{x}{a_{e}cos(\theta)} } } { d\, (cos(\theta )) }$ --- (*)

This however, is not the final solution.  We have to consider the fact that the charge is in orbit of radius  $r_e$  and  the loop defining  $B$ is centered at the circumference along a radial line.

We are concerned with change in  $E$ perpendicular to the loop.

$\sigma=\cfrac{\pi-\phi}{2}=\cfrac{\pi}{2}-\cfrac{\phi}{2}$

$E_{\bot}=Esin(\sigma)=Esin(\cfrac{\pi}{2}-\cfrac{\phi}{2})=Ecos(\phi/2)$

Consider,

$x^2=r^2_e+r^2_e-2r^2_ecos(\phi)=2r^2_e(1-cos(\phi))$,  

$x=r_e\sqrt{2(1-cos(\phi))}=2r_{ e }sin(\phi /2)$

$\cfrac { dx }{ dt }=r_{ e }cos(\phi/2)\cfrac{d\,\phi}{d\,t}$

and,

$x_\bot=xcos(\phi/2)$

$\cfrac { dx_\bot }{ dt }=\cfrac { dx }{ dt }cos(\phi/2)-xsin(\phi/2).\cfrac{1}{2}\cfrac { d\phi }{ dt }$

$\cfrac { dx_{ \bot  } }{ dt } =\left( r_{ e }cos^{ 2 }(\phi /2)-xsin(\phi /2).\cfrac { 1 }{ 2 }  \right) \cfrac { d\phi  }{ dt }$

Substitute $x$ into the above.

$\cfrac { dx_{ \bot  } }{ dt } =r_{ e }cos(\phi )\cfrac { d\phi  }{ dt }$

Consider the term,

$e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }=e^{ -\cfrac { r_{ e }\sqrt { 2(1-cos(\phi )) }  }{ a_{ e }cos(\theta ) }  }=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }$

Let

$h(\phi,\theta)=e^{ -\cfrac { 2r_{ e }sin(\phi /2) }{ a_{ e }cos(\theta ) }  }$

which has both  $\phi$  and  $\theta$  dependence.

From (*),

$\cfrac { \partial \, E_{ A } }{ \partial t } =2\pi E_{ o }\cfrac { d\, x }{ d\, t } \cfrac { x^{ 2 } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac{1}{cos^{ 4 }(\theta )}h(\phi,\theta) } { d\, (cos(\theta )) }$

Adjusted for circular orbit,

$\cfrac { \partial \, E_{ A } }{ \partial t } =2\pi E_{ o }cos(\phi/2 )\cfrac { d\, x_{ \bot  } }{ d\, t } \cfrac { x^{ 2 }_{ \bot  } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta) } { d\, (cos(\theta )) }$

The direction of  $E$  is adjusted by  $cos(\phi/2)$.  The the distance term in the exponential remains as  $x$ as we see from the above diagram.

Assuming that  $B$  is constant around the loop,

$\oint{B}d\,r=2\pi rB_o$

$2\pi rB_o=\mu _{ o }\varepsilon _{ o }2\pi E_{ o }cos(\phi /2)\cfrac { d\, x_{ \bot  } }{ d\, t } \cfrac { x^{ 2 }_{ \bot  } }{ a_{ e } } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }$

This is wrong. $r\ne x_\bot tan(\theta _{ o })$ refer to post "Not To Be taken Too Seriously, Please" dated 14 May 2016 

$x_\bot tan(\theta _{ o })B_{ o }=\cfrac {\mu _{ o }\varepsilon _{ o } E_{ o }  }{ a_{ e } }cos(\phi/2 )\cfrac { d\, x_{ \bot  } }{ d\, t } x^{ 2 }_{ \bot  } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }$

$B_{ o }=\cfrac { \mu _{ o }\varepsilon _{ o }E_{ o } }{ a_{ e } }cos(\phi/2 )\cfrac { d\, x_{ \bot  } }{ d\, t } x_{ \bot  } \cfrac {cos(\theta_o) }{ sin(\theta _{ o }) } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta) } { d\, (cos(\theta )) }$

Consider,

$\cfrac { d\, x_{ \bot  } }{ d\, t } x_{ \bot  }=2r^{ 2 }_{ e }cos(\phi /2)sin(\phi/2)cos(\phi )\cfrac { d\phi  }{ dt }$

Let

$f(\phi)=cos(\phi /2).2cos(\phi /2)sin(\phi/2)cos(\phi )=\cfrac{1}{2}sin(2\phi)cos(\phi /2)$

such that,

$cos(\phi /2)\cfrac { d\, x_{ \bot  } }{ d\, t } x_{ \bot  } =r^2_ef(\phi)\cfrac { d\phi  }{ dt }$

And we let,

$g(\phi,\theta_o)= \cfrac { cos(\theta _{ o }) }{ sin(\theta _{ o }) } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }$

and

$\omega=\cfrac{d\,\phi}{d\,t}$

$B_o$  becomes,

$B_{ o }=\cfrac {\mu _{ o }\varepsilon _{ o }E_{ o } }{ a_{ e } } cos(\phi /2)\cfrac { d\, x_{ \bot  } }{ d\, t }x_{ \bot  }  \cfrac {cos(\theta_o) }{ sin(\theta _{ o }) } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta) } { d\, (cos(\theta )) }$

$B_{ o }=\cfrac { \mu _{ o }\varepsilon _{ o }\omega r^{ 2 }_{ e }E_{ o } }{ a_{ e } } f(\phi )g(\phi ,\theta _{ o })$

where the term  $g(\phi,\theta_o)$  share the variable  $\phi$  with  $f(\phi)$  through  $h(\phi,\theta_o)$  and the two are not independent.

A plot of  $f(\phi)$  is shown below.  $f(\phi)$ is periodic and bounded.

This shows that the used of the exponential form is appropriate for  $E$.

Electron Orbit B Field..BLEWUP

What is the  $B$  field established by an electron in orbit.

From Maxwell,

$\oint { B } \,dr=\varepsilon _{ o }\mu _{ o }\cfrac { \partial \,  }{ \partial t } \left\{ \oint { E } \,dA \right\}$

$E=\cfrac{e}{4\pi\varepsilon_o d^2}$

Total  $E$  over flat circular x-section when the charge is at a distance  $x$  from the loop,

$E_A=\int_{0}^{r} {2\pi xtan(\theta).\cfrac{e}{4\pi\varepsilon_o}\left(\cfrac{cos(\theta)}{x}\right)^2}d\,r$

$\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } } \cfrac { \partial \,  }{ \partial t } \left\{ \int _{ 0 }^{ r }{ sin(\theta )cos(\theta ).\cfrac { 1 }{ x }  } d\, r \right\}$

$\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 4\varepsilon _{ o } } \cfrac { \partial \,  }{ \partial t } \left\{ \int _{ 0 }^{ r }{ sin(2\theta )\cfrac { 1 }{ x }  } d\, r \right\}$

$\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 4\varepsilon _{ o } } \left\{ \int _{ 0 }^{ r }{ 2cos(2\theta )\cfrac { 1 }{ x } \cfrac { d\, \theta  }{ d\, t }  } -sin(2\theta )\cfrac { 1 }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } d\, r \right\}$

$\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } } \left\{ \int _{ 0 }^{ r }{ (2cos^{ 2 }(\theta )-1)\cfrac { 1 }{ x } \cfrac { d\, \theta  }{ d\, t }  } -sin(\theta )cos(\theta )\cfrac { 1 }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } d\, r \right\}$

From geometry,

$\cfrac { r }{ x } =tan(\theta )$

$-\cfrac { r }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } =sec^{ 2 }(\theta )\cfrac { d\, \theta  }{ d\, t }$

$\cfrac { d\, \theta  }{ d\, t } =-\cfrac { rcos^{ 2 }(\theta ) }{ x^{ 2 } } \cfrac { d\, x }{ d\, t }$

and

$\cfrac { 1 }{ x } =sec^{ 2 }(\theta )\cfrac { d\, \theta  }{ dr }$

$dr=xsec^{ 2 }(\theta ){ d\, \theta  }$

So we have,

$\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } } \left\{ \int _{ 0 }^{ r }{ -(2cos^{ 2 }(\theta )-1)\cfrac { 1 }{ x } \cfrac { rcos^{ 2 }(\theta ) }{ x^{ 2 } } \cfrac { d\, x }{ d\, t }  } -sin(\theta )cos(\theta )\cfrac { 1 }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } d\, r \right\}$

$\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } } \cfrac { 1 }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } \left\{ \int _{ 0 }^{ r }{ -(2cos^{ 2 }(\theta )-1)tan(\theta )cos^{ 2 }(\theta )-sin(\theta )cos(\theta ) } d\, r \right\}$

$\cfrac { \partial \, E_{ A } }{ \partial t } =-\cfrac { e }{ 2\varepsilon _{ o } } \cfrac { 1 }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } \left\{ \int _{ 0 }^{ r }{ 2cos^{ 2 }(\theta )sin(\theta )cos(\theta ) } d\, r \right\}$

$\cfrac { \partial \, E_{ A } }{ \partial t } =-\cfrac { e }{ 2\varepsilon _{ o } } \cfrac { 1 }{ x } \cfrac { d\, x }{ d\, t } \left\{ \int _{ 0 }^{ \theta _{ o } }{ cos^{ 2 }(\theta )sin(2\theta ) } sec^{ 2 }(\theta )d\, \theta  \right\}$

$\cfrac { \partial \, E_{ A } }{ \partial t } =-\cfrac { e }{ 2\varepsilon _{ o } } \cfrac { 1 }{ x } \cfrac { d\, x }{ d\, t } \left\{ \int _{ 0 }^{ \theta _{ o } }{ sin(2\theta ) } d\, \theta  \right\}$

$\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 4\varepsilon _{ o } } \cfrac { 1 }{ x } \cfrac { d\, x }{ d\, t } \left\{ cos(2\theta _{ o })-1 \right\}$ --- (*)

We have also to consider the fact that the charge is in orbit of radius  $r_e$  and  the loop defining  $B$ is centered at the circumference along a radial line.

$\sigma=\cfrac{\pi-\phi}{2}=\cfrac{\pi}{2}-\cfrac{\phi}{2}$

$sin(\sigma)=sin(\cfrac{\pi}{2}-\cfrac{\phi}{2})=cos({\phi}/{2})$

$E_\bot=Esin(\sigma)=Ecos(\phi/2)$

$x_\bot=xsin(\sigma)=xcos(\phi/2)$

We are concerned with the change in $E$ perpendicular to the smaller orbit in the loop.   From (*)

$\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } } \cfrac { d\, x }{ d\, t } \cfrac { 1 }{ x } \left\{ cos(2\theta _{ o })-1 \right\}$

 So the equation (*) becomes,

$\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } } sin(\sigma )\cfrac { d\, x_{ \bot  } }{ d\, t } \cfrac { 1 }{ x }  \left\{ cos(2\theta _{ o })-1 \right\}$

 $\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } }cos(\phi/2)\cfrac { d\, x_{ \bot  } }{ d\, t }\cfrac { 1 }{ x } \left\{ cos(2\theta _{ o })-1 \right\}$  --- (1)

The perpendicular $E$ field component is given by $E_\bot=Ecos(\phi/2)$, not by changing  $x$  to $x_\bot$.

Consider,

$x^2=r^2_e+r^2_e-2r^2_ecos(\phi)=2r^2_e(1-cos(\phi))$,  

$x=r_e\sqrt{2(1-cos(\phi))}=2r_{ e }sin(\phi /2)$

$\cfrac { dx }{ dt }=r_{ e }cos(\phi/2)\cfrac{d\,\phi}{d\,t}$

and,

$x_\bot=xcos(\phi/2)$

$\cfrac { dx_\bot }{ dt }=\cfrac { dx }{ dt }cos(\phi/2)-xsin(\phi/2).\cfrac{1}{2}\cfrac { d\phi }{ dt }$

$\cfrac { dx_{ \bot  } }{ dt } =\left( r_{ e }cos^{ 2 }(\phi /2)-xsin(\phi /2).\cfrac { 1 }{ 2 }  \right) \cfrac { d\phi  }{ dt }$

Substitute $x$ into the above.

$\cfrac { dx_{ \bot  } }{ dt } =r_{ e }cos(\phi )\cfrac { d\phi  }{ dt }$

$\cfrac { dx_{ \bot  } }{ dt } \cfrac { 1 }{ x } =r_{ e }\cfrac { 1 }{ 2r_{ e }sin(\phi /2) } cos(\phi )\cfrac { d\phi  }{ dt }$

$\cfrac { dx_{ \bot  } }{ dt } \cfrac { 1 }{ x } =\cfrac { 1 }{ 2 } \cfrac { cos(\phi ) }{ sin(\phi /2) } \cfrac { d\phi  }{ dt }$

From (1),

 $\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 2\varepsilon _{ o } }cos(\phi/2)\cfrac { d\, x_{ \bot  } }{ d\, t }\cfrac { 1 }{ x }  \left\{ cos(2\theta _{ o })-1 \right\}$

$\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { e }{ 4\varepsilon _{ o } } cos(\phi /2)\cfrac { 1 }{ 2 } \cfrac { cos(\phi ) }{ sin(\phi /2) } \cfrac { d\phi  }{ dt }  \left\{ cos(2\theta _{ o })-1 \right\}$

We know at this point that $B$ is going to blowup when integrated over  $\phi$.  This suggests that  $\cfrac{1}{x^2}$ is not approprite here.

Assuming that  $B$  is constant around the loop,

$\oint{B}d\,r=2\pi r_pB_o$

where $r_p$ is the radius of the $B$ orbit.

$2\pi r_pB_o=\mu _{ o }\varepsilon _{ o }\cfrac { \partial \, E_{ A } }{ \partial t } =\cfrac { \mu _{ o }e }{ 4 } cos(\phi /2)\left\{ \cfrac { 1 }{ sin(\phi /2) } -2sin(\phi /2) \right\} \cfrac { d\phi  }{ dt } \left\{ cos(2\theta _{ o })-1 \right\}$

$B_{ o }=\cfrac { \mu _{ o }e }{ 8\pi r_{ p } } cos(\phi /2)\left\{ \cfrac { 1 }{ sin(\phi /2) } -2sin(\phi /2) \right\} \cfrac { d\phi  }{ dt } \left\{ cos(2\theta _{ o })-1 \right\}$

This is a time varying $B$  field, changing as the electron travels at  $\omega=\cfrac{d\phi}{d\,t}$ around its orbit.  We can average $B^2$ over one period to obtain the energy in needed to establish the orbit.

 $\cfrac { 1 }{ T } \int _{ 0 }^{ T }{ B^{ 2 }_{ o } } d\, t=\cfrac { 1 }{ T } \int _{ 0 }^{ T }{ \left( \cfrac { \mu _{ o }e }{ 8\pi r_{ p } } sin^{ 2 }(\theta _{ o }) \right) ^{ 2 }\left( cos(\phi /2)\left\{ \cfrac { 1 }{ sin(\phi /2) } -2sin(\phi /2) \right\}  \right) ^{ 2 }\cfrac { d\phi  }{ dt } \cfrac { d\phi  }{ dt }  }$

$=\cfrac { 1 }{ T } \left( \cfrac { \mu _{ o }e }{ 8\pi r_{ p } } sin^{ 2 }(\theta _{ o }) \right) ^{ 2 }\int _{ 0 }^{ 2\pi }{ \left( cos(\phi /2)\left\{ \cfrac { 1 }{ sin(\phi /2) } -2sin(\phi /2) \right\}  \right) ^{ 2 }\omega  } d\phi$

as $\omega=\cfrac{d\phi}{d\,t}$.

Consider,

$\int _{ 0 }^{ 2\pi}{ \left( cos(\phi /2)\left\{ \cfrac { 1 }{ sin(\phi /2) } -2sin(\phi /2) \right\}  \right) ^{ 2 }\omega  } d\phi$

KABOOM!!