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.