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.