\(g(\phi,\theta_o)= \cfrac { cos(\theta _{ o }) }{ sin(\theta _{ o }) } \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }\)
\(g(\phi,\theta_o)=cot(\theta_o)\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } h(\phi,\theta)} { d\, (cos(\theta )) }\)
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 \(r_e\),
\(\cfrac { \partial \, B_{ o } }{ \partial \, r_{ e } } =\cfrac { \mu _{ o }\varepsilon _{ o }\omega r_{ e }E_{ o } }{ a_{ e } } f(\phi )\\\left[ 2g(\phi ,\theta _{ o })-r_{ e }\cfrac { 2sin(\phi /2) }{ a_{ e } } cot(\theta _{ o })\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 5 }(\theta ) } h(\phi ,\theta )\, } { d\, (cos(\theta )) } \right]\)
Once again the conclusion is, the orbit does not collapse.
