For
θo such that
Bo is at the extrema, we look at
g(ϕ,θo)=cos(θo)sin(θo)∫θo01cos4(θ)h(ϕ,θ)d(cos(θ))
g(ϕ,θo)=cot(θo)∫θo01cos4(θ)h(ϕ,θ)d(cos(θ))
where,
h(ϕ,θ)=e−2resin(ϕ/2)aecos(θ)
Consider,
∂g∂θo=∂∂θo{cot(θo)∫θo01cos4(θ)h(ϕ,θ)d(cos(θ))}
∂g∂θo=−csc2(θo)∫θo01cos4(θ)h(ϕ,θ)d(cos(θ))+cot(θo)∂∂θo∫θo01cos4(θ)h(ϕ,θ)(−sin(θ))d(θ)
∂g∂θo=−csc2(θo)∫θo01cos4(θ)h(ϕ,θ)d(cos(θ))−cos(θo)sin(θo)1cos4(θo)h(ϕ,θo)sin(θo)
∂g∂θo=−csc2(θo)∫θo01cos4(θ)h(ϕ,θ)d(cos(θ))−1cos3(θo)h(ϕ,θo)
Consider the extrema,
∂g∂θo=0
1sin2(θo)∫θo01cos4(θ)h(ϕ,θ)d(cos(θ))+1cos3(θo)h(ϕ,θo)=0
Since,
h(ϕ,θ)>0 and
g(ϕ,θo)>0 for
0<θo<π/2 and all
ϕ
A plot of (1/sin(x))^2 and (1/cos(x))^3 shows that,
which means by itself
g(ϕ,θo) has no extrema in the range
0<θo<π/2. Since, its first derivative of
g(ϕ,θo) is negative
Bo decreasse with
(θo). 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
Br orbit (Refer to post "Orbit In Blue").
rp do not effect
re in reciprocal.
rp is the result of a centripetal force provided for by the B field established by the electron in orbit
re. The nucleus will be accelerated to terminal speed and reaches a minimum
rp value such that the centripetal force balances the Lorentz's force (B field) and drag at terminal speed. The radius
re affects
rp;
rp does not affects
re.
From the post "Electron Orbit B Field II",
Bo=μoεoωr2eEoaef(ϕ)g(ϕ,θo)
We consider
Bo at its maximum,
∂Bo∂re=∂∂re{μoεoωr2eEoaef(ϕ)g(ϕ,θo)}
∂Bo∂re=2μoεoωreEoaef(ϕ)g(ϕ,θo)+μoεoωr2eEoaef(ϕ)∂g(ϕ,θo)∂re
∂Bo∂re=μoεoωreEoaef(ϕ)[2g(ϕ,θo)+re∂g(ϕ,θo)∂re]
∂g(ϕ,θo)∂re=cot(θo)∫θo01cos4(θ)∂h(ϕ,θ)∂red(cos(θ))
∂h(ϕ,θ)∂re=e−2resin(ϕ/2)aecos(θ).−2sin(ϕ/2)aecos(θ)=−h(ϕ,θ)2sin(ϕ/2)aecos(θ)
Therefore,
∂g(ϕ,θo)∂re=−cot(θo)∫θo01cos4(θ)h(ϕ,θ)2sin(ϕ/2)aecos(θ)d(cos(θ))
∂g(ϕ,θo)∂re=−2sin(ϕ/2)aecot(θo)∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))
So,
∂Bo∂re=μoεoωreEoaef(ϕ)[2g(ϕ,θo)−re2sin(ϕ/2)aecot(θo)∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))]
For extrema,
∂Bo∂re=0
When
aesin(ϕ/2)g(ϕ,θo)=recot(θo)∫θo01cos5(θ)h(ϕ,θ)d(cos(θ)) ---
(*)
re=aesin(ϕ/2)∫θo01cos4(θ)h(ϕ,θ)d(cos(θ))∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))
Consider the second derivative,
∂2Bo∂r2e=2μoεoωEoaef(ϕ)g(ϕ,θo)+2μoεoωreEoaef(ϕ)∂g(ϕ,θo)∂re+2μoεoωreEoaef(ϕ)∂g(ϕ,θo)∂re+μoεoωr2eEoaef(ϕ)∂2g(ϕ,θo)∂r2e
∂2Bo∂r2e=2μoεoωEoaef(ϕ)[g(ϕ,θo)+2re∂g(ϕ,θo)∂re+r2e2∂2g(ϕ,θo)∂r2e]
∂2g(ϕ,θo)∂r2e=(2sin(ϕ/2)ae)2cot(θo)∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))
So,
∂2Bo∂r2e=2μoεoωEoaef(ϕ)[g(ϕ,θo)−re4sin(ϕ/2)aecot(θo)∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))+2r2e(sin(ϕ/2)ae)2cot(θo)∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))]
∂2Bo∂r2e=2μoεoωEoaef(ϕ)cot(θo)[∫θo01cos4(θ)h(ϕ,θ)d(cos(θ))−re4sin(ϕ/2)ae∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))+2r2e(sin(ϕ/2)ae)2∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))]
∂2Bo∂r2e=2μoεoωEoaef(ϕ)cot(θo){Ar2e−Bre+C}
where,
C=∫θo01cos4(θ)h(ϕ,θ)d(cos(θ))
B=4sin(ϕ/2)ae∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))
A=2(sin(ϕ/2)ae)2∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))
A>0 and
C>0 for
0<θ<π/2 but
B depends on
ϕ/2.
B2−4AC=16(sin(ϕ/2)ae)2∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))[∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))−12cot(θo)∫θo01cos4(θ)h(ϕ,θ)d(cos(θ))]
If
B2−4AC>0 then
∂2Bo∂r2e can be negative, since
A>0 (part of the graph of
Ar2e−Bre+C is below the x-axis). However this depends on the value of
θo and
cot(θo).
cot(θo) is a monotonously decreasing function, that means for larger values of
θo or
rp,
Bo is maximum.
Applying the condition for extrema, from
(*)
aesin(ϕ/2)g(ϕ,θo)=recot(θo)∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))
aeresin(ϕ/2)∫θo01cos4(θ)h(ϕ,θ)d(cos(θ))=∫θo01cos5(θ)h(ϕ,θ)d(cos(θ))
For a possible maxima in
Bo,
B2−4AC>0 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
\thetas 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.