\(v^2=-\cfrac { 1 }{ m } \psi _{ c } \cfrac { \psi _{ c }+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\) --- (1)
\(v\) is complex. Where,
\(u=ln(cosh(x- x_{ z }))\), \(x_z=\cfrac{1}{2}x_a\) and
\(\psi_c=\psi_n-\psi_{max}\)
A plot of which,
The scaling factor \(\cfrac{\psi_c}{m}\) plays a minor role in the shape of \(v^2\) vs \(x\). \(v^2\) and so the orbiting particle \(KE\), is confined to a narrow range about \(\psi_{max}\), at \(x=x_z\) of the particle. At values of \(x\) closer \(x=x_z\), \(v^2\) has a minimum, \(v^2_{min}\).
On approach to \(x_z\) from higher values of \(x\), since,
\(\lim\limits_{x\to x_z^{+}}\cfrac { \psi _{ c }+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\to\infty\)
This means \(v\) increases rapidly as the particle fall from higher orbit but because of the speed limit \(c\), at light speed, the particle does not reach \(x=x_z\).
Also, on approach to \(x_z\) from lower values of \(x\),
\(\lim\limits_{x\to x_z^{-}}\cfrac { \psi _{ c }+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\to(-i)\infty\)
As \(v\) is already complex, \(-i\) rotates \(v\) back along the radial line and the particle is expelled beyond \(x=x_z\) with great velocity. The particle jumps to higher orbit, \(x\gt x_z\) at light speed \(c\) (assuming that it has reach light speed, \(c\) before passing through \(x=x_z\)). A plot of \(v^2\) around \(x_z\) is shown below,
To obtain \(v_{min}\) consider this, when \(x\to x_z\pm\delta\), where \(\delta=small\),
\(\lim\limits_{x\to x_z}\cfrac{d(v^2)}{dx}=\lim\limits_{x\to x_z}\cfrac{d(v^2)}{du}\cfrac{du}{dx}=\epsilon\)
as
\(\delta\to small\), not necessarily zero, \(\epsilon\to0\). Substitute equation (1) for \(v^2\),
\(\cfrac{d(v^2)}{dx}=-\cfrac{\psi_c}{m}\cfrac{d}{du}\left\{\cfrac{1}{u}\left(\psi_c+u\right)e^{u}\left(e^{2u}-1\right)^{1/2}\right\}\cfrac{du}{dx}=\epsilon\)
Consider,
\(\cfrac { e^{ u }\left\{\psi _{ c }(-u+e^{ 2u }(2u-1)+1)+u^{ 2 }(2e^{ 2u }-1)\right\} }{ u^2\sqrt { e^{ 2u }-1 } } \cfrac{du}{dx}=\epsilon\)
\( \cfrac { e^{ u }\left\{u^{ 2 }(2e^{ 2u }-1)+\psi _{ c }(u(2e^{ 2u }-1)-(e^{ 2u }-1))\right\} }{ u^{ 2 }\sqrt { e^{ 2u }-1 } } \cfrac { du }{ dx } =\epsilon \)
\( \cfrac { e^{ u }\left\{u(\psi _{ c }+u)(2e^{ 2u }-1)-\psi _{ c }(e^{ 2u }-1)\right\} }{ u^2\sqrt { e^{ 2u }-1 } } \cfrac { du }{ dx } =\epsilon \)
\( e^{ u }\left\{ \cfrac { (\psi _{ c }+u) }{ u } e^{ u }-\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1) }{ u^{ 2 }(2e^{ 2u }-1) } \right\} \cfrac { du }{ dx } =\cfrac { e^{ u }(e^{ 2u }-1)^{ 1/2 } }{ (2e^{ 2u }-1) } \epsilon \)
\( \cfrac { (\psi _{ c }+u) }{ u } e^{ 2u }=\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1)\cfrac { du }{ dx } } \epsilon +\psi _{ c }\cfrac { e^{ 2u }(e^{ 2u }-1) }{ u^{ 2 }(2e^{ 2u }-1) } \)
\(\because \cfrac{\epsilon}{\cfrac { du }{ dx }}=\cfrac { dv^{ 2 } }{ du } \)
\( \cfrac { (\psi _{ c }+u) }{ u } e^{ 2u }\cfrac { du }{ dx } =\cfrac { e^{ u }(e^{ 2u }-1) ^{1/2}}{ (2e^{ 2u }-1) } \cfrac { du }{ dx } \cfrac { dv^{ 2 } }{ du } +\psi _{ c }\cfrac { e^{ 2u }(e^{ 2u }-1)}{ u^{ 2 }(2e^{ 2u }-1) } \cfrac { du }{ dx } \)
\( \cfrac { du }{ dx }=tanh(x)=\cfrac{(e^{2u}-1)^{1/2}}{e^u}\),
\( \cfrac { (\psi _{ c }+u) }{ u } e^{ 2u }\cfrac { (e^{ 2u }-1)^{ 1/2 } }{ e^{ u } } =\cfrac { e^{ u }(e^{ 2u }-1) ^{1/2}}{ (2e^{ 2u }-1) } \cfrac { dv^{ 2 } }{ dx } +\psi _{ c }\cfrac { e^{ 2u }(e^{ 2u }-1) }{ u^{ 2 }(2e^{ 2u }-1) } \cfrac { (e^{ 2u }-1)^{ 1/2 } }{ e^{ u } } \)
\( \cfrac { (\psi _{ c }+u) }{ u } e^{ u }(e^{ 2u }-1)^{ 1/2 }=\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1) } \cfrac { dv^{ 2 } }{ dx } +\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) } \)
Wrong! Never divide by zero or the possibility of zero. But \(\cfrac { du }{ dx }\) was not actually divided with!
So,
\(v^2=-\cfrac{\psi_c}{m}\left\{\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1) } \cfrac { dv^{ 2 } }{ dx } +\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) } \right\}\)
So, when \(\delta\) is sufficiently small, from the diagram above, the velocity is minimum \(v=v_{min}\) when,
\(\cfrac{d(v^2)}{dx}=0\)
which occurs before \(x=x_z\). This leads to,
\(\require{cancel}\)
\(v^2_{min}=-\cfrac{\psi_c}{m}\left\{\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1) } \cancelto{0}{\cfrac { dv^{ 2 } }{ dx }} +\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1) ^{3/2}}{ u^{ 2 }(2e^{ 2u }-1) } \right\}_{x=x_{min}}\)
\(v^2_{min}=-\cfrac{\psi^2_c}{m}\left\{\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) } \right\}_{x=x_{min}}\)
Since,
\(u=log(cosh(x-x_z))\)
\(\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) }=\cfrac { cosh(x-x_z)sinh^{ 3 }(x-x_z) }{ ln^{ 2 }(cosh(x-x_z))(2cosh^{ 2 }(x-x_z)-1) } =\cfrac { sinh(2(x-x_z))sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z))cosh(2(x-x_z)) } =tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) } \)
All \(v^2_{min}\) for \(\psi_c\gt0\) lie on this line,
\(y=tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) } \)
For each elevation \(\psi_c\) in energy, the particle orbits at different radius with different velocity. The minimum velocities squared of all particles elevated to different \(\psi_c\)s are given by the curve above. If energy above this minimum \(KE\) represented by the curve is emitted, then this curve is the baseline of the emission spectrum. This is emission across a continuous spectrum not in packets of discrete frequencies.
How?? In packets of \(\psi\), at low speed, as the particle's photon equivalent, but across a continuous spectrum as v^2 is continuous.
The particle is in circular motion around with velocity, \(v_{min}\) in a direction perpendicular to \(x\). Only when \(x\to x_z^{-}\) does the particle switch direction and travel along \(x\) with great velocity.
When \(x\to x_z^{+}\), the particle is in circular motion with radius, \(x_{radius}=x\). When \(x\to x_z^{-}\), the particle travels along the radial line, \(x\).
The steep values of \(v\) on either side of \(x=x_z\) restrict the location of the particle to small values around \(x=x_z\). This is a Quantum Well except that \(x=x_z\) is also forbidden. They are two Quantum Wells side by side. It is however possible to escape from the lower Quantum Well in the region \(x\lt x_z\) by approaching \(x=x_z\). At that point \(iv\) is rotated by the \(-i\) factor and the particle speed beyond \(x=x_z\) along the radial line, in the direction of \(x\) into the adjacent Quantum Well at very high speed. This is quantum tunneling, but it is possible only from the well at lower region \(x\lt x_z\), to the well at higher region \(x\gt x_z\).
\(\lim\limits_{x\to x_z}\cfrac{d(v^2)}{dx}=\lim\limits_{x\to x_z}\cfrac{d(v^2)}{du}\cfrac{du}{dx}=\epsilon\)
as
\(\delta\to small\), not necessarily zero, \(\epsilon\to0\). Substitute equation (1) for \(v^2\),
\(\cfrac{d(v^2)}{dx}=-\cfrac{\psi_c}{m}\cfrac{d}{du}\left\{\cfrac{1}{u}\left(\psi_c+u\right)e^{u}\left(e^{2u}-1\right)^{1/2}\right\}\cfrac{du}{dx}=\epsilon\)
Consider,
\(\cfrac { e^{ u }\left\{\psi _{ c }(-u+e^{ 2u }(2u-1)+1)+u^{ 2 }(2e^{ 2u }-1)\right\} }{ u^2\sqrt { e^{ 2u }-1 } } \cfrac{du}{dx}=\epsilon\)
\( \cfrac { e^{ u }\left\{u^{ 2 }(2e^{ 2u }-1)+\psi _{ c }(u(2e^{ 2u }-1)-(e^{ 2u }-1))\right\} }{ u^{ 2 }\sqrt { e^{ 2u }-1 } } \cfrac { du }{ dx } =\epsilon \)
\( \cfrac { e^{ u }\left\{u(\psi _{ c }+u)(2e^{ 2u }-1)-\psi _{ c }(e^{ 2u }-1)\right\} }{ u^2\sqrt { e^{ 2u }-1 } } \cfrac { du }{ dx } =\epsilon \)
\( e^{ u }\left\{ \cfrac { (\psi _{ c }+u) }{ u } e^{ u }-\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1) }{ u^{ 2 }(2e^{ 2u }-1) } \right\} \cfrac { du }{ dx } =\cfrac { e^{ u }(e^{ 2u }-1)^{ 1/2 } }{ (2e^{ 2u }-1) } \epsilon \)
\( \cfrac { (\psi _{ c }+u) }{ u } e^{ 2u }=\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1)\cfrac { du }{ dx } } \epsilon +\psi _{ c }\cfrac { e^{ 2u }(e^{ 2u }-1) }{ u^{ 2 }(2e^{ 2u }-1) } \)
\(\because \cfrac{\epsilon}{\cfrac { du }{ dx }}=\cfrac { dv^{ 2 } }{ du } \)
\( \cfrac { (\psi _{ c }+u) }{ u } e^{ 2u }\cfrac { du }{ dx } =\cfrac { e^{ u }(e^{ 2u }-1) ^{1/2}}{ (2e^{ 2u }-1) } \cfrac { du }{ dx } \cfrac { dv^{ 2 } }{ du } +\psi _{ c }\cfrac { e^{ 2u }(e^{ 2u }-1)}{ u^{ 2 }(2e^{ 2u }-1) } \cfrac { du }{ dx } \)
\( \cfrac { du }{ dx }=tanh(x)=\cfrac{(e^{2u}-1)^{1/2}}{e^u}\),
\( \cfrac { (\psi _{ c }+u) }{ u } e^{ 2u }\cfrac { (e^{ 2u }-1)^{ 1/2 } }{ e^{ u } } =\cfrac { e^{ u }(e^{ 2u }-1) ^{1/2}}{ (2e^{ 2u }-1) } \cfrac { dv^{ 2 } }{ dx } +\psi _{ c }\cfrac { e^{ 2u }(e^{ 2u }-1) }{ u^{ 2 }(2e^{ 2u }-1) } \cfrac { (e^{ 2u }-1)^{ 1/2 } }{ e^{ u } } \)
\( \cfrac { (\psi _{ c }+u) }{ u } e^{ u }(e^{ 2u }-1)^{ 1/2 }=\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1) } \cfrac { dv^{ 2 } }{ dx } +\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) } \)
Wrong! Never divide by zero or the possibility of zero. But \(\cfrac { du }{ dx }\) was not actually divided with!
So,
\(v^2=-\cfrac{\psi_c}{m}\left\{\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1) } \cfrac { dv^{ 2 } }{ dx } +\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) } \right\}\)
So, when \(\delta\) is sufficiently small, from the diagram above, the velocity is minimum \(v=v_{min}\) when,
\(\cfrac{d(v^2)}{dx}=0\)
which occurs before \(x=x_z\). This leads to,
\(\require{cancel}\)
\(v^2_{min}=-\cfrac{\psi_c}{m}\left\{\cfrac { e^{ u }(e^{ 2u }-1)^{1/2} }{ (2e^{ 2u }-1) } \cancelto{0}{\cfrac { dv^{ 2 } }{ dx }} +\psi _{ c }\cfrac { e^{ u }(e^{ 2u }-1) ^{3/2}}{ u^{ 2 }(2e^{ 2u }-1) } \right\}_{x=x_{min}}\)
\(v^2_{min}=-\cfrac{\psi^2_c}{m}\left\{\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) } \right\}_{x=x_{min}}\)
Since,
\(u=log(cosh(x-x_z))\)
\(\cfrac { e^{ u }(e^{ 2u }-1)^{3/2} }{ u^{ 2 }(2e^{ 2u }-1) }=\cfrac { cosh(x-x_z)sinh^{ 3 }(x-x_z) }{ ln^{ 2 }(cosh(x-x_z))(2cosh^{ 2 }(x-x_z)-1) } =\cfrac { sinh(2(x-x_z))sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z))cosh(2(x-x_z)) } =tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) } \)
All \(v^2_{min}\) for \(\psi_c\gt0\) lie on this line,
\(y=tanh(2(x-x_z))\cfrac { sinh^{ 2 }(x-x_z) }{ 2ln^{ 2 }(cosh(x-x_z)) } \)
For each elevation \(\psi_c\) in energy, the particle orbits at different radius with different velocity. The minimum velocities squared of all particles elevated to different \(\psi_c\)s are given by the curve above. If energy above this minimum \(KE\) represented by the curve is emitted, then this curve is the baseline of the emission spectrum. This is emission across a continuous spectrum not in packets of discrete frequencies.
How?? In packets of \(\psi\), at low speed, as the particle's photon equivalent, but across a continuous spectrum as v^2 is continuous.
The particle is in circular motion around with velocity, \(v_{min}\) in a direction perpendicular to \(x\). Only when \(x\to x_z^{-}\) does the particle switch direction and travel along \(x\) with great velocity.
The steep values of \(v\) on either side of \(x=x_z\) restrict the location of the particle to small values around \(x=x_z\). This is a Quantum Well except that \(x=x_z\) is also forbidden. They are two Quantum Wells side by side. It is however possible to escape from the lower Quantum Well in the region \(x\lt x_z\) by approaching \(x=x_z\). At that point \(iv\) is rotated by the \(-i\) factor and the particle speed beyond \(x=x_z\) along the radial line, in the direction of \(x\) into the adjacent Quantum Well at very high speed. This is quantum tunneling, but it is possible only from the well at lower region \(x\lt x_z\), to the well at higher region \(x\gt x_z\).
How is \(iv\) applied as the particle approachs \(x_z^{-}\)? Does the particle turn slowly transcribing a spiral or does it turn abruptly near \(x_z^{-}\).
Note:
Do not try to take the square root of \(-i\), rotate \(v\) by \(i\) for the first negative sign, then rotate \(v\) by \(-i\) as the result of taking the limit to approach \(x_z^{-}\).
