Neil Bohr's $n\lambda$ Constraint

Consider the Lagrangian,

$L=T-V$

$T=\cfrac{1}{2}m_p\dot{x}^2$ --- (1)

And the Euler-Lagrange Equation,

$\cfrac{d\,}{d\,t}\left(\cfrac{\partial\, L}{\partial\,\dot{x}}\right)=\cfrac{\partial\, L}{\partial\,x}$

Because of (1),

$\cfrac{d\,}{d\,t}\left(m_p\dot{x}-\cfrac{\partial\, V}{\partial\,\dot{x}}\right)=m_p\dot{x}\cfrac{d\,\dot{x}}{d\,x}-\cfrac{\partial\, V}{\partial\,x}$

But,

$\cfrac{d\,\dot{x}}{d\,x}=\cfrac{d\,}{d\,x}\left\{\cfrac{d\,x}{d\,t}\right\}=\cfrac{d\,c}{d\,x}=0$

So,

$\cfrac{\partial\, V}{\partial\,\dot{x}}=\int{\cfrac{\partial\, V}{\partial\,x}}dt+m_p\dot{x}$

And,

$\cfrac{\partial\, V}{\partial\,x}=\cfrac{\partial\, V}{\partial\,t}\cfrac{d\,t}{d\,x}+\cfrac{\partial\, V}{\partial\,\dot{x}}\cfrac{d\,\dot{x}}{d\,x}=\cfrac{\partial\, V}{\partial\,t}\cfrac{d\,t}{d\,x}$

So,

$\cfrac{\partial\, V}{\partial\,\dot{x}}=\int{\cfrac{\partial\, V}{\partial\,t}\cfrac{d\,t}{d\,x}}dt+m_p\dot{x}$

If we take the integral over one period around the orbit,

$\cfrac{\partial\, V}{\partial\,\dot{x}}=\int^T_0{\cfrac{\partial\, V}{\partial\,t}\cfrac{d\,t}{d\,x}}d\,t+m_p\dot{x}$

$\cfrac{\partial\, V}{\partial\,\dot{x}}=\cfrac{1}{c}\Delta V+m_p\dot{x}$

where $\dot{x}=c$

$\cfrac{\partial\, V}{\partial\,\dot{x}}=\cfrac{1}{c}\left(\Delta V+m_pc^2\right)$

where $\Delta V$ is due to the phase difference in $\psi$, as the wave not of integral multiple of $n_z$ completes one revolution.  The following diagram shows the phase difference when the particle completes one revolution.  It is at the same projected position on the circumference of the orbit, but at a different height $x$ above the orbit.

$\Delta V=\cfrac { \partial \, V }{ \partial \, \dot { x }  } c-m_{ p }c^{ 2 }$

The particle is now in the same plane perpendicular to the circumference of the orbit where it started but at a different position along $x$.  The direction of $x$ is also perpendicular to the circumference of the orbit.  The velocity component of the particle along the orbit remains at $c$.

We know however, without the action of any external forces,

$V=\cfrac{1}{2}m_p\dot{x}^2$

any change in $V$ is the result of a corresponding change in $T$,  So,

$\Delta V=\cfrac { \partial \,  }{ \partial \, \dot { x }  }\left\{\cfrac{1}{2}m_p\dot{x}^2\right\} c-m_{ p }c^{ 2 }$

As $\dot{x}=c$,

$\Delta V=0$

A zero difference in $V$ implies that the phase difference in $\psi$ is also zero;  that when there is no external force,

$n=n_z=1,2,3,...$

$L_{cir}=n_z\lambda_{or}$

The orbital circumference, $L_{cir}$ is an integer multiples of wavelength, $\lambda_{or}$ such that the phase of the particle on completion of one revolution around its orbit is the same, and that it is at the same relative position above the orbital circumference when its projected position on the circumference is the same.

Without any external forces, $\psi$ is a standing wave.  Good Night!