Energy Band Gap....Gap...Gap.Gap

In the case of a body subjected to heat, its heat content or temperature changes by,

$\cfrac{d\,T}{d\,t}$

Given an electron orbital radius,  $r_e$  vs  $T$  curve,

$\cfrac{d\, r_e}{d\,T}=\cfrac{d\,r_e}{d\,t}.\cfrac{d\,t}{d\,T}$

A discontinuity in  $\cfrac{d\, r_e}{d\,T}$  will result in a similar break in  $\cfrac{d\,r_e}{d\,t}$.

In this case the time information need to formulate velocity along  $r_e$  is provided for by  the rate of temperature change, $\cfrac{d\,T}{d\,t}$  which is continuous under normal circumstances.

We can also use Hamilton's principle of least action in place of the time information needed.

A electron at an orbital radius  $r_e$  from the nucleus is depicted below,

Is is assume here that the tangential orbital velocity, $v$ is near terminal speed as a result of drag and does not change with distance $x$ from the nucleus and time  $t$.  It is excluded from the Lagrangian below.

If we formulate the Lagrangian along the radial line,

$L=T-V$

$L=\cfrac{1}{2}m_ev^2_{re}-(-PE_{re})$

$PE_{re}$  is negative as zero potential is defined at  $x\rightarrow\infty$;  in this system the forces are attractive.

we know that,

$\cfrac{d}{d\,t}\left\{\cfrac{\partial\,L}{\partial\,\dot x}\right\}=\cfrac{d}{d\,t}\left\{\cfrac{\partial\,}{\partial\,\dot x}\left\{\cfrac{1}{2}m_ev^2_{re}+PE_{re}\right\}\right\}$

$=\cfrac{d}{d\,t}\left\{\cfrac{\partial\,}{\partial\,\dot x}\left\{\cfrac{1}{2}m_e(\cfrac { d\, r_{ e } }{ d\, T } \cfrac { d\, T }{ d\, x } \cfrac { d\, x }{ d\, t })^2+PE_{re}\right\}\right\}$

$=\cfrac{d}{d\,t}\left\{m_e(\cfrac { d\, r_{ e } }{ d\, T } \cfrac { d\, T }{ d\, x })^2{\dot x}+\cfrac{\partial\,PE_{re}}{\partial\,\dot x}\right\}$

$=\cfrac{\partial\,L}{\partial\,x}=\cfrac{\partial}{\partial\,x}\left\{\cfrac{1}{2}m_ev^2_{re}+PE_{re}\right\}$

$=\cfrac{\partial}{\partial\,x}\left\{\cfrac{1}{2}m_e(\cfrac { d\, r_{ e } }{ d\, T } \cfrac { d\, T }{ d\, x } \cfrac { d\, x }{ d\, t })^2+PE_{re}\right\}$

$=m_{ e }(\cfrac { d\, r_{ e } }{ d\, T } \cfrac { d\, x }{ d\, t } )^{ 2 }\cfrac { d\, T }{ d\, x } \cfrac { d^{ 2 }T }{ d\, x^{ 2 } } +\cfrac { \partial \, PE_{ re } }{ \partial \, x }$

$\because \cfrac{d^2\,r_e}{d\,xd\,t}=\cfrac{d\,(\cfrac{d\,r_e}{d\,x})}{d\,t}=0$,  $\cfrac{d^2\,x}{d\,xd\,t}=\cfrac{d\,(\cfrac{d\,x}{d\,x})}{d\,t}=0$

So,

$\cfrac { d }{ d\, t } \left\{ m_{ e }(\cfrac { d\, r_{ e } }{ d\, T } \cfrac { d\, T }{ d\, x } )^{ 2 }{ \dot { x }  }+\cfrac { \partial \, PE_{ re } }{ \partial \, \dot { x }  }  \right\} =m_{ e }(\cfrac { d\, r_{ e } }{ d\, T } \cfrac { d\, x }{ d\, t } )^{ 2 }\cfrac { d\, T }{ d\, x } \cfrac { d^{ 2 }T }{ d\, x^{ 2 } } +\cfrac { \partial \, PE_{ re } }{ \partial \, x }$

We know that  $x$  is a proxy for  $r_e$,  ie  $x=r_e$

$\cfrac { d }{ d\, t } \left\{ m_{ e }(\cfrac { d\, r_{ e } }{ d\, T } \cfrac { d\, T }{ d\, r_{ e } } )^{ 2 }{ \dot { r_{ e } }  }+\cfrac { \partial \, PE_{ re } }{ \partial \, \dot { r_{ e } }  }  \right\} =m_{ e }(\cfrac { d\, r_{ e } }{ d\, T } \cfrac { d\, r_{ e } }{ d\, t } )^{ 2 }\cfrac { d\, T }{ d\, r_{ e } } \cfrac { d^{ 2 }T }{ d\, r_{ e }^{ 2 } } +\cfrac { \partial \, PE_{ re } }{ \partial \, r_{ e } }$

since  $\cfrac { \partial \, PE_{ re } }{ \partial \, r_e }=m_e\ddot r_e=F$;  this force does work against an opposing force  $-\cfrac { \partial \, PE_{ re } }{ \partial \, r_e }$.  The result is an increase in potential energy, as the expression suggests.

$m_{ e }{ \ddot { r_{ e } }  }+\cfrac { d }{ d\, t } \left\{ \cfrac { \partial \, PE_{ re } }{ \partial \, \dot { r_{ e } }  }  \right\} =m_{ e }(\cfrac { d\, r_{ e } }{ d\, t } )^2\cfrac { d\, r_{ e } }{ d\, T }\cfrac { d^{ 2 }T }{ d\, r^{ 2 }_e } +\cfrac { \partial \, PE_{ re } }{ \partial \,r_e }$

Using Schwartz Theorem,

$\cfrac{\partial}{\partial\,\dot r_e}\left\{\cfrac{d\,PE_{re}}{d\,t}\right\}=m_e\dot r^2_{ e }\cfrac { d\, r_{ e } }{ d\, T }\cfrac { d^2 T }{ d\, r^2_e }$

As when  $\dot r_e=0$,  at turning point for,  $r_e$,  $\cfrac{d\,PE_{re}}{d\,t}=0$,  we have

$\cfrac{d\,PE_{re}}{d\,t}=\cfrac{1}{3}m_e\dot r_{ e }^3\cfrac{ d\, r_{ e } }{ d\, T }\cfrac { d^2 T }{ d\, r^2_e }$

The unit dimension of this expression is consistent.  The expression implies that, as the electron moves away from the nucleus, $PE_{re}$ although always minimum (ie.  at  $r_e$,  $PE_{re}$ is a local minima ),  is moving from curve to curve with different mimimum value.  More importantly, the term  $\cfrac { d\, r_{ e } }{ d\, T }\cfrac { d^2 T }{ d\, r^2_e }$  suggests that there is a discontinuity in  $PE_{re}$ as the electron perform SHM about a mean orbit  $r_e$ over the kink in the profile.

Notice that in the derivation of  $\cfrac{d\,PE_{re}}{d\,t}$  we used the proxy relationship,

$v_{re}=\cfrac { d\, r_{ e } }{ d\, T } \cfrac { d\, T }{ d\, x } \cfrac { d\, x }{ d\, t }$

where  $x$  is actually  $r_e$.   This method delays the immediate cancellation of  certain terms and allows the derivation to evolve until it is convenient to apply the substitution,  $x=r_e$.