Pag.Pag...Pag....Pag Dnab Ygrene

From the post "Energy Band Gap....Gap...Gap.Gap",

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

It is possible to rearrange the terms to give,

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

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

$\cfrac{d\,PE_{re}}{d\,t}=\cfrac{2}{3}KE_{re}\cfrac{ d\, r_e } { d\, T }\cfrac { d^2 T }{ d\, r^2_e }\cfrac{d\, r_e}{d\,t}$

We will formulate the change of total energy,

$U=T+V=KE+(-PE)$

$\cfrac{d\,U}{d\,t}=\cfrac{d\,KE_{re}}{d\,t}-\cfrac{d\,PE_{re}}{d\,t}$

So the total change in energy is,

$\Delta U=\int{1.}d\,KE_{re}-\cfrac{2}{3}\int{KE_{re}\cfrac{ d\, r_e } { d\, T }\cfrac { d^2 T }{ d\, r^2_e }d\,r_e}$

$\Delta U= KE_{re}-\cfrac{2}{3}\int{KE_{re}\cfrac{ d\, r_e } { d\, T }\cfrac { d^2 T }{ d\, r^2_e }d\,r_e}$

$\Delta U= KE_{re}-\cfrac{2}{3}\int{KE_{re}\cfrac{ d\, r_e } { d\, T }\cfrac{d(\cfrac { d T }{ d\, r_e })}{d\,r_e}d\,r_e}$

$\Delta U= KE_{re}-\cfrac{2}{3}\int{KE_{re}\cfrac{ d\, r_e } { d\, T }d(\cfrac { d T }{ d\, r_e })}$

$KE_{re}$ is an expression in $\dot r_e$  which  in  the use of the Lagrangian is treated as an independent variable.

$\Delta U= KE_{re}\left\{1-\cfrac{2}{3}ln(\cfrac { d T }{ d\, r_e })\right \}+C$

If we focus on the kink point  $r_{e1}$  to  $r_{e1}+\varepsilon$  where  $\varepsilon$  is very small, and
let  $r_{e2}=r_{e1}+\varepsilon$.  $r_{e1}$  is just before the kink and  $r_{e2}$  is just after the kink in the  $r_e$  vs  $T$  profile.

$\Delta U= \left\{KE_{re}\left\{1+\cfrac{2}{3}ln(\cfrac { d\, r_e }{ d T })\right \}\right\}^{b}_{a}$

where  $a=\cfrac { d\, r_e }{ d T }|_{re2}$  and   $b=\cfrac { d\, r_e }{ d T }|_{re1}$

Note that the direction of integration is from higher orbit to lower orbit.

$\Delta U= \left\{KE_{re}+\cfrac{2}{3}KE_{re}ln(\cfrac { d\, r_e }{ d T })\right\}^{b}_{a}$

$\Delta U= KE_{re}|_{r{e1}}-KE_{re}|_{r{e2}}+\cfrac{2}{3}KE_{re}|_{r{e1}}ln(\cfrac{ d\, r_e } { d T }|_{r{e1}})-\cfrac{2}{3}KE_{re}|_{r{e2}}ln(\cfrac { d\, r_e } { d T }|_{r{e2}})$

 Since,  $r_e$  did not change,  $PE_{re}$  does not change.  $PE$ after all is defined as energy stored by virtue of position. All the energy of the band gap is then due to a change in  $KE_{re}$ alone.

$\Delta U=\Delta KE_{re}=KE_{re}|_{re1}-KE_{re}|_{re2}$

So,

$-\cfrac{2}{3}KE_{re}|_{r{e2}}ln(\cfrac { d\, r_e }{ d T }|_{r{e2}})+\cfrac{2}{3}KE_{re}|_{r{e1}}ln(\cfrac { d\, r_e }{ d T }|_{r{e1}})=0$

and

$KE_{re}|_{r{e2}}=KE_{re}|_{r{e1}}{ln(\cfrac{ d\, r_e }{ d T }|_{r{e1}})\over ln(\cfrac { d\, r_e }{ d T }|_{r{e2}})}$

Moving from lower to higher orbit,

$\Delta U=KE_{re}|_{re2}-KE_{re}|_{re1}=KE_{re}|_{r{e1}}\left\{{ln(\cfrac { d\, r_e }{ d T }|_{r{e1}})\over ln(\cfrac { d\, r_e }{ d T }|_{r{e2}})}-1\right\}$

$E_{BG}=\Delta U=KE_{re}|_{r{e1}}\left\{{ln(\cfrac{ d\, r_e } { d T }|_{r{e1}})\over ln(\cfrac { d\, r_e }{ d T }|_{r{e2}})}-1\right\}$

This is a much more involved quantification of Energy Band Gap.  We know that,

$\cfrac{ d\, r_e } { d T }<0$

But,  $\int{\cfrac{1}{-x}}d(-x)=ln(-x)+c$

So,   $\int{\cfrac{1}{x}}d(x)=ln(|x|)+c$

and so, knowing that we have negative gradients,

$E_{BG}=KE_{re}|_{r{e1}}\left\{{ln(|(\cfrac{ d\, r_e } { d T }|_{r{e1}})|)/ ln(|(\cfrac { d\, r_e }{ d T }|_{r{e2}})|)}-1\right\}$

we use their absolute values only.  And since,

$|(\cfrac{ d\, r_e } { d T }|_{r{e1}})|$ is large

$E_{BG}<0$,    when

$|(\cfrac { d\, r_e }{ d T }|_{r{e2}})|<1$

This is the condition for quantum emission on transit to a higher orbit.  Very strange indeed.  For most material  $r_e$  is small, the change of  $r_e$  with  $T$  is even smaller.  So this is a common phenomenon, the top part of the  $r_e$ vs  $T$ graph has a very gentle slope less then  $1$.

Losing energy going from a lower orbit to a higher one, is as if the electron is repulsed by the nucleus.