What Am I Doing? Pressure Lamp

If we formulate the Lagrangian along the radial line, since/assuming all  perpendicular axes have no effects

$L=T-V$

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

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

we know that,

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

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

$\cfrac { d }{ d\, t } \left\{ m_{ e }\dot { r } _{ re }+\cfrac { \partial PE_{ re }\,  }{ \partial \, \dot { r } _{ e } }  \right\}$

$\because \cfrac { d }{ d\, t } \left\{ \cfrac { \partial PE_{ re }\,  }{ \partial \, \dot { r } _{ e } }  \right\} =\cfrac { \partial \, (d\, PE_{ re })\,  }{ \partial \, \dot { r } _{ e }dt } =\cfrac { \partial \, (\cfrac { dPE_{ re } }{ dt } )\,  }{ \partial \, (\cfrac { d\, { r }_{ e } }{ d\, t } ) } =\cfrac { \partial \, PE_{ re }\,  }{ \partial \, { r }_{ e } }$

$\cfrac { d }{ d\, t } \left\{ m_{ e }\dot { r } _{ re }+\cfrac { \partial PE_{ re }\,  }{ \partial \, \dot { r } _{ e } }  \right\} \\ =m_{ e }\ddot { r } _{ re }+\cfrac { \partial PE_{ re }\,  }{ \partial \, { r }_{ e } }$

$=2\cfrac { \partial (PE_{ re })\,  }{ \partial \, { r }_{ e } }$,    since  $m_e\ddot{r_e}=\cfrac { \partial PE_{ re }\,  }{ \partial \, { r }_{ e } }$

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

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

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

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

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

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

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

So,

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

$m_{ e }\cfrac { d\, r_{ e } }{ d\, T } \cfrac { d\, r_{ e } }{ d\, t } \cfrac { d }{ dt } (\cfrac { d\, T }{ d\, r_{ e } } )=\cfrac { \partial (PE_{ re })\,  }{ \partial \, { r }_{ e } }$

${ m_{ e } } \cfrac { d\, r_{ e } }{ d\, T } \cfrac { d^{ 2 }T }{ dt^{ 2 } } =\cfrac { \partial (PE_{ re })\,  }{ \partial \, { r }_{ e } }$

where $\cfrac { d\, r_{ e } }{ d\, T }$ is negative.

This means at the kink in the  $r_e$  vs  $T$ curve, resonance occurs at  (from the post "I like SHM, Death Rays Again, Way Cool...".),

$\omega^2_o={\cfrac { \partial ^{ 2 }(PE_{ re })\,  }{ \partial \, { r }^{ 2 }_{ e } }}_{kink}= \cfrac{\partial}{\partial\,r_e}\left\{-\cfrac { d\, r_{ e } }{ d\, T } \cfrac { d^{ 2 }T }{ dt^{ 2 } }\right\} =-\cfrac { d\, r_{ e } }{ d\, T }_{kink}\cfrac{d^2}{d\,t^2} \left\{\cfrac{d\,T }{ d\,r_e}_{kink} \right\}$

per unit mass

The temperature gradient is changing as a result of applying $\cfrac{d\,T}{d\,t}$.  Resonance is obtained when the applied frequency is the value of the gradient at the kink point, times the second time derivative of the change in the reciprocal of the gradient at the kink point.

Since the material is radiating packets of energy,  it is expected to be damped,

We used the model,    $\ddot { r } _{ e }+2p\dot { r } _{ e }+\omega ^{ 2 }_or_{ e }$

When the driving force is sinusoidal,

$\omega^2_n=\omega^2_o-2p^2$

$\omega^2_o=  {-\cfrac { d\, r_{ e } }{ d\, T }}_{kink} \cfrac{d^2}{d\,t^2} \left\{\cfrac{d\,T }{ d\,r_e}_{kink}  \right\}$

where  $p$  is the damping factor.  If  $T=T_oe^{iw_at}T(r_e)$  then at maximum, ${\cfrac { d^{ 2 }T }{ dt^{ 2 }}}_{extrema} =-T_ow^2_a$  and its driving frequency = $w_a$.   ie ${\cfrac { d^{ 2 }T }{ dt^{ 2 }}}_{extrema} =-T_ow^2_a$ occurs $w_a$ times per seconds when  $T$ is applied.

At resonance, driving frequency = damped resonance frequency,

$A.T_ow^2_a-2p^2=w^2_a$

$(AT_o-1).w^2_a=2p^2$

$w^2_a=\cfrac{2p^2}{(AT_o-1)}$

provided  $AT_o>1$

where $A=  {\cfrac { d\, r_{ e } }{ d\, T }}_{kink}\cfrac{d\,T }{ d\,r_e}_{kink}=1$

So, it is possible for resonance provided,

 $T_o>1$

In a kerosene pressure lamp increasing pressure increases chemical reaction rate that in turns increases  ${\cfrac { d^{ 2 }T }{ dt^{ 2 } }_{kink}}$,  a bright glow results when resonance frequency is reached. The material in resonance glow is thorium oxide in a pressure lantern.