Rectified Waveform To The Rescue

Once again, at resonance, for

$p=\xi\omega_o$

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

$\omega^2_n=\omega^2_o-2\xi^2\omega^2_o=\omega^2_a$

$\omega^2_o(1-2\xi^2)=\omega^2_a$

From the post "What Am I Doing? Pressure Lamp"

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


when $T=time(t)T(r_e)$ is separable in time and $r_e$,

$\cfrac{d^2}{d\,t^2} \left\{\cfrac{d\,T }{ d\,r_e} \right\} =\cfrac{d\,T(r_e) }{ d\,r_e}_{kink} \cfrac{d^2time(t)}{d\,t^2}=\cfrac{d\,T }{ d\,r_e}_{kink}\cfrac{d^2time(t)}{d\,t^2}$

Therefore,

${ {-\cfrac { d\, r_{ e } }{ d\, T }}_{kink} \cfrac{d^2}{d\,t^2} \left\{\cfrac{d\,T }{ d\,r_e}_{kink}  \right\}(1-2\xi^2)=\omega^2_a}$ --- (*)

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

Since the driving force is electrical, it is possible to apply rectified waveform instead of a pure sinusoidal.

In the both cases the applied frequencies doubles,  $\omega_a\rightarrow2\omega_a$,  For a normal rectified waveform,

$\cfrac{d^2T}{d\,t^2}$ is always negative

$\omega_a=2\omega$

and

$\cfrac{d^2}{d\,t^2} \left\{\cfrac{d\,T }{ d\,r_e}_{kink}\right\}=-T_o\omega^2\cfrac{d\,T }{ d\,r_e}_{kink}=-\cfrac{1}{4}T_o\omega^2_a\cfrac{d\,T }{ d\,r_e}_{kink}$

So,  expression (*) becomes,

$-\cfrac { 1 }{ 4 } T_o\omega^2_a(1-2\xi^2)=\omega^2_a$

$-\cfrac { 1 }{ 4 } T_o (1-2\xi^2)=1$

$\xi ^{ 2 }=\cfrac{1}{2} (1+\cfrac { 4 }{T_{ o } } )$

it is possible that  $\xi=1$.

When the inverted rectified waveform is applied,

$\cfrac{d^2T}{d\,t^2}$ is always positive

$\omega_a=2\omega$

and

$\cfrac{d^2}{d\,t^2} \left\{\cfrac{d\,T }{ d\,r_e}_{kink}\right\}=T_o\omega^2\cfrac{d\,T }{ d\,r_e}_{kink}=\cfrac{1}{4}T_o\omega^2_a\cfrac{d\,T }{ d\,r_e}_{kink}$

$\cfrac { 1 }{ 4 } T_o\omega^2_a(1-2\xi^2)=\omega^2_a$

$\cfrac { 1 }{ 4 }T_o (1-2\xi^2)=1$

$\xi ^{ 2 }=\cfrac{1}{2} (1-\cfrac { 4 }{T_{ o } } )$

It is not possible that,  $\xi=1$, since $T_o\gt 1$

To achieve $\xi=1$ would depend on the material and may not be achievable.  If it is achievable then the material can be made very conductive with a rectified voltage at resonance frequency applied to it at low temperature.

This is because, at critical damping, the electrons are pushed once over the kink.  A short pulse of this rectified waveform at the resonance frequency will increase the number of electrons in the conduction band and make the material more conductive.  Imagine an antenna pulse with this waveform before transmitting or receiving.

If not at critical damping, the reverse voltage on the negative cycle of the applied waveform will pull some of the electron back into the valence band, reducing conductivity.

On the practical side, simply switch the waveform to the inverted rectified wave and reduce  $\omega_a$  to half of resonance frequency at a suitable material temperature.  The glow should disappear immediately as the system is now critically damped.

When we consider in terms of current,  $I$   and voltages,  $V$,  of the oscillating electrical voltage system used to drive the system, things are even simplier.  Consider,

$\cfrac{d\,T}{d\,t}=VIcos(\theta)=\cfrac{1}{2}V_oI_oe^{i2\omega_d t}cos(\theta)$

$\cfrac{d^2T}{d\,t^2}=i\omega_dV_oI_oe^{i2\omega_d t}cos(\theta)$

where  $cos(\theta)$  is the power factor, and the applied voltage is a sinusoidal

$V=V_oe^{i\omega_dt}$

The system is actually driven at twice the dial frequency,  $\omega_d$.  ie

$\omega_a=2\omega_d$

We let,

$T_o=V_oI_ocos(\theta)$    then

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

A negative sign was introduced previously because the gradient was negative in order to oscillate, in this case, the negative sign appears as a phase lag.

$\omega^2_o= \cfrac{1}{2}\omega_aT_oe^{-i\pi/2}$

At resonance,

$\omega^2_o(1-2\xi^2)=\omega^2_a$

$\cfrac{1}{2}\omega_aT_oe^{-i\pi/2}(1-2\xi^2)=\omega^2_a$

$\xi^2=\cfrac{1}{2}(1-\cfrac{\omega_a}{\cfrac{1}{2}T_oe^{-i\pi/2}})$

$\xi^2=\cfrac{1}{2}(1+\cfrac{\omega_a}{\cfrac{1}{2}T_o})$

when,

$\omega_a=\cfrac{1}{2}T_o$,    $\xi=1$

And so, the electrons moved to the conduction band and jammed.

The advantage is  $w_a$ can be low but applied over a longer time to ensure that all electrons has moved to the conduction band.

When,

$\omega_a<\cfrac{1}{2}T_o$,

$\xi<1$

and the system will glow in resonance.  This suggest that when a piece of wire is heated with a current, the wire will stop glowing when the frequency of the applied voltage is increased beyond the point,  $\cfrac{1}{2}T_o$.  All power input is then just heat, no light.

When the waveform is rectified,  $\omega_a=4\omega_d$.

Since,

$\omega^2_o=-\cfrac{d^2T}{d\,t^2}$

$\omega^2_o(1-2\xi^2)=\omega^2_a=-\cfrac{d^2T}{d\,t^2}(1-2\xi^2)=\omega^2_a$

When ${\cfrac { d^2 T }{ d\, t^2 }}$ is always negative,

  $\xi^2=\cfrac{1}{2}(1-\cfrac{\omega_a}{\cfrac{1}{4}T_o})$

$\xi$  can be reduced to zero when

$\omega_a=\cfrac{1}{4}T_o$ for a rectified waveform.

When ${\cfrac { d^2 T }{ d\, t^2 }}$ is always positive for the inverted rectified waveform,

$\xi^2=\cfrac{1}{2}(1+\cfrac{\omega_a}{\cfrac{1}{4}T_o})$

it will serve the purpose of pushing electrons into the conduction band.

$\xi=1$,    when

$\omega_a=\cfrac{1}{4}T_o$ for a inverted rectified waveform.

The factor  $e^{i2\omega_d t}$  is not involved in the algebra because we are concern only with the extrema value,  a point on the waveform and how often this point occurs, ie. its frequency.

Unfortunately, all these are hypotheses subjected to human failings and dumb mathematical mistakes.