\(\cfrac{d\,r_e}{d\,T}\cfrac{d\,T}{d\,(d_s)}=\cfrac{d\,r_e}{d\,(d_s)}\)
A discontinuity in \(\cfrac{d\,r_e}{d\,T}\) will also result in a discontinuity in the rate of change of space density along the \(r_e\) direction. \(d_s\) is still continuous with \(r_e\) but a similar kink occurs in the graph \(d_s\) vs \(r_e\).
The following illustrative plot shows the discontinuity in red, (the actual function plotted is 1/x*(1-(1-1/x)^(1/2)).)
If drag force is directly proportional to density, \(d_s\) and velocity squared, \(v^2\)
\(F=A\,d_s v^2\) where \(A\) is a constant of proportionality.
The work done against this force (by an electron) is,
\(\int{F}d\,r_e=\int{A\,d_sv^2}d\,r_e\)
\(=A .\int { (d_{ s }) } d\, r_{ e }.v^{ 2 }-\int { A.\int { (d_{ s }) } d\, r_{ e }.2v }\, d\, r_{ e }\)
If we were to define,
\(m_{sA}=\int { (d_{ s }) } d\, r_{ e }\)
as mass per unit area of space and,
\(A_sm_s=Am_{sA}=A .\int { (d_{ s }) } d\, r_{ e} \)
where \(m_s\) is the mass of a packet of space, then the term,
\(A .\int { (d_{ s }) } d\, r_{ e }.v^{ 2 }=A_sm_sv^{ 2 }\)
is just a multiple of the of the \(KE_s\) of a packet of space with mass \(m_s\) at velocity \(v\). So, the work done against drag, in part, increases the velocity of this packet of space,
Work done against drag \(F\), \(\int{F}d\,r_e=A_{ s }\, m_{ s }v^{ 2 }-2A_{ s }\int { \, m_{ s }.v } \, d\, r_{ e }\)
At the kink point, integrating from \(re1\) to \(re2=re1+\varepsilon\) where \(\varepsilon\) is small.
\(\int^{re2}_{re1}{F}\,d\,r_e=\left\{A_{ s }\, m_{ s }v^{ 2 }\right\}^{re1+\varepsilon}_{re1}-2A_{ s }\left \{ \, m_{ s }.v \right\}^{re1+\varepsilon}_{re1}.\varepsilon\)
Work done in passing over the kink is,
\(\int{F}\,d\,r_e|_{kink}=lim_{\varepsilon\rightarrow0}\left\{\left\{A_{ s }\, m_{ s }v^{ 2 }\right\}^{re1+\varepsilon}_{re1}-2A_{ s }\left \{ \, m_{ s }.v \right\}.\varepsilon\right\}\)
\(\int{F}\,d\,r_e|_{kink}=A_{ s }m_{ s }v^{ 2 }_{re2}-A_{ s }m_{ s }v^{ 2 }_{re1}=2A_s\left\{\cfrac{1}{2}{m_{ s }v^{ 2 }_{re2}-\cfrac{1}{2}m_{ s }v^{ 2 }_{re1}}\right\} =2A_s\Delta KE_s\)
This discontinuity in velocity across the kink point is consistent with the treatment of electrons across the kink point. For the case of electrons, \(r_e \) is a constant at the kink point, so \(PE_e\) is constant and so the change in total energy is attributed wholly to a change in \(KE_e\). ie velocity is discontinuous across the kink point.
Furthermore, this work done against the drag force is equal to the change in total energy of the electron on passing the kink point, because the \(r_e\) vs \(T\) curve on which the kink reside was derived by considering the effect of temperature on drag at high velocity, in the first place. Therefore the \(KE_s\) gained by the packet of space at the kink point is equal to the band gap.
\(E_{BG}=2A_s\Delta KE_s\)
Metaphorically, the electron collided with a packet of space with mass \(m_s\) at the kink point. The packet of space gained kinetic energy, \(\Delta KE_s\) as given by the expression above, from the collision.
If we set \(2A_s=1\), then the quantum is just a packet of space with kinetic energy. This awkwardness result in part from considering electrons as point particles without the second and third dimension, but density as mass per unit volume in 3D. And, in part from the undetermined constant of proportionality in the expression for drag force.
This packet of space has a boundary delimited by non zero velocities. If space in made up of particles, then the extend over which the particles have non zero velocities constitute the body of this packet.
The unknow factor \(2A_s\) actually allows for multiple packets of space/quanta to be created at the kink from one transition of an electron across the band gap. All quanta have the same velocity but their masses that was derived from an integral of space density, can be broken down into a sum of multiple terms. This means the mass of individual quantum may not be equal, but their total sum is constrained by the integral of space density over \(r_e\).
\(\int { (d_{ s }) } d\, r_{ e }=m_{sA1}+m_{sA2}+...\)
And Alice broke the looking glass. This is counter intuitive, because we would expect a collisions involving different masses to have different velocities. In this case, part of the kinetic energy of the electron \(\Delta KE_s\), (the work done against drag), is divided directly in proportion to their masses among the quanta created. They all have the same velocity.
Does these packets of energy travel as waves?
