Why is the variance of the absorbed photons greater than the emitted photons of either peaks?
The variance around the absorption peak is due to the spread of particle velocities around mode/mean \(v_m\).
The variance at the two fluorescence peaks are due to the variance in \(\theta\) as both \(x_p\) and \(x_v\) in ratio, distribute themselves around their respective mean values. \(c^2\) is a constant. Although, from the post "Emmy Noether" dated 6 Jul 2015,
\(299792458=\cfrac{1}{2}(\cfrac{\bar\alpha N_p}{\bar\alpha N_p-1})=c\)
where \(\bar\alpha\) is the average fraction energy shared, and \(N_p\) total number of particles in entanglement. Both \(\bar\alpha\) and \(N_p\) has a mean and a variance, for a specific instance of a particle.
Bullshit is otherwise fiction, if not for bulls.
