The expected \(m\) in,
\(P(energy\,state\,m)=e^{-\cfrac{E_a}{k.T_{boom}}}.\cfrac{\cfrac{1}{n^m}}{m!}\)
is,
\(E[m]=\sum^\infty_1{me^{-\cfrac{E_a}{k.T_{boom}}}.\cfrac{\cfrac{1}{n^m}}{m!}}\)
\(E[m]=e^{-\cfrac{E_a}{k.T_{boom}}}\sum^\infty_1{\cfrac{m\cfrac{1}{n^m}}{m!}}\)
From,
\(\sum^\infty_0 {k\cfrac{z^k}{k!}}=ze^z=\sum^\infty_1 {k\cfrac{z^k}{k!}}\)
because the term when \(k=0\) is zero.
\(E[m]=e^{-\cfrac{E_a}{k.T_{boom}}}\sum^\infty_1{\cfrac{m\cfrac{1}{n^m}}{m!}}=e^{-\cfrac{E_a}{k.T_{boom}}}.\cfrac{1}{n}e^{\cfrac{1}{n}}\)
But,
\(\cfrac{E_a}{kT_{boom}}=\cfrac{1}{n}\)
\(E[m]=\cfrac{1}{n}\)
What happened? \(P(energy\,state\,m)\) should have been \(P(energy\,state\,m,\,\cfrac{1}{m})\), but is the result still valid.
\(P(energy\,state\,m)\) is the probability of a particle being in the \(m\) energy state, that is, with thermal energy \(m\) times its kinetic energy.
\({kT_{boom}}={m}.{E_a}\)
In this result, \(E_a\) is not the energy state of the particle. But, the sum of thermal and kinetic energy, \(E\) is,
\(E=thermal\,energy+kinetic\,energy={kT_{boom}}+E_a\)
\(E=(m+1)E_a\)
Compared to Boltzmann Distribution,
\(F({\rm {state}})\propto e^{-{\frac {E}{kT}}}\)
\(P(energy\,state\,m)\) has an explicit discrete random variable \(m\), but is to be interpreted as the reciprocal of a energy multiple and its total energy is \(m+1\) multiple of its kinetic energy at a given temperature \(T_{boom}\).
In this dream world, all matter with temperature is nuclear, to a greater or less extent, depending on the fraction of particles with the most probable speed \(v_p=v_{boom}\) given \(T_{boom}\). All matter with temperature is radiating.
Temperature particles, \(T^{+}\) and \(T^{-}\) and charged particles, \(p^{+}\) and \(e^{-}\) are likely candidates in solids, in water and other polar liquid. Gravity particles, \(g^{+}\) and \(g^{-}\) in the case of crystalline solids.
