The Problem With M

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.