Is there an approximation to the expression for the probability of a particle being in energy partition \(E_j\) ,
\(p(E_j)=\cfrac{(\eta E_j)^{j/2}e^{-\sqrt{\eta E_j}}}{j!}\)?
we can rewrite,
\(p(E_j)=e^{\large{jln\left\{\sqrt{\eta E_j}\right\}-\sqrt{\eta E_j}}}.\cfrac{1}{\Gamma(j+1)}\)
But where's \(T\), the temperature of the system?
\(T\) is in \(E_j\). In this equation there is only one expression for energy. The energy of a particle is not divided into \(E\) and \(T\), where \(T\) is part of \(E\).
Is this still useful?
\(T\) is often fixed at 298.15 K in calculations involving Boltzmann or Fermi-Dirac Distributions.
How does \(E\) related to \(T\) as we measure it? Is \(T\propto E\)?
