\(\bar{E}=\cfrac{nRT}{N_A}\)
then,
\(n=\cfrac{\bar{E}}{kT}\)
where \(k=\cfrac{R}{N_A}\)
When we rewrite,
\(\bar{E}=E_1+E_2+...E_i...+E_{\small{N}}=\cfrac{R}{N_A}(n_1T_1+n_2T_2+...n_iT_i...+n_{\small{N}}T_{\small{N}})\)
If \(E\) is partitioned into \(N\) values ( \(N\) states) with corresponding average \(\bar{E_i}\) then
\(n_i=\cfrac{E_i}{kT_i}\) --- (!)
which is assuming,
\(n=\sum^n_1{n_i}\) and
\(\bar{E}=\sum^n_1{E_i}\)
This would not work when the particles behave as waves or partially as waves. AND when the mean value \(\bar{E_i}\) is not estimated by a simple average within the population of \(n_i\) number of particles, in the partition \(i\).
If \(pV=nT\) where \(T\) is interpreted as energy per mole, then \(T_i\) is the average energy within the partition \(i\) of number \(n_i\) with total energy \(E_i\). Equation (1) reduces to,
\(n_i=\cfrac{E_i}{T_i}\)
which is consistent. But fails when the total energy of the partition, \(E_i\) is not estimated by the \(n_iT_i\), where \(T_i\) is the average energy of the system measured, supposedly the temperature measured in kelvin of the partition. (No, the equation state one thing, the experimenter does another.)
Then comes Poisson,
\(p(\lambda,j)=\cfrac{\lambda^je^{-\lambda}}{j!}\) or its alternate/approx. form,
\(p(\lambda,j)=e^{\large{\{jln(\lambda)-\lambda-ln(\Gamma(j+1))\}}}\)
where \(\Gamma(n-1)=n!\) when \(n\) is an integer.
If particles with the same number of entanglement encounters over the a fixed time interval have the same energy. (This is not collision in the physical world, but collision in the time dimension.) The assertion here stated differently is that entanglement is the reason for energy partitioning in a system. That the energy of a system of particles occupies discrete states because of different degree of entanglement. The degree of entanglement depends on the discrete (not average) number of collisions in the time dimension over an observation interval.
We note that previously, the index \(i\) was arbitrary. But here, for \(j\),
\(p(\lambda_j,n_{j})=\cfrac{\lambda^je^{-\lambda}}{j!}\)
\(n_j\) denotes a partition where the particles have \(j\) number of collisions in the fixed time interval.
If \(T\propto\lambda\), that the average energy of a particle is proportional to the number of collision it encounters,
\(\lambda_j=a_jT_j\) --- (1)
\(p(\lambda_j,n_{j})=p(j)=p(E_j)\)
where we assume that particles with the same number of entanglement encounters over the a fixed time interval have the same energy.
\(p(E_j)=\cfrac{\lambda^je^{-\lambda}}{j!}\) --- (*)
But, \(\lambda\), the average number of collisions a particle encounters is proportional to its population size, \(n\),
\(\lambda_j=b_j.n_j\)
So, from (1),
\(\lambda^2_j=a_jb_j.T_jn_j\)
where \(a_j\) and \(b_j\) are constants of proportionallity (for the time being, in general \(\lambda(T,n)\)).
but
\(T_jn_j=E_j\)
so, \(\lambda_j=\sqrt{a_jb_j E_j}=\sqrt{\eta E_j}\)
where \(\eta=a_jb_j\)
We find that \(j=0\),
\(p(E_0)=\cfrac{(\eta E_0)^{0/2}e^{-\sqrt{\eta E_0}}}{0!}=e^{-\sqrt{\eta E_0}}\)
Luckily, when \(T=0\), \(E_0=0\) then
\(p(E_0)=e^{-\sqrt{\eta .0}}=1\)
Which is true, because there is no partitioning possible when all particles have energy zero (negative energy not allowed). The converse is also true, that when \(E_0=0\), \(p(E_0)=1\), so
\(p(E_0)=1\iff E_0=0\)
and,
\(p(E_0)=1\iff T=0\), \(\because T_i=\cfrac{E}{n_i}\)
But \(\lambda\) need not be zero, because there can still be collisions in the time dimension without net exchange of energy within the observation interval. \(\lambda\) is in time not in space.
In might seem that it is collision in space that partition a system into energy levels. This is true for particles mobile in space, as in a gas. But electrons in a semiconductor bounded to the conduction band around a nucleus do not physically collide with other electrons in space. It is entanglement, that partitions such bounded electrons. After they break free from the nucleus, they collide with the surrounding lattice (holes) and other electrons, at this point, their energy redistribute/re-partition due to collisions in space.
Have a nice day.
