Fermi–Dirac Statistics But Twisted

From the post "Beam Me Up Scotty!", $x_z$ is continuous as long as,

$m_{\rho}c^2-\int^{x_a}_{0}{\psi}dx\ge0$

where $x_a=2x_z$

If particles are distributed evenly along $x_z$, and if Pauli Exclusion Principle (Please note the next post "Discrete And Quietly, Please") is about mass not occupying the same space at the same time (and, space and time fully determine state of the particle) then, the distribution of $\psi$ along $x_z$,

$\cfrac{\partial\,\psi}{\partial\,x}=\cfrac{\partial\,\psi}{\partial\,x_z}.\cfrac{\partial\,x_z}{\partial\,x}$

As, $\cfrac{\partial\,\psi}{\partial\,x}=-F_\rho$

$\cfrac{\partial\,\psi}{\partial\,x_z}=-F_\rho.(\cfrac{\partial\,x_z}{\partial\,x})^{-1}=-i\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{z}) \right).(-1)$

$\cfrac{\partial\,\psi}{\partial\,x_z}=-i\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x_z-x) \right)$

is the distribution of $\psi$ among the masses.  Imagine all the masses line-up along state space, $x_z$ each occupying $\Delta x$, each with energy density $\psi$ around it, then the distribution of energies associated with such particles is just $\cfrac{\partial\,\psi}{\partial\,x_z}$.

This looks like "Fermi–Dirac statistics", but is not, it is the $\psi$ distribution of particles as the parameter $x_z$ of each particle varies uniformly about a mean $x_{ave}$.  If the size of such particles, ($x_a=2x_z$) is uniformly distributed around a mean value $x_{ave}$ and $0\le x_z\lt x_{z\,max}$, where $x_{z\,max}$ is given by,

$m_{\rho\,p}=m_{\rho}c^2-\int^{x_{max}}_{0}{\psi}dx=0$

since, $x_{a}=2x_{z\,max}=x_{max}$, equivalently we have,

$m_{\rho\,p}=m_{\rho}c^2-\int^{2x_{z\,max}}_{0}{\psi}dx=0$

that the particle has some mass $m_{\rho\,p}\ne0$;

then "Fermi–Dirac statistics" is a good fit for the distribution of energy density among the population as $x_z$ varies, because,

$-tanh(x)=-\cfrac{e^x-e^{-x}}{e^x+e^{-x}}=\left({1-e^{2x}}\right)\cfrac{1}{e^{2x}+1}$

where $\left({1-e^{2x}}\right)$ is the energy density of state $x$, ($x$ is the one single index to the energy states of the particle) and

$\cfrac{1}{e^{2x}+1}$, the Fermi function.

However,  given a population of particles closely distributed where $\psi$ of each particle equals to the $\psi$ around it.  We can define a quasi center of one such population $c_p$, then the distribution of $\psi(x)$ around this center is the energy distribution of the particles around it.

From the post "Not Quite The Same Newtonian Field",

$\psi_p=i{ 2{ mc^{ 2 } } }\left\{-ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }(x-x_{zp})))+ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }x_{zp}))\right\}$

$\psi_p=\psi_p(x_{zp})=\psi(x)$

The number of particles with energy $\psi(x)$ around $c_p$ on a circle of radius $x$ is,

$n(x)=2\pi x.\cfrac{1}{x_a}=\pi\cfrac{x}{x_{zp}}$

since $x_a=2x_{zp}$

What is $\psi(x)$?  $\psi(x)$ is the effect of the total $\psi_T$ within a confining circle of radius $x$ just below point $x$.

 $\psi_T=\int^x_{x_{zp}}{n(x)\psi_p(x_{zp})}\,dx=\int^x_{x_{zp}}{\pi\cfrac{x}{x_{zp}}\psi_p(x_{zp})}\,dx$

If we assume that the effect of $\psi(x)$ is simply the average of $\psi_T$ below $x$, then

$\psi(x)=\cfrac { 1 }{ 2\pi x^{ 2 } } \int _{ x_{ zp } }^{ x }{ \pi \cfrac { x }{ x_{ zp } } \psi_p (x_{ zp }) } \, dx$

Let's consider the change in $\psi(x)$ with $x$,

$\cfrac{\partial\,\psi(x)}{\partial\,x}=-\cfrac { 2 }{ 2\pi x^{ 3 } } \int _{ x_{ zp } }^{ x }{ \pi \cfrac { x }{ x_{ zp } } \psi_p (x_{ zp }) } \, dx+\cfrac { \psi _p(x_{ zp }) }{ x_{ zp } } \cfrac { 1 }{ 2x }$

$\cfrac{\partial\,\psi(x)}{\partial\,x}=-\cfrac { 2 }{ 2\pi x^{ 3 } } \left[ \pi \cfrac { x^{ 2 } }{ 2x_{ zp } } \psi_p (x_{ zp }) \right]^{x}_{x_{zp}}+\cfrac { \psi _p(x_{ zp }) }{ x_{ zp } } \cfrac { 1 }{ 2x }$

$\cfrac{\partial\,\psi(x)}{\partial\,x}=-\cfrac { 1 }{ 2x } \left[ \cfrac { 1 }{ x_{ zp } } \psi_p (x_{ zp }) \right] +\cfrac { \psi _p(x_{ zp }) }{ x_{ zp } } \cfrac { 1 }{ 2x }+\cfrac { x_{ zp } }{ 2x^{ 3 } } \psi_p (x_{ zp })$

$\cfrac{\partial\,\psi(x)}{\partial\,x}=\cfrac { x_{ zp } }{2x^{ 3 } } \psi _p(x_{ zp })$

as $\psi$ is per unit volume, over the total volume of radius $x$,

$\cfrac{\partial\,\psi_V(x)}{\partial\,x}=\cfrac { x_{ zp } }{ 2x^{ 3 } } \psi _p(x_{ zp })\cfrac{4}{3}\pi x^3$

and so, the change in energy along $x$, $\psi_V(x)$ becomes, 

$\cfrac{\partial\,\psi_V(x)}{\partial\,x}=\cfrac {2\pi }{ 3 } x_{ zp } \psi_p (x_{ zp })$

which is independent of $x$.  This implies, as we travel along $x$, the change in $\psi_V$ as we pass each particle is a constant; which is obvious.  Integrating, we have energy, $\psi_V$ about a quasi-center in a population of particles,

$\psi_V(x)=\cfrac {2\pi x }{ 3 } x_{ zp } \psi _p(x_{ zp })=\cfrac {2n(x) }{ 3 } x^2_{ zp } \psi_p (x_{ zp })$

This means, given a population of particles and the constraint that $\psi$ of each particle (taken to be at $\psi_p(x_{zp})$, centered locally) is equal to its surrounding $\psi$ (the average of $\psi$ in a confining circle just below point $x$, where the particle is at), does not restrict the values of $x_{zp}$.

The precarious point is that the value of $\psi$ just beyond a circle, ($x_{zp}$ being small), is the average of $\psi$ within the circle.  If this is true, then there is no restriction on the values of $x_{zp}$ for $\psi_V$ being non-zero, continuous, real, etc.  That, given a population of particles, each individual particle is free to take on different values of $\psi_p(x_{zp})$.  More clearly, the constraint of being in a population of particles does not affect the values of individual $\psi_p(x_{zp})$ and its corresponding $x_{zp}$.  If $x_{zp}$ is a random variable from a large population, we can assume that it follows a Normal Distribution.

$\phi (x)=\cfrac { 1 }{ \sqrt { 2\pi  }  }{ e }^{ -\cfrac { x^{ 2 } }{ 2 }  }$

from which we exclude the case of 

$m_{\rho}c^2-\int^{x_{max}}_{0}{\psi}dx=0$

since, $x_{a}=2x_{z\,max}=x_{max}$, equivalently we exclude,

$m_{\rho}c^2-\int^{2x_{z\,max}}_{0}{\psi}dx=0$

which would corresponds to a mass-less boson, $m_{\rho\,p}=0$ (cf. post "It All Adds Up"), ie

$\phi (x-x_{z\,max}=0)=\cfrac { 1 }{ \sqrt { 2\pi  }  }$

But excluding one point does not change the density distribution function of $x_{zp}$, suppose $x_{zp}$ is distributed about some average value, $x_{ave}$

$\phi (x_{zp}-x_{ave})=\cfrac { 1 }{ \sqrt { 2\pi  }  }{ e }^{ -\cfrac { (x_{zp}-x_{ave})^{ 2 } }{ 2 }  }$

We have close to a Fermi Dirac Statistics description of such particles.

$E(x_{zp})=\phi (x_{zp}-x_{ave}).\cfrac{\partial\,\psi}{\partial\,x_z}$

where the distribution of energy is,

$E(x_{zp})=\cfrac { 1 }{ \sqrt { 2\pi  }  }{ e }^{ -\cfrac { (x_{zp}-x_{ave})^{ 2 } }{ 2 }  } B.tanh\left( {A(x_{zp}-x_{ave}) }\right)$

where $x=x_{ave}$,

$A=\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }$

$B=-i\sqrt { 2{ mc^{ 2 } } }\,G$  and,

$G$ is derived from a constant of proportionality after the integration for $F_{\rho}$ from the post "Not Exponential, But Hyperbolic And Positive Gravity!".

Let $x_j=x_{zp}-x_{ave}$

As the random variable $x_j$ can be appropriately scaled without any change to the Gaussian probability density function.

 $E(x_{zp})=\cfrac { AB }{ \sqrt { 2\pi  }  }{ e }^{ -\cfrac { (Ax_j)^{ 2 } }{ 2 }  }\left({e^{2Ax_j}-1}\right)\cfrac{1}{e^{2Ax_j}+1}$

 $E(x_{zp})=i\cfrac { { G^{ 2 } } }{ \sqrt { 2\pi  }  }{ e }^{ -\cfrac { (Ax_j)^{ 2 } }{ 2 }  }\left({1-e^{2Ax_j}}\right)\cfrac{1}{e^{2Ax_j}+1}$

with $A=\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }$

We see a part similar to the Fermi Function,

$\cfrac{1}{e^{2Ax_j}+1}$

but, this is an expression for the energy distribution of a large population of particles for which $x_{zp}$ is randomly distributed by the Normal Distribution.  To obtain $E(x_{zp})$ per unit energy,

 $N(x_{zp})=\cfrac{E(x_{zp})}{\psi_p(x_{zp})}$

 $N(x_{zp})=\cfrac { { G^{ 2 } } }{2{ mc^{ 2 } } \sqrt { 2\pi  }  }{ e }^{ -\cfrac { (Ax_j)^{ 2 } }{ 2 }  }\left({1-e^{2Ax_j}}\right)\cfrac{1}{e^{2Ax_j}+1}.\cfrac{1}{ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }x_{zp}))}$

where $N(x_{zp})$ is the average number of fermions per unit energy at $x_{zp}$ per unit volume (ie. per unit $\psi_p(x_{zp})$).  And the density of states, $g(x_{zp})$ is given by,

$g(x_{zp})=\cfrac { { G^{ 2 } } }{2{ mc^{ 2 } } \sqrt { 2\pi  }  }{ e }^{ -\cfrac { (Ax_j)^{ 2 } }{ 2 }  }\left({1-e^{2Ax_j}}\right).\cfrac{1}{ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }x_{zp}))}$

with $A=\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }$, $x_j=x_{zp}-x_{ave}$ and $G$ a constant of proportionality.

It is more likely that $x_{zp}$, not affected by a large population of particles for which it is part, follows a Normal Distribution rather than being uniformly distributed over the range $0\le x_{zp}\lt x_{z\,max}$.  In both cases, we can obtained an expression similar to the Fermi Function.

$\cfrac{1}{e^{2Ax_j}+1}$

where $2A=\cfrac{1}{kT}=2\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }$

In both cases, this expression is derived readily from considering the change of $\psi$ along state space $x_z$,

$\cfrac{\partial\,\psi}{\partial\,x_z}$

neither the density of states $g(x_{zp})$, nor the Fermi Function need to be defined separately.  A plot of -e^(-(x^2)/2)*tanh(x) is show below,

Compared to when $x_{zp}$ is uniformly distributed, the tail ends of this distribution tend towards zero at extreme values of $x_{zp}$.  This is more reasonable.

Note:  In both cases, uniform or normal distribution, $x_{ave}=\cfrac{x_{z\,max}}{2}$