# Degrees of freedom in M-B distributions used for solving Poisson equation

I am having some trouble to understand the Maxwell-Boltzmann probability functions or more precisely how to use it in the Poisson equation.

The problem is the following: consider the case of an ion gas (single-species) trapped in an electron beam with cylindrical symmetry. The electron beam is supposed well described through a Gaussian or Normal radial density distribution and it dominates the total charge density of the plasma. The system is bounded axially, in a potential well for instance.

Once the ionic species inside the electron beam has reached a collision equilibrium (several "thermalization time"), its density distribution may be described through a Maxwell-Boltzmann energy-density distribution. I would like to consider the three different cases: 1, 3 or 5 degrees of freedom at which the total energy of the ions is redistributed per DoF (equipartition theorem). The probability density functions for the three different cases of DoF can be expressed in terms of density per energy volume (likeliness between $$E$$ and $$E + d E$$, here in [$$m^{-3}/eV$$]):

$$f_{1 D o F}(E) d E=n_{o} \sqrt{\frac{1}{\pi E k T}} e^{\frac{-E}{k T}} dE$$

$$f_{3 D o F}(E) d E=2 n_{o} \sqrt{\frac{E}{\pi}}\left(\frac{1}{E k T}\right)^{\frac{3}{2}} e^{\frac{-E}{k T}} d E$$

$$f_{5 D o F}(E) d E=\frac{4}{3} n_{0}\left(\frac{1}{\pi}\right)^{\frac{1}{2}}(E)^{\frac{3}{2}}\left(\frac{1}{k T}\right)^{\frac{5}{2}} e^{\frac{-E}{k T}}d E$$

Now, if ones wants to solve the Poisson equation, to obtain the evolution of the potential in the radial coordinate:

$$-\nabla^{2} \varphi=\frac{e\left(q n_{i}+n_{e}\right)}{\varepsilon_{0}}$$

The density of electron $$n_{e}$$ depends on the radial coordinate and it is not a problem to solve the Poisson equation. The equation becomes non-linear notably due to the dependence of $$n_{i}$$ in $$E = q\varphi$$.

In most of the literature concerning trapped ions in electron beams, the Poisson equation is reduced to that form:

$$-\nabla^{2} \varphi=\frac{e}{\varepsilon_{0}}\left(q n_{0} e^{\left(\frac{-q\varphi}{k T}\right)}+n_{e}(r)\right)$$

I tend to think that this equation is well fitted to a Boltzmann description of the energy-density distribution. Is it possible to use M-B distribution with different DoF instead of the Boltzmann distribution? In such case the density at $$E = 0$$ [$$eV$$] is $$n_{i}(0) = 0$$ [$$m^{-3}$$]. I am trying to understand the validity of this equation and how in such case express $$n_{i}$$.

$$n_{i} \propto n_{0} E^{\alpha} e^{\frac{-E}{k T}}$$

Thank you in advance for any remark and comment!