I'm a math student and I studies PDEs which come from plasma physics, and especially the Vlasov-Poisson-Fokker-Planck system. The system can be found in this article, page 2:


The system describes the evolution of electron density $f$ in a plasma where the ion density $n_{i}$ is treated as a data and is stationary.

It is constituted by 2 equations : the first is the Fokker-Planck equation

$$ \partial_{t}f + v\partial_{x}f - E\partial_{v}f = \partial_{v}(vf+\partial_{v}f) $$

where $E$ is the electric field. $E$ is generated by the particles, and is the derivative of a potential $E=-\partial_{x}\phi$. Using the Maxwell-Gauss equation, we get :

$$ -\Delta\phi = \dfrac{n_{i}-n_{e}}{\varepsilon_{0}} $$

where $n_{e}(t,x) = \int_{\mathbb{R}} f(t,x,v) dv$ is the electron density. In my opinion, this should be the second equation of the VPFP system. But in the article, the second equation is :

$$ -\delta^{2}\Delta\phi = n_{i} - n_{e} $$ where $\delta$ is the Debye length. I don't understand which step is used to derive the Poisson equation with the Debye length (the one from the article) from mine (the one derive from Maxwell-Gauss).

Moreover, in the article above, the Debye length $\delta$ is a fixed parameter, independent of the space variable and of the densities. But the wikipedia page (https://en.wikipedia.org/wiki/Debye_length) gives the following formula :

$$ \lambda_{D} = \sqrt{\dfrac{\varepsilon_{0}k_{b}/q_{e}^{2}}{n_{e}T_{e} + \sum_{j}z_{j}^{2}n_{j}/T_{j}}} $$

which seems to depend on the ion and electron densities. I don't understand how the Debye length, which depends on the time and space, can be chosen uniform in space and in time.

How to establish this Poisson equation? And what expression should be used for the Debye length?


3 Answers 3


The Debye length is derived from the assumption of thermodynamic equilibrium. In thermodynamic equilibrium the density of ions is given by a Boltzmann distribution $$n_j(x)=n_j(r) \exp(-z_j(\phi(x)-\phi(r)),$$ with $n_j(x)$ denoting the local density of ions with valency $z_j$. Note that here $n_j(x)$ has units of concentration here, in contrast with your definition. The location $r$ is an arbitrarily chosen, static, reference point, and it is custom to choose $\phi(r)=0$ and to denote the reference concentration $n_j(r)=n_{b,j}$ as a bulk concentration. Substituting the Boltzmann distribution into the Poisson equation then yields the aptly named Poisson-Boltzmann equation $$ −\delta^2 \Delta \phi=\sum_{j=0}^N \frac{z_jn_j}{n_{b,j}}.$$ The Debye length is then given by $$ \delta=\bigg(\frac{\epsilon k_BT}{\sum_{j=0}^N n_{b,j}z_j^2}\bigg)^{1/2},$$ where the crucial distinction with your definition is in the subscript $b$ of the bulk concentration $n_{b,j}$ being a defined quantity fixed in space. This quantity is also independent of time, as we are considering thermodynamic equilibrium where time does not alter ensemble averages. Thus the derivation from thermodynamics makes the Debye length a well-defined quantity. I dont see an explicit expression for the Debye length in arXiv paper, so it seems you overlooked the subscript on the Wiki page.

While the Debye length is strictly only defined in thermodynamic equilibrium, it seems to work surprisingly well in for near-equilibrium transport theory and beyond. However, tricky issues arise when your chosen reference (bulk) state changes with time. Usually this problem is circumvented by choosing your reference state deep into a homogeneous bulk reservoir.

Note. I have no experience with Debye length in the plasma context so I do not immediately see how to derive it using two different temperatures (it also seems you are missing division factor in front of $T_e$); I assume the derivation goes along the same lines. Finally, I tried to follow your notation but it wasn't immediately clear which factors were unitless or unitful. There may be factors $\epsilon$, $k_BT$ and $e$ missing.

  • $\begingroup$ If I understand correctly, the Poisson Boltzmann equation is valid at thermodynamic equilibrium, but it is used in a dynamic, nonstationary context on the arXiv paper. How can one justify the use of the Poisson Boltzmann equation instead of the classic Poisson equation? $\endgroup$ Commented Aug 2, 2023 at 7:20
  • 1
    $\begingroup$ This may be unsatisfactory for a math student, but I dont think there is a rigorous derivation showing the range of validity for the Debye length. Most transport theory is based on the (non-explicit) assumption of near-equilibrium state parameters, where the material is assumed to be fully thermalised. A simple example breaking the assumption of a single Debye length can be given by net charge recombination: spatially homogeneous charge recombination results in a time dependent Debye length, spatially heterogeneous charge recombination results in a position and time dependent Debye length. $\endgroup$
    – Willem
    Commented Aug 3, 2023 at 19:15

I believe that equation is non-dimensionalized, and $x$ is in units of the Debye length $\delta$. If you start with the original Vlasov-Poisson equation and nondimensionalize the variables (i.e., $v$ in units of $v_{th}$ (thermal velocity), $t$ in units of plasma frequency, $x$ in units of Debye length), you end up with a $\delta^2$ in Poisson's equation, which comes from the Laplace operator. Yes, the Debye length, rigorously speaking, depends on position and time, but we're just looking for a typical length-scale to nondimensionalize the equation with. We could, for instance, average out the Debye length over space and time and choose that as the "default" value by which we nondimensionalize. Hope that makes sense. Or, we could only average it out over space at time $t=0$ and use that value.


The Debye length is the result of the equilibrium between thermal fluctuations and electrostatic forces. It defines the length scale at which charge densitiy correlations "typically" decay.

Formally, it is derived by solving Poisson equation with the charge density set equal to the Boltzmann distribution. Thus, it is a self-consistent equation for the potential.

Alternatively, it is also the stationary solution of Poisson-Nernst-Planck system of equations.

  • $\begingroup$ Thank you for your answer, but I think that my question was not clear and that there was a misunderstanding about the Poisson equation. I edited the question. $\endgroup$ Commented Jul 27, 2023 at 20:20

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