# Criteria for electrostatic approximation in microscopic/kinetic plasmas

I started pondering about the electrostatic approximation in relation to plasmas, where a particle's motion $$\mathbf x(t)$$ is governed by the Lorentz force:

$$m\ddot{\mathbf x}=q(\mathbf E+\dot{\mathbf x}\times\mathbf B).$$

It is clear that when the $$\mathbf B$$-field is constant, or at least slowly varying, Maxwell's equations de-couple such that the $$\mathbf E$$-field can be solved independently of the $$\mathbf B$$-field,

$$\nabla\times\mathbf E =\mathbf 0, \\ \nabla\cdot\mathbf E = \frac{\rho}{\varepsilon_0},$$

Usually, of course, one introduces the electric potential and solves the Poisson equation. So far all fine. But this does not mean, however, that $$\mathbf B=\mathbf 0$$. Indeed, the $$\mathbf B$$-field is given from the remaining Maxwell equations once the $$\mathbf E$$-field is solved for:

$$\nabla\times\mathbf B=\mu_0\mathbf J + \mu_0\varepsilon_0\frac{\partial\mathbf E}{\partial t} \\ \nabla\cdot\mathbf B=0$$

Clearly, if there is a non-negligible $$\mathbf B$$-field, this should be included in the Lorentz force when calculating the trajectory of the particle. However, I usually see that when the electrostatic approximation is applied, the $$\mathbf B$$-field is usually ignored in the Lorentz force.

Note that I do not talk about external fields, which may come in addition, but fields self-consistently determined from the particles in the plasma, i.e., $$\rho$$ and $$\mathbf J$$ given by the collection of particles.

I suppose there must be some condition for neglecting $$\mathbf B$$ in the Lorentz force, and I would expect this condition to follow from considering magnitudes of the various terms in the equations, but I'm not sure what this condition is. Slow particles such that the current is small? But slow compared to what? Speed of light?

Any help appreciated.

Edit:

Since this is a kinetic plasma the sources can be expressed as a sum of contribution due to each particle in the plasma:

$$\rho=\sum_p q_p \delta(\mathbf x-\mathbf x_p) \\ \mathbf J=\sum_p q_p \dot{\mathbf x}_p \delta(\mathbf x-\mathbf x_p)$$

• There are two potential sources for the B field (neglecting a nearby magnet of current carrying device): the one from an incoming radiation, or the one created by another charge moving under the influence of the Lorentz force. In both cases, you can estimate the ratio between E and B and thus see if ignoring B in the Lorentz force is justified. – EigenDavid Jul 15 '19 at 12:02
• Incoming radiation would be an external field not created by the plasma itself, so that is irrelevant. I was initially thinking perhaps I can do some dimensional reasoning on Ampére's law, showing the B-field to be negligible, but if I could show that the magnetic part of the Lorentz force is negligible in comparison to the electric part, that would work too. But how can I do that? And what is the necessary assumption? I suppose slow particles would apply to most of the literature I read. – sigvaldm Jul 17 '19 at 7:53

I started pondering about the electrostatic approximation in relation to plasmas...

First, the term electrostatic has a specific definition in terms of macroscopic phenomena and is not the same as that for an electrostatic wave. In the former, one assumes that both $$\partial_{t} \mathbf{B} = 0$$ and $$\partial_{t} \mathbf{E} = 0$$. Generally, one also assumes no local current sources, which drops the $$\mathbf{j}$$-dependent terms. This is not necessarily required, it just happens to be a typical assumption. In the case of a plasma, this is equivalent to requiring all motion to be parallel to $$\mathbf{B}$$ which then precludes the need for the magnetic field in the Lorentz force equation.

In the case of electrostatic waves, the only thing assumed is that $$\partial_{t} \mathbf{B} = 0$$. If the wave is a linear plane wave, then Faraday's law reduces to something like $$\mathbf{k} \times \mathbf{E} = 0$$, where $$\mathbf{k}$$ is the wave vector. This is the same as saying that the wave phase is propagating along the electric field fluctuations, which means the wave is a linearly polarized, longitudinal oscillation.

I suppose there must be some condition for neglecting $$\mathbf{B}$$ in the Lorentz force, and I would expect this condition to follow from considering magnitudes of the various terms in the equations, but I'm not sure what this condition is. Slow particles such that the current is small? But slow compared to what? Speed of light?

There are a few conditions when you can neglect $$\mathbf{B}$$ in the Lorentz force equation:

• All motion is parallel to $$\mathbf{B}$$, thus $$\mathbf{v} \times \mathbf{B} = 0$$;
• $$\mathbf{B} = 0$$, that is, there is no local or external magnetic field; or
• $$\lvert \mathbf{B} \rvert \ll \lvert \mathbf{E} \rvert$$ (in Gaussian units).

In the last case, the difference in magnitude allows one to transform to a reference frame where the magnetic field is effectively zero.

• Thank you for your reply. It is well written. I do disagree, however, that $\partial_t \mathbf E=\mathbf 0$ is necessary for the electrostatic approximation even for the macroscopic case. That is the magnetostatic approximation. – sigvaldm Dec 5 '19 at 13:23
• I think I have found the answer in one of my textbooks. The usual electrostatic approximation only requires $\partial_t \mathbf B=\mathbf 0$. Hockney&Eastwood is careful enough to refer to the electrostatic plasma approximation, as when the current is small enough to not induce any significant $\mathbf B$-field. This fulfills your third condition for ignoring the magnetic part in the Lorentz force, and is a stricter condition than the usual electrostatic approximation, where a constant $\mathbf B$-field would do. Not all plasma phenomena needs this restricted definition, of course. – sigvaldm Dec 5 '19 at 13:29
• @sigvaldm - $\partial_{t} \mathbf{E} = 0$ is not strictly required, but it is generally assumed or taking in the limit when the value is small. The word static implies a lack of change and electro specifically refers to electric fields, so it's not really a stretch to find that one assumes there is no time variation in the electric field. – honeste_vivere Dec 5 '19 at 14:42
• Ah, yes. The word electrostatic is indeed often used for constant (static) electric field. I suppose the terminology is slightly ambiguous then, and I accept your point of view, although I still prefer the other one. – sigvaldm Dec 6 '19 at 7:51