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I'm learning about $\mathbf{E}\times \mathbf{B}$ drifts in plasmas. The drift speed is given by $\mathbf{v}=\mathbf{E}\times \mathbf{B}/B^2$. Chen (in Introduction to Plasma Physics and Controlled Fusion) derives this by considering $\ddot{\mathbf{v}}=\mathbf{F}$ where $\mathbf{F}$ is the Lorentz force on a single particle of charge $q$, and sets this equal to zero for steady velocities. $$\ddot{\mathbf{v}}=0=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})\Rightarrow\mathbf{E}+\mathbf{v}\times\mathbf{B}=0.$$ He then uses vector identities to get the final answer, but it's clear that solving such an equation will indeed lead to a velocity independent of everything but $\mathbf{E}$ and $\mathbf{B}$.

This implies that a neutral particle with $q=0$ must move with this velocity in such a field. This should be wrong because a neutral particle should not interact with the field. What is going on?

honeste_vivere
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xln
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2 Answers2

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  1. By inspection, you may verify that $\mathbf{v}_E=\frac{\mathbf{E}\times\mathbf{B}}{B^2}$ is one valid solution to the differential equation for an electrically neutral particle $$m\dot{\mathbf{v}}=0(\mathbf{E}+\mathbf{v}\times\mathbf{B})=0.$$ But any constant $\mathbf{v}_0$ is a solution to this equation! This is in agreement with the uniqueness theorem for a first order linear differential equation in $\mathbf{v}$, when there are no specified boundary conditions. In order to get this equation to enforce $\mathbf{v}=\mathbf{v}_E$ for a particle with $q=0$, you'd need to divide by zero.

    The important point that your derivation uses is that they look for values of $\mathbf{v}$ that, when placed in the (nontrivial) Lorentz force equation for a nonzero charge, give $\dot{\mathbf{v}}=0$.

  2. Plasmas are defined to have low populations of non-ionized particles (you may read about the Saha equation, though it isn't valid when the Debaye length decreases). It's often convenient to say that most particles are moving with that velocity (remember that both positive and negative charges have the same $\mathbf{v}_E$).

Rishi
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The generalized equation is actually charge-dependent and given by: $$ \mathbf{V}_{F \perp} = \frac{ \mathbf{F} \times \mathbf{B} }{ q \ B^{2} } \tag{0} $$ where $\mathbf{F}$ is any external force, $\mathbf{B}$ is the magnetic field vector, and $q$ is the charge of the particle. For the Lorentz force, the $\mathbf{F}$ term goes to $q \mathbf{E}$.

This implies that a neutral particle with q = 0 must move with this velocity in such a field. This should be wrong because a neutral particle should not interact with the field. What is going on?

If the charge is zero, then the Lorentz force is zero so there will be no perpendicular drift or drift at all associated with this.

There is a slight caveat to this in a weakly ionized plasma though. Charge-neutral collisions can induce their own type of drifts that include neutral particle drifts (e.g., see discussions at https://physics.stackexchange.com/a/363523/59023 and https://physics.stackexchange.com/a/312306/59023). But again, this requires the influence/presence of charged particles.

As an aside, the ExB-drift arises from the frozen-in approximation (see derivation at https://physics.stackexchange.com/a/452325/59023), which is an indirect way of expressing a Lorentz transformation to a frame with no net electric field. That is, the particles have bulk motions transverse to the magnetic field only because of an electric field. Or conversely, there exists a frame of reference where the magnetic flux is conserved because there is no net electric field.

honeste_vivere
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