A charged particle will rotate in a circular path within a magnetic field if it has no velocity component in the direction of the field. If there is a component in the direction of the field, it forms a helical motion, known as electromagnetic gyration. On top of this gyration, electric drift may occur if the electric field changes over time or is not uniform in space. Due to drift, the helical path of the gyrating particle shifts in a specific direction. We will discuss two types of drift: E-cross-B drift (cross-field drift) and polarization drift.

Suppose a charge is gyrating within a magnetic field, and an external electric field is also present. Then the two components of the particle’s equation of motion can be separated: one parallel to the magnetic field and the other perpendicular. The parallel component is

$$ m\dot{v}_\parallel = qE_\parallel $$

which results in an acceleration along the magnetic field. However, in most plasmas of the solar system, a parallel electric field cannot be maintained along the magnetic field because the fast-moving electrons along the magnetic field lines cancel out this electric field. If the perpendicular E-field to the B-field is along the x-axis, the two perpendicular components of the equation of motion will be

$$\begin{align*} & \dot{v}_x = \omega_g v_y + \frac{q}{m} E_x \\ & \dot{v}_y = - \omega_g v_x \end{align*}$$ because $\dot{v}_x = (qB/m)v_y+qE_x/m$ and $\omega_g=qB/m$; there will be no E-field in the y-component. Differentiating the above two equations yields

$$\begin{align*} & \ddot{v}_x = - \omega_g^2 v_x \\ & \ddot{v}_y = - \omega_g^2 \left(v_y+\frac{E_x}{B}\right) \end{align*}$$ which looks identical to [[em-gyration|the gyration equation]] if we assume $v_y+E_x/B\equiv v_y’$. Therefore, these two equations also describe a gyration, though in this case, the guiding center of the gyration drifts in the -y direction. This drift of the guiding center is called E-cross-B drift, with a velocity of

$$ \mathbf{v}_E = \frac{\mathbf{E}\times\mathbf{B}}{B^2} $$

which does not depend on the electric charge, meaning that the drift for all types of charges is in the same direction. The explanation is provided with a video below.

Here, the E-field is perpendicular to the B-field, and we see the gyration planes of an ion and an electron. When moving in the direction of the E-field, ions accelerate, while in the opposite direction, they decelerate. In the case of electrons, the opposite occurs: they decelerate in the direction of the E-field and accelerate in the opposite direction. But since the gyration of electrons is also opposite, the final drift of both types of charges ends up in the same direction. This can be clarified further with an example of an ion. When an ion moves in the direction of the E-field, its gyro-radius increases due to acceleration, and in the opposite direction, the radius decreases. Because of this increase and decrease in radius, the center of the entire gyro-orbit, i.e., the guiding center, begins to drift in one direction. The drift direction is along the cross product of the E and B fields, perpendicular to both fields.

The actual explanation of drift can be derived from the Lorentz transformation of the electric field in the moving system of a moving charge. In the moving system, the transformed field is

$$ \mathbf{E}' = \mathbf{E} + \mathbf{v}\times\mathbf{B} = 0 $$

because the charge is assumed to be free. Therefore,

$$ \mathbf{E} = -\mathbf{v}\times\mathbf{B} $$

Taking the cross product of both sides with the B-field gives the drift velocity.

From the equation of motion of a charge in E and B fields, the above equation and another drift relation can be derived directly. Taking the cross product of both sides of the equation of motion with $\mathbf{B}/B^2$, we get

\begin{align*} & m\frac{d\mathbf{v}}{dt}\times\frac{\mathbf{B}}{B^2} = q\frac{\mathbf{E}\times\mathbf{B}}{B^2} + \frac{q}{B^2} (\mathbf{B}\times\mathbf{v}\times\mathbf{B}) \\ & \Rightarrow \frac{m}{q}\frac{d\mathbf{v}}{dt}\times\frac{\mathbf{B}}{B^2} = \frac{\mathbf{E}\times\mathbf{B}}{B^2} + \frac{q}{B^2}[(\mathbf{B}\cdot\mathbf{B})\mathbf{B} - (\mathbf{B}\cdot\mathbf{v})\mathbf{B}] \\ & \Rightarrow \mathbf{v} - \frac{\mathbf{B(\mathbf{v}\cdot\mathbf{B})}}{B^2} = \frac{\mathbf{E}\times\mathbf{B}}{B^2} - \frac{m}{q}\frac{d\mathbf{v}}{dt}\times\frac{\mathbf{B}}{B^2} \end{align*}

where the two terms on the left give a perpendicular velocity vector, and the first term on the right is the E-cross-B drift. Assuming any temporal changes within the gyro-period are negligible, this perpendicular velocity can be considered as the perpendicular drift velocity $\mathbf{v}_d$. Then,

$$ \mathbf{v}_d = \mathbf{v}_E - \frac{m}{qB^2} \frac{d}{dt}(\mathbf{v}\times\mathbf{B}) = \mathbf{v}_E + \frac{1}{\omega_g B} \frac{d\mathbf{E}_\perp}{dt} $$

because we have already seen in the Lorentz transformation that $\mathbf{v}\times\mathbf{B}=-\mathbf{E}$ and gyro-frequency $\omega_g = qB/m$; the last term on the right side of this equation is called polarization drift

$$ \mathbf{v}_p = \frac{1}{\omega_g B} \frac{d\mathbf{E}_\perp}{dt} $$

which differs from E-cross-B drift in at least two ways. This drift is proportional to the particle’s mass and also depends on its charge; both terms are found in the gyro-frequency. The direction of polarization drift is along the electric field, but it works in opposite directions for electrons and ions. The resulting current is

$$ \mathbf{j}_P = n_e e (\mathbf{v}_{Pi}-\mathbf{v}_{Pe}) = \frac{n_e(m_i+m_e)}{B^2} \frac{d\mathbf{E}_\perp}{dt} $$

which drives electrons and ions in opposite directions and ionizes the plasma. Since the mass of ions is much greater than that of electrons, polarization current primarily arises through ions, as the mass of electrons can be considered negligible.

If the electric field is inhomogeneous in both time and space, corrections might be needed in the two terms of the drift velocity $\mathbf{v}_d$. In that case, the complete relation becomes

\begin{align*} \mathbf{v}_d &= \frac{\mathbf{E}\times\mathbf{B}}{B^2} + \frac{1}{\omega_g B} \frac{d\mathbf{E}_\perp}{dt} \\ &= \left(1+\frac{1}{4}r_g^2\nabla^2\right) \frac{\mathbf{E}\times\mathbf{B}}{B^2} + \frac{1}{\omega_g B} \left(\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla\right) \mathbf{E}_\perp \end{align*}

where the first term takes the second spatial derivative ($\nabla^2$) of the cross product of the electric and magnetic fields, and the second term takes both temporal and spatial derivatives of the perpendicular electric field.

The second-order effect is called finite Larmor radius effect, which only applies when the electric field is inhomogeneous within the gyration orbit. The notation $\mathbf{v}\cdot\nabla$ indicates a convective derivative, measuring how the field changes due to the motion of the electron within the electric field. These two corrections to the electric field are used here without derivation.