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--- un:electric-drift [2024/10/29 23:35] – created asad
+++ un:electric-drift [2024/11/06 11:16] (current) – asad
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 ====== Electric Drift ======
-A charged particle will move in a circular path in a magnetic field if there is no component of its velocity along the field. If there is a component along the field, it results in helical motion, known as [[em-gyration|electromagnetic gyration]]. On top of this gyration, electric drift occurs if the electric field changes with time or is non-uniform in space. Due to the drift, the helical path of the gyrating particle shifts in a specific direction. We will discuss two types of drifts: E-cross-B drift (cross-field drift) and polarization drift.
+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 [[em-gyration|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.
 ===== - E-cross-B Drift =====
-Consider a charged particle gyrating in a magnetic field with an external electric field also present. The two components of the particle's velocity equation can be written separately: one parallel to the magnetic field and one perpendicular to it. The parallel component is
+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 acceleration along the magnetic field. However, in most plasmas of the solar system, it is difficult to maintain an electric field parallel to the magnetic field because fast-moving electrons along the magnetic field lines cancel out the electric field. If the perpendicular electric field to the B-field lies along the x-axis, the two perpendicular components of the velocity equation become
+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
 $$\begin{align*}
@@ Line 14: / Line 16: @@
 \end{align*}$$
-where  $\dot{v}_x = (qB/m)v_y+qE_x/m$  and  $\omega_g = qB/m$ . For the y-component, there is no electric field. By differentiating the two equations above, we get
+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
 $$\begin{align*}
 & \ddot{v}_x = - \omega_g^2 v_x \\
-& \ddot{v}_y = - \omega_g^2 \left(v_y + \frac{E_x}{B}\right)
+& \ddot{v}_y = - \omega_g^2 \left(v_y+\frac{E_x}{B}\right)
 \end{align*}$$
-These equations resemble the [[em-gyration|gyration equations]] if we assume  $v_y + E_x/B \equiv v_y'$ . Thus, these two equations also describe a gyration, although 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 velocity
+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
- $\mathbf{v}_E = \frac{\mathbf{E} \times \mathbf{B}}{B^2}$
+ $\mathbf{v}_E = \frac{\mathbf{E}\times\mathbf{B}}{B^2}$
-which does not depend on the electric charge, so the drift is the same for all types of charges. The video below explains this:
+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.
 {{https://svs.gsfc.nasa.gov/vis/a000000/a004200/a004265/EBvOrthogonal.HD1080.webm?600}}
-Here, the E-field is perpendicular to the B-field, and the plane of gyration for both an ion and an electron is shown. As ions accelerate along the direction of the E-field and decelerate in the opposite direction, the opposite happens for electrons. However, since electrons gyrate in the opposite direction, both ions and electrons ultimately drift in the same direction. Using an ion as an example makes this clearer: when the ion moves along the E-field, its gyro-radius increases due to acceleration, and decreases when moving in the opposite direction. As a result of this alternating increase and decrease, the center of the gyro-orbit, i.e., the guiding center, starts to drift. The drift direction is along the cross product of the E and B fields, i.e., perpendicular to both fields.
+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 true explanation of the drift lies in the Lorentz transformation of the electric field in the frame of the moving charge. In the moving frame, the transformed field is
+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
- $\mathbf{E}' = \mathbf{E} + \mathbf{v} \times \mathbf{B} = 0$
+ $\mathbf{E}' = \mathbf{E} + \mathbf{v}\times\mathbf{B} = 0$
-because the charge is considered free. Therefore,
+because the charge is assumed to be free. Therefore,
- $\mathbf{E} = -\mathbf{v} \times \mathbf{B}$
+ $\mathbf{E} = -\mathbf{v}\times\mathbf{B}$
-Taking the cross product of both sides with the B-field yields the drift velocity.
+Taking the cross product of both sides with the B-field gives the drift velocity.
 ===== - Polarization Drift =====
-Another drift relation can be derived directly from the equation of motion of a charge in the presence of E and B fields. Taking the cross product of both sides of the equation of motion with  $\mathbf{B}/B^2$ , we get
-$$\begin{align*}
+From the equation of motion of a charge in E and B fields, the above equation and another drift relation can be derived directly. Taking the cross product of both sides of the equation of motion with $\mathbf{B}/B^2$, we get
-& m\frac{d\mathbf{v}}{dt} \times \frac{\mathbf{B}}{B^2} = q\frac{\mathbf{E} \times \mathbf{B}}{B^2} + \frac{q}{B^2} (\mathbf{B} \times \mathbf{v} \times \mathbf{B}) \\
-& \Rightarrow \frac{m}{q} \frac{d\mathbf{v}}{dt} \times \frac{\mathbf{B}}{B^2} = \frac{\mathbf{E} \times \mathbf{B}}{B^2} + \frac{q}{B^2}[(\mathbf{B} \cdot \mathbf{B})\mathbf{B} - (\mathbf{B} \cdot \mathbf{v})\mathbf{B}] \\
-& \Rightarrow \mathbf{v} - \frac{\mathbf{B} (\mathbf{v} \cdot \mathbf{B})}{B^2} = \frac{\mathbf{E} \times \mathbf{B}}{B^2} - \frac{m}{q} \frac{d\mathbf{v}}{dt} \times \frac{\mathbf{B}}{B^2}
-\end{align*}$$
-where the two terms on the left together represent a perpendicular velocity vector, and the first term on the right is the E-cross-B drift. If we assume that all temporal changes during a gyro-period are negligible, this perpendicular velocity can be thought of as the perpendicular drift velocity $\mathbf{v}_d$. Thus,
+\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*}
-$$ \mathbf{v}_d = \mathbf{v}_E - \frac{m}{qB^2} \frac{d}{dt}(\mathbf{v} \times \mathbf{B}) = \mathbf{v}_E + \frac{1}{\omega_g B} \frac{d\mathbf{E}_\perp}{dt} $$
+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,
-Since we know from the Lorentz transformation that  $\mathbf{v} \times \mathbf{B} = -\mathbf{E}$  and the gyro-frequency  $\omega_g = qB/m$ , the last term on the right side of the equation is known as polarization drift:
+ $\mathbf{v}_d = \mathbf{v}_E - \frac{m}{qB^2} \frac{d}{dt}(\mathbf{v}\times\mathbf{B}) = \mathbf{v}_E + \frac{1}{\omega_g B} \frac{d\mathbf{E}_\perp}{dt}$
+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
  $\mathbf{v}_p = \frac{1}{\omega_g B} \frac{d\mathbf{E}_\perp}{dt}$
-This drift differs from E-cross-B drift in at least two ways: it is proportional to the particle’s mass and also depends on its charge; both of these terms are related to the gyro-frequency.
+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.