\subsection{E × B Drift}

This subsection shows how a uniform electric field crossed with a uniform magnetic field produces a constant guiding-centre drift, derivable from the Hamilton--Jacobi equation by recognizing the harmonic nature of the transverse motion.

\dfn{Crossed-field Hamiltonian in the Landau gauge}{
A particle of mass $m$ and charge $q$ moves in a uniform electric field $\vec{E} = E_0\,\hat{\bm{y}}$ and a uniform magnetic field $\vec{B} = B_0\,\hat{\bm{z}}$. We choose the Landau gauge for the vector potential, $\vec{A} = (-B_0 y, 0, 0)$, and the scalar potential $\varphi = -E_0 y$. The curl of the vector potential,
\[
\nabla\times\vec{A}
= \left(\pdv{A_z}{y} - \pdv{A_y}{z}\right)\hat{\bm{x}}
+ \left(\pdv{A_x}{z} - \pdv{A_z}{x}\right)\hat{\bm{y}}
+ \left(\pdv{A_y}{x} - \pdv{A_x}{y}\right)\hat{\bm{z}}
= B_0\,\hat{\bm{z}},
\]
reproduces the magnetic field, and the gradient of the scalar potential gives $\vec{E} = -\nabla\varphi = E_0\,\hat{\bm{y}}$. The electromagnetic Hamiltonian for a charged particle,
\[
\mcH = \frac{1}{2m}\bigl|\vec{p} - q\vec{A}\bigr|^2 + q\varphi,
\]
becomes explicitly
\[
\mcH = \frac{1}{2m}\Bigl[\bigl(p_x + qB_0 y\bigr)^2 + p_y^2 + p_z^2\Bigr] - qE_0 y.
\]
The coordinates $x$ and $z$ do not appear in $\mcH$, so they are cyclic and their conjugate momenta $p_x$ and $p_z$ are conserved.}

\nt{Gauge choice for crossed fields. The Landau gauge $\vec{A} = (-B_0 y, 0, 0)$ differs from the gauge used in the pure magnetic-field problem (subsection A.11), where cylindrical symmetry guided the choice. Here, placing the electric field along the $y$-direction makes this particular Landau gauge algebraically convenient: because $\vec{A}$ depends only on $y$, the coordinate $x$ remains cyclic and $p_x$ is automatically conserved. This conserved momentum $\alpha_x$ couples directly into the guiding-centre position, allowing the drift velocity to emerge cleanly from the Hamilton--Jacobi formalism without solving coupled differential equations.}

\thm{$E \times B$ drift velocity from the guiding centre}{
For crossed fields $\vec{E} = E_0\,\hat{\bm{y}}$ and $\vec{B} = B_0\,\hat{\bm{z}}$, the guiding centre of the orbit lies at
\[
y_c = \frac{mE_0/B_0 - \alpha_x}{qB_0},
\]
where $\alpha_x$ is the conserved canonical $x$-momentum. The time-averaged $x$-velocity (drift velocity) is
\[
v_d = \frac{\alpha_x + qB_0 y_c}{m} = \frac{E_0}{B_0},
\]
pointing in the $\hat{\bm{x}} = \hat{\bm{E}} \times \hat{\bm{B}}$ direction. The drift velocity is universal: every particle, regardless of charge or mass, drifts at this same speed.}

\pf{Derivation of the drift velocity from the Hamilton--Jacobi equation}{
Because the Hamiltonian has no explicit time dependence, energy is conserved: $\mcH = E$. The time variable separates as $\mcS = W(x,y,z) - Et$. The coordinates $x$ and $z$ are cyclic in $\mcH$, so their conjugate momenta are constants:
\[
\pdv{W}{x} = \alpha_x,
\qquad
\pdv{W}{z} = \alpha_z.
\]
The full action takes the additive form
\[
\mcS(x,y,z,t) = \alpha_x x + W_y(y) + \alpha_z z - Et.
\]
The time-independent Hamilton--Jacobi equation $\mcH(\vec{r},\nabla\mcS) = E$ becomes
\[
\frac{1}{2m}\Bigl[\bigl(\alpha_x + qB_0 y\bigr)^2 + \bigl(\der{W_y}{y}\bigr)^2 + \alpha_z^2\Bigr] - qE_0 y = E.
\]
The term involving the only remaining unknown is isolated by solving for $\bigl(\der{W_y}{y}\bigr)^2$:
\[
\bigl(\der{W_y}{y}\bigr)^2 = 2mE - \alpha_z^2 + 2mqE_0 y - \bigl(\alpha_x + qB_0 y\bigr)^2.
\]
Expand the square $\bigl(\alpha_x + qB_0 y\bigr)^2 = \alpha_x^2 + 2\alpha_x qB_0 y + q^2B_0^2 y^2$ and collect terms by powers of $y$:
\[
\bigl(\der{W_y}{y}\bigr)^2
= -q^2B_0^2 y^2 + 2mqE_0 y - 2\alpha_x qB_0 y + 2mE - \alpha_x^2 - \alpha_z^2.
\]
Factor the linear-$y$ terms:
\[
\bigl(\der{W_y}{y}\bigr)^2
= -q^2B_0^2 y^2 + 2qB_0\bigl(mE_0/B_0 - \alpha_x\bigr)y + 2mE - \alpha_x^2 - \alpha_z^2.
\]
Complete the square on the right-hand side. Factor out $-q^2B_0^2$ from the quadratic and linear terms in $y$:
\[
-q^2B_0^2\Biggl[y^2 - \frac{2}{qB_0}\Bigl(\frac{mE_0}{B_0} - \alpha_x\Bigr)y\Biggr].
\]
Add and subtract the square of half the coefficient of $y$ inside the bracket:
\[
-q^2B_0^2\Biggl[y - \frac{1}{qB_0}\Bigl(\frac{mE_0}{B_0} - \alpha_x\Bigr)\Biggr]^2
+ q^2B_0^2\Biggl[\frac{1}{qB_0}\Bigl(\frac{mE_0}{B_0} - \alpha_x\Bigr)\Biggr]^2.
\]
The shift in brackets is the guiding-centre coordinate:
\[
y_c = \frac{mE_0/B_0 - \alpha_x}{qB_0}.
\]
Completing the square has a clear physical meaning: it reveals the equilibrium position $y_c$ at which the electric force $qE_0\,\hat{\bm{y}}$ is exactly balanced by the magnetic force arising from the guiding-centre's $x$-velocity. At $y = y_c$, the canonical momentum combination $\alpha_x + qB_0 y_c = mE_0/B_0$ gives precisely the velocity needed for the magnetic Lorentz force to cancel the electric force. Deviations from $y_c$ are therefore harmonic oscillations about this equilibrium.
Substituting back, the full equation becomes
\[
\bigl(\der{W_y}{y}\bigr)^2 + q^2B_0^2\bigl(y - y_c\bigr)^2 = 2mE - \alpha_z^2 - \alpha_x^2 + q^2B_0^2 y_c^2.
\]
The right-hand side is a constant determined by the energy and separation constants. Define $C = 2mE - \alpha_z^2 - \alpha_x^2 + q^2B_0^2 y_c^2$. Then
\[
\bigl(\der{W_y}{y}\bigr)^2 + q^2B_0^2\bigl(y - y_c\bigr)^2 = C.
\]
This is precisely the Hamilton--Jacobi equation for a harmonic oscillator in the shifted variable $Y = y - y_c$, with frequency
\[
\omega_c = \frac{|q|B_0}{m}.
\]
The $y$-motion oscillates sinusoidally about $y_c$ with the cyclotron frequency. The explicit time dependence of the coordinate follows from inverting the generating function:
\[
y(t) = y_c + A\sin\bigl(\omega_c t + \varphi\bigr),
\]
where $A$ is the oscillation amplitude determined by the total energy and $\varphi$ is a phase set by initial conditions. Trigonometric averaging over one cyclotron period $T_c = 2\pi/\omega_c$ gives
\[
\langle y \rangle = \frac{1}{T_c}\int_0^{T_c} y(t)\,\dd t
= y_c + \frac{A}{T_c}\int_0^{T_c} \sin\bigl(\omega_c t + \varphi\bigr)\,\dd t
= y_c,
\]
because the sine function integrates to zero over a complete period. The guiding-centre position is thus the exact time-average of the $y$-coordinate.

Now compute the kinematic $x$-velocity. From Hamilton's equation, $\dot{x} = \pdv{\mcH}{p_x}$:
\[
v_x = \pdv{\mcH}{p_x}
= \frac{1}{m}\bigl(p_x + qB_0 y\bigr).
\]
The canonical momentum $p_x = \pdv{\mcS}{x} = \alpha_x$ is constant. With the trigonometric average $\langle y \rangle = y_c$ established, the drift velocity follows from substituting $y_c$ into the expression for $v_x$:
\[
\langle v_x \rangle = \frac{\alpha_x + qB_0 y_c}{m}.
\]
Substitute the explicit expression for the guiding centre:
\[
qB_0 y_c = qB_0\cdot\frac{mE_0/B_0 - \alpha_x}{qB_0}
= \frac{mE_0}{B_0} - \alpha_x.
\]
The separation constant $\alpha_x$ cancels out:
\[
\alpha_x + qB_0 y_c = \alpha_x + \frac{mE_0}{B_0} - \alpha_x = \frac{mE_0}{B_0}.
\]
Dividing by $m$ gives the drift velocity:
\[
\langle v_x \rangle = \frac{E_0}{B_0}.
\]
The drift is positive in the $+x$ direction, equal to $(E_0/B_0)\,\hat{\bm{x}} = (\vec{E}\times\vec{B})/B_0^2$, and depends on neither the particle's mass nor its charge.}

\cor{Comparison with the Lorentz-force prediction}{
The steady-state solution of the Lorentz-force equation $m\dot{\vec{v}} = q(\vec{E} + \vec{v}\times\vec{B})$ for zero acceleration, $\dot{\vec{v}} = 0$, requires
\[
q\vec{E} + q\vec{v}\times\vec{B} = 0,
\qquad\text{or}\qquad
\vec{v}\times\vec{B} = -\vec{E}.
\]
Take the cross product of both sides with $\vec{B}$ from the right:
\[
(\vec{v}\times\vec{B})\times\vec{B} = -\vec{E}\times\vec{B}.
\]
Using the vector identity $(\vec{a}\times\vec{b})\times\vec{c} = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{b}\cdot\vec{c})\vec{a}$:
\[
(\vec{v}\cdot\vec{B})\vec{B} - B^2\vec{v} = -\vec{E}\times\vec{B}.
\]
Since $\vec{E} \perp \vec{B}$, the velocity component parallel to $\vec{B}$ does not contribute to the drift, and choosing $\vec{v}\cdot\vec{B} = 0$ gives
\[
\vec{v}_d = \frac{\vec{E}\times\vec{B}}{B^2}.
\]
With $\vec{B} = B_0\,\hat{\bm{z}}$ and $\vec{E} = E_0\,\hat{\bm{y}}$:
\[
\vec{v}_d = \frac{E_0 B_0\,\hat{\bm{x}}}{B_0^2}
= \frac{E_0}{B_0}\,\hat{\bm{x}}.
\]
This matches the Hamilton--Jacobi result exactly. The cross product $\vec{E}\times\vec{B}$ determines the drift direction and division by $B^2$ converts the magnitude into a velocity. Both formalisms predict the same drift regardless of the particle's charge or mass.}

\nt{Universality of the $E \times B$ drift velocity. The drift velocity $v_d = E_0/B_0$ is independent of both the particle's mass and its charge sign. A charge-sign reversal changes the sense of Larmor rotation and shifts the guiding centre $y_c$, but these two effects exactly compensate in the time-averaged $x$-velocity. An electron and a proton spiralling in the same crossed fields share the same guiding-centre drift, even though their gyroradii and cyclotron frequencies differ enormously. This universality is precisely why the result is called the $E\times B$ drift velocity: the expression $\vec{v}_d = \vec{E}\times\vec{B}/B^2$ contains no mass or charge, so every charged species drifts together. In plasma physics this means bulk plasma moves as a coherent fluid rather than separating into counter-streaming components.}

\mprop{Properties of the $E \times B$ drift}{
The drift velocity $\vec{v}_d = \vec{E}\times\vec{B}/B^2$ satisfies the following properties:

\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item Independence of charge sign. Positive and negative charges drift in the same direction at the same speed. The sign of $q$ cancels between the force $q\vec{E}$ and the Lorentz deflection $q\vec{v}\times\vec{B}$.
\item Independence of mass. Heavy ions and light electrons drift side by side at the same velocity. The mass appears in neither $\vec{E}\times\vec{B}$ nor $B^2$.
\item Direction perpendicular to both fields. The drift points along $\hat{\bm{E}}\times\hat{\bm{B}}$, orthogonal to the plane containing the two fields.
\item Magnitude depends on the field ratio. $v_d = E_0/B_0$ grows with stronger electric field and weaker magnetic field. Doubling both fields leaves the drift unchanged.
\item The guiding centre $y_c$ itself depends on $q$, $m$, and $\alpha_x$, but these dependencies cancel in the drift velocity $\langle v_x\rangle$.
\item The transverse $y$-motion is a harmonic oscillation with cyclotron frequency $\omega_c = |q|B_0/m$ about $y_c$. The full trajectory is a trochoid: the superposition of circular Larmor motion and uniform drift.
\end{enumerate}
}

\ex{Trochoidal orbit geometry}{
When the transverse kinetic energy is large compared to the electric-field energy scale $E_0$ times the gyroradius, the particle traces a cycloid-like path in the $xy$ plane while drifting in $x$. If the drift speed exceeds the thermal speed, the orbit is a prolate trochoid with open loops; at equal speeds it is a common cycloid with cusps; and at slower drift the orbit folds back on itself as a curtate trochoid. In every case the guiding centre advances uniformly at $v_d = E_0/B_0$.}

\qs{Electron in crossed electric and magnetic fields}{
An electron travels through a region with crossed fields $\vec{E} = (500\,\mathrm{V/m})\,\hat{\bm{y}}$ and $\vec{B} = (0.01\,\mathrm{T})\,\hat{\bm{z}}$. The electron has mass $m_e = 9.11\times 10^{-31}\,\mathrm{kg}$ and charge $q_e = -e = -1.60\times 10^{-19}\,\mathrm{C}$.

\begin{enumerate}[label=(\alph*)]
\item Compute the drift velocity $\vec{v}_d = \vec{E}\times\vec{B}/B^2$ and show it equals $50\,000\,\mathrm{m/s}$ in the $\hat{\bm{x}}$ direction.
\item Using the guiding-centre formula $y_c = (mE_0/B_0 - \alpha_x)/(qB_0)$, verify that the drift $\langle v_x\rangle = (\alpha_x + qB_0 y_c)/m$ is independent of both $m$ and $q$.
\item For a proton ($m_p = 1.67\times 10^{-27}\,\mathrm{kg}$, $q_p = +e = +1.60\times 10^{-19}\,\mathrm{C}$) in the same fields, confirm that the drift velocity is identical to the electron's.
\end{enumerate}}

\sol \textbf{Part (a).} The cross product of the field vectors is
\[
\vec{E}\times\vec{B}
= \bigl(500\,\mathrm{V/m}\bigr)\bigl(0.01\,\mathrm{T}\bigr)\,\hat{\bm{y}}\times\hat{\bm{z}}
= 5.00\,\mathrm{V\!\cdot\!T/m}\,\hat{\bm{x}}.
\]
The unit check: $1\,\mathrm{V\!\cdot\!T/m} = 1\,(\mathrm{V/m})/(\mathrm{T}) = 1\,\mathrm{m/s}$, because $\mathrm{T} = \mathrm{V\!\cdot\!s/m^2}$. The squared magnetic field strength is
\[
B^2 = (0.01\,\mathrm{T})^2 = 1.00\times 10^{-4}\,\mathrm{T^2}.
\]
The drift velocity is
\[
\vec{v}_d = \frac{\vec{E}\times\vec{B}}{B^2}
= \frac{5.00\,\mathrm{V\!\cdot\!T/m}}{1.00\times 10^{-4}\,\mathrm{T^2}}\,\hat{\bm{x}}.
\]
Equivalently, using the scalar ratio directly:
\[
\frac{E_0}{B_0} = \frac{500\,\mathrm{V/m}}{0.01\,\mathrm{T}}
= 50\,000\,\mathrm{m/s}.
\]
Therefore,
\[
\vec{v}_d = 50\,000\,\mathrm{m/s}\,\hat{\bm{x}}.
\]
The drift is in the $\hat{\bm{x}}$ direction, perpendicular to both $\vec{E}$ ($\hat{\bm{y}}$) and $\vec{B}$ ($\hat{\bm{z}}$), consistent with the $\hat{\bm{E}}\times\hat{\bm{B}}$ rule.

\textbf{Part (b).} The guiding-centre position is
\[
y_c = \frac{mE_0/B_0 - \alpha_x}{qB_0}.
\]
Substitute into the drift formula for the time-averaged $x$-velocity:
\[
\langle v_x\rangle = \frac{\alpha_x + qB_0 y_c}{m}.
\]
First evaluate the product $qB_0 y_c$:
\[
qB_0 y_c = qB_0\cdot\frac{mE_0/B_0 - \alpha_x}{qB_0}
= \frac{mE_0}{B_0} - \alpha_x.
\]
The factors of $qB_0$ cancel cleanly. Now the numerator of the drift is
\[
\alpha_x + qB_0 y_c = \alpha_x + \Bigl(\frac{mE_0}{B_0} - \alpha_x\Bigr)
= \frac{mE_0}{B_0}.
\]
Dividing by $m$:
\[
\langle v_x\rangle = \frac{mE_0/B_0}{m} = \frac{E_0}{B_0}.
\]
Both the charge $q$ and the mass $m$ have cancelled algebraically. The drift depends only on the field ratio $E_0/B_0$. For the numerical values of this problem:
\[
\frac{E_0}{B_0} = \frac{500\,\mathrm{V/m}}{0.01\,\mathrm{T}} = 50\,000\,\mathrm{m/s},
\]
which agrees with part~(a).

\textbf{Part (c).} For the proton, the guiding centre is
\[
y_c^{\text{(p)}} = \frac{m_p E_0/B_0 - \alpha_x^{\text{(p)}}}{q_p B_0}.
\]
The proton mass $m_p = 1.67\times 10^{-27}\,\mathrm{kg}$ is about $1837$ times the electron mass, and the proton charge $q_p = +e$ has the opposite sign. The numerical value of $y_c^{\text{(p)}}$ therefore differs substantially from the electron's guiding centre. However, the cancellation in the drift formula is purely algebraic and does not depend on the numerical values of $m$ or $q$. The proton drift is
\[
\langle v_x^{(\text{p})}\rangle = \frac{E_0}{B_0}
= \frac{500\,\mathrm{V/m}}{0.01\,\mathrm{T}}
= 50\,000\,\mathrm{m/s}.
\]
This is identical to the electron drift velocity. Every charged particle in the same crossed fields drifts at the same speed in the same direction, as predicted by both the Hamilton--Jacobi and Lorentz-force formalisms.

Therefore,
\[
\vec{v}_d = (50\,000\,\mathrm{m/s})\,\hat{\bm{x}}
\qquad\text{for both electron and proton, independent of charge and mass.}
\]