fix(HJ): A.12 — Explain gauge choice, guiding center physics, averaging, universality

concepts/advanced/crossed-fields-hj.tex

@@ -21,7 +21,7 @@ becomes explicitly
 \]
 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{Choice of gauge does not affect physical observables. The Landau gauge $\vec{A} = (-B_0 y, 0, 0)$ breaks translational symmetry in $y$ but preserves it in $x$, making $p_x$ the conserved momentum. An alternative symmetric gauge $\vec{A} = \tfrac12 B_0(-y, x, 0)$ would be natural for purely rotational problems but obscures the drift structure we want to expose here.}
+\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
@@ -76,6 +76,7 @@ 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.
@@ -88,14 +89,24 @@ This is precisely the Hamilton--Jacobi equation for a harmonic oscillator in the
 \[
 \omega_c = \frac{|q|B_0}{m}.
 \]
-The $y$-motion oscillates sinusoidally about $y_c$ with the cyclotron frequency. The time average of the position is the guiding centre, $\langle y \rangle = y_c$.
+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. Averaging over one cyclotron orbit, $\langle y \rangle = y_c$, so the drift velocity is
+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 $v_x$:
 \[
 \langle v_x \rangle = \frac{\alpha_x + qB_0 y_c}{m}.
 \]
@@ -140,7 +151,7 @@ With $\vec{B} = B_0\,\hat{\bm{z}}$ and $\vec{E} = E_0\,\hat{\bm{y}}$:
 \]
 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{A charge-sign reversal changes both the sense of Larmor rotation and the location of the guiding centre $y_c$, but these two effects exactly compensate in the averaged $x$-velocity. An electron and a proton spiralling in the same crossed fields therefore share the same guiding-centre drift, even though their individual gyroradii and rotation frequencies differ enormously. This universality is what makes the $E\times B$ drift so important in plasma physics: bulk plasma drifts as a coherent fluid.}
+\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: