153 lines
8.1 KiB
TeX
153 lines
8.1 KiB
TeX
\subsection{Charged Particle in Uniform Electric Field}
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This subsection solves the Hamilton--Jacobi equation for a charged particle in a uniform electric field, showing that Jacobi's theorem reproduces the parabolic motion dictated by the constant electric force $\vec{F} = q\vec{E}$. The problem is formally identical to the projectile motion treatment in~A.06: the separation ansatz, the characteristic function, and the Jacobi inversion follow exactly the same algebra, with the gravitational acceleration $g$ replaced by $-qE_0/m$. Likewise, the uniform field between parallel plates studied in Unit~9 (e9-3) produces a constant electric force that accelerates the particle uniformly; the HJ solution presented here applies directly to that configuration as well.
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\dfn{Hamiltonian for a charged particle in a uniform electric field}{
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A particle of mass $m$ and charge $q$ in a uniform electric field $\vec{E} = E_0\,\hat{\bm{z}}$ (with $\vec{B} = 0$) is described by the scalar potential $\varphi = -E_0 z$ and zero vector potential. The Hamiltonian is
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\[
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\mcH = \frac{p_x^2 + p_y^2 + p_z^2}{2m} - qE_0 z.
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\]
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The coordinates $x$ and $y$ are absent from $\mcH$, so they are cyclic and the conjugate momenta $p_x$, $p_y$ are constants of the motion. The Hamilton--Jacobi equation for $\mcS(\vec{r},t)$ is
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\[
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\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y}\right)^2 + \left(\pdv{\mcS}{z}\right)^2\right] - qE_0 z + \pdv{\mcS}{t} = 0.
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\]}
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\nt{This problem is the electromagnetic analogue of projectile motion. The gravitational acceleration $g$ is replaced by the electric acceleration $qE_0/m$ and the direction of $\hat{\bm{y}}$ by $\hat{\bm{z}}$. The two problems are formally equivalent under the substitution $g \to -qE_0/m$.}
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\thm{Complete integral and trajectory from Jacobi's theorem}{
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The complete integral of the Hamilton--Jacobi equation for a charged particle in the uniform field $\vec{E} = E_0\,\hat{\bm{z}}$ is
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\[
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\mcS(x,y,z,t) = p_x x + p_y y + \frac{2\sqrt{2m}}{3qE_0}\left(E_z + qE_0 z\right)^{3/2} - Et,
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\]
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where $p_x$ and $p_y$ are the conserved transverse momenta and $E_z = E - (p_x^2 + p_y^2)/(2m)$. Jacobi's theorem with respect to the energy, $\pdv{\mcS}{E} = \beta_E$, yields the trajectory along the field direction:
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\[
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z(t) = v_{0z}\,t + \frac{1}{2}\,\frac{qE_0}{m}\,t^2,
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\]
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a parabola identical in form to the kinematic equation for constant acceleration.}
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\pf{Separation of the HJ equation and extraction of $z(t)$}{
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Because the Hamiltonian has no explicit time dependence, use the ansatz $\mcS = p_x x + p_y y + W_z(z) - Et$, where $p_x$, $p_y$, and $E$ are the separation constants. The partial derivatives are
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\[
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\pdv{\mcS}{x} = p_x,
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\qquad
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\pdv{\mcS}{y} = p_y,
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\qquad
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\pdv{\mcS}{z} = \der{W_z}{z},
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\qquad
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\pdv{\mcS}{t} = -E.
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\]
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Substitute into the HJ PDE:
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\[
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\frac{1}{2m}\Bigl(p_x^2 + p_y^2 + \left(\der{W_z}{z}\right)^2\Bigr) - qE_0 z = E.
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\]
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Define the energy associated with $z$-motion, $E_z = E - (p_x^2 + p_y^2)/(2m)$, and solve for the derivative:
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\[
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\der{W_z}{z} = \sqrt{2m\left(E_z + qE_0 z\right)}.
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\]
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Integrate with respect to $z$. Set $u = E_z + qE_0 z$, so $\dd u = qE_0\,\dd z$:
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\[
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W_z(z) = \frac{\sqrt{2m}}{qE_0}\int\sqrt{u}\,\dd u
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= \frac{2\sqrt{2m}}{3qE_0}\left(E_z + qE_0 z\right)^{3/2}.
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\]
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Reassemble the principal function:
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\[
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\mcS(x,y,z,t) = p_x x + p_y y + \frac{2\sqrt{2m}}{3qE_0}\left(E_z + qE_0 z\right)^{3/2} - Et.
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\]
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Jacobi's theorem with respect to $E$ gives
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\[
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\pdv{\mcS}{E} = \frac{2\sqrt{2m}}{3qE_0}\cdot\frac{3}{2}\left(E_z + qE_0 z\right)^{1/2} - t = \beta_E,
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\]
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since $\pdv{E_z}{E} = 1$ with $p_x$ and $p_y$ held fixed. Simplify:
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\[
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\frac{\sqrt{2m}}{qE_0}\sqrt{E_z + qE_0 z} - t = \beta_E.
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\]
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The square root equals $p_z(z)/\sqrt{2m} = m v_z$ divided by $\sqrt{2m}$, so multiplying by $qE_0/\sqrt{2m}$ gives
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\[
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v_z(t) = \frac{qE_0}{m}\,(t + \beta_E).
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\]
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At $t = 0$, set $v_z(0) = v_{0z}$. Then $\beta_E = v_{0z}\,m/(qE_0)$ and
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\[
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v_z(t) = v_{0z} + \frac{qE_0}{m}\,t.
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\]
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Integrating once more with $z(0) = 0$:
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\[
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z(t) = v_{0z}\,t + \frac{1}{2}\,\frac{qE_0}{m}\,t^2.
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\]
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This is the trajectory along the field. The transverse coordinates evolve uniformly, with Jacobi's theorem applied to $p_x$ and $p_y$ giving $x(t) = (p_x/m)t + \beta_x$ and $y(t) = (p_y/m)t + \beta_y$.}
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\nt{Verification against the Lorentz force}{Newton's second law with $\vec{F} = q\vec{E} = qE_0\,\hat{\bm{z}}$ gives the component equation $m\,\dv[2]{z}{t} = qE_0$. Integrating twice subject to $z(0) = 0$ and $\dot{z}(0) = v_{0z}$ yields
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\[
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z(t) = v_{0z}\,t + \frac{1}{2}\,\frac{qE_0}{m}\,t^2,
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\]
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which matches the Hamilton--Jacobi trajectory exactly. The constant acceleration $a_z = qE_0/m$ depends on the charge-to-mass ratio and the field strength. For an electron ($q < 0$) the acceleration opposes the field direction, just as a positively charged particle accelerates along the field. The equivalence between HJ and the force-law approach holds for any time-independent potential.}
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\qs{Electron in a uniform electric field}{
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An electron ($q = -e = -1.60\times 10^{-19}\,\mathrm{C}$, $m = 9.11\times 10^{-31}\,\mathrm{kg}$) moves in a uniform electric field $\vec{E} = 1000\,\mathrm{N/C}$ directed along $+\hat{\bm{z}}$. The electron is released from rest at the origin at $t = 0$.
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\begin{enumerate}[label=(\alph*)]
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\item Write the Hamilton--Jacobi equation for this system. Show that $x$ and $y$ are cyclic coordinates and identify the corresponding separation constants.
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\item For an electron with $p_x = p_y = 0$ and $v_{0z} = 0$, find $z(t)$ using Jacobi's theorem and give the canonical $z$-momentum $p_z$ as a function of time.
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\item At $t = 1.0\,\mathrm{ns} = 1.0\times 10^{-9}\,\mathrm{s}$, compute the position $z(t)$ and kinetic energy. Compare to the $F = ma$ prediction.
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\end{enumerate}}
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\sol \textbf{Part (a).} The Hamiltonian is
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\[
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\mcH = \frac{p_x^2 + p_y^2 + p_z^2}{2m} - qE_0 z,
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\]
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with $E_0 = 1000\,\mathrm{N/C}$. The Hamilton--Jacobi equation reads
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\[
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\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y}\right)^2 + \left(\pdv{\mcS}{z}\right)^2\right] - qE_0 z + \pdv{\mcS}{t} = 0.
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\]
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Neither $x$ nor $y$ appears explicitly in the Hamiltonian, so both are cyclic. Their conjugate momenta
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\[
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\pdv{\mcS}{x} = \alpha_x,
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\qquad
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\pdv{\mcS}{y} = \alpha_y,
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\]
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are conserved separation constants. The complete integral is $\mcS = \alpha_x x + \alpha_y y + W_z(z) - Et$.
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\textbf{Part (b).} With $p_x = p_y = 0$ we have $\alpha_x = \alpha_y = 0$ and the action reduces to $\mcS = W_z(z) - Et$. The HJ equation gives
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\[
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\frac{1}{2m}\left(\der{W_z}{z}\right)^2 - qE_0 z = E.
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\]
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Solve for the canonical momentum:
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\[
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p_z(z) = \der{W_z}{z} = \sqrt{2m\bigl(E + qE_0 z\bigr)}.
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\]
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Jacobi's theorem yields the velocity $v_z(t) = \dfrac{qE_0}{m}(t + \beta_E)$. The particle starts from rest, so $v_z(0) = 0$ fixes $\beta_E = 0$ and
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\[
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v_z(t) = \frac{qE_0}{m}\,t,
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\qquad
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p_z(t) = qE_0\,t.
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\]
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Integrating $v_z(t)$ with $z(0) = 0$:
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\[
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z(t) = \frac{1}{2}\,\frac{qE_0}{m}\,t^2.
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\]
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Because $q = -e < 0$ and $E_0 > 0$, the acceleration is negative and the electron moves in the $-z$ direction.
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\textbf{Part (c).} Compute the acceleration:
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\[
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a = \frac{qE_0}{m} = \frac{(-1.60\times 10^{-19})(1000)}{9.11\times 10^{-31}}\,\mathrm{m/s^2}
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= -1.76\times 10^{14}\,\mathrm{m/s^2}.
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\]
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At $t = 1.0\times 10^{-9}\,\mathrm{s}$ the position is
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\[
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z = \frac{1}{2}\,a\,t^2
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= \frac{1}{2}\,(-1.76\times 10^{14})(1.0\times 10^{-18})\,\mathrm{m}
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= -8.8\times 10^{-5}\,\mathrm{m}.
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\]
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The speed is $|v_z| = |a|\,t = (1.76\times 10^{14})(1.0\times 10^{-9})\,\mathrm{m/s} = 1.76\times 10^{5}\,\mathrm{m/s}$. The kinetic energy is
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\[
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K = \tfrac{1}{2}\,m\,v_z^2 = \tfrac{1}{2}\,(9.11\times 10^{-31})(1.76\times 10^{5})^2\,\mathrm{J}
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= 1.4\times 10^{-20}\,\mathrm{J}.
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\]
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In electron volts, $K = (1.4\times 10^{-20})/(1.60\times 10^{-19})\,\mathrm{eV} = 0.088\,\mathrm{eV}$. From the $F = ma$ approach, $\vec{F} = q\vec{E} = (-1.60\times 10^{-19})(1000)\,\hat{\bm{z}}\,\mathrm{N} = -1.60\times 10^{-16}\,\hat{\bm{z}}\,\mathrm{N}$. The resulting acceleration $a = -1.76\times 10^{14}\,\mathrm{m/s^2}$ is identical and the integrated kinematics $z = \tfrac{1}{2}at^2$ reproduce both the position and energy exactly.
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Therefore,
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\[
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z(1.0\,\mathrm{ns}) = -8.8\times 10^{-5}\,\mathrm{m},
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\qquad
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K = 1.4\times 10^{-20}\,\mathrm{J} = 0.088\,\mathrm{eV}.
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\]
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