Introduction to Electrodynamics · Problem 11.32

Electrodynamics and Relativity — David J. Griffiths

Physics · radiation-reaction

Problem

A charged particle, traveling in from −∞-\infty along the xx axis, encounters a rectangular potential energy barrier U(x)={U0,if 0<x<L,0,otherwise.U(x) = \begin{cases} U_0, & \text{if } 0 < x < L, \\ 0, & \text{otherwise.} \end{cases} Show that, because of the radiation reaction, it is possible for the particle to tunnel through the barrier—that is, even if the incident kinetic energy is less than U0U_0, the particle can pass through. [Hint: Your task is to solve the equation a=τa˙+Fm,a = \tau\dot{a} + \frac{F}{m}, subject to the force F(x)=U0[−δ(x)+δ(x−L)].F(x) = U_0[-\delta(x) + \delta(x-L)]. Refer to Probs. 11.19 and 11.31, but notice that this time the force is a specified function of xx, not tt. There are three regions to consider: (i) x<0x < 0, (ii) 0<x<L0 < x < L, (iii) x>Lx > L. Find the general solution for a(t)a(t), v(t)v(t), and x(t)x(t) in each region, exclude the runaway in region (iii), and impose the appropriate boundary conditions at x=0x = 0 and x=Lx = L. Show that the final velocity (vfv_f) is related to the time TT spent traversing the barrier by the equation L=vfT−U0mvf(τe−T/τ+T−τ),L = v_f T - \frac{U_0}{mv_f}\left(\tau e^{-T/\tau} + T - \tau\right), and the initial velocity (at x=−∞x = -\infty) is vi=vf−U0mvf[1−11+U0mvf2(e−T/τ−1)].v_i = v_f - \frac{U_0}{mv_f}\left[1 - \frac{1}{1 + \frac{U_0}{mv_f^2}\left(e^{-T/\tau} - 1\right)}\right]. To simplify these results (since all we're looking for is a specific example), suppose the final kinetic energy is half the barrier height. Show that in this case vi=vf1−(L/vfτ).v_i = \frac{v_f}{1 - (L/v_f\tau)}. In particular, if you choose L=vfτ/4L = v_f\tau/4, then vi=(4/3)vfv_i = (4/3)v_f, the initial kinetic energy is (8/9)U0(8/9)U_0, and the particle makes it through, even though it didn't have sufficient energy to get over the barrier!]

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