EM II Problem 27Energy Loss of a Charged Particle Moving in a Dielectric Chris Mueller Dept. of Physics, University of Florida 1 March, 2010 27. Jackson 7.26 A charged particle (charge Ze) moves at constant velocity v through a medium descried by a dielectric function (q, ω )/ 0 or, equivalently, by a conductivity function σ (q, ω ) = iω [ 0 − (q, ω )]. It is desired to calculate the energy loss per unit time by the moving particle in terms of the dielectric function (q, ω ) in the approximation that the electric field is the negative gradient of the potential and current flow obeys Ohm’s law, J (q, ω ) = σ (q, ω )E (q, ω ). (a) Show that with suitable normalization, the Fourier transform of the particle’s charge density is Ze δ (ω − q · v ) ρ(q, ω ) = (2π )3 (b) Show that the Fourier components of the scalar potential are φ(q, ω ) = ρ(q, ω ) q 2 (q, ω ) (c) Starting from dW/dt = J · Ed3 x show that the energy loss per unit time can be written as dW Z 2 e2 d3 q ∞ 1 − = δ (ω − q · v ) dω ω dt 4π 3 q2 0 (q, ω ) [This shows that [ (q, ω )]−1 is related to energy loss and provides, by studying characteristic energy losses in thin foils, information on (q, ω ) for solids.] Instructor’s Notes: This problem employs Fourier transformation in both space and time. The normalizations chosen for this problem seem to be X (r, t)= d3 qdωX (q, ω )eiq·r−iωt & X (q, ω ) = d3 rdt X (r, t)e−iq·r+iωt (2π )4 where X is any of the variables E, J, D, ρ, φ. If we were to follow the pattern of (7.104), there would have been a factor of 1/(2π )2 in both transforms instead of 1/(2π )4 in only one of them. The space time analog of Jackson (7.105) is D(r, t) = where G(r − r , t − t ) = d3 r dt G(r − r , t − t )E (r , t ) d3 qdω (q, ω )eiq·(r−r) −iω(t−t ) (2π )4 which is the natural 4D generalization of (7.106). Part a The charge density of a moving point charge can be expressed in terms of a delta function. ρ(r, t) = Zeδ (r − vt) 1 ω ) · E (−q. ω ) = −iq 2 q (q. t) · E (r. ω ) = ρ(q. ω ) q 2 (q. ω ) · E (q . ω ) (2π ) (2π )3 d3 q ∞ Z 2 e2 0 − (q. Laplace’s equation is valid. ω )e−i(ω +ω d qdω dω Since we are working under the approximation that E can be expressed as the gradient of the scaler potential ρ(q. ω ) (q. ω ) Part c We begin with the power equation dW = dt = = J (r. ω ) = ρ(q. We begin by breaking up the integral into two parts. ω ) Substituting into the integral gives dW = −(2π )3 i dt ω ρ(q. t)d3 r d3 q dω J (q . ω ) · E (q . ω ) = −∇φ(q. ω ) e−i(ω +ω )t d3 qdω dω [ 0 − (q. ω ) −∞ )t 3 d qdω dω What we now want to do is symmetrize the integral. ω )eeq d q d3 q dω dω d3 r )t 3 ·r −iω t d3 r )·r −i(ω +ω )t 3 = (2π )3 = (2π )3 = (2π )3 J (q . ω ) ω [ 0 − (q. ω ) (q. ω ) = Ze d3 rdtδ (r − vt)e−iq·r+iωt (2π )4 Ze = dtei(ω−q·v)t (2π )4 Ze = δ (ω − q · v ) (2π )3 Part b Since we are working under the approximation that the electric field is the gradient of a scaler potential. ρ(q. −∇2 φ(q. ω )ei(q +q · d3 q dw E (q . ω )] Ze Ze = −(2π )3 i δ (ω − q · v ) δ (ω + q · v ) e−i(ω +ω 2 3 q (−q. ω ) = ωdω δ (ω − q · v ) 3 2 i(2π ) q (−q. ω )]E (q.We want to Fourier transform this charge density. ω )e−i(ω +ω iω [ 0 )t 3 d q d3 q dω dω d qdω dω )t 3 − (q. ω )δ (q + q )e−i(ω +ω J (q. ω ) q 2 (−q. ω ) ρ(−q. ω ) · E (−q. ω )eiq ·r−iω t J (q . We then reverse the bounds on the 2 . ω )]q 2 2 2 q q (q. ω ) One of the nice features of Fourier space is that spatial and temporal derivatives are simply algebraic ∂ manipulations by letting ∇ → iq and ∂t → −iω . one from −∞ to 0 and the other from 0 to ∞. This equation therefore simplifies nicely to φ(q. −ω ) (q. ω ) E (q. since the integral is taken over all q . ω ) (−q. ω ) − 0 − (−q. ω ) = (−q. −ω ) = Hence. ω ) Since we are working in Fourier space (−q. we can simultaneously send q to q → −q . ω ) 3 . −ω ) − (q. Also. ω ) 1 (q. ω ) = −2iIm dω ω Im 0 1 δ (ω − q · v ) (q. −ω ) (q. ω ) (−q. −ω ) (q. Since the integrand is odd in ω we can switch the bounds on the first to be from 0 to ∞ and let all of the ω ’s in the integrand go to ω → ω . −ω ) (q. −ω ) (q. −ω ) − (q. ω ) (−q.first integral obtaining a minus sign. −ω ) (−q. Combining the two separate integrands we find 0 − (q. (−q. ω ) Putting all of this back into the integrand gives − dW Z 2 e2 = dt 4π 3 d3 q q2 ∞ ∗ (q.