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EM II Problem 27Energy Loss of a Charged Particle Moving in a Dielectric

Chris MuellerDept. of Physics, University of Florida

1 March, 2010

27. Jackson 7.26 A charged particle (charge Ze) moves at constant velocity v through amedium descried by a dielectric function (q, )/0 or, equivalently, by a conductivityfunction (q, ) = i[0 (q, )]. It is desired to calculate the energy loss per unit timeby the moving particle in terms of the dielectric function (q, ) in the approximation thatthe electric field is the negative gradient of the potential and current flow obeys Ohmslaw, J(q, ) = (q, ) E(q, ).

(a) Show that with suitable normalization, the Fourier transform of the particles chargedensity is

(q, ) =Ze

(2)3( q v)

(b) Show that the Fourier components of the scalar potential are

(q, ) =(q, )

q2(q, )

(c) Starting from dW/dt =

J Ed3x show that the energy loss per unit time can bewritten as

dWdt

= Z2

e2

43

d3

qq2

0

d

1(q, )

( q v)

[This shows that [(q, )]1 is related to energy loss and provides, by studyingcharacteristic energy losses in thin foils, information on (q, ) for solids.]

Instructors Notes: This problem employs Fourier transformation in both space andtime. The normalizations chosen for this problem seem to be

X(r, t)=

d3qdX(q, )eiqrit & X(q, ) =

d3rdt

(2)4X(r, t)eiqr+it

where X is any of the variables E, J, D, , . If we were to follow the pattern of (7.104),there would have been a factor of 1/(2)2 in both transforms instead of 1/(2)4 in onlyone of them. The space time analog of Jackson (7.105) is

D(r, t) =

d3rdtG(r r, t t) E(r, t)

where

G(r r, t t) =

d3qd

(2)4(q, )eiq(rr)

i(tt)

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.

(r, t) = Ze(r vt)

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We want to Fourier transform this charge density.

(q, ) =Ze

(2)4

d3rdt(r vt)eiqr+it

=Ze

(2)4

dtei(qv)t

=Ze

(2)3( q v)

Part bSince we are working under the approximation that the electric field is the gradient of a scaler

potential, Laplaces equation is valid.

2(q, ) =

(q, )

(q, )

One of the nice features of Fourier space is that spatial and temporal derivatives are simply algebraicmanipulations by letting iq and

t i. This equation therefore simplifies nicely to

(q, ) =(q, )

q2 (q, )

Part cWe begin with the power equation

dW

dt=

J(r, t) E(r, t)d3r

=

d3qd J(q, )ei

qrit

d3qdw E( q, )ee

qrit

d3r

=

J(q, ) E( q, )ei(

q+ q)ri(+)td3qd3qddd3r

= (2)3

J(q, ) E( q, )(q + q)ei(+)td3qd3qdd

= (2)3

J(q, ) E(q, )ei(+)td3qdd

= (2)3

i[0 (q, )] E(q, ) E(q, )ei(

+)td3qd d

Since we are working under the approximation that E can be expressed as the gradient of thescaler potential

E(q, ) = (q, ) = iq(q, )

q2(q, )

Substituting into the integral gives

dW

dt= (2)3i

q2[0 (q,

)]q2

(q, )

q2(q, )

(q, )

q2(q, )

ei(

+)td3qdd

= (2)3i

[0 (q,

)]

q2(q, )(q, )

Ze

(2)3( q v)

Ze

(2)3( + q v)

ei(

+)td3qdd

=Z2e2

i(2)3

d3q

q2

d

0 (q, )

(q,)(q, )

( q v)

What we now want to do is symmetrize the integral. We begin by breaking up the integral intotwo parts, one from to 0 and the other from 0 to . We then reverse the bounds on the

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first integral obtaining a minus sign. Since the integrand is odd in we can switch the bounds onthe first to be from 0 to and let all of the s in the integrand go to . Also, since theintegral is taken over all q, we can simultaneously send q to qq. Combining the two separateintegrands we find

0 (q, )

(q,)(q, )

0 (q,)

(q,)(q, )

=

(q,) (q, )

(q,)(q, )

Since we are working in Fourier space

(q,) = (q, )

Hence, (q,) (q, )

(q,)(q, )

= 2iIm

1

(q, )

Putting all of this back into the integrand gives

dW

dt=

Z2e2

43

d3q

q2

0

d Im

1

(q, )

( q v)

3