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