Citation preview

PHYSICS 880.06 (Fall 2005)

Proble Set 1 Solution

(1.1) A&M Problem 1.4 ³ ´ p dp p = −e E + ×H − , dt mc τ ˆ, H = Hz z E(t) = Re(E(ω)e−iωt ). (a)

Seek steady-state solutoin of this form p(t) = −iωp(ω) =

Re(p(ω)e−iωt ), µ ¶ p(ω) p(ω) −e E(ω) + ×H − . mc τ

µ ¶ 1 −iω + px (ω) = τ µ ¶ 1 −iω + py (ω) = τ µ ¶ 1 −iω + pz (ω) = τ

µ ¶ 1 −e Ex (ω) + py (ω)Hz , mc µ ¶ 1 px (ω)Hz , −e Ey (ω) − mc −eEz (ω).

E(ω) = Ex (ω)ˆ x + Ey (ω)ˆ y, Ey = ±iEx , Ez = 0. The solution is px

=

py pz

= =

−eτ Ex , 1 − i(ω ∓ ωc )τ ±ipx , 0,

where ωc

=

eHz . mc

The current density is j jx jy jz

p , m σ0 Ex , = 1 − i(ω ∓ ωc )τ = ±ijx , = 0, = −ne

where σ0

ne2 τ . m

=

1

(b)

From Maxwell equations, ¶ 4πiσ 1+ E, ω σ0 . 1 − i(ω ∓ ωc )τ ω2 c2

2

−∇ E = σxx = σyy

=

µ

Look for a solution of this form Ex (k, t) = E0 e−i(kz−ωt) . Plugging in, Ã ! 2 ω 1 p k 2 c2 = ω 2 1 − = ω 2 ²(ω), ω ω ∓ ωc + i/τ where

(c)

²(ω)

=

ωp2

=

²(ω)

=

ωp2 1 , ω ω ∓ ωc + i/τ 4πne2 . m

1−

For polarization Ey = iEx , 1−

ωp2 1 . ω ω − ωc + i/τ

(SKETCH/PLOT?...) Assuming ωp /ωc À 1 and ωc τ À 1, for large ω, one can rewrite the above eq. as ²(ω) = 1 −

ωp ω

1 ω ωp

ωc ωp

−

+

≈1−

i τ ωp

ωp2 ωp 1 = 1 − , ω ωωp ω2

which is positive for ω > ωp , and real solutions for k exist. For small but positive ω, one has ²(ω) = 1 +

ωp2 1 , ω ωc − ω − i/τ

which, if τ is larger and therefore the i/τ term is ignored, is positive for ω < ωc , and consequently real solutions for k exist. (d)

For ω ¿ ωc (but still > 0), ²(ω) k 2 c2 ω

ωp2 1 ωp2 ≈ , ω −ωc ωωc ωp2 2 ωp2 = ²ω 2 ≈ ω = ω, ωωc ωc k 2 c2 = ωc 2 . ωp ≈ 1−

λ = 1 cm, T = 10 kilogauss. c = 3 × 1010 cm/s, e = 4.8 × 10−10 esu. Taking a typical metalic electron density of 1023 /cm3 , the helicon frequency is f=

eH k 2 c2 1 Hc 2 Hc ω = = k = 2 2 ne 2 ne 2π mc 4πne 2π 8π 8π m

µ

2π λ

¶2 =

2

Hc 1 (104 )(3 × 1010 ) 1 = = 3.1 Hz. 2 2ne λ (2)(1023 )(4.8 × 10−10 ) (1)2