PHYSICS 880.06 (Fall 2005)
Proble Set 1 Solution
(1.1) A&M Problem 1.4
dp p p = −e E + ×H − , dt mc τ H = H z zˆ, E(t) = Re(E(ω)e−iωt ).
(a) Seek steady-state steady-state solutoin of this form p(t) = Re(p(ω )e−iωt ), p(ω) p(ω) iω p(ω ) = −e E(ω) + . −iωp ×H − mc τ
1 −iω + px (ω) = −e E x (ω) + τ 1 py (ω) = −e E y (ω ) − −iω + τ 1 −iω + pz (ω) = −eE z (ω ). τ
ˆ + E y (ω)y ˆ, E(ω ) = E x (ω )x E y = ±iE x , E z = 0. The solution is px
−eτ
=
py
1 − i(ω ∓ ωc )τ = ±ipx ,
pz
= 0,
E x ,
where ωc
=
eH z . mc
The current density is p , m σ0 = E x , 1 − i(ω ∓ ωc )τ = ±ijx , = 0,
j = −ne jx jy jz where σ0
ne2 τ . m
=
1
1 py (ω)H z , mc 1 px (ω)H z , mc
(b)
From Maxwell equations, 2
−∇ E
=
σxx = σyy
=
ω2 c2
4πiσ 1+ E, ω σ0 . 1 − i(ω ∓ ωc )τ
Look for a solution of this form E x (k, t) = E 0 e−i(kz −ωt) . Plugging in, k 2 c2
= ω2
ω p2 1 1− ω ω ∓ ωc + i/τ
= ω 2 (ω),
where ω p2 1 (ω ) = 1 − , ω ω ∓ ωc + i/τ 4πne2 2 ω p = . m (c) For polarization E y = iE x , (ω) = 1 −
ω p2 1 . ω ω − ωc + i/τ
(SKETCH/PLOT?...) Assuming ω p /ωc 1 and ωc τ 1, for large ω , one can rewrite the above eq. as ω p (ω ) = 1 − ω
1 ω ωp
ωc ωp
−
+
ω p 1 ≈1− ω ωω
i τ ωp
p
ω p2 = 1− 2, ω
which is positive for ω > ω p , and real solutions for k exist. For small but positive ω, one has (ω) = 1 +
ω p2 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), ω p2 1 ω p2 ≈ , ω −ωc ωω c ω p2 2 ω p2 = ω2 ≈ ω = ω, ωω c ωc k 2 c2 = ωc 2 . ω p
(ω ) ≈ 1 − k 2 c2 ω
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 ω eH k 2 c2 1 Hc 2 Hc = = = f = k 2π mc 4πne 2π 8π 2 ne 8π 2 ne m 2
2π λ
2
=
2
Hc 1 (104 )(3 × 1010 ) 1 = = 3.1 Hz. Hz . 2ne λ2 (2)(10 23 )(4. )(4.8 × 10−10 ) (1)2
