Download Energy densities in a free electron gas

Energy densities in a free electron gas
A plane, monochromatic, transverse electromagnetic wave propagates in a medium containing a
density ne of free electrons, which oscillate in the EM fields with negligible friction.
Calculate
a) the dispersion relation of the wave, the phase (vp ) and group (vg ) velocities and the relation
between the amplitudes of the electric (E0 ) and magnetic (B0 ) fields;
b) the EM energy density uEM (averaged over an oscillation period) as a function of E0 ;
c) the kinetic energy density uK (averaged over an oscillation period), defined as uK = ne me hv2 i /2
where v is the electron oscillation velocity, and the total energy density u = uEM + uK .
d) Consider the transmission of an EM wave incident from vacuum on the plane boundary of the
free electron medium, in conditions for which vg and vp are real quantities. Use the above results to
verify the conservation of the energy flux expressed by the relation
c(ui − ur ) = vg ut ,
(1)
where ui , ur and ut are the total energy densities for the incident, reflected and transmitted waves,
respectively.
1
Solution
a) We use the complex representation for all fields, i.e. A(x, t) = Re(Ãeikx−iωt ). For the electric field
of the wave we have Ẽ = E0 and E0 can be taken as a real number. The equation of motion for an
electron (neglecting the nonlinear magnetic term) is
me
dv
d2 r
= me
= −eE ,
2
dt
dt
(2)
which brings for the velocity and displacement
ṽ = −
ie
E0 ,
me ω
r̃ =
e
E0 .
me ω 2
(3)
The polarization density is
P̃ = −ene r̃ = −
ωp2
n e e2
E
=
−ε
0
0 2 E0 ≡ ε0 χ(ω)E0 ,
me ω 2
ω
(4)
from which we obtain for the dielectric permittivity function
εr (ω) = 1 + χ(ω) = 1 −
ωp2
.
ω2
(5)
The dispersion relation is
ω2 =
k 2 c2
= ωp2 + k 2 c2 .
εr (ω)
The phase and group velocities are
−1/2
ωp2
ω
vp = = c 1 − 2
,
k
ω
1/2
ωp2
∂ω
=c 1− 2
vg =
,
∂k
ω
(6)
(7)
so that both vp and vg are real if ω > ωp , and vp vg = c2 holds. Finally, using the equation ∇ × E =
−∂t B i.e. ik Ẽ = iω B̃, we obtain E0 = vp B0 .
c) From the definition of the EM energy density
c2
ε0 2
ε0 2
1 2
ε0 2
2 2
uEM =
E +
B = (E0 + c B0 ) = E0 1 + 2
2
2µ0
4
4
vp
2
ωp
ε0 2
=
(8)
E0 2 − 2 .
4
ω
d) From the definition of the kinetic energy density
uK
2
me 1 eE0 1 ne e 2 2
me 2 E
E
=
= ne v = ne
2
2 2 me ω 4 me ω 2 0
ε0 ωp2 2
=
E .
4 ω2 0
D
2
(9)
Thus
ε0 2
E ,
2 0
(10)
vg Et2 = c (E0 2 − Er2 ) .
(11)
u = uEM + uK =
indipendently on ne .
e) In our case Eq.(1) can be rewritten as
1/2
Using Fresnel formulas as a function of the phase velocity vp = c/n with n = εr
Er =
vp − c
E0 ,
vp + c
Et =
2vp
E0
vp + c
(12)
we obtain
4vg vp2 = 4c2 vp ,
which is equivalent to vg vp = c2 .
3
(13)