Physics 417G : Solutions for Problem set 7
Due : March 25, 2016
Please show all the details of your computations including intermediate steps.
1
Problem 1
Consider the following spherical wave. The plane wave can be obtained when the radius of the spherical
wave is very big.
sin θ
sin(kr − ωt)
~
E(r, θ, φ, t) = A
cos(kr − ωt) −
φ̂ ,
r
kr
where ω = ck.
~ obeys all four of vacuum Maxwell’s equations, and find the associated magnetic field.
a) Show that E
~ over a full cycle to get the intensity vector I.
~ (Does it
b) Calculate the Poynting vector. Average S
−2
point in the expected direction? Does it fall off like r ?)
~ over a spherical surface A
~ to determine the total power radiated.
c) Integrate I~ · dA
Sol: a) It is straight forward to show that the electric field satisfy the Maxwell’s equations. For
~ ·E
~ = 1 ∂Eφ = 0. The corresponding magnetic field can be
example, Gauss’s law is satisfied as ∇
r sin θ ∂φ
obtained from Faraday’s law as
~
1 ∂
∂
∂B
~ ×E
~ = 1
=∇
(sin θEφ )r̂ −
(rEφ )θ̂
∂t
r
sin
θ
∂θ
r ∂r
2 sin θ cos θE0
1
sin θE0
1
1
=
cos(kr − ωt) −
sin(kr − ωt) r̂ −
[ 2 − k] sin(kr − ωt) − cos(kr − ωt) θ̂ .
r2 sin θ
kr
r
kr
r
−
By integrating with respect to t, we get
2 cos θE0
sin θE0
1
1
1
~
B=
cos(kr − ωt) r̂ +
sin(kr − ωt) +
[ 2 − k] cos(kr − ωt) + sin(kr − ωt) θ̂ .
ωr2
kr
ωr
kr
r
~ can be shown to vanish as well as the Ampere’s law.
The divergence of B
b) The Poynting vector can be computed directly as
2
~= 1E
~ ×B
~ = E0 sin θ 2 cos θ [1 − 1 ] sin(kr − ωt) cos(kr − ωt) + 1 (cos2 (kr − ωt) − sin2 (kr − ωt)) θ̂
S
µ0
µ0 ωr2
r
k2 r2
kr
1
1
2
+ sin θ [ 2 3 − ] sin(kr − ωt) cos(kr − ωt) + k cos2 (kr − ωt) + 2 (cos2 (kr − ωt) − sin2 (kr − ωt)) r̂ .
k r
r
kr
By averaging over the full cycle, we use hsin(kr−ωt) cos(kr−ωt)i = 0, hsin2 (kr−ωt)i = hcos2 (kr−ωt)i =
1/2, we get
2
2
2
~ = E0 sin θ sin θk 1 r̂ = E0 sin θ r̂ .
I~ = hSi
µ0 ωr2
2
2µ0 cr2
It points in the r̂ direction, and falls as 1/r2 , as expected for a spherical wave.
c) The total power radiated is
Z
P =
2
~ = E0
I~ · dA
2µ0 c
Z
dφdθr2 sin θ
sin2 θ
4π E02
=
.
r2
3 µ0 c
2
Problem 2
~
We view the atomic polarizablity formula, p~(ω) = αE(ω),
as a response function. Here we would like
to find some physical consequences of causality of response function using the Fourier Transformation
and its inverse
Z ∞
Z ∞
dω ~
iωt
~
~
~
E(t)
=
E(ω)e−iωt ,
E(ω)
=
dtE(t)e
.
2π
−∞
−∞
a) Using a Fourier transform for p~(t) obtain an expression of the response function in terms of α̃(t − t0 )
~ 0 ).
and E(t
~
~
Hint you can use p~(ω) = αE(ω),
and do the inverse Fourier transform of E.
b) Now statement of ’response’ and thus
can be transferred as α̃(t) = 0 for t < 0.
R ∞requiring causality
−iωt
α(ω)e
,
show
that the statement of causality can be
Using the Fourier transform of α̃(t) = −∞ dω
2π
expressed as the statement “α(ω) is analytic in the upper half plane in the complex ω plane.”
To have a relation between the real part and imaginary part of α, we have some digression here.
c) For a general function α2 , consider the following integral
Z ∞
1
α2 (ω 0 )dω 0
g(ω) =
.
iπ −∞ ω 0 − ω
When ω lies in the real axis, we have a divergent integral. To avoid this divergence, we deform our
pole in two different ways, one by using ω → ω ± i. Then the pole is not directly on the real axis and
we can perform the integral. Perform the integral to obtain the following integrals
1
[g(ω + i) − g(ω − i)] ,
2
1
[g(ω + i) + g(ω − i)] .
2
Note that you can use the contour integral to evaluate the first, while Principal value for the second.
d) Now note that the α given in the problem 3 has only poles in the lower half in ω plane. By considering
a linear combination of the results in c) and setting α2 = α, evaluate g(ω − i) for the closed contour
including the entire upper half of the ω plane. What is the expression of α(ω) in terms of poles and
Principal values? Using this connect the real and imaginary part of α(ω). This is Kramers-Kronig
relations.
Sol: a) The vector p~(t) can have the Fourier transformation as
Z ∞
Z ∞
dω
dω
−iωt
−iωt
~
p~(t) =
p~(ω)e
=
α(ω)E(ω)e
2π
2π
−∞
−∞
Z ∞
Z ∞
Z ∞
dω
0 ~ 0 −iω(t−t0 )
~ 0) ,
=
α(ω)
dt E(t )e
=
dt0 α̃(t − t0 )E(t
2π
−∞
−∞
−∞
~ 0 ) through
which relates the time dependence of the quantity p~(t) to that of the electric field E(t
Z ∞
0
dω
α̃(t − t0 ) =
α(ω)e−iω(t−t ) .
−∞ 2π
R∞
−iωt
. The integral can be changed into a full contour integral with
b) Consider α̃(t) = −∞ dω
2π α(ω)e
the contour circling over the upper half plane as in the figure. The reason to do in the upper half
plane is because we want to have a vanishing integral over the integral for the angle (0, π). Thus the
statement of the causality, meaning that there will not be response for t < 0 because there is no cause,
translates into the statement that the function α̃(t) = 0 for t < 0, which is again translated into the
statement α(ω) is analytic in the upper half plane in ω space.
c) Assuming that α2 decay rapidly at the infinity in ω plane, the first expression can be evaluated
using a contour integral of going right below the real axis and coming back right above the real axis
and including the pole ω = ω 0 . Thus it gives the residue of the pole. Thus
1
1
1
[g(ω + i) − g(ω − i)] = 2πiα2 (ω) = α2 (ω) .
2
2
iπ
Thus the integral in real axis is actually discontinuous by the residue. The other expression can be
expressed as a principle value, which is averaging over the two functions
Z ∞
1
α2 (ω 0 )dω 0
1
[g(ω + i) + g(ω − i)] = P
.
2
iπ
ω0 − ω
−∞
d) Now by subtracting these two expressions and identifying α2 = α, we get
Z ∞
1
α(ω 0 )dω 0
g(ω − i) = P
− α(ω) .
0
iπ
−∞ ω − ω
The left hand side can be evaluate with the same integral performed in b) using the contour given in
the figure. The result vanishes. Thus we get
Z ∞
1
α(ω 0 )dω 0
α(ω) = P
.
0
iπ
−∞ ω − ω
Thus we have two independent expressions that connecting the real and imaginary parts of the function
α as
Z ∞
dω 0 Im[α(ω 0 )]
Re[α(ω)] = P
,
ω0 − ω
−∞ π
Z ∞
dω 0 Re[α(ω 0 )]
Im[α(ω)] = −P
.
ω0 − ω
−∞ π
Note that the relation is non-local. To know Re[α(ω)], we need information of Im[α(ω)] for all frequency
space.
3
Problem 3
Consider a nucleus of charge −q with a cloud of electrons with a charge q of radius a. In this problem,
~ In
we assume that the electron cloud is a uniform sphere even in the presence of
field E.
external electric
q
~
r
2
~
~−
the Midterm 1, we obtained that the force acting on the electrons as F~ = q E
r.
4π0 a3 ≡ q E −mω0 ~
The second part is restoring force.
~
~ 0 e−iωt (the spatial part ei~k·~r is assumed to vary
Now we use the oscillating electric field E(t)
=E
slowly) and damping term F~damping = mη~r˙ . η will have its physical meaning depending on the physical
situation. Combining them together, we have the force equation for a mass m,
~
m~r¨ = q E(t)
− mω02~r − mη~r˙ .
(1)
~
~ 0 e−iωt , solve the equation for ~r.
a) For given electric field E(t)
=E
b) The electric dipole moment is p~ = q~r. Express this in terms of the external electric field to get
~ What is the atomic polarizability α?
p~ = αE.
c) Now Polarization is average over the dipole moment per unit volume. Thus we can show P~ = nh~
pi =
~ where n is the density. Compute the dielectric constant r = = 1 + χe .
0 χe E,
0
d) We know the permittivity is complex. Obtain 1 , 2 by writing = 1 + i2 .
Observe that the atomic polarizability is frequency dependent and complex. As a direct consequence, so is the dielectric constant (ω). We are trying to get physical consequences of the fact.
Consider the source free Maxwell’s equations
~
~ ×E
~ = − ∂B ,
(ii) ∇
∂t
~
~ ×H
~ = ∂D ,
(iv) ∇
∂t
~ ·D
~ =0,
(i) ∇
~ ·B
~ =0,
(iii) ∇
~ = (ω)E,
~ B
~ = µH
~ and we assume µ is real and constant. We use the wave form
where D
~
~ x, t) = B
~ 0 (ω)ei(~k·~r−ωt) .
B(~
~ 0 (ω)ei(k·~r−ωt) ,
E(~x, t) = E
(2)
~ 0, B
~ 0 without derivatives.
e) Write down the Maxwell’s equations in terms of ~k and ω and E
f) What is the dispersion relation (the relation between ~k and ω)? You can derive it from the Maxwell’s
equations.
g) You notice that ~k should be complex! Denote ~k = (k1 + ik2 )ẑ. Rewrite the electric and magnetic
fields. What are the consequences of the complexified wave vector ~k?
h) Using the result of d), compute k1 , k2 in terms of 1 , 2 .
i) For very high frequency and very low frequency, show that the wave has a transparent propagation.
j) For ω ≈ ω0 η, and |1 | 2 , show that the wave is decay rapidly.
k) For 1 < 0 and |1 | 2 , show that the wave is reflected.
Sol: a) By trying ~r = ~r0 e−iωt , we get
~r(t) =
~0
qE
1
e−iωt .
m ω02 − ω 2 − iηω
b) The atomic polarizability gives
p~ = q~r =
q2
1
~ ,
E
m ω02 − ω 2 − iηω
α=
q2
1
.
m ω02 − ω 2 − iηω
c) The dielectric constant reads
r = 1 + n
q2
1
.
m ω02 − ω 2 − iηω
d) The complex permittivity is
1 = 0 +
nq 2
ω02 − ω 2
,
2
m (ω0 − ω 2 )2 + η 2 ω 2
2 =
nq 2
ηω
m (ω02 − ω 2 )2 + η 2 ω 2
e) The Maxwell’s equations are
~ =0,
(i) (ω)~k · E
~ =0,
(iii) ~k · B
~ = ωB
~ ,
(ii) ~k × E
~ = −ωµ(ω)E
~ ,
(iv) ~k × B
f) By taking ~k× to (ii) and use (i)&(iv) we get
~k · ~k = ω 2 µ(ω) .
g) Now the field look like
~ x, t) = E
~ 0 (ω)e−k2 z ei(k1 z−ωt) ,
E(~
~ 0 (ω)e−k2 z ei(k2 z−ωt) = |k| ẑ × E
~ 0 (ω)e−k2 z ei(k2 z−ωt+φ) ,
~ x, t) = k1 + ik2 ẑ × E
B(~
ω
ω
where φ = tan−1 (k2 /k1 ).
Thus we can see at least two consequences, (A) The fields decay as they propagate along the z direction,
(B) there are phase difference between the electric and magnetic field by the phase φ.
h) The wave vector decomposes as
r
k1 = ±ω
µ
2
q
1/2
2
2
1 + 2 + 1
,
r
k2 = ±ω
µ
2
q
21
+
22
1/2
− 1
i) For very high frequency and very low frequency, we see 1 > 0, 1 2 . Thus
s
p
µ22
2
k2 ≈ ±ω
=
k1 k1 .
k1 ≈ ±ω µ|1 | ,
41 |
21
Thus the imaginary part is small compared to the real part, showing that the wave has a transparent
propagation.
j) For ω ≈ ω0 η, and |1 | 2 , then the resonance peak for 2 is very sharp. For ω ≈ ω0 ± η/2, we
have
1 ≈ 0 ,
2 ≈
nq 2 1
|1 | .
m ω0 γ
q
µ22
Then we have k1 ≈ k2 ≈ ±ω
2 . This means that the wave decay rapidly as it propagates! The
frequency is close to the resonance frequency and the electrons are easily excited to absorb the energy
from the wave.
k) With the condition one can see that
r
p
2
µ
|1 |
k1 ≈ ±ω
,
k2 ≈ ±ω µ|1 | =
k1 k1
2 1 |
22
The wave number is almost pure imaginary that indicate that the wave will rapidly decay inside the
medium, not absorbed, but reflected.
4
Problem 4
Let us consider the electromagnetic waves in conducting matter. Using the oscillating electric field
~
~ 0 e−iωt (the spatial part ei~k·~r is assumed to vary slowly) and damping term F~damping = mη~r˙ .
E(t)
=E
The motion of electrons can be described by the force equation
m
~
m~r¨ = q E(t)
− ~r˙ .
τ
(3)
where m is an electron mass and τ is the scattering time, representing the average time traveling before
an electron bounces off.
a) Write the equation (3) in terms of ~v = ~r˙ , and solve the equation for the velocity ~v .
b) Using the two equations
~ x, t) = nq~v (~x, t) ,
J(~
~ 0 (ω) ,
J~0 (ω) = σ(ω)E
~ x, t) = J~0 (ω)e−iωt , n is the density of electrons, and the first equation is Ohm’s law, derive
where J(~
the conductivity σ(ω) in term of τ and the other parameters.
~ · J~ + ∂ρ = 0, and the result of b) to compute J(~
~ x, t), compute the
c) Using the continuity equation, ∇
∂t
density ρ(~x, t).
~ 0, B
~0
d) Write down the following Maxwell’s equations with the sources in terms of ~k and ω and E
without derivatives.
~
~ ×E
~ = − ∂B ,
(ii) ∇
∂t
~
~ ×H
~ = J~f + ∂ D ,
(iv) ∇
∂t
~ ·D
~ = ρf ,
(i) ∇
~ ·B
~ =0,
(iii) ∇
~ = (ω)E,
~ B
~ = µH
~ and we assume µ is real and constant. We use the same wave form (2).
where D
e) By applying i~k on the equation (iv) you obtained in d), find the modified wave equations, for both
~ and B.
~ What is the dispersion relation?
E
f) By writing ~k = (k1 + ik2 )ẑ, evaluate k1 , k2 in terms of ω, µ, .
g) Put the wave vector obtained in f) into the the wave form (2), to obtain the skin depth.
h) Compute the skin depth in a poor conductor (σ ω). Find the skin depth for (pure) water by
using Tabless 4.2, 6.1 and 7.1.
i) Compute the skin depth in a good conductor (σ ω). Find the skin depth (in nanometers) for a
typical metal (σ ≈ 107 (Ωm)−1 ) in the visible range (ω ≈ 1015 /s), assuming ≈ 0 and µ ≈ µ0 . Why
are metals opaque?
Let us discuss the physical consequences of the modification compared to the previous case due to
the change
ef f (ω) = 1 (ω) + i2 (ω) + i
σ(ω)
.
ω
(4)
Note that σ(ω) can be complex as we obtained in h). We observe that the Maxwell’s equations are the
same as the previous case in c) except (ω) is changed into ef f (ω). If ef f (ω) is non-zero the physical
consequences are similar before.
j) If ef f (ω) = 0, and ωτ 1 and 1 2 (low frequency limit), What happen to the electric field?
k) If ef f (ω) = 0, and ωτ 1 and 1 ≈ 0 2 (high frequency limit), What happen to the electric
& magnetic fields?
Sol a) Trying ~v (t) = ~v0 (ω)e−iωt , we get
~v0 (ω) =
~ 0 (ω)
qτ E
.
m 1 − iωτ
b) Thus the conductivity has the form
~ 0 (ω)
nq 2 τ E
~
~
J(ω)
= σ(ω)E(ω)
=
,
m 1 − iωτ
σ(ω) =
nq 2 τ
1
.
m 1 − iωτ
i(~
k·~
r −ωt)
~ x, t) = σ(ω)E(ω)e
~
c) From b) we know that form of current J(~
. The density is
ρ(~x, t) =
σ(ω) ~ ~
~
k · E(ω)ei(k·~r−ωt) .
ω
d) The Maxwell’s equations are
σ(ω) ~ ~
(i)
(ω) + i
k·E =0,
ω
~ =0,
(iii) ~k · B
~ = ωB
~ ,
(ii) ~k × E
~ = −ωµ (ω) + i σ(ω) E
~ ,
(iv) ~k × B
ω
e) The wave equations are
~k 2 B
~ = ω 2 µ (ω) + i σ(ω) B
~ ,
ω
~k 2 E
~ = ω 2 µ (ω) + i σ(ω) E
~ .
ω
The dispersion relation is
~k 2 = ω 2 µ (ω) + i σ(ω) .
ω
f) The k1 , k2 are
r
k1 = ω
s
1/2
2
µ(ω) 
σ
+ 1
,
1+
2
ω(ω)
r
k2 = ω
s
1/2
2
µ(ω) 
σ
− 1
.
1+
2
ω(ω)
g) The electromagnetic fields have the following wave forms
~ t) = E
~ 0 (ω)e−k2 z ei(~k1 z−ωt) ,
E(z,
~ t) = B
~ 0 (ω)e−k2 z ei(~k1 z−ωt) .
B(z,
(5)
Thus the skin depth is
1
1
d=
=
k2
ω
s
s
−1/2
2
2 
σ
1+
− 1
.
µ(ω)
ω(ω)
h) The expansion of the skin depth at poor conduction goes
r
k2 = ω
s
µ(ω) 
1+
2
σ
ω(ω)
1/2
2
r
≈ω
− 1
µ(ω)
2
"
σ
ω(ω)
#1/2
2
/2
=
σ
2
r
µ
.
σ(ω)
Thus the skin depth has
s
r
2 σ(ω)
80.1 × (8.85 × 10−12 )
5
d=
= 2 × (2.5 × 10 )
= 1.2 × 104 m .
σ
µ
4π × 10−7
where we use the numerical values for pure water = 80.10 , µ ≈ µ0 , σ = 1/(2.5 × 105 ).
i) For a good conductor, we get
r
k2 = ω
s
µ(ω) 
1+
2
σ
ω(ω)
1/2
2
− 1
r
≈ω
1/2 r
µ(ω)
σ
µωσ(ω)
.
=
2
ω(ω)
2
Thus the skin depth has
s
s
2
2
d=
=
= 1.3 × 10−8 m = 13 nano meter .
µωσ(ω)
1015 × (4π × 10−7 ) × 107
So the fields do not penetrate far into a metal. This is what accounts for their opacity.
2
2
1
j) In the low frequency limit, σ(ω) = nqm τ 1−iωτ
≈ nqm τ = σDC . And we demand ef f (ω) = 1 (ω) +
σDC
i ω = 0. Thus ω = −iσDC /1 . Then the electric field becomes
~ 0 (ω)ei~k·~r e−σDC t/1 ,
E(~x, t) = E
which tells that the electric field decays in time. Actually there is a longitudinal component of the
electric field.
2
2
1
1
DC
≈ i nqm τ ωτ
= i σωτ
. And we demand ef f (ω) =
k) For high frequency limit, σ(ω) = nqm τ 1−iωτ
0 (ω) −
become
σDC
ω2 τ
= 0. In this case ω = ωp =
nq 2
m0 ,
plasma frequency. Then the electric and magnetic fields
~ 0 (ωp )ei(~k·~r−ωp t) ,
E(~x, t) = E
~ = 0. Thus there exist a longitudinal component of the electric
while the magnetic field vanishes, B
~
~
field because k · E 6= 0.