Classical Electrodynamics
Gabriel Barello
Jackson 7.1
Recall the definitions of a1 , a2 , a+ and a− as
E1 = a1 eiδ1 ,
iδ+
E+ = a+ e
E2 = a2 E iδ2 (Linear Basis)
,
E− = a− e
iδ−
(1)
(Circular Basis)
(2)
Recall the equation relating the stokes parameters to the polarization and amplitude in the linear and circularly polarized
bases
Linear
Circular
2
2
2
2
s0 = |1 · E| + |2 · E| =
s1 = |1 · E| − |2 · E| =
a21
a21
+
−
a22
a22
s0 =
s2 =
∗
s3 =
∗
s1 =
s2 = 2Re[(1 · E) (2 · E)] = 2a1 a2 cos(δ2 − δ1 )
s3 = 2Im[(1 · E) (2 · E)] = 2a1 a2 sin(δ2 − δ1 )
|∗+
(3)
· E| + |∗− · E|2 = a2+ + a2−
2Re[(∗+ · E)∗ (∗− · E)] = 2a+ a− cos(δ−
2Im[(∗+ · E)∗ (∗− · E)] = 2a+ a− sin(δ−
|∗+ · E|2 − |∗− · E|2 = a2+ − a2−
2
(4)
− δ+ )
(5)
− δ+ )
(6)
(7)
From which we can deduce, taking δ1 = δ+ = 0
Linear
r
s0 + s1
,
a1 =
2
r
s0 − s1
,
a2 =
2
s
δ2 = arccos
s22
s20 − s21
Circular
r
s0 + s3
a+ =
2
r
s0 − s3
a− =
2
s
!
,
δ− = arccos
(8)
(9)
(10)
s22
s20 − s23
!
Now I can just make a fun little mathematica notebook which takes in stokes parameters and spits out plots!
a.
b.
Linear
Amplitude
√
3
5
Phase
π
4
24
arccos( 25
)
Circular
Aplitude
√
3
5
Phase
arccos( √25 )
0
3
1.0
2
0.5
1
-1.0
-0.5
0.5
-3
1.0
-2
-1
1
-1
-0.5
-2
-1.0
-3
Where the left figure depicts the field for part a. and the right figure for part b.
1
2
3
(11)
Zangwill 16.11
Consider two antipodal points on the Poincare Sphere describing fields with electric fields E and E 0 . We can choose to
represent our fields in either the linear or circular polarization bases, I choose to use the linear basis. Then, saying that two
points are antipodes on the poincare sphere means that
s0 = s00 −→
s1 =
s2 =
s3 =
−s01
−s02
−s03
|1 · E 0 |2 + |2 · E 0 |2 = |1 · E|2 + |2 · E|2
0 2
−→
−→
0 2
2
(12)
2
(13)
(1 · E 0 )∗ (2 · E 0 ) = −(1 · E)∗ (2 · E)
(14)
|1 · E | − |2 · E | = |2 · E| − |1 · E|
Armed with these relations, we press onwards. To determine the relative orientation of the E fields, in particular whether
they are orthogonal, we shall copute their inner product. Recall that for complex vectors, the inner product is E 0∗ · E. The
analysis is as follows
E 0∗ · E = ((1 · E 0 )1 + (2 · E 0 )2 )∗ · ((1 · E)1 + (2 · E)2 ) = (1 E 0 )∗ (1 · E) + (2 · E 0 )∗ (2 · E)
(1 · E)∗ (2 · E)
= (2 · E 0 )∗ (2 · E) − (1 · E)
(2 · E 0 )
2 · E
= (2 · E 0 )∗ (2 · E) − |1 · E|2
2 · E 0
2 · E
(|1 · E 0 |2 − |1 · E|2 − |2 · E 0 |2 )
= (2 · E 0 )∗ (2 · E) +
2 · E 0
By equation 14
By equation 12
Note here that the first term cancels with the final part of the second term. That is,
2 · E
2 · E
0 2
|2 · E | =
(2 · E 0 )∗ (2 · E 0 ) = (2 · E 0 )∗ (2 · E)
2 · E 0
2 · E 0
So that the term evaluated above and the first term in the expression of E 0∗ · E cancel. We now have the expression
2 · E
(|1 · E 0 |2 − |1 · E|2 )
(15)
E 0∗ · E =
2 · E 0
However, by adding equations 12 and 13 we get the expression
2|1 · E 0 |2 = 2|2 · E|2 → |1 · E 0 |2 − |1 · E|2 = 0
(16)
0∗
So, in fact, it is true that E · E = 0, or in other words antipodes on the poincare sphere represent orthogonal polarization
states.
Zangwill 16.7
a.
Consider the angular momentum of electromagnetic fields
Z
~ EM = 0 d3 x ~r × (E
~ × B)
~
L
(17)
~ and introducing explicit indices we can write
By introducing the vector potential in place of B
~ EM i = 0
L
Z
Z
= 0
Z
= 0
d3 x ijk rj klm El mno ∂n Ao
(18)
d3 x (δkn δlo − δko δln )ijk rj El ∂n Ao
(19)
d3 x (El ijk rj ∂k Al − ijk rj El ∂l Ak )
(20)
(21)
2
The first term is exactly the Lorbital we are looking for. We can integrate the second term by parts, use the fact that
these are sourceless fields so that ∂l El = 0 (this was stated in the original statement in Zangwill’s book) and use the fact
that ∂l rj = δlj to write the second term as
Z
Z
Z
3
3
~ spin
−0 d x ijk rj El ∂l Ak = 0 d x ijk δlj El Ak = 0 d3 x ijk Ej Ak = L
(22)
Thus, indeed the decomposition stated in the problem is correct.
b.
Consider a gauge tranformation, A → A0 = A + ∇φ with φ a scalar function. Since the electric potential does not appear,
we need not worry ourselves with it. Plugging this in
~0 EM = 0
L
Z
~ × (A
~ + ∇φ))
d3 x Ek (~r × ∇)(Ak + ∇k φ) + (E
Z
~ EM + 0 d3 x Ek (~r × ∇)∇k φ + (E
~ × ∇φ)
=L
(23)
(24)
It is obvious that the second term will be nonzero in general, for example if we take φ to be a linear function of the
coordinates, the first term is zero, but the second term will not in general be zero even with a linear function φ.
c.
Consider a circularly polarized plane wave in the coloumb gauge
± iŷ
~ ± = E0 x̂ √
exp[i(kz − ωt)]
E
2
(25)
In the coloumb gauge
E = −∂t A → A = −i
E0 x̂ ± iŷ
√ exp[i(kz − ωt)]
ω
2
(26)
Consider the object
~ spin i = ±ωẑh0
±ωẑ · hL
Z
d3 x<(E × H ∗ )i = 0
E0
T
Z
T
Z
dt
d3 x cos[kz − ωt]2 = 0
0
|E0 |2
2
Z
d3 x = hui
Z
d3 x = huitot
(27)
So indeed the identity follows. In the case that hui = ~ω this says that a left/right circularly polarized plane wave of
energy has spin angular momentum in the ẑ direction of ±~. Cute!
Zangwill 16.3
Consider an evanescent wave
~ = ŷE0 exp[i(hz − ωt) − κx]
E
(28)
a.
Since the components of the electric field obey the wave equation, we have that
κ2 − h2 = ω 2
(29)
b.
~ = 1 n̂ × E
~ and n̂ =
The B field is related to the E field of an EM wave by the equation B
c
~ =− √ 1
B
(x̂h + iκẑ) E0 exp[i(hz − ωt) − κx]
c h2 − κ2
3
hz−iκx
√
h2 −κ2
(because n̂ · n̂ = 1). Thus
(30)
c.
The B field will be circularly polarized when h = κ.
d.
The time average of the poynting vector is given by
1
1
1
Re[E ∗ × H] = Re[ E ∗ × B]
2
2
µ0
1
1
−(hz−ωt)−κx
∗ −i(hz−ωt)−κx
(hx̂ + iκẑ)E0 e
Re ŷE0 e
=
× √
2µ0
c h2 − κ2
|E0 |2
√
=−
e−2κx Re [ŷ × (hx̂ + iκẑ)]
2cµ0 h2 − κ2
|E0 |2
√
=−
e−2κx Re [−hẑ + iκx̂]
2cµ0 h2 − κ2
r
1
h
0
= √
|E0 |2 e−2κx ẑ
2 h2 − κ2 µ0
hSi =
(31)
(32)
(33)
(34)
(35)
Jackson 7.2
Consider a plane wave incident normally on an a setup of three nonpermiable materials.
a.
Since these plane waves are incident to the planar interface we can choose to only analize one particular linear polarization
since the perpendicular one will behave exactly the same, and we are free to combine them since this whole problem is linear.
Thus from the solution to a particular polarization, the general solution follows.
We choose to orient the ẑ axis normal to the interface and pointing to the right. The incident light wave is chosen to be
linearly polarized in the x̂ direction with wave vector k. The x component of the E fields in the three regions are given as
follows
Region I
EI = E0 ei(kz−ωt) + E00 ei(−kz−ωt) ,
k = ωn
(36)
k 0 = ωn0
(37)
k 00 = ωn00
(38)
Region II
0
0
EII = E1 ei(k z−ωt) + E10 ei(−k z−ωt) ,
Region III
EIII = E2 ei(k
00
z−ωt)
,
Where, by the symmetry of the problem, we have inferred that the left moving wave vectors in the first and second region
are the negative of the right moving wave vectors. The y component of the B field is given by dividing these fields by c, and
putting a − sign in front of the leftmoving waves, since their wave vectors point in the opposite direction.
Now we must simply impose the boundary conditions on the E, D, B and H fields at each interface. That is, the normal
component of D and B are continuous, which is trivial since there is no normal component of the fields, and the parallel
components of E and H are continuous. These conditions, at t = 0 give
n0
(E1 − E10 )
n
0
0
00
n00
E 1 eik d − E1 e−ik d = 0 E2 eik d
n
E0 + E00 = E1 + E10 ,
0
0
E 0 − E00 =
00
E1 eik d + E10 e−ik d = E2 eik d ,
Solving these equations gives total transmission and reflection coefficients
4
(39)
(40)
|E00 |2
−4nn02 n00 + n2 (n02 + n002 ) + n02 (n02 + n002 ) + (n2 − n02 )(n02 − n002 ) cos(2dn0 ω)
=
=R
2
|E0 |
4nn02 n00 + n2 (n02 + n002 ) + n02 (n02 + n002 ) + (n2 − n02 )(n02 − n002 ) cos(2dn0 ω)
16nn02 n00
|E2 |2
=
<
=T
|E0 |2
(n + n0 )2 (n0 + n00 )2 + (n − n0 )(n0 − n00 )((n − n0 )(n0 − n00 ) + 2(n + n0 )(n0 + n00 ) cos[2dn0 ω])
<
(41)
(42)
(43)
The coefficients are plotted below for the case d = 1.
Figure 1: This plot shows the transmission and reflection coefficients for d = 1 (in each case, the transmission coefficient is the
one that starts out the larger of the two). The horizontal axis is the frequency which is plotted from ω = 0 to ω = 2πd = 2π
so as to show a full period for both R and T . From left to right this gives plots where (n, n0 , n00 ) = (1, 2, 3), (3, 2, 1) and
(2, 4, 1).
b.
When n00 = 1 the reflection coefficient reduces to
R=1−
n2
+ (1 + n(4 +
n))n02
+
n04
8nn02
+ (n − n0 )(n + n0 )(n02 − 1) cos[2dn0 ω]
(44)
If we can get cos(2n0 ω) = −1 the requirement that R = 0 reduces to
√
(45)
n2 + (1 + n(4 + n))n02 + n04 − (n − n0 )(n + n0 )(n02 − 1) − 8nn02 = 0 → n0 = n
√
Thus, if we choose n0 = n and d = (m − 1/2) √πnω with m an integer, the reflection coefficient will be zero for the
frequency ω.
Problem 6
a.
Ohms law gives
~
J~ = σ E
(46)
Thus, even in the sourceless maxwell equations, we get nonzero cources for B and thus E
∇·E =0
(47)
∇·B =0
(48)
∂H
∇×E =−
∂t
∇ × H = σE + (49)
∂E
∂t
(50)
Taking the curl of the third and fourth equations, using a vector identity from the front of Jackson, and plugging in for
the divergence and curl temrs, we arrive at
5
∂2E
∂E
+ µ 2
∂t
∂t
2
∂B
∂
B
∇2 B = σ
+ 2
∂t
∂t
∇2 E = σ
(51)
(52)
just as we wanted.
b.
Consider now a monocromatic field of the form
~ =E
~ 0 ei(kz−ωt)
E
~ =H
~ 0 ei(kz−ωt)
H
(53)
(54)
Plugging these into the above we arrive at
∇2 E = −k 2 E = −σiωE − ω 2 E
2
2
(55)
2
∇ H = −k H = −σiωH − ω H
(56)
However, if we allow the permitivity to be ˆ = + iσ/ω we can write the first of these equations as
−k 2 E = −σiωE − (ˆ
− iσ/ω)ω 2 E = −ˆ
ω 2 E
(57)
And similarly for the H equation, which is just as we would expect from the maxwell equations with no conductivity.
c.
From √
the last part we have that k 2 = ˆω 2 , but ˆ is complex, so k must be as well. Intriducing the compelx index of refraction
n̂ = c ˆµ this relation can be written
k 2 = n̂2 ω 2 /c2
(58)
d.
If we assume that µ ∼
= µ0 then we have that n0 =
q
0
=
q
0 +iσ/ω
0
=
p
σ
1 + i ω
. Finally, recall that for an EM wave, k = nω,
in order to satisfy the maxwell equations, so that we can write the E and H fields for a wave propogating in the ẑ direction
as
p
E = x̂E0 exp[iω( 1 + i(σ/ω)z + t)]
p
p
+ i ωσ E0
exp[iω( 1 + i(σ/ω)z + t)]
H = ŷ
√
µ0
(59)
(60)
So the fields are in fact transverse to the propogation direction. Also, the magnitude of H is
s
+ iσ/ω
H0 =
E0
µ0
(61)
e.
Recall that k satisfies the relation k 2 = ˆµω 2 = ( + iσ/ω)µω 2 . This is a complex number with magnitude µω 2
and argument φ = arctan(σ/ω). It is easy now to solve for k, it is
p
k = µω 2 (2 + (σ/ω)2 )1/4 exp[i arctan(σ/ω)/2]
p
2 + (σ/ω)2
(62)
The imaginary part is then
=(k) =
p
µω 2 (2 + (σ/ω)2 )1/4 sin(arctan(σ/ω)/2)
6
(63)
In the limit where the imaginary part of ˆ dominates, the phase of k is just π/4 and the magninute is
say that the imaginary component of k is
r
r
µσω
2
=(k) =
→δ=
2
µσω
7
√
µσω. That is to
(64)