Classical Electrodynamics. Gabriel Barello. Jackson 7.1. Recall the definitions of
a1,a2,a+ and a− as. E1 = a1eiδ1 ,. E2 = a2Eiδ2 (Linear Basis). (1). E+ = a+eiδ+ ...
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 ˆ ± iˆ y ~ ± = E0 x √ exp[i(kz − ωt)] E 2
(25)
In the coloumb gauge E = −∂t A → A = −i
E0 x ˆ ± iˆ y √ exp[i(kz − ωt)] ω 2
(26)
Consider the object
~ spin i = ±ωˆ ±ωˆ z · hL z h0
Z
d3 x
