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Liénard-Wiechert Potentials
Point charge q with given trajectory w(t).�Source-to-�eld� vector
R = r−w
and velocity v(t)
v(t) =dw(t)
dt
retarded time tr implicitly de�ned:
∣r−w(tr)∣ = c(t− tr)⇒
tr = t− R(tr)
c
1
Wave equation for potential paravector:
Φ = � + A with source J = c� + J
□2Φ = Z0Jfor Lorenz gauge
⟨∂Φ⟩s
= 0
Φ =1
□2Z0J =
1
cG(x ) ★ J (x )
where:
□2G (x) = 4��(4) (x) x = (r, t) →
retarded Green's function:
G (x) =1
∣r∣�
(t− ∣r∣
c
)≡ 1
r�(t− r
c
)↑
1-d delta function
Solution:
Φ (x) =1
c
∫G (x− x′)J (x′) d4x′ =
1
c
∫ 1
∣r− r′∣�
(t− t′ − ∣r− r′∣
c
)J (x′) dt′d3r′
2
For a point charge:
�(x) = q �3(r−w(t))
J (x′) = q(c + v(t′))�3(r′ −w(t′))
Φ (x) =∫ q
R(t′)�
(t′ − t +
R(t′)
c
)(1 +
v(t′)
c)dt′
Using the identity
�
(t′ − t +
R(t′)
c
)=
� (t′ − tr)
∣1 +1
cR(tr)∣
and from
RR = R ⋅ R = −R⋅w = −R⋅v = −Rn ⋅ v
Φ (x) = q
[1 +−→�
R(1− n ⋅−→� )
]tr
where−→� =
v
c
3
Point source-to-�eld paravector
Q = cT + R = c(t− t′) + (r−w(t′))
QQ = c2T 2 −R2
retarded time condition:
QQ∣t′=tr = 0
Q∣tr = R(1 + n)
velocity paravector
u = (c + v) = c1 +−→�√
1− �2
with uu = c2 and
⟨Qu⟩ s =⟨R(1 + n) c(1−
−→� )⟩s
= Rc(1−n⋅−→� )
Potentials (covariant form):
Φ (x) = q
[u
⟨Qu⟩s
]tr
4
Fields: ∂=1
c∂t −∇
we need ∂G(x) = ∂
(� (�)
r
)for � = t− r
c:
∂
(1
r
)= −∇1
r=r
r2(Coulomb)
∂ (� (�)) = (∂�) �′ (�) =1
c(1 + r) �′ (�)
Hence
∂
(� (T −R/c)
R
)=
n
R2� (s)+
1
cR(1 + n) �′ (s)
where
s = t− t′ − R
c= t− t′ − ∣r− r′∣
c
5
In terms of the potential
ℱ = E + IB =⟨∂Φ⟩v
for the general case:
J = c� + J
Je�menko solution
E (r, t) =∫ [ nR2� (r′, tr) +
n
cR� (r′, tr)−
1
c2RJ (r′, tr)
]d3r′
B (r, t) =∫ [(J (r′, tr)
cR2+J (r′, tr)
c2R
)× n
]d3r′
6
Field of a moving point charge
ℱ = ℱc + ℱrad
Coulomb radiativev a
∼ 1
r2∼ 1
r
unit vector �point source-to-�eld�
n(t) =R
R=
r−w(t)
∣r−w∣
a) ℱ2rad = 0 & ℱrad = (1 + n)Erad
Brad = n× Erad
b)ℱ2c ∕= 0
Bc = n× Ec
Rest frame: Erf = qn
R2(no radiation)
7
Green function (revisited)
dv
et
c2t2 − r2 = (ct + r) (ct− r)
G (x) =1
r�(t− r
c
)= 2c�
(c2t2 − r2
)∣r
so
G (x− x′) = 2c�(QQ)∣r = 2c� (�) ∣r
whereQ = cT + R
and� = QQ = c2T 2 −R2
8
To calculate �elds for point charge,
∂G (x− x′) = 2c∂(�(QQ))
= 2c∂(�(�))
= 2c(∂�)dt′
d�
d
dt′� (�)
Factors:
∂� = 2Q = 2(cT + R)
[d�
dt′
]tr
= −2cT 2−2RR = −2Rc(1−n⋅−→� )
so
∂G (x− x′) = − 2Q
R(1− n ⋅−→� )
d
dt′�(�)
9
ℱ =1
c
∫[∂G(x− x′)]J (x′)d4x′∣v =
−2q∫dt′
Q
R(1− n ⋅−→� )
(1−−→� )∣v
d
dt′�(�)
Integrating by parts:
ℱ = 2q
∫d
dt′
⎡⎢⎣⟨Q(1−
−→� )⟩v
R(1− n ⋅−→� )
⎤⎥⎦ �(�)dt′ =
=q
c
∫d
dt′
⎡⎢⎣⟨Q(1−
−→� )⟩v
R(1− n ⋅−→� )
⎤⎥⎦ �(t′ − tr)R(1− n ⋅
−→� )dt′ =
ℱ = q1
cR(1− n ⋅−→� )
d
dt′
⎡⎢⎣⟨Q(1−
−→� )⟩v
R(1− n ⋅−→� )
⎤⎥⎦tr
Finally:
ℱ (x) =
⎡⎢⎣ q
cR(1− n ⋅−→� )
d
dt′
⎛⎜⎝⟨
(1 + n)(1−−→� )⟩v
1− n ⋅−→�
⎞⎟⎠⎤⎥⎦tr
10
a) Coulomb term
ℱc =
⎡⎢⎣ q
2R2
⟨(1 + n)(1−
−→� )⟩v⟨
(1 + n)(1−−→� )⟩3s
⎤⎥⎦tr
Ec =
[q
2R2
n−−→�
(1− n ⋅−→� )3
]tr
where−→� ≡ 1
c
dw
dt
Bc = n× Ec∣tr
b) Radiation term
acceleration:
a =dv
dt=d2w
dt2
Erad =
[ 2
c2R× (Ec × a)
]tr
Brad = n× Erad∣tr
ℱrad = (1 + n)Erad
11
c) Non relativistic: Ec ≃ qn
R2
and Erad ≃ −[ qc2a⊥R
]tr
a⊥ ≡ n× (a× n)
Radiated Power
ℱrad = (1 + n)Erad
Srad =c
8�
⟨ℱradℱ †rad
⟩v⇒
Srad =c
4�E2
radn
Nonrelativistic:
Srad ≃q2
4�c3∣a⊥∣2
R2n
≃ q2a2 sin2 �
4�R2c3n
12
Power per solid angle:
dP
dΩ= R2(n ⋅ Srad)
and∫sin2 �dΩ = 2� ⋅ 4
3Non-relativistic Larmor formula
P ≃ 2
3
q2a2
c3
13
Charge in uniform motion
a = 0 ℱrad = 0
ℱc = ⟨(1 + n)Ec⟩v
Ec =
⎡⎢⎣ q
R2 2n−−→�(
1− n ⋅−→�)3⎤⎥⎦r
where n =R (tr)
R (tr)
R(n−−→�)∣r = R−
−→� R ≡ r
tret ←→ t n−−→� =
r
R
14
Ec =q
2r
(n ⋅ r)3
with n ⋅ r = r cosA = r√
1− sin2A
Law of sines:sinA
�R=
sin �
R
sinA = � sin �
and n ⋅ r = r√
1− �2 sin2 �
E (r, t) =q
2r
r3(1− �2 sin2 �
)32
in terms of the present position r
15
Dipole radiation:
p (t) = qr(t) = p0 cos(!t)
qa(t) = −!2p0 cos(!t)
Magnetic �eld for dipole radiation:
Brad = −[q(n× a)
c2R
]tr
Bdip ≃q!2
c2rr× p0 cos
(!t− !r
c
)Electric �eld for dipole radiation:
Erad = −r×Brad
Edip ≃q!2
c2r[p0 − (p0 ⋅ r)r] cos
(!t− !r
c
)
16
ℱdip = (1 + r)Edip
ℱ (r, t) ≃ q!2
c2r(1 + r)p0⊥ cos!(t− r
c)
r× p0 = p0 sin � (−e�)
E (r, t) ≃ −!2
c2rp0 sin � cos(!t− !r
c) e�
B (r, t) ≃ −!2
c2rp0 sin � cos(!t− !r
c) e�
17
Characteristics of dipole radiation
outgoing spherical waves
! = kc dispersionless
E0 = B0
E ⊥ B and in phase
E,B plane tangent to spherical wave front
Radiated power
⟨S⟩ave ≃!4p20 sin2 �
8�R2c3r
⟨P ⟩ave ≃p20!
4
3c3
or ∣� ⋅ p∣2for polarization �
18
Antenna sin2 �
Spatial regions:
near(static) zone d≪ r ≪ �
intermediate d≪ r ∼ �
far(radiation) d≪ �≪ r
d - dimension of source
� =2�
k
19
Multipole Radiation
static:
1
∣r− r′∣=
1
r>
∑(r<r>
)lPl (r ⋅ r′)
plane wave:
k ⊥ propagating plane
eik⋅r so k → ∇eik⋅r = ik
n ∼ ∇�
and r×∇� ⊥ ∇� M=r×∇
choose E or H ∝M�
then the other one ∝ ∇× (M�)
20
De�ne
Elong ∝ ∇�lmEmag ∝M�lm
Eel ∝ ∇× (M�lm)
where �lm(k, r) is the solution to Helmholtzequation
21
Separation of variables in sphericalcoordinates r ≡ (r,Ω)
Hemlholtz eq(∇2 + k2
) = 0
i) Separate radial and angular parts:
p2 = (r ⋅ p)2 + (r× p)2 = p2r + l2
ii) w/∇operator
∇2=∇2r +
1
r2(r×∇)2 =
1
r2∂2
∂r2r2 +
M 2
r2
iii) Assume
lm (r, k) =ul (r)
rYlm (Ω)(
d2
dr2+ k2 − l (l + 1)
r2
)ul (r) = 0
angular: M 2Ylm (Ω) = −l (l + 1)Ylm (Ω)
MzYlm (Ω) = imYlm (Ω)
22
Properties:
M×M=−M (as commutators)
r ⋅M=0[M,∇2
]=0 [r,M] =0
Spherical Bessel functions:
jl (x) nl (x)
ℎ±l (x) = jl (x)± inl (x)
Rodrigues: jl (x) = (−x)l(
1
x
d
dx
)lsinx
x
First few:
j0 (x) =sinx
xj1 (x) =
sinx
x2− cosx
x
n0 (x) = −cosx
xn1 (x) = −cosx
x2− sinx
x
ℎ+0 (x) =1
i
eix
xℎ+1 (x) = −e
ix
x
(1 +
i
x
)Bessel eq:[
1
x
d2
dx2x− l (l + 1)
x2+ 1
]Zl (x) = 0
x→ kr
23
Recursion relations:
2l + 1
xZl (x) = Zl−1 (x) + Zl+1 (x)
Asymptotics:
jl (x)→ 1
xsin
(x− l�
2
)nl (x)→ −1
xcos
(x− l�
2
)
Green function G (r, r′) =eik∣r−r
′∣
4� ∣r− r′∣
G (r, r′) = ik∑
jl (kr<)ℎ+l (kr>)∑m
Y ∗lm (Ω′)Ylm (Ω)
discontinuity @ r = r′
24
Solution Helmholtz scalar equation:
�lm (r) =
{fl (kr) Ylm (Ω)
gl (kr) Ylm (Ω)
M� = r×∇� vector solution:
M 2 (M�) = M(M 2�
)= −�M�
i) pure magnetic multipoles
r ⋅ E(m)lm = 0 so
E(m)lm = gl (kr)MYlm (Ω)
B(m)lm = − i
k∇× E
(m)lm from Maxwell's
25
ii) pure electric multipoles
r ⋅B(e)lm = 0 so
B(e)lm = fl (kr)MYlm (Ω)
E(e)lm =
i
k∇×B
(e)lm
General solution:
E =∑
lm
[i
kAlm∇× (flMYlm) + Blm glMYlm
]H =
∑lm
[Alm flMYlm −
i
kBlm∇× (glMYlm)
]
26
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