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