### Electromagnetic Potentials

```Electromagnetic Potentials
E = -f
Scalar Potential f and Electrostatic Field E
 x E = -∂B/∂t
 x -f = 0 ≠ -∂B/∂t
Substitute E = -f in Faraday’s law
 x E =  x (-f - ∂A/∂t) = 0 - ∂( x A)/∂t = -∂B/∂t
E = -f - ∂A/∂t
Generalize to include Vector Potential A
B=xA
Identify B in terms of Vector Potential
E = -f - ∂A/∂t
B=xA
Electromagnetic Potentials
(A,f) 4-vector generates E, B 3-vectors ((A,f) redundant by one degree)
Suppose (A,f) and (A’,f’) generate the same E, B fields
E = -f - ∂A/∂t = -f’ - ∂A’/∂t
B =  x A =  x A’
Let A’ = A + f
 x A’ =  x (A + f) =  x A +  x f =  x A
What change must be made to f to generate the same E field?
E = -f’ - ∂A’/∂t = -f’ - ∂ (A + f )/∂t = -f - ∂A/∂t
A’ = A + f
f’ = f - ∂f/∂t
Gauge Transformation
Electromagnetic Potentials
A = AL + AT
L and T components of A
A’ = AL + AT + f
Change of gauge
. A’ = . AL + . AT + . f = . AL + 2 f
Choose . A’ = 0
f’ = f - ∂f/∂t
. AT = 0
f = -AL
f’ = f - ∂f/∂t
A’ = AT
f’ = f + ∂AL/∂t
E = -f - ∂A/∂t = (-f) - ∂(AL+ AT)/∂t
 x E =  x (-f - ∂A/∂t) =  x -∂AT/∂t
E = -f’ - ∂A’/∂t = (-f - ∂AL/∂t) - (∂AT/∂t)
 x E =  x -∂AT/∂t
 x f =  x ∂AL/∂t = 0
Electromagnetic Potentials
Coulomb Gauge
Choose . A = 0
Represent Maxwell laws in terms of A,f potentials and j, r sources
 x B = mo j + moeo ∂E/∂t
Maxwell-Ampère Law
 x ( x A) = mo j + moeo ∂ (-f - ∂A/∂t)/∂t
 (. A) - 2 A = mo j – 1/c2 ∂f/∂t - 1/c2 ∂ 2A/∂t2
1 ∂2
1 ∂f
2
 − c2 ∂t2 A = – mo j + c2 ∂t
. E = r /eo
Gauss’ Law
. (-f - ∂A/∂t) = -. f - ∂. A/∂t = - 2f = r /eo
Electromagnetic Potentials
. A = 0
1 ∂2
1 ∂f
2 − c2 ∂t2 A = − mo j + c2 ∂t
-2f = r /eo
Coupled equations for A, f
Coulomb or Transverse Gauge
Electromagnetic Potentials
Lorentz Gauge
Choose . A = – 1/c2 ∂f/∂t
 x B = mo j + moeo ∂E/∂t
Maxwell-Ampère Law
 (. A) - 2 A = mo j – 1/c2 ∂f/∂t - 1/c2 ∂ 2A/∂t2
1 ∂2
2 − c2 ∂t2 A = – mo j
. E = r /eo
Gauss’ Law
. (-f - ∂A/∂t) = -. f - ∂. A/∂t = -. f + 1/c2 ∂2f/∂t2 = r /eo
1 ∂2
r
2 − c2 ∂t2 f = –
e
o
Electromagnetic Potentials
. A = – 1/c2 ∂f/∂t
1 ∂2
2 − c2 ∂t2 A = – mo j
1 ∂2
r
2 − c2 ∂t2 f = –
e
o
mo j
= r
f
eo
□
A
□2
1 ∂2
= c2 ∂t2 − 2
2
Lorentz Gauge
Electromagnetic Potentials
mo j
= r
f
eo
□2
A
□2
1 ∂2
= c2 ∂t2 − 2
Each component of A, f obeys wave equation with a source
□2G(r - r’, t - t’) = d(r - r’) d(t - t’)
d(r - r’) d(t - t’)
G(r - r’, t - t’) =
Defining relation for Green’s function
Represents a point source in space and time
q(t − t’) d(t − t’ − r − r’ /c)
4p r − r’
Proved by substitution
d(t − t’ − r − r’ /c) is non-zero for t − t’ = r − r’ /c i.e. time taken for signal
to travel from r’ to r at speed c (retardation of the signal)
Unit step function q(t − t’) ensures causality (no response if t’ > t)
Electromagnetic Potentials
f(r, t) = dr’dt’ G(r − r’, t − t’)
=
r r’, t’
eo
Solution in terms of G and source
q(t − t’) d(t − t’ − r − r’ /c) r r’, t’
dr’dt’
4p r−r’
eo
Let t = r − r’ /c be the retardation time, then there is a contribution to f(r, t)
from r r’, t’ at t’ = t - t. Hence we can write, more simply,
1
f(r, t) = 4pe
dr’
r r’,t − t
r−r’
c.f. GP Eqn 13.11
mo
j r’,t − t
dr’
4p
r−r’
c.f. GP Eqn 13.12
o
Similarly
A(r, t) =
These are retarded vector and scalar potentials
r = (x, y, z) Field Point
z
+q
Using retarded potentials, calculate E(r,t), B(r,t) for dipole at origin
l
x
y
r' = (0, 0, z’) Source Point
-q
Charge
Current
Dipole Moment
Current Density
q(t) = qo Re {eiwt}
I(t) = dq/dt = qo Re {iw eiwt}
p(t) = po Re {eiwt} = qo l Re {eiwt}
a
j(t) = I(t) / p a2
Retarded Electric Vector Potential
A(r, t) =
mo j (r’, t − t)
dr’
4p
r – r’
A || ez because j || ez
Retardation time t = |r - ezz’| / c if l << c t then t ≈ |r| / c = r / c
Az(r, t) ≈
mo l I (t − r / c )
for distances r >> l
r
4p
. A = – 1/c2 ∂f/∂t
. A = ∂Az(r, t) / ∂z =
=
Obtain f from Lorentz Gauge condition
mo l ∂ I (t − r / c )
r
4p ∂z
mo l
∂I (t − r / c) z
z
−
− I (t − r / c) 3 = – moeo ∂f/∂t
2
r
4p
∂(t − r / c) cr
∂f/∂t =
∂I (t − r / c) z
l
z
+
I
(t
−
r
/
c)
r3
4peo ∂(t − r / c) cr2
Differentiate wrt z and integrate wrt t to obtain
l
z
z
f(r, z, t) =
q(t − r / c) 3 + I(t − r / c) 2
4peo
r
cr
Az(r, t) =
mo l I (t − r / c )
r
4p
I (t − r / c) dt =
dq
(t − r / c) dt = q(t − r/c)
dt
∂I (t − r / c)
dt = I (t − r / c)
∂(t − r / c)
since d(t - r/c) = dt
Charge
q(t) = qo Re {eiwt}
Current
I(t) =
dq
= qo Re {iw eiwt}
dt
Electric Field E(r, t) = - f -
∂A
∂t
Switch to spherical polar coordinates
l
cos q
cos q
f(r, q, t) =
q(t − r / c) 2 + I(t − r / c)
4peo
r
cr
=
∂ 1 ∂
1
∂
,
,
∂r r ∂q r sinq ∂
po eiw(t – r / c)
2 2ik
1 ik
-f(r, q, t) =
cos q 2 +
− k2 , sin q 2 +
,0
r
r
r
r
r
4peo
k=w/c
po = qol is the dipole amplitude
Obtain part of E field due to A vector
Az(r, t) =
mo l I (t − r / c )
r
4p
mo l qo w eiw(t – r / c)
A(r, q, t) =
cos q, − sin q , 0
r
4p
Cartesian representation
Spherical polar rep’n
mo l qo w 2 eiw(t – r / c)
∂A(r, q, t)
=−
cos q, − sin q , 0
r
4p
∂t
∂A(r, q, t)
l qo w 2 eiw(t – r / c)
=−
cos q, − sin q , 0
c2 r
4peo
∂t
∂A(r, q, t)
po eiw(t – r / c) 2
=
k cos q, −k2 sin q, 0
r
4peo
∂t
Total E field
po eiw(t – r / c)
2 2ik
1 ik
E(r, q, t) =
cos q 2 +
, sin q 2 + − k2 , 0
r
r
r
r
r
4peo
Long range (radiated) electric field, proportional to
1
r
po eiw(t – r / c)
q, t) = 4pe
0 , − k2 sin q, 0
r
o
sin q, sin2q polar plots
Short range, electrostatic field
w = 0 i.e. k = w / c → 0
Total E field
po eiw(t – r / c)
2 2ik
1 ik
E(r, q, t) =
cos q 2 +
, sin q 2 + − k2 , 0
r
r
r
r
r
4peo
Eelectrostat. = -f(r, q) =
po
2 cos q, sin q , 0
4peor3
Classic field of electric point dipole
Obtain B field from  x A
1
∂
∂Aq 1 1 ∂Ar
∂
xA=
A sinq −
,
−
rA
r sinq ∂q
∂ r sinq ∂ ∂r
1 ∂
∂Ar
,
rAq −
r ∂r
∂q
mo l qo w eiw(t – r / c)
A(r, q, t) =
cos q, − sin q , 0
r
4p
po w eiw(t – r / c)
=
cos q, − sin q , 0
c 2r
4peo
po eiw(t – r / c)
ik
xA=
0, 0, sin q
− k2
4peo
cr
r
po eiw(t – r / c)
t)= −
0, 0, k2sin q
cr
4peo
Power emitted by Hertz Dipole
The Poynting vector, N, gives the flux of radiated energy Jm-2s-1
The flux N = E x H depends on r and q, but the angle-integrated flux is constant
N = E x H = Eq B r / mo
po eiw(t – r / c) 2
Eq = −
k sin q
4peo
r
Eq
po eiw(t – r / c) 2
B=
=−
k sin q
c
cr
4peo
W=
1
N . dS =
mo
2p
p
dq r2 sin q Eq B
d
0
0
1
N . dS =
mo
W=
2p
p
dq r2 sin q Eq B
d
0
2p
p
d
dq
0
0
0
=
1
moc
=
2p po 2 4 4
k cos2 w(t – r / c)
moc 4pe
3
o
r2 sin q
po 1 2
k sin q
4peo r
p
0
2
dq sin3q =
cpo2k4
po2w4
w2 Io2 l 2 p Io2 l 2
W =
=
=
=
12peo 12peoc3 12 p eo c3 3eo c λ 2
Average power over one cycle
I o = w qo
po = qo l
po w = Io l
4
3
<cos2 w(t – r / c)> =
1
2
Half Wave Antenna
r = (x, y, z) Field Point
z
r’’
r
l/2
q’
r' = (0, 0, z’)
q
Source Point
y
x
r′′
r
r
=
=
sinq sin(p – q′) sinq′
r′′2 = z′2 + r2 – 2 r z’ cosq
r′′ = r2 – 2 r z’ cosq + z’2
z’
z’2
cosq + 2
r
r
r′′ ⋍ r – z’ cosq
r′′ = r
Current distribution
I(z’, t) = Io cos (2p z’/ l) eiwt
Current distribution on wire is
half wavelength and harmonic in time
1 –2
r′′
r z’ cosq
t – c ⋍ t – c+
c
Single Hertz Dipole
Eq
po eiw(t – r / c) 2
−
k sin q
r
4peo
Eq
Iow l sin q eiw(t – r / c)
−
r
4peoc2
Current distribution in antenna
k2 po = w2/c2 qo l = w Io l / c2
I(z’, t) = Io cos
I o = w qo
2p z’ iwt
e
l
Radiation from antenna is equivalent to sum of radiation from Hertz dipoles
Io w dz’ sin q ’
2p z’ eiw(t – r′′ / c)
dEq = −
cos
l
4peoc2
r′′
t–
r′′
r z’ cosq
⋍ t– +
c
c
c
Io w sin q
2p z’ iw(t – r / c + z’ cosq / c) sin q ’
r
⋍−
dz’ cos
e
= ′′
2
r
r′′
4peoc
l
r
2 sin
r
q
sin q
⋍ r
Io w sin q
2p z’ iw(t – r / c + z’ cosq / c)
dEq ⋍ −
dz’ cos
e
4peoc2 r
l
Io w sin q iw(t – r / c)
Eq ⋍ −
e
2
r
4pe c
o
+l/4
dz’ cos
−l/4
2p z’ iw z’ cosq / c
e
l
2p z’ 1 i2pz’/l
cos
= e
+ e−i2pz’/l
l
2
+l/4
−l/4 dz’ cos
2p z’ iw z’ cosq / c 2 cos p cosq /2
e
=
l
k
sin2 q
k=
2p
l
Half Wave Antenna electric field
e
I
w
o
4peoc2
Io e
Eq ⋍ −
2peoc
iw(t – r / c)
r
iw(t – r / c)
r
2 cos p cosq /2
sin q
k
sin2q
cos p cosq /2
sinq
c.f. GP 13.24 NB phase difference
Hertz Dipole electric field
Eq
l sin q eiw(t – r / c)
Io
l w eiw(t – r / c)
=−
sin q
r
r
4peoc2
2peoc 2c
l w πl
= ≪1
λ
2c
In general, for radiation in vacuum B = k x E / c, hence for antenna
e
I
o
2peoc c r
iw(t – r / c)
cos p cosq /2
sin q
Io e
Eq ⋍ −
2peoc
iw(t – r / c)
r
cos p cosq /2
sin q
Io e
B ⋍ −
2peoc c r
cos p cosq /2
sin q
2p
p
iw(t – r / c)
W=
1
N . dS =
mo
1
Io2
=
mo 4p2eo2 c3
dq r2 sin q Eq B
d
0
0
2p
p
d
dq
0
0
cos2 p cosq /2
sinq
sin2q
cos2(t − r/c)
cos2 p cosq /2
I o2 p
W =
dq
4peo c 0
sinq
I o2
W ⋍ 73
2
Average power over one cycle
Half Wave Antenna
cos2 p cosq /2
I o2 p
W =
dq
4peo c 0
sinq
I o2
W ⋍ 73
2
p
0
cos2 p cosq /2
dq
= 1.21883
sinq
W = 5 kW if Io ⋍ 12 A
Polar plot for half wave antenna
Hertz Dipole
2 2
2 2
Io2 w2 l
I o2 k l
W=
=
=
4 p eo c c 3
4 p eo c 3
12 p eo c3
ω2Io2 l 2
Polar plot for Hertz dipole
```