1. Electromagnetic Potentials
E = -f Scalar Potential f and Electrostatic Field E
 x E = -∂B/∂t Faraday’s Law
 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 =  x A Identify B in terms of Vector Potential
E = -f - ∂A/∂t B =  x A

2. 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

3. 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 . AT = 0
Choose . A’ = 0 f = -AL A’ = AT
f’ = f - ∂f/∂t f’ = f - ∂f/∂t f’ = f + ∂AL/∂t
E = -f - ∂A/∂t = (-f) - ∂(AL+ AT)/∂t
 x E =  x (-f - ∂A/∂t) =  x -∂AT/∂t  x f =  x ∂AL/∂t = 0
E = -f’ - ∂A’/∂t = (-f - ∂AL/∂t) - (∂AT/∂t)
 x E =  x -∂AT/∂t

4. 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
2 − 1 c2 ∂2 ∂t2 A = – mo j + 1 c2 ∂f ∂t
. E = r /eo Gauss’ Law
. (-f - ∂A/∂t) = -. f - ∂. A/∂t = - 2f = r /eo

5. Electromagnetic Potentials
2 − 1 c2 ∂2 ∂t2 A = − mo j + 1 c2 ∂f ∂t
-2f = r /eo Coulomb or Transverse Gauge
Coupled equations for A, f

6. 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
2 − 1 c2 ∂2 ∂t2 A = – mo j
. E = r /eo Gauss’ Law
. (-f - ∂A/∂t) = -. f - ∂. A/∂t = -. f + 1/c2 ∂2f/∂t2 = r /eo
2 − 1 c2 ∂2 ∂t2 f = – r eo

7. Electromagnetic Potentials
. A = – 1/c2 ∂f/∂t
2 − 1 c2 ∂2 ∂t2 A = – mo j Lorentz Gauge
2 − 1 c2 ∂2 ∂t2 f = – r eo
□2 A f = mo j r eo
□2 = 1 c2 ∂2 ∂t2 − 2

8. Electromagnetic Potentials
□2 A f = mo j r eo Each component of A, f obeys wave equation with a source
□2 = 1 c2 ∂2 ∂t2 − 2
□2G(r - r’, t - t’) = d(r - r’) d(t - t’) Defining relation for Green’s function
d(r - r’) d(t - t’) Represents a point source in space and time
G(r - r’, t - t’) = 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)

9. Electromagnetic Potentials
f(r, t) = dr’dt’ G(r − r’, t − t’) r r’, t’ e o Solution in terms of G and source
= dr’dt’ q(t − t’) d(t − t’ − r − r’ /c) 4p r−r’ r r’, t’ e o
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,
f(r, t) = 1 4p e o dr’ r r’,t − t r−r’ c.f. GP Eqn 13.11
Similarly
A(r, t) = m o 4p dr’ j r’,t − t r−r’ c.f. GP Eqn 13.12
These are retarded vector and scalar potentials

10. Radiation by Hertz Electric Dipole+q -q l x y z
r = (x, y, z) Field Point
r' = (0, 0, z’) Source Point
Charge q(t) = qo Re {eiwt}
Current I(t) = dq/dt = qo Re {iw eiwt}
Dipole Moment p(t) = po Re {eiwt} = qo l Re {eiwt}
Wire Radius a
Current Density j(t) = I(t) / p a2
Using retarded potentials, calculate E(r,t), B(r,t) for dipole at origin

11. Radiation by Hertz Electric Dipole
Retarded Electric Vector Potential
A(r, t) = mo 4p j (r’, t − t) r – r’ dr’ 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 4p I (t − r / c ) r for distances r >> l
. A = – 1/c2 ∂f/∂t Obtain f from Lorentz Gauge condition
. A = ∂Az(r, t) / ∂z = mo l 4p ∂ ∂z I (t − r / c ) r
= mo l 4p − ∂I (t − r / c) ∂(t − r / c) z cr2 −I (t − r / c) z r3 = – moeo ∂f/∂t
∂f/∂t = l 4peo ∂I (t − r / c) ∂(t − r / c) z cr2 +I (t − r / c) z r3

12. Radiation by Hertz Electric Dipole
Differentiate wrt z and integrate wrt t to obtain
f(r, z, t) = l 4peo q(t − r / c) z r3 +I(t − r / c) z cr2
Az(r, t) = mo l 4p I (t − r / c ) r
I (t − r / c) dt = dq dt (t − r / c) dt =q(t − r/c)
∂I (t − r / c) ∂(t − r / c) dt = I (t − r / c) since d(t - r/c) = dt
Charge q(t) = qo Re {eiwt}
Current I(t) = dq dt = qo Re {iw eiwt} Electric Field E(r, t) = - f - ∂A ∂t

13. Radiation by Hertz Electric Dipole
Switch to spherical polar coordinates
f(r, q, t) = l 4peo q(t − r / c) cos q r2 +I(t − r / c) cos q cr
 = ∂ ∂r , 1 r ∂ ∂q , 1 r sinq ∂ ∂𝜑
-f(r, q, t) = po 4peo eiw(t – r / c) r cos q 2 r2 + 2ik r −k2 ,sin q 1 r2 + ik r , 0
k = w / c po = qol is the dipole amplitude

14. Radiation by Hertz Electric Dipole
Obtain part of E field due to A vector
Az(r, t) = mo l 4p I (t − r / c ) r Cartesian representation
A(r, q, t) = mo l 4p qo 𝑖w eiw(t – r / c) r cos q, − sin q , 0 Spherical polar rep’n
- ∂A(r, q, t) ∂t = − mo l 4p qo 𝑖w 2 eiw(t – r / c) r cos q, − sin q , 0
- ∂A(r, q, t) ∂t = − l 4peo qo 𝑖w 2 eiw(t – r / c) c2 r cos q, − sin q , 0
- ∂A(r, q, t) ∂t = po 4peo eiw(t – r / c) r k2 cos q, −k2 sin q, 0

15. Radiation by Hertz Electric Dipole
Total E field
E(r, q, t) = po 4peo eiw(t – r / c) r cos q 2 r2 + 2ik r ,sin q 1 r2 + ik r −k2 , 0
Long range (radiated) electric field, proportional to 1 r
Erad(r, q, t) = po 4peo eiw(t – r / c) r 0 ,− k2 sin q, 0
Radiated E field lines
sin q, sin2q polar plots

16. Radiation by Hertz Electric Dipole
Short range, electrostatic field w = 0 i.e. k = w / c → 0
Total E field
E(r, q, t) = po 4peo eiw(t – r / c) r cos q 2 r2 + 2ik r ,sin q 1 r2 + ik r −k2 , 0
Eelectrostat. = -f(r, q) = po 4peor3 2 cos q, sin q , 0
Classic field of electric point dipole

17. Radiation by Hertz Electric Dipole
Obtain B field from  x A
 x A= 1 r sinq ∂ ∂q A𝜑 sinq − ∂Aq ∂𝜑 , 1 r 1 sinq ∂Ar ∂𝜑 − ∂ ∂r rA𝜑 , 1 r ∂ ∂r rAq − ∂Ar ∂q
A(r, q, t) = mo l 4p qo 𝑖w eiw(t – r / c) r cos q, − sin q , 0
= po 4peo 𝑖w eiw(t – r / c) c2r cos q, − sin q , 0
 x A= po 4peo eiw(t – r / c) cr 0, 0, sin q ik r −k2
Radiated part of B field
Brad(r,q, t)=− po 4peo eiw(t – r / c) cr 0, 0,k2sin q Brad = Erad / c

18. Radiation by Hertz Electric Dipole
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 B𝜑 r / mo
Eq = − po 4peo eiw(t – r / c) r k2sin q
B𝜑= Eq c =− po 4peo eiw(t – r / c) cr k2sin q
W = N . dS = 1 mo 0 2p d𝜑 0 p dq r2 sin q Eq B𝜑

19. Radiation by Hertz Electric Dipole
W = N . dS = 1 mo 0 2p d𝜑 0 p dq r2 sin q Eq B𝜑
= 1 moc 0 2p d𝜑 0 p dq r2 sin q po 4peo 1 r k2sin q 2
= 2p moc po 4peo 2k4 4 3 cos2 w(t – r / c) p dq sin3q = <cos2 w(t – r / c)> = 1 2
W = cpo2k4 12peo = po2w4 12peoc3 = w2 Io2 l 2 12 p eo c Average power over one cycle

20. Radiation by Half-wave Antennax y z
r = (x, y, z) Field Point
r' = (0, 0, z’) Source Point q’ q
Half Wave Antenna r’ r
r‘ sinq = r sin(p – q’) = r sinq’
r‘2 = z’2 + r2 – 2 r z’ cosq
r‘ = r2 – 2 r z’ cosq + z’2
r‘ = r 1 – 2 z’ r cosq + z’2 r2
r‘ ⋍ r – z’ cosq
t – r’ c ⋍ t – r c + z’ cosq c
Current distribution
I(z’, t) = Io cos (2p z’/ l) eiwt
Current distribution on wire is
half wavelength and harmonic in time

21. Radiation by Half-wave Antenna
Single Hertz Dipole
Eqrad = − po 4peo eiw(t – r / c) r k2sin q k2 po = w2/c2 qo l = w Io l / c2 Io = w qo
Eqrad = − Iow l sin q 4peoc2 eiw(t – r / c) r
Current distribution in antenna I(z’, t) = Io cos 2p z’ l eiwt
Radiation from antenna is equivalent to sum of radiation from Hertz dipoles
dEq = − Io w dz’ sin q ’ 4peoc2 cos 2p z’ l eiw(t – r′ / c) r′ t – r’ c ⋍ t – r c + z’ cosq c
⋍− Io w 4peoc2 sin q r dz’ cos 2p z’ l eiw(t – r / c + z’ cosq / c) sin q ’ r’ = r r’ 2 sin q r ⋍ sin q r

22. Radiation by Half-wave Antenna
dEq ⋍− Io w 4peoc2 sin q r dz’ cos 2p z’ l eiw(t – r / c + z’ cosq / c)
Eq ⋍− Io w 4peoc2 sin q r eiw(t – r / c) −l/4 +l/4 dz’ cos 2p z’ l eiw z’ cosq / c
cos 2p z’ l = 1 2 ei2pz’/l + e−i2pz’/l
−l/4 +l/4 dz’ cos 2p z’ l eiw z’ cosq / c = 2 k cos p cosq /2 sin2 q k = 2p l

23. Radiation by Half-wave Antenna
Half Wave Antenna electric field
Eqrad⋍− Io w 4peoc2 eiw(t – r / c) r sin q 2 k cos p cosq /2 sin2q
Eqrad⋍− Io 2peoc eiw(t – r / c) r cos p cosq /2 sinq c.f. GP NB phase difference
Hertz Dipole electric field
Eqrad = − Iow l sin q 4peoc2 eiw(t – r / c) r = − Io 2peoc l w 2c eiw(t – r / c) r sin q l w 2c = π l λ ≪1
In general, for radiation in vacuum B = k x E / c, hence for antenna
B𝜑rad⋍− Io 2peoc eiw(t – r / c) c r cos p cosq /2 sin q

24. Radiation by Half-wave Antenna
Eq ⋍− Io 2peoc eiw(t – r / c) r cos p cosq /2 sin q
B𝜑⋍− Io 2peoc eiw(t – r / c) c r cos p cosq /2 sin q
W = N . dS = 1 mo 0 2p d𝜑 0 p dq sin q Eq B𝜑
= 1 mo Io2 4p2eo2 c p d𝜑 0 p dq sinq cos2 p cosq /2 sin2q cos2(t − r/c)
W = Io2 4peo c 0 p dq cos2 p cosq /2 sinq
W ⋍73 Io Average power over one cycle

25. Radiation by Half-wave Antenna
W = Io2 4peo c 0 p dq cos2 p cosq /2 sinq p dq cos2 p cosq /2 sinq =
W ⋍73 Io W =5 kW if Io⋍12 A radiation resistance = 73 W
Polar plot for half wave antenna
Hertz Dipole
W = ω2Io2 l 2 12 p eo c3 = Io2 4 p eo c w2 c2 l 2 3 = Io2 4 p eo c k2l 2 3
Polar plot for Hertz dipole

26. Radiation by Half-wave Antenna
Full Wave Antenna
Eq ⋍−𝑖 Io 2peoc eiw(t – r / c) r sin p cosq sin q
0 p dq sin2 p cosq sinq =
W ⋍73 Io W =5 kW if Io⋍12 A radiation resistance = 73 W
Hertz Dipole
W = Io2 l 2 12 p eo c3 = Io2 4 p eo c w2 c2 l 2 3 = Io2 4 p eo c k2l 2 3