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

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 PotentialsA = 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

Electromagnetic PotentialsCoulomb 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) - A = mo j – 1/c2 ∂f/∂t - 1/c2 ∂ 2A/∂t2

. E = r /eo Gauss’ Law

. (-f - ∂A/∂t) = -. f - ∂. A/∂t = - r /eo

Electromagnetic Potentials

. A = 0

-2f = r /eo Coulomb or Transverse Gauge

Coupled equations for A, f

Electromagnetic PotentialsLorentz Gauge

Choose . A = – 1/c2 ∂f/∂t

x B = mo j + moeo ∂E/∂t Maxwell-Ampère Law

(. A) - A = mo j – 1/c2 ∂f/∂t - 1/c2 ∂ 2A/∂t2

. E = r /eo Gauss’ Law

. (-f - ∂A/∂t) = -. f - ∂. A/∂t = -. f + 1/c2 ∂2f/∂t2 = r /eo

Electromagnetic Potentials. A = – 1/c2 ∂f/∂t

Lorentz Gauge

□2

□2 =

Electromagnetic Potentials□2 Each component of A, f obeys wave equation with a source

□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’) = Proved by substitution

) is non-zero for i.e. time taken for signalto travel from r’ to r at speed c (retardation of the signal)

ensures causality (no response if t’ > t)

Electromagnetic Potentials Solution in terms of G and source

Let be the retardation time, then there is a contribution to from at t’ = t - . Hence we can write, more simply,

c.f. GP Eqn 13.11

Similarly

c.f. GP Eqn 13.12

These are retarded vector and scalar potentials

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 aCurrent Density j(t) = I(t) / p a2

Using retarded potentials, calculate E(r,t), B(r,t) for dipole at origin

Radiation by Hertz Electric Dipole Retarded Electric Vector Potential

A(r, t) A || ez because j || ez

Retardation time t = |r - ezz’| / c if l << c t then t ≈ |r| / c = r / c

Az(r, t) for distances r >> l

. A = – 1/c2 ∂f/∂t Obtain f from Lorentz Gauge condition

. A = ∂Az(r, t) / ∂z =

= –∂f/∂t

∂f/∂t =

Radiation by Hertz Electric Dipole Differentiate wrt z and integrate wrt t to obtain

Az(r, t)

since d(t - r/c) = dt

Charge q(t) = qo Re {eiwt}

Current I(t) = = qo Re {iw eiwt} Electric Field E(r, t) = - f -

Radiation by Hertz Electric Dipole Switch to spherical polar coordinates

-

k = w / c is the dipole amplitude

Radiation by Hertz Electric Dipole Obtain part of E field due to A vector

Az(r, t) Cartesian representation

A(r, t) Spherical polar rep’n

-

-

-

Radiation by Hertz Electric Dipole Total E field

E

Long range (radiated) electric field, proportional to

Erad

Radiated E field lines

, polar plots

Radiation by Hertz Electric Dipole Short range, electrostatic field = 0 i.e. k = / c → 0

Total E field

E

Eelectrostat. = -

Classic field of electric point dipole

Radiation by Hertz Electric Dipole Obtain B field from x A

A(r, t)

Radiated part of B field

t) = / c

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 = / mo

Radiation by Hertz Electric Dipole

= = > =

=

Average power over one cycle

= w qo po = qo po w =

Radiation by Half-wave Antenna

l/2

x

y

zr = (x, y, z) Field Point

r' = (0, 0, z’) Source Pointq’

q

r ′′sin q =

rsin ( p  –  q ′ )=

rsin q ′

t – t –

I(z’, t) = Io cos (2p z’/ l) eiwt

Current distribution on wire is half wavelength and harmonic in time

Half Wave Antenna

r

r’’

Current distribution

Radiation by Half-wave AntennaSingle Hertz Dipole

= = / =

Current distribution in antenna (z’, t) = cos

Radiation from antenna is equivalent to sum of radiation from Hertz dipoles

t – t –

Radiation by Half-wave Antenna

k =

Radiation by Half-wave AntennaHalf Wave Antenna electric field

c.f. GP 13.24 NB phase difference

Hertz Dipole electric field

1

In general, for radiation in vacuum B = k x E / c, hence for antenna

Radiation by Half-wave Antenna

=

=

Average power over one cycle

Radiation by Half-wave AntennaHalf Wave Antenna =

Polar plot for half wave antenna

Hertz Dipole

Polar plot for Hertz dipole

0.4

0.2

0.20.4

1.0

0.50.51.0