electromagnetic potentials
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
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