E f Scalar Potential f and Electrostatic Field E x E B t Faradays Law x f 0 B t ID: 618840
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Slide1
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 ASlide2
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 Slide3
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
-
∂
A
L
/
∂
t) -
(∂
A
T
/
∂
t)
x
E
=
x -
∂
A
T
/
∂
tSlide4
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) - A = mo j – 1/c2 ∂f/∂t - 1/c2 ∂ 2A/
∂
t
2
.
E = r /eo Gauss’ Law . (-f - ∂A/∂t) = -. f - ∂. A/∂t = - r /eo
Slide5
Electromagnetic Potentials
. A = 0 -2f = r /eo Coulomb or Transverse Gauge Coupled equations for A, fSlide6
Electromagnetic Potentials
Lorentz 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
/
e
o
Slide7
Electromagnetic Potentials
. A = – 1/c2 ∂f/∂t Lorentz Gauge □2 □2 =
Slide8
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 functiond(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 signal
to travel from
r
’ to
r
at speed c (retardation of the signal) ensures causality (no response if t’ > t) Slide9
Electromagnetic Potentials
Solution in terms of G and sourceLet
be the retardation time, then there is a contribution to
from
at t’ = t - . Hence we can write, more simply,
c.f.
GP
Eqn
13.11Similarly
c.f
.
GP
Eqn
13.12
These are
retarded vector and scalar potentials
Slide10
Radiation by Hertz Electric Dipole
+q-qlxyzr = (x, y, z) Field Pointr' = (0, 0, z’) Source PointCharge q(t) = qo Re {eiwt}Current I(t) = dq/dt = qo Re {iw eiwt}Dipole Moment p(t) = po Re {eiwt} = qo l Re {
e
i
wt} Wire Radius aCurrent Density j(t) = I(t) / p a2Using retarded potentials, calculate E(r,t), B(r
,t
) for dipole at originSlide11
Radiation by Hertz Electric Dipole
Retarded Electric Vector PotentialA(r, t) A || ez because j || ez Retardation time t = |r - ezz’| / c if l << c t then t ≈ |r| / c = r / cAz(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 =
Slide12
Radiation by Hertz Electric Dipole
Differentiate wrt z and integrate wrt t to obtainAz(r, t)
since d(t - r/c) =
dt
Charge q(t
) =
q
o
Re {
e
i
w
t
}
Current
I
(t
) =
=
q
o
Re {
i
w
eiwt
} Electric Field
E
(
r
, t) = -
f
-
Slide13
Radiation by Hertz Electric Dipole
Switch to spherical polar coordinates
-
k =
w
/ c
is the dipole amplitude
Slide14
Radiation by Hertz Electric Dipole
Obtain part of E field due to A vectorAz(r, t) Cartesian representationA(r, t) Spherical polar rep’n-
-
-
Slide15
R
adiation by Hertz Electric Dipole Total E fieldELong range (radiated) electric field, proportional to
E
rad
Radiated
E
field lines
,
polar plots
Slide16
Radiation by Hertz Electric Dipole
Short range, electrostatic field = 0 i.e. k = / c → 0Total E fieldE
E
electrostat
. = -
Classic field of electric point
dipole
Slide17
Radiation by Hertz Electric Dipole
Obtain B field from x A
A
(r,
t)
Radiated part of
B
field
t)
=
/ c
Slide18
Radiation by Hertz Electric Dipole
Power emitted by Hertz DipoleThe Poynting vector, N, gives the flux of radiated energy Jm-2s-1The flux N = E x H depends on r and q, but the angle-integrated flux is constantN = E x H = / mo
Slide19
Radiation by Hertz Electric Dipole
=
=
> =
Average
power over one cycle
Slide20
Radiation by Half-wave Antenna
l/2xyzr = (x, y, z) Field Pointr' = (0, 0, z’) Source Pointq’q
t –
t
–
I
(z’, t) =
I
o
cos
(2
p
z’/
l
)
e
i
w
t
Current distribution on wire is
half wavelength and harmonic in time
Half
W
ave
A
ntenna
r
r’
Current distributionSlide21
Radiation by Half-wave Antenna
Single Hertz Dipole = = / =
Current distribution in antenna
(z’, t) =
cos
Radiation from antenna is equivalent to sum of radiation from Hertz dipoles
t
–
t –
Slide22
Radiation by Half-wave Antenna
k =
Slide23
Radiation by Half-wave Antenna
Half Wave Antenna electric field
c.f.
GP 13.24 NB phase differenceHertz Dipole electric field
1
In general, for radiation in vacuum
B
=
k
x
E
/ c, hence for antenna
Slide24
Radiation by Half-wave Antenna
=
=
Average power over one cycle
Slide25
Radiation by Half-wave Antenna
Half Wave Antenna =
Polar plot for half wave antenna
Hertz Dipole
Polar
plot for
Hertz dipole
Slide26
Radiation by Half-wave Antenna
Full Wave Antenna
Hertz Dipole