Electromagnetic Potentials

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Electromagnetic Potentials - PowerPoint PPT Presentation

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

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Electromagnetic Potentials

E= -fScalar Potential f and Electrostatic Field E

 x E = -∂B/∂t Faraday’s Law

 x -f = 0 ≠ -∂B/∂t Substitute E = -fin 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 + ff’= 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 . 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 Potentials

Coulomb Gauge

Choose . A = 0

Represent Maxwell laws in terms of A,f potentials and j, r sources

 x B = moj + moeo∂E/∂t Maxwell-Ampère Law

 x ( x A) = moj + moeo∂(-f - ∂A/∂t)/∂t

 (. A)- A = moj – 1/c2 ∂f/∂t - 1/c2∂2A/∂t2

. E = r /eoGauss’ Law

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

Electromagnetic Potentials

. A = 0

-2f = r /eoCoulomb or Transverse Gauge

Coupled equations for A, f

Electromagnetic Potentials

Lorentz Gauge

Choose . A = – 1/c2∂f/∂t

 x B = moj + moeo∂E/∂t Maxwell-Ampère Law

 (. A)- A = moj – 1/c2 ∂f/∂t - 1/c2∂2A/∂t2

. E = r /eoGauss’ 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, fobeys 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 signal

to 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 Eqn13.12

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

y

l

x

r' = (0, 0, z’) Source Point

-q

Charge q(t) = qo Re {eiwt}

Current I(t) = dq/dt = qoRe {iweiwt}

Dipole Moment p(t) = poRe {eiwt} = qolRe {eiwt}

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

• 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 =
• Differentiate wrt z and integrate wrtt to obtain
• Az(r, t)
• since d(t - r/c) = dt
• Charge q(t) = qo Re {eiwt}
• Current I(t) = = qoRe {iweiwt} Electric Field E(r, t) = - f -
• Switch to spherical polar coordinates
• -
• k = w / c is the dipole amplitude
• Obtain part of E field due to A vector
• Az(r, t) Cartesian representation
• A(r, t) Spherical polar rep’n
• -
• -
• -
• Total E field
• E
• Long range (radiated) electric field, proportional to

, polar plots

Short range, electrostatic field = 0 i.e. k = / c → 0

Total E field

E

Eelectrostat. = -

Classic field of electric point dipole

• Obtain B field from  x A
• A(r, t)
• Radiated part of B field
• t)= / c
• 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
• = = > =
• =
• Average power over one cycle
• = wqopo = qo po w =

Half Wave Antenna

r = (x, y, z) Field Point

z

r’’

r

q’

r' = (0, 0, z’) Source Point

t – t –

y

q

l/2

x

Current distribution

I(z’, t) = Iocos (2p z’/ l) eiwt

Current distribution on wire is

half wavelength and harmonic in time

• Single Hertz Dipole
• = = / =
• Current distribution in antenna (z’, t) = cos
• Radiation from antenna is equivalent to sum of radiation from Hertz dipoles
• t – t –
• Half 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