Electromagnetism - Lecture 17

Radiation Fields

• The Lorentz Gauge

• Hertzian Dipole

• Radiation Fields

• Antennas

• Synchrotron Radiation

1

The Lorentz Gauge

Electromagnetism (Maxwell’s Equations) are unchanged by:

V → V −∂Λ

∂tA → A + ∇Λ

The gauge transformation Λ is a scalar satisfying:

∇2Λ −1

c2

∂2Λ

∂t2= −

(

∇.A +1

c2

∂V

∂t

)

In electrostatics we use the Coulomb gauge:

∇.A = 0∂V

∂t= 0 ∇2V = −

ρ

ε0∇2A = −µ0J

In electrodynamics we use the Lorentz gauge:

∇.A = −1

c2

∂V

∂t∇2V −

1

c2

∂2V

∂t2= −

ρ

ε0∇2A−

1

c2

∂2A

∂t2= −µ0J

2

Retarded Potentials

Variations in charge density ρ or current density J at (r′, t′) lead to

changes in the potentials V and A at (r, t):

V (r, t) =1

4πε0

∫

V

ρ(r′, t − ∆t)

|r− r′|dτ ′

A(r, t) =µ0

4π

∫

V

J(r′, t − ∆t)

|r− r′|dτ ′

These are known as retarded potentials

The propagation time of the changes is:

∆t = t − t′ =|r− r′|

c

The effect of the changes in ρ or J propagate from r′ to r as an

electromagnetic wave with speed c

3

Notes:

Diagrams:

4

The Hertzian Dipole

An electric dipole has its charges oscillating with frequency ω:

Q = Q0 sin ωt p = Qa = p0 sinωt

This is a simple model for atomic and molecular vibrations

Corresponds to oscillating current between the ends of the dipole:

I =dQ

dt= I0 cosωt I0 = ωQ0

The changes in Q and I are propagated as electromagnetic waves

radiated outwards from the centre of the dipole

EM waves are produced by oscillating charges and currents

Reverse process: Absorption of EM waves creates oscillating Q,I

5

Potentials of Hertzian Dipole

There is a retarded magnetic vector potential parallel to the

current, which is along the direction of the dipole:

Az(r, t) =µ0

4π

∫

I(r′, t − ∆t)

|r− r′|dz′

We take the far-field limit r a and ignore the variation in I along

the dipole, which is equivalent to assuming λ a

Az(r, t) =µ0a

4πrI(t − ∆t) =

µ0

4πr

dp

dt

The scalar potential is obtained from the Lorentz gauge condition:

∇.A = −1

c2

∂V

∂t= −µ0ε0

∂V

∂t

V (r, t) =cos θ

4πε0r2

(

p +r

c

dp

dt

)

6

Fields of Hertzian Dipole

The magnetic field is obtained from B = ∇×A:

B =µ0 sin θ

4π

(

1

r2

dp

dt+

1

rc

d2p

dt2

)

φ

B is in the φ direction (around the dipole axis)

The electric field is obtained from M4 after some tedious

manipulation using ∇×B in spherical polars:

E =1

4πε0

(

2p cos θ

r3r +

p sin θ

r3θ +

2 cos θ

r2c

dp

dtr +

sin θ

r2c

dp

dtθ +

sin θ

rc2

d2p

dt2θ

)

E has components in the r and θ directions

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Notes:

Diagrams:

8

Interpretations of Hertzian Fields

1. Static field is proportional to 1/r3 and depends on p

E =p

4πε0r3

(

2 cos θ r + sin θ θ)

2. Induction fields are proportional to 1/r2 and dp/dt

E =1

4πε0r2c

dp

dt

(

2 cos θ r + sin θ θ)

B =µ0 sin θ

4πr2

dp

dtφ

3. Radiation fields are proportional to 1/r and d2p/dt2

E =sin θ

4πε0rc2

d2p

dt2θ B =

µ0 sin θ

4πrc

d2p

dt2φ

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Properties of Radiation Fields

At large distances r a the radiation fields dominate

• Bφ and Eθ are perpendicular to r and to each other

• The Poynting vector N = E×H points radially outwards

• The amplitudes of the fields vary with sin θ and the Poynting

vector is proportional to sin2 θ

• The ratio of the amplitudes is B0 = E0/c, which is a

characteristic of electromagnetic waves

• The fields can be written in the form of plane waves:

Bφ =µ0p0 sin θ

4πrcω2ei(ωt−kr) Eθ =

p0 sin θ

4πε0rc2ω2ei(ωt−kr)

10

Power radiated by Hertzian Dipoles

The Poynting vector is:

Nr =sin2 θ

16π2ε0r2c3

(

d2p

dt2

)2

Most power is radiated in the midplane of the dipole, and none is

radiated along the dipole axis!

Integrating over a spherical surface of radius r, the total power is:

P =

∮

A

N.dS =

(

d2p

dt2

)2 ∫

sin2 θ

4πε0c3d(cos θ) =

1

6πε0c3

(

d2p

dt2

)2

Radiated power is conserved since this is independent of r

The time-averaged radiated power is proportional to ω4:

< P >=p20ω

4

12πε0c3

11

Notes:

Diagrams:

12

Half-Wave Antennas

Reception and transmission of EM waves by antennas is the basis

of TV, mobile phones, satellite communication ...

Practical antennas do not satisfy λ a

⇒ have to include variation of I along length of antenna

For a half-wave antenna a = λ/2 the integral over Idz′ gives:

Eθ =I0

2πε0cr

cos[(π cos θ)/2]

sin θei(ωt−kr)

which is still peaked at θ = 90 and zero at 0

The power radiated can be expressed in terms of an impedance:

< P >=< I >2 Rrad Rrad = 73Ω

13

Full Wave Antennas

For a full-wave antenna a = λ, the factor cos[(π cos θ)/2] from the

integral over Idz′ is replaced by sin(π cos θ):

Eθ =I0

2πε0cr

sin(π cos θ)

sin θei(ωt−kr)

The angular distribution has four lobes

(see Grant & Phillips P.447 for pictures)

The impedance of a full-wave antenna is Rrad ≈ 100Ω

14

Accelerated Charges

A Hertzian dipole is an example of an accelerated charge

All accelerated charges radiate EM waves

General form for the radiation field is:

Eθ(t) =Qa(t′) sin θ

4πε0rc2

where a(t′) is the acceleration at t′ = t − r/c

For a non-relativistic charge Q in circular motion a = Rω2:

E =QRω2 sin θ

4πε0rc2

and the total power radiated from the charge is:

P =2Q2R2ω4

3c3

15

Synchrotron Radiation

An important application of circularly accelerated charges is to

produce beams of synchrotron radiation

For a relativistic charge the angular distribution of the radiation is

boosted into a narrow cone around the direction of the particle:

dP

dθ=

Q2a2

4πc3

sin2 θ

(1 − β cos θ)5

This has a peak at θmax = 1/2γ with width θrms = 1/γ

As β → 1, γ 1 the radiated beam becomes tangential

See Jackson Pp. 669-670 for pictures

The total power radiated is proportional to γ4 as β → 1:

P =2Q2cγ4β4

3R2

These energy losses limit the maximum energy of circular accelerators

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