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:
∂Λ
V →V − A → A + ∇Λ
∂t
The gauge transformation Λ is a scalar satisfying:
2
 
1 ∂ Λ 1 ∂V
∇2 Λ − 2 2 = − ∇.A + 2
c ∂t c ∂t

In electrostatics we use the Coulomb gauge:

∂V 2 ρ
∇.A = 0 =0 ∇ V =− ∇2 A = −µ0 J
∂t 0
In electrodynamics we use the Lorentz gauge:
1 ∂V 2 1 ∂2V ρ 2 1 ∂2A
∇.A = − 2 ∇ V− 2 2 =− ∇ A− 2 2 = −µ0 J
c ∂t c ∂t 0 c ∂t

2
 Retarded Potentials
Variations in charge density ρ or current density J at (r0 , t0 ) lead to
changes in the potentials V and A at (r, t):
1 ρ(r0 , t − ∆t) 0
Z
V (r, t) = dτ
4π0 V |r − r0 |
µ0 J(r0 , t − ∆t) 0
Z
A(r, t) = dτ
4π V |r − r0 |
These are known as retarded potentials
The propagation time of the changes is:

0 |r − r0 |
∆t = t − t =
c

The effect of the changes in ρ or J propagate from r0 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 sin ωt

This is a simple model for atomic and molecular vibrations

Corresponds to oscillating current between the ends of the dipole:
dQ
I= = I0 cos ωt I0 = ωQ0
dt

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:
µ0 I(r0 , t − ∆t) 0
Z
Az (r, t) = dz
4π |r − r0 |

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

the dipole, which is equivalent to assuming λ  a
µ0 a µ0 dp
Az (r, t) = I(t − ∆t) =
4πr 4πr dt
The scalar potential is obtained from the Lorentz gauge condition:
1 ∂V ∂V
∇.A = − 2 = −µ0 0
c ∂t ∂t
 
cos θ r dp
V (r, t) = p+
4π0 r 2 c dt

6
 Fields of Hertzian Dipole
The magnetic field is obtained from B = ∇ × A:
2
 
µ0 sin θ 1 dp 1 d p
B= 2
+ 2
φ̂
4π r dt rc dt
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:
2
 
1 2p cos θ p sin θ 2 cos θ dp sin θ dp sin θ d p
E= r̂ + θ̂ + 2 r̂ + 2 θ̂ + θ̂
4π0 r3 r3 r c dt r c dt rc2 dt2

E has components in the r and θ directions

7
Notes:

Diagrams:

8
 Interpretations of Hertzian Fields

1. Static field is proportional to 1/r 3 and depends on p

p  
E= 3
2 cos θ r̂ + sin θ θ̂
4π0 r

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

1 dp  
E= 2
2 cos θ r̂ + sin θ θ̂
4π0 r c dt
µ0 sin θ dp
B= 2
φ̂
4πr dt
3. Radiation fields are proportional to 1/r and d2 p/dt2
sin θ d2 p µ0 sin θ d2 p
E= θ̂ B= φ̂
4π0 rc2 dt2 4πrc dt2

9
 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:
µ0 p0 sin θ 2 i(ωt−kr) p0 sin θ 2 i(ωt−kr)
Bφ = ω e Eθ = ω e
4πrc 4π0 rc2

10
 Power radiated by Hertzian Dipoles
The Poynting vector is:
2  2
2
sin θ d p
Nr =
16π 2 0 r 2 c3 dt2
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:

 2 2 Z 2  2 2
d p sin θ 1 d p
I
P= N.dS = 2 3
d(cos θ) =
A dt 4π 0 c 6π0 c3 dt2
Radiated power is conserved since this is independent of r
The time-averaged radiated power is proportional to ω 4 :
p20 ω 4
< P >=
12π0 c3

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 0 gives:

I0 cos[(π cos θ)/2] i(ωt−kr)
Eθ = e
2π0 cr sin θ
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 0 is replaced by sin(π cos θ):
I0 sin(π cos θ) i(ωt−kr)
Eθ = e
2π0 cr sin θ

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:
Qa(t0 ) sin θ
Eθ (t) =
4π0 rc2
where a(t0 ) is the acceleration at t0 = t − r/c
For a non-relativistic charge Q in circular motion a = Rω 2 :
QRω 2 sin θ
E=
4π0 rc2
and the total power radiated from the charge is:
2Q2 R2 ω 4
P=
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 Q 2 a2 sin2 θ
=
dθ 4πc3 (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 γ 4 as β → 1:
2Q2 cγ 4 β 4
P=
3R2
These energy losses limit the maximum energy of circular accelerators

16