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PHAS3201 EMT Radiation Emission

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					Electromagnetic Theory: PHAS3201, Winter 2008
9. Emission of Radiation
1    Retarded Potentials
We will consider the emission of electromagnetic waves from sources. To start with, we will solve for the scalar
and vector potentials in terms of charge and current densities.

Fields

    • How are fields determined by potentials ?
    • B is easy (from     · B = 0):
                                                          B=     ×A                                          (1)

    • We get E from Faraday’s law:
                                                                   ∂A
                                                     E=− φ−                                                  (2)
                                                                   ∂t
    • Use these to relate A and φ to J and ρ

    TAKE NOTES

Summary

    • We wish to solve for fields E(r, t) and B(r, t)
    • It is easier to work in terms of potentials:

                                                     B =         ×A                                          (3)
                                                                      ∂A
                                                     E = − φ−                                                (4)
                                                                      ∂t

    • From Maxwell’s equations, we find:

                                       2           ∂             ρ
                                           φ   +      ( · A) = −                                             (5)
                                                   ∂t            0

                                      2               ∂2A                      ∂φ
                                  −       A    +   µ 2 + ( · A) + µ               = µJ                       (6)
                                                      ∂t                       ∂t

    • We can use gauges to simplify

Lorenz Condition

    • We will impose a condition on our potentials:

                                                                 ∂φ
                                                       ·A+ µ        =0                                       (7)
                                                                 ∂t

    • This is known as the Lorenz condition.
    • Notice that we can always write A → A +          Λ and φ → φ − ∂Λ/∂t
    • So:
                                                      2          ∂2Λ
                                                          Λ− µ       =0                                      (8)
                                                                 ∂t2
    • All potentials belonging to the Lorenz gauge satisfy this condition


PHAS3201 Winter 2008                       Section IX. Emission of Radiation                                  1
PHAS3201: Electromagnetic Theory

Wave Equations

    • The vector potential now satisfies:
                                                       2               ∂2A
                                                  −        A+ µ            = µJ             (9)
                                                                       ∂t2
    • The scalar potential can be shown to satisfy:

                                                           2           ∂2φ   ρ
                                                   −           φ+ µ        =               (10)
                                                                       ∂t2

    • How do we solve these equations ?

    TAKE NOTES

Retarded Time

    • We write t = t − r/c
    • This is called retarded time
    • We choose:
                                                                       q(t − r/c)
                                                  f (r − ct) =                             (11)
                                                                          4π 0
    • This means that we can write:
                                                                      q(t − r/c)
                                                   φ(r, t) =                               (12)
                                                                        4π 0 r
    • This solves for a point charge at origin.

    If we now apply this solution to Eq. (10) we can write:

Form

    • Retarded scalar potential:
                                                                1          ρ (r , t )
                                             φ(r, t) =                                dv   (13)
                                                               4π 0    V   |r − r |

    • We have t = t − |r − r |/c
    • By considering the components of A we can write:

                                                                µ0         J (r , t )
                                              A(r, t) =                               dv   (14)
                                                                4π    V    |r − r |

    TAKE NOTES


2    Hertzian Dipole
Geometry

    • Two small spheres connected by a short wire
    • Each sphere has charge q(t)
    • Wire length l, no capacitance
    • Spherical polars shown

    TAKE NOTES


PHAS3201 Winter 2008                    Section IX. Emission of Radiation                    2
PHAS3201: Electromagnetic Theory




                                                       Figure 1:


Potentials

   • The vector potential:
                                                            µ0 l   I(t − r/c)
                                          Az (r, t) =                               (15)
                                                            4π          r

   • The scalar potential:
                                               l       z    q(t − r/c) I(t − r/c)
                                 φ(r, t) =                            +             (16)
                                             4π    0   r2        r          c

   • Choose:
                                          q(t − r/c) = q0 cos ω(t − r/c)            (17)

   TAKE NOTES

Important Points

   • These do not depend on φ
   • E and B are perpendicular
   • The power is radially outwards

   • The average power is:
                                                       ¯    l2 ω 2 I02
                                                       P =        3 2
                                                                                    (18)
                                                           6π 0 c
   • The radial components of E and B are zero

   TAKE NOTES




PHAS3201 Winter 2008                  Section IX. Emission of Radiation               3

				
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