Chapter Nine Radiation

Heinrich Rudolf Hertz (1857 - 1894)

October 12, 2001

Contents

1 Introduction 1

2 Radiation by a localized source 3 2.1 The Near Zone ...... 6 2.2 The Radiation or Far Zone ...... 6

3 Multipole Expansion of the Radiation Field 9 3.1 Electric Dipole ...... 10 3.1.1 Example: Linear Center-Fed Antenna ...... 13 3.2 Magnetic Dipole ...... 14 3.3 Comparison of Dipoles ...... 16 3.4 Electric Quadrupole ...... 18 3.4.1 Example: Oscillating Charged Spheroid ...... 21 3.5 Large Radiating Systems ...... 23 3.5.1 Example: Linear Array of Dipoles ...... 24

1 4 Multipole expansion of sources in waveguides 26 4.1 Electric Dipole ...... 26

5 Scattering of Radiation 31 5.1 Scattering of Polarized Light from an Electron ...... 31 5.2 Scattering of Unpolarized Light from an Electron ...... 34 5.3 Elastic Scattering From a Molecule ...... 35 5.3.1 Example: Scattering O a Hard Sphere ...... 38 5.3.2 Example: A Collection of Molecules ...... 40

6 Diraction 43 6.1 Scalar Diraction Theory: Kircho Approximation ...... 45 6.2 Babinet’s Principle ...... 51 6.3 Fresnel and Fraunhofer Limits ...... 53

7 Example Problems 55 7.1 Example: Diraction from a Rectangular Aperture ...... 55 7.2 Example: Diraction from a Circular Aperture ...... 57 7.3 Diraction from a Cross ...... 60 7.4 Radiation from a Reciprocating Disk ...... 61

1 Introduction

An electromagnetic wave, or electromagnetic radiation, has as its sources electric accelerated charges in motion. We have learned a great deal about waves but have not given much thought to the connection between the waves and the sources that produce them. That oversight will be rectied in this chapter. The scattering of electromagnetic waves is produced by bombarding some object (the scatterer) with an electromagnetic wave. Under the inuence of the elds in that

2 wave, charges in the scatterer will be set into some sort of coherent motion1 and these moving charges will produce radiation, called the scattered wave. Hence scattering phenomena are closely related to radiation phenomena. Diraction of electromagnetic waves is similar. One starts with a wave incident on an opaque screen with holes, or aperture, in it. Charges in the screen, especially around the apertures, are set in motion and produce radiation which in this case is called the diracted wave. Thus radiation, scattering, and diraction are closely related. We shall start our investigation by considering the radiation produced by some specied localized distribution of charges and currents in harmonic motion. We delay until Chap. 14, the discussion of non-harmonic sources.

1The response to a harmonic excitation is of the same frequency, and thus coherent

3 Diffraction Scattering

2 Radiation by a localized source

Suppose that we are given some charge and current densities (x, t) and J(x, t)2. These produce potentials which, in the Lorentz gauge (Chap. 6), can be found im- (+) mediately using the retarded Green’s function G (x, t; x0, t0) which we shall write simply as G(x, t; x0, t0):

1 3 A(x, t) = d x0dt0G(x, t; x0, t0)J(x0, t0) (1) c Z 2In this chapter, we assume = = 1

4 3 (x, t) = d x0dt0G(x, t; x0, t0)(x0, t0). (2) Z The Green’s function itself is given by

(t t0 x x0 /c) G(x, t; x0, t0) = | | . (3) x x | 0| Because all of the equations we shall use to compute elds are linear in the elds themselves, we may conveniently treat just one Fourier component (frequency) of the eld at a time. To this end we write

1 ∞ iωt J(x, t) = dωJ(x, ω)e (4) 2 Z∞ where

J(x, ω) = J(x, ω) (5) is required in order that J(x, t) be real; Eq. (5) is known as a “reality condition.” We may equally well, and more conveniently, replace Eqs. (4) and (5) by

∞ iωt J(x, t) = dωJ(x, ω)e . (6) < Z0 We will do this and will in general not bother to write in front of every complex < expression whose real part must be taken. We will just have to remember that the real part is the physically meaningful quantity. Similarly, we introduce the Fourier transform of the charge density,

∞ iωt (x, t) = dω(x, ω)e . (7) Z0 In the following we shall suppose that the sources have just a single frequency component,

J(x, ω0) = J(x)(ω ω0), ω0 > 0 (8) (x, ω0) = (x)(ω ω0), ω0 > 0. (9) Thus, assuming ω > 0,

iωt J(x, t) = J(x)e and (10)

iωt (x, t) = (x)e . (11)

5 From the continuity condition on the sources,

∂(x, t) + J(x, t) = 0, (12) ∂t ∇ we nd that (x) and J(x) are related by

J(x) (x) = i∇ . (13) ω Using Eq. (10) in Eq. (1) and employing Eq. (3) for the Green’s function, we nd, upon completing the integration over the time, that

1 3 J(x0) ik x x iωt A(x, t) = d x0 e | 0|e (14) c x x Z | 0| where, as usual, k ω/c. Dene A(x) by

iωt A(x, t) = A(x)e ; (15) comparison with Eq. (14) gives

ik x x0 1 3 e | | A(x) = d x0 J(x0) . (16) c x x Z | 0| From here the recipe is to nd B(x, t) from the curl of A(x, t); then the electric eld 3 1 is found from B(x, t) = c ∂E(x, t)/∂t, which holds in regions where the current ∇ density vanishes. These elds have the forms

iωt iωt B(x, t) = B(x)e E(x, t) = E(x)e (17) where

i B(x) = A(x) E(x) = B(x). (18) ∇ k ∇ We have reduced everything to a set of straightforward calculations - integrals and derivatives. Doing them exactly can be tedious, so we should spend some time

3Notice that we never have to evaluate the scalar potential.

6 thinking about whether there are any simplifying approximations that may have general validity or at least validity in some cases of interest. There are approximations based on expansions in powers of some small parameter. We can see what may be possible by realizing that there are three relevant lengths in any radiating system. Provided the origin of coordinates is taken to be somewhere within the source in the integrals above, these are the size of the source, r0 = x0 ; the distance of the | | observer from the source, represented by r = x ; and the wavelength of the radiation, | | = 2/k. The magnitude r0 is never larger than d, the size of the source. Focusing on just the relative size of and r, we identify the three traditional regimes below. λ d

source r

d << r << λ Near or static zone

d << r ~ λ Intermediate or induction zone d << λ << r Far or radiation zone

Here we have specied also that d be much smaller than the other two lengths. That simplies the discussion of the three regimes and so is a convenience but it is not always met in practice nor is it always necessary. In particular, the elds far away from the source (in the radiation zone) have characteristic forms independent of the relative size of and d provided r is large enough. Also, man-made sources such as antennas (and antenna arrays) are often intentionally constructed to have d and even << d in which case the inequalities above are not always satised. On the other hand, natural radiating systems, such as atoms and nuclei, typically do satisfy

7 the condition d << and d << r for any r at which it is practical to detect the radiation.

2.1 The Near Zone

Consider rst the near zone. Here d << and r << , so there is a simple expansion of the exponential factor,

ik x x e | 0| = 1 + ik x x0 + ... (19) | | which leads to

1 3 J(x0) iωt A(x, t) = d x0 [1 + ik x x0 + ...]e . (20) c x x | | Z | 0| The leading term in this expansion is

1 3 J(x0) iωt A(x, t) = d x0 e . (21) c x x Z | 0| Aside from the harmonic time dependence, this is just the vector potential of a static current distribution J(x), and that is the origin of the name “static” zone; the magnetic induction here has a spatial dependence which is the same as what one would nd for a static current distribution with the spatial dependence of the actual oscillating current distribution. We nd this result for the simple reason that in the near zone the exponential factor can be approximated as unity.

2.2 The Radiation or Far Zone

In the radiation or far zone (r d), the story is completely dierent because in this regime the behavior of the exponential dominates the integral. We can most easily see what will happen if we rst expand the displacement x x0 in powers of | | r0/r (d/r):

1/2 2 2 1/2 x x0 = [(x x0) (x x0)] = (r 2x x0 + r0 ) | | 2 1/2 2 1/2 2x x0 r0 n x0 r0 = r 1 + = r 1 2 + (22)  r2 r !   r r !      8 where n = x/r is a unit vector in the direction of x. Carrying the expansion to second order in r0/r, we have

2 2 n x0 1 r0 1 n x0 x x0 = r 1 + . (23) | |  r 2 r ! 2 r !    Similarly, 2 2 1 1 n x0 1 r0 n x0 = 1 + + 3 1 . (24) x x r  r 2 r !  r !  | 0|  0    Using the rst of these expansions we can write 

2 2 kr r n x i 0 0 ik x x ikr ik(n x ) 2 r r e | 0| = e e 0 e 2 kr 2 (n x ) i 0 1 0 ikr ik(n x ) 2r r 2 = e e 0 e 0 . (25) h i

2 Given r >> r0 and kr0 /r << 1, it is clear that this exponential function can be approximated by just the rst two factors; the third represents a change of phase by an amount small compared to a radian. Further, in the far zone it is sucient to approximate 1 1 = . (26) x x r | 0| Putting these pieces together we have

i(kr ωt) e 3 ik(n x ) 2 A(x, t) = d x0 J(x0)e 0 (r d and r d /) . (27) cr Z

This expression is always valid for r “large enough” which means r >> r0 and r >> 2 kr0 . The relative size of and d is unimportant. The behavior of A(x, t) on r and t is explicitly given by the factor in front of the integral; the integral depends on the direction of x but not on its magnitude. Hence in the far zone the vector potential always takes the form ei(kr ωt) A(x, t) = f(, ) (28) r where 1 3 ik(n x ) f(, ) d x0 J(x0)e 0 ; (29) c Z 9 and specify, in polar coordinates, the direction of x (or n). Knowing the form of the potential so precisely makes it easy to see what must be the form of the elds E and B in the far zone. First, ei(kr ωt) ei(kr ωt) B(x, t) = A(x, t) = f(, ) ik [n f(, )] (30) ∇ ∇ r ! r where we have discarded terms of relative order /r or d/r. Further, from Eqs. (18) and (30), we can nd E(x,t); to the same order as B, it is ei(kr ωt) E(x, t) = ik [n (n f(, ))] = B(x, t) n. (31) r There are two essential features of these equations.

First, both E and B in the radiation zone is that the eld strengths are pro- 1 portional to r ; this is very dierent from the case for static elds which fall 2 o at least as fast as r (consider the static zone).

Second, the radiation elds are transverse, meaning they are perpendicular to x or n; they are also perpendicular to each other.

The Poynting vector in the far zone also has a simple basic form:

c c ik 2 < S >= (E B) = [n (n f(, ))] [n f (, )] 8 8 r ! ck2 = [n(n f(, )) f(, )] [n f (, )] 8r2 { } 2 ck 2 = (n f(, ))[n(n f (, )) f (, )] n f(, ) + f (, )(n f(, )) 8r2 | | n ck2 o = n[ f(, ) 2 n f(, ) 2].(32) 8r2 | | | | This is presumably the time-averaged energy current density. Because it points radi- ally outward4, it also gives directly the angular distribution of radiated power: d ck2 P = [ f(, ) 2 n f(, ) 2]. (33) d 8 | | | |

4In the near or intermediate zones, there are non-zero components in other directions as well.

10 If we integrate over the solid angle, we nd the total power radiated: ck2 = d2x < S > n = d [ f(, ) 2 n f(, ) 2]. (34) P IS 8 Z | | | | Notice that the radiated power is, appropriately, independent of r.

3 Multipole Expansion of the Radiation Field

Thus far, we have only assumed that r , d. Now we will consider the d/. Consider two limits:

If d/ 1, then all elements of the source will essentially be in phase, and thus an observer at r cannot learn about the structure of the source from the emitted radiation. In this limit, we need only consider the rst nite moment in d/ (if the series is convergent).

If d/ > 1, then the elements of the source will not radiate in phase, and an observer at r may learn about the details of the structure of the source by analyzing the interference of the radiation pattern (i.e. Bragg diraction). In this case, to be discussed in sec. V, we need to retain more terms in the series.

Let us now go back and attempt to evaluate f(, ). If kd << 1, or 2d/ << 1, then it is not unreasonable to proceed with the evaluation by expanding the expo- ik(n x ) nential function e 0 ,

1 3 1 2 2 f(, ) = d x0 J(x0) 1 ik(n x0) k (n x0) + ... . (35) c Z 2

3.1 Electric Dipole

The rst term in the expansion is just the volume integral of J(x0); one can write it as 1 3 1 3 d x0 J(x0) = d x0 [ 0 J(x0)]x0 (36) c Z c Z ∇ 11 which one can show by doing integration by parts starting from the right-hand side of this equation. Now employ Eq. (13) to have

1 3 3 d x0 J(x0) = ik d x0 x0(x0). (37) c Z Z The right-hand side can be recognized as the electric dipole moment of the amplitude (x) of the harmonically varying charge distribution. Let us dene the electric dipole moment p in the usual way, 3 p d x0 x0(x0). (38) Z The electric dipole contribution fed(, ) to f(, ) is thus

f (, ) = ikp; (39) ed it is in fact independent of and . The corresponding contribution to the vector potential, Aed(x, t), is ei(kr ωt) A (x, t) = ikp . (40) ed r We have used only d << and d << r; no assumption about the relative size of and r has been made. It is somewhat tedious, but nevertheless instructive, to evaluate the elds without making any assumptions so that we can see their form in the near, intermediate, and far zones. First, the magnetic induction is

eikr Bed(x) = Aed(x) = ik p ∇ ∇ r ! 1 eikr 1 eikr = ik ik (n p) = k2 1 (n p). (41) r r ikr r Further, the electric eld is i E (x) = ( B (x)) ed k ∇ ed eikr 1 eikr 1 = ik 1 (n p) + ik 1 (n p) ∇ " r ikr # r ikr ∇

ik 1 1 1 ikr = ik 2 1 + 3 e n (n p) ( r r ! ikr ikr )

12 eikr 1 x x +ik 1 (p ) p r ikr ∇ r ∇ r 2 k 2 2 ikr = 1 + + 2 2 e n (n p) r ikr k r ik 1 p (p n)n 3p p + eikr 1 + r ikr r r r r ! k2 1 = n (n p) + (1 ikr)[3n(p n) p] eikr. (42) ( r r3 )

The electric eld is divided up in this fashion to bring out, rst, the form in the radiation zone which is the rst term and, second, the form in the near zone r << which is the second term. The spatial dependence of the latter is the same as the eld of a static dipole, [3n(p n) p]/r3, but do not forget that it oscillates with angular frequency ω. In the intermediate zone where all contributions are comparable, the eld is complex indeed.5 In the radiation zone, where the elds become quite simple, they are

k2 k2 B (x) = eikr(n p) and E = eikrn (n p). (43) ed r ed r The same conclusion may be reached much more simply from Eqs. (30), (31), and (39). As remarked earlier, the elds in the far zone are transverse to n. If we let p dene the polar axis, then Bed is azimuthal, i.e., in the direction of . This is a special feature of electric dipole radiation. Further, Eed(x) is in the direction of .

E S p r B

5And this is only the electric dipole part of the eld which is, along with the magnetic dipole part of the eld, by far the simplest.

13 FromEqs. (32) and (39) we nd that the time-averaged Poynting vector in the radiation zone is

ck4 ck4 < S >= [ p 2 n p 2]n = p 2 sin2 n, (44) 8r2 | | | | 8r2 | | where is the usual polar angle, i.e., the angle between the dipole moment and the direction at which the radiation is observed. The radiated power per unit solid angle is d ck4 P = r2 < S > n = p 2 sin2 , (45) d 8 | | and the total power radiated is

d ck4 p 2 ck4 p 2 = d P = | | dd sin sin2 = | | . (46) P Z d 8 Z 3

14 3.1.1 Example: Linear Center-Fed Antenna

Consider the short, linear, “center-fed” antenna shown below. z

d/2 n

-d/2 x Fig. 1 Short, center-fed, linear antenna.

For such an antenna, the current density can be crudely approximated by

z iωt J(x, t) = 3I0(x)(y) 1 2| | e (47) d ! for z < d/2; for z > d/2, it is zero. Given this current density, we can evaluate the | | | | divergence and so nd the charge density,

2I z J(x) = 0 (x)(y) = iω(x) (48) ∇ d z | | or 2iI z (x) = 0 (x)(y) , z < d/2. (49) ωd z | | | | Hence, 3 iI0d p = d x x(x) = 3. (50) Z 2ck 15 Now we have only to plug Eq. (50) into Eqs. (45) and (46) to nd d I2 I2k2d2 P = 0 (kd)2 sin2 and = 0 . (51) d 32c P 12c The calculation in this example may be expected to provide a good approximation to the total radiated power provided kd << 1 so that the electric dipole term dominates, unless it vanishes. There are many instances where this happens.

3.2 Magnetic Dipole

When this happens it is necessary to look at higher-order terms in the expansion ik x x of the phase factor e | 0|. Let’s look now at the next one. Start from the exact expression for A(x),

ik x x0 1 3 e | | A(x) = d x0 J(x0) c Z x x0 ikr | | e 3 n x0 = Aed(x) + d x0 J(x0) ik(n x0) [1 + (d/r, d/)]. (52) cr Z r ! O where the second term in the () comes from the exponential, and the rst comes from the corresponding denominator. The integral we wish to examine is

1 3 d x0 J(x0)(n x0) c Z 1 3 = d x0 [J(x0)(n x0) + x0(n J(x0))] + [J(x0)(n x0) x0(n J(x0))] 2c Z { } 1 3 1 3 = d x0 [J(x0)(n x0) + x0(n J(x0))] + d x0 n (J(x0) x0) (53) 2c Z 2c Z What is the point of breaking the integral into two pieces, symmetric and antisymmetric under interchange of x0 and J(x0)? There are several related points. One is that in the near zone the second term on the right-hand side produces a magnetic induction which has the form of the induction of a static magnetic dipole while the rst term produces an electric eld which has the form of the eld of a static electric quadrupole. Hence the radiation from the former is called magnetic dipole radiation while that from the latter is known as electric quadrupole radiation. Also, the

16 magnetic dipole part produces a purely transverse electric eld in all zones while the electric quadrupole part gives a purely transverse magnetic induction. Recall that for electric dipole radiation, B(x, t) is purely transverse in all zones as well. Let us look at the magnetic dipole elds rst. From Eqs. (52) and (53) we have

ikr e 1 3 1 Amd(x) = ik n d x0 (J(x0) x0) r r Z 2c eikr 1 ik 1 (n m) (54) r ikr where 1 3 m d x0 [x0 J(x0)] (55) 2c Z is the magnetic dipole moment, familiar from our study of magnetostatics. Rather than plow ahead with with the evaluation of the curl to nd B, etc., let us recall the electric dipole results

eikr A (x) = ik p (56) ed r

ikr 2 e 1 Bed(x) = k 1 (n p). (57) r ikr We see that Bed is the same in functional form as Amd; consequently, Eed, which is the curl of Bed, must be the same in form as the curl of Amd, or Bmd. Hence we can immediately write, using Eq. (43),

eikr eikr B (x) = k2 n (n m) + (1 ikr)[3n(n m) m]. (58) md r r3

Notice that in the near zone Bmd(x) is the same as that of a static dipole. As for the corresponding electric eld, we could work through the tedious derivatives of the magnetic induction, but it happens that this is one time when it is much easier to evaluate the the eld from the potentials. We know that

1 ∂A(x, t) E(x, t) = (x, t) . (59) ∇ c ∂t

17 Also, in the Lorentz gauge

1 ∂(x, t) A(x, t) + = 0 or A(x) ik(x) = 0. (60) ∇ c ∂t ∇ Using this relation for (x), we nd that

i E(x) = ( A(x)) + ikA(x) (61) k ∇ ∇ for any electric eld which is harmonic in time. From our expression for Amd(x), one can easily see that A = 06 and so E is simply proportional to the vector ∇ md md potential, ikr 2 e 1 Emd(x) = k 1 (n m). (62) r ikr

3.3 Comparison of Dipoles

To summarize:

The electric dipole and magnetic dipole elds are the same with E and B interchanged, E B and B E when p m. md ⇔ ed md ⇔ ed ⇔ In the near zone E and B have the form of static dipole elds, while in all ed md zones, Bed and Emd are purely azimuthal in direction.

The Poynting vector in the far zone has the same form for both electric dipole and magnetic dipole elds; in the latter case it is

ck4 < S >= r2n m 2 sin2 , (63) 8 | | leading to results for the power distribution and total power which are the same as for electric dipole radiation, Eqs. (45) and (46), with m in place of p. Notice particularly that if one measures d /d, and nds the sin2 pattern, then one P 6An equivalent statement is that a magnetic dipole is always charge neutral, so that = 0.

18 knows that the radiation is indeed7 dipole radiation; however, it is impossible to distinguish electric dipole radiation from magnetic dipole radiation without examining its polarization.

Magnetic Dipole Electric Dipole E B

n x = r n x = r E B m p 2 i kr 2 i kr B =k e ( n x mp ) / r E = -k e (n x m) / r E = -n x B B = n x E

p m B - E E B

These formulas suggest that for a given set of moving charges, one should get as much power out of an oscillating magnetic dipole as an oscillating electric dipole. This is not so, since the magnetic dipole moment is

1 3 m d x0 [x0 J(x0)] (64) 2c Z thus v m Qd (65) | | c where v is the speed of the moving charge, Q is the magnitude of this charge, and d is the size of the current loop. The size of the corresponding electric dipole moment

7Higher-order multipoles produce radiation with distinctive angular distributions which are never proportional to sin2 .

19 is roughly Qd. Thus, v m p (66) | | c | | We see that v 2 md ed . (67) P c P Thus, in a system with both an electric and magnetic dipole moment, the latter is usually a relativistic correction to the former.

3.4 Electric Quadrupole

Let’s go now to the other contribution to the vector potential in Eq. (53). This is called the electric quadrupole term; knowing that, we naturally expect to nd that it produces an electric eld in the near zone which has the characteristic form of a static electric quadrupole eld. The vector potential is

ikr e 1 1 3 Aeq(x) = ik d x0 [J(x0)(n x0) + x0(n J(x0))] r r 2c Z ikr e 1 1 3 = ik d x0 (n x0)( 0 J(x0))x0. (68) r r 2c Z ∇ To demonstrate this algebraic step, consider the ith component of the nal expression:

1 3 ∂Jj 1 3 ∂ d x0 (n x0) xi0 = d x0 [xi0 (n x0)]Jj 2c  ∂xj0  2c ∂xj0 Z Xj Xj Z   1 3 = d x0 [ (n x0) + x0 n ]J 2c ij i j j Z Xj 1 3 = d x0 [Ji(n x0) + xi0 (n J)] (69) 2c Z which matches the ith component of the original expression in Eq. (68). Making use of Eq. (13) for J(x) and also using ω = ck, we nd, from Eq. (68), that ∇ 2 ikr k e 1 3 Aeq(x) = 1 d x0 x0(n x0)(x0). (70) 2 r ikr Z

20 There are nine components to the integral since the factor of n in the integrand can be used to project out three numbers by, for example, letting n be each of the Carte- sian basis vectors. That is, the basic integral over the source, (x), which appears here is a dyadic, d3x xx(x). It is symmetric and so has at most six independent components. NoticeR also that Aeq depends on the direction of n, which was not the case for either Aed or Amd; the evaluation of the elds is further complicated as a consequence. The electric quadrupole vector potential can be written in terms of the components

Qij of the electric quadrupole moment tensor which we dened long ago. Recall that

3 2 Qij d x (3xixj r ij)(x). (71) Z

Take combinations of these to construct the components Qi of a vector Q:

3 Q (n) Q n . (72) i ij j jX=1 From these denitions one can show quite easily that

1 3 n Q(n) n d x0 x0(n x0)(x0) . (73) 3 Z To see this, note that the ith component of Q/3 is

1 3 2 Q = d x0 [x0 ( n x0 )(x0) r0 n (x0)/3]. (74) 3 i i j j i Z Xj The rst term on the right-hand side of this relation produces the ith component of the integral in Eq. (73); the second term is some i-independent quantity multiplied by ni; the three terms in Q of this form give something which is directly proportional to n and so they do not contribute to n Q. Thus have we established the validity of Eq. (73). But what good is it? It tells us that we can write n A in terms of n Q, but does not allow us to write A eq eq itself in terms of Q. However, if we restrict our attention to the radiation zone, then

21 n A is all we will need, because in this zone, B = ik(n A) and E = n B. Thus, in the far zone, ik3 eikr B (x) = [n Q(n)] (75) eq 6 r and ik3 eikr E (x) = n [n Q(n)]. (76) eq 6 r The time-averaged power radiated is

d 2 c 2 c 2 P = r < S n > = r [E(x) B(x)] n = r [B(x) n] E(x) d 8 8 c k6 ck6 = n [n Q(n)] 2 = n Q(n) 2. (77) 8 36| | 288 | | The right-hand side does not have any single form as a function of and that we can extract because Q(n) depends in an unknown way (in general) on these angles. We can proceed to a general result only up to a point:

2 n Q = (n Q) (n Q) = [(n Q) n] Q | | 2 2 = Q [n(Q n) Q] = Q n Q ; (78) | | | | this is a brilliant derivation of the statement that sin2 = 1 cos2 . Further,

Q Q (n Q)(n Q) = Q n Q n n Q n n Q n . (79) ij j ik k i ij j k kl l Xijk Xijkl The ni’s are direction cosines and so obey the identities 4 d ninj = ij (80) Z 3 and 4 d ninjnknl = (ijkl + ikjl + iljk). (81) Z 15 These enable us at least to get a simple result for the total radiated power: d = d P P Z d 6 ck 4 2 4 2 = Q [Q Q + Q + Q Q ] 288  3 | ij| 15 ii kk | ik| ij ji   Xij Xik  ck6 =  Q 2  (82) 360 | ij| Xij 22 where we have used the facts that Qij = Qji and i Qii = 0. By choosing appropriate axes (the principal axesPof the quadrupole moment matrix or tensor) one can always put the matrix of quadrupole moments into diagonal form. Further, only two of the diagonal elements can be chosen independently because the trace of the matrix must vanish. Hence any quadrupole moment matrix is a linear combination of two basic ones.

3.4.1 Example: Oscillating Charged Spheroid

A commonly occurring example is an oscillating spheroidal charge distribution leading to Q = Q and Q = Q = Q /2. (83) 33 0 22 11 0 Then

Qi = Qijnj = Qiini (84) Xj or 1 Q = Q0 cos 3 sin (cos 1 + sin 2) . (85) 2 From this we have

2 2 2 1 2 Q = Q0 cos + sin , | | 4 2 1 2 n Q = Q0 cos sin , 2 2 2 4 2 2 1 4 n Q = Q0 cos cos sin + sin , (86) | | 4 and so

2 2 2 2 1 2 4 2 2 1 4 Q n Q = Q0 cos + sin cos + sin cos sin | | | | 4 4 2 2 2 1 2 2 2 2 = Q0 cos sin + cos sin + cos sin 4 9 = Q2 cos2 sin2 . (87) 4 0

23 Hence, d ck6 P = Q2 sin2 cos2 . (88) d 128 0 This is a typical, but not uniquely so, electric quadrupole radiation distribution.

dP q dW

Radiation Pattern of an oscillating charged spheriod

Higher-order multipole radiation (including magnetic quadrupole radiation) is ik(n x ) found by expanding the factor e 0 in Eq. (27) to higher order in powers of d/. One can do this in a complete and systematic fashion after rst developing some appropriate mathematical machinery by generalizing the spherical harmonics to vector elds,8 the purpose being to construct an orthonormal set of basis functions for the electromagnetic elds.

8Thereby producing the so-called vector spherical harmonics. This is the subject matter of Chapter 16.

24 3.5 Large Radiating Systems

Before abandoning the topic of simple radiating systems, let us look at one example of an antenna which is not small compared the wavelength of the emitted radiation. Our example is an array of antennas, each of which is itself small compared to and each of which will be treated as a point dipole. We take the current density of this array to be J(x) = I a (x x )eij . (89) 0 j 3 Xj One can easily see that this is an array of point antennas located at positions xj; they all have the same current but are not necessarily in phase, the phase of the jth iωt antenna being given by j (the additional phase factor e , common to all antennas, has been omitted, as usual). Antenna farm p D d << λ r >> D No matter how large the array, we can specify that x is large enough that the observation point is in the far zone in which case the vector potential can be taken as eikr A(x) = f(, ) (90) r where

I0a 3 ik(n x ) i f(, ) = d x0 (x0 x )e 0 e j 3 c j Z Xj I0a i[ k(n x )] = e j j . (91) 3 c Xj Referring back to Eq. (33) we nd that the distribution of radiated power is

d k2I2a2 P = 0 sin2 w 2 (92) d 8c | |

25 where i( kn x ) w e j j . (93) Xj The factor of sin2 arises because each element of this array is treated in the dipole approximation. The factor w 2 contains all of the information about the relative | | phases of the amplitudes from the dierent elements, i.e., about the interference of the waves from the dierent elements.

3.5.1 Example: Linear Array of Dipoles

As a special and explicit example, suppose that

j = j0 and xj = aj3, (94)

meaning that the elements are equally spaced in a line along the z-axis and that they have relative phases that increase linearly along the array. z a q n^ dipoles x

Then

i(0 ka cos )j j w = e x (95) Xj Xj where

i(0 ka cos ) x = e . (96)

The sum is easy to evaluate if we know where j begins and ends:

j2 j2 j1+1 1 x xj = xj1 , (97) 1 x jX=j1 so 2 i()(j2 j1+1) 1 e w 2 = (98) | | 1 ei() 26 where () ka cos . Let j = n and j = n, corresponding to an array of 0 1 2 2n + 1 elements centered at the origin. Then

2 1 ei(2n+1) 1 cos[(2n + 1)] w 2 = = . (99) | | 1 ei 1 cos This is a function which is in general of order unity and which oscillates as a function of . It has, however, a large peak of size (2n + 1)2 when () is an integral multiple 9 of 2. The peak occurs at that angle 0 where (0) = 0 or cos 0 = 0/ka. If we choose 0 = 0, the peak is at 0 = /2; further, if ka < 2 or a < , there is no other such peak. The width of the peak can be determined from the fact that the factor w goes to zero when (2n + 1)() = 2. Assuming that n >> 1, one nds that the corresponding angle diers from 0 by where

2 1 = (2n + 1) . (100) (2n + 1)ka[1 ( /ka)2] 0 Hence the antenna becomes increasingly directional with increasing n. The accompa- 2 2 2 nying gure shows the power distribution in units of k I0 a /8c for 5 elements with ka = and 0 = 0. dP/dΩ

-10

-30 0 50 100 150

9 More generally, (0) = 2m where m is an integer; because we control 0, we can make it small enough that the peak corresponds to the particular case m = 0.

27 4 Multipole expansion of sources in waveguides

We saw in Chapter 8 that a eld in a waveguide could be expanded in the normal modes of the waveguide with an integral over the sources, consisting of a current distribution and apertures, determining the coecients in the expansion. If the sources are small in size compared to distances over which elds in the normal modes vary, then one can do the integrals in an approximate fashion by making a multipole expansion of the sources.

4.1 Electric Dipole

Consider rst the part of the amplitude which is produced by some explicit current distribution. It is ( ) 2Z 3 ( ) A = d x0 J(x0) E . (101) c Z The eld in the mode is given by

( ) ik z E (x0) = [E (x0, y0) E (x0, y0)]e 0 . (102) 3 z

The origin of the coordinate x0 is at some appropriately chosen point which is probably not near the center of the source distribution. Let us use a dierent coordinate x having as origin a point near the center of the source. Also, let’s let the electric eld in the mode be called E(x) as a matter of convenience. Then we have to do an integral of the form

3 3 ∂Ei d x J(x) E(x) = d x J(x) E(0) + ixj + ... (103)  ∂x  Z Z i.j j 0 X   where we are assuming that the eld varies little over the size of the source. In this integral the rst term can be converted to an integral over the charge density,

d3x J(x) = iω d3x x(x) (104) Z Z

28 provided the integration by parts can be done without picking up a contribution from the surface of the integration volume. That is not automatic here because the surface includes some points on the wall of the waveguide including those points where the current is fed into the guide, so one has to exercise some care in applying this formula. Assuming that it is okay, we see that the leading term in the expansion ( ) of the integrand produces to a term in the coecient A which is proportional to p E(0). The next term in the expansion can, as we have seen, be divided into symmetric and antisymmetric parts:

∂Ei 3 1 ∂Ei ∂Ej 3 d x Ji(x)xj = d x [Jixj Jjxi] ∂x 4 ∂x ∂x ! i,j j 0 Z i,j j i 0 Z X X 1 ∂Ei 3 + d x [Jixj + Jjxi]. (105) 2 ∂x j 0 Z The rst term on the right-had side is set up in such a way as to display explicitly a component of E, which is i(ω/c)B, and the same component of the magnetic ∇ dipole moment. Hence this term is proportional to m B(0). The remaining term can be handled in the way that we treated the electric quadrupole part of the vector potential earlier; it becomes

iω ∂Ei 3 iω ∂Ei d x xixj(x) = Qij (106) 2 ∂x 6 ∂x i,j j 0 Z i,j j 0 X X provided one can throw away the contributions that come from the surface when the integration by parts is done. The nal step is achieved by making use of the fact that E = 0. ∇ In the next order, not shown in Eq. (103), the antisymmetric part provides the magnetic quadrupole contribution. Without delving into the algebra of the derivation, we state that the result is of the same form as Eq. (106) but with the magnetic M quadrupole moment tensor Qij in place of the electric quadrupole moment tensor, B in place of E, and an overall relative ( ). The components of the magnetic quadrupole

29 moment tensor are dened in the same way as those of the electric quadrupole moment tensor except that in place of (x) there is 1 M (x) = [x J(x)]. (107) 2c∇ ( ) The nal result for the amplitude A with all indices in place is ( ) ( ) ( ) 2ω ( ) ( ) 1 ∂Ei M ∂Bi A = i p E (0) m B + Q Q + ... .  ij ij   c 6 i,j ∂xj ∂xj n X 0 0    (108)  We can see immediately some interesting features of this result. For example, if one wants to produce TM modes, which have z components of E but not of B, then this is most eciently done by designing the source to have a large pz. At the same time, hardly any TE mode will be generated if there is only a z component of p because the TE mode has no Ez to couple to p. Hence this expression gives one a good idea how to design a source to produce, or not produce, modes of a given kind. Now let’s look at the same expansion if the source of radiation is an aperture rather than an explicit current distribution. We derived in chapter 8 that in this case

( ) Z 2 ( ) A = d x0 n E(x0) H (x0) 2 Sa Z h i Z 2 ( ) = d x0 [n E(x0)] H (x0) (109) 2 ZSa where the integral is over the aperture Sa, E(x0) is the electric eld actually present at point x0, and n is the inward directed normal at the aperture. Notice that only the tangential component of the electric eld contributes to this integral. Assuming that the aperture is small compared to distances over which the elds in the normal modes vary, we can expand the latter and nd, after changing the origin to a point in the aperture,

( ) ( ) Z 2 ( ) ∂Hi A = d x (n E ) H (0) + x + ... tan  i j  2 ZSa i,j ∂xj X 0 ( )  ( )  Z B (0) ∂H = d2x (n E ) + d2x (n E ) i x + ... .(110)  tan tan i j  2 ZSa i.j ZSa ∂xj  X 0    30 The rst term here has already an appropriate form. As for the second one, note that we can break up the integrand into even and odd pieces,

( ) ∂H 1 ∂H (n E ) i x = [(n E ) x (n E ) x ] i tan i j tan i j tan j i ∂xj 2 ∂xj 0 ( ) 1 ∂H + [(n E ) x + (n E ) x ] i . (111) tan i j tan j i 2 ∂xj 0 Take just the antisymmetric part of this expression and complete the integral over x. The contribution to the amplitude, aside from a factor of Z /2, is ( ) ( ) 1 ∂H ∂H d2x[(n E ) x (n E ) x ] i j tan i j tan j i   4 i,j ZSa ∂xj ∂xi X 0   1 2 iω ( ) = d x [(n Etan)ixj (n Etan)jxi]kij Ek (0) 4 Sa c i,jX,k Z iω 2 ( ) = d x E (0) [2x (n Etan)] 4c ZSa iω 2 ( ) iω 2 ( ) = d xE (0) [x (n Etan)] = d x E (0) n(x Etan). (112) 2c ZSa 2c ZSa In this string of algebra we have made use of the fact that H = i(ω/c)E and ∇ ( ) that Etan is orthogonal to E in the aperture because of the boundary conditions on the latter eld. ( ) Putting this piece into the expression for A along with the leading one, we nd

( ) Z 1 ( ) 2 iω ( ) 2 A = B (0) d x n Etan(x) E (0) n d xx Etan(x) 2 ( ZSa 2c ZSa ) (113) Dening 2 peff n d x x Etan(x) (114) 4 ZSa and c 2 meff d x n Etan(x (115) 2iω ZSa we can write

( ) iωZ ( ) ( ) A = [p E (0) m B (0) + ...] (116) c eff eff

31 Thus we nd that in the small wavelength limit, apertures are equivalent to dipole sources. The eective dipole moments are found by solving for, or using some simple approximation for, the elds in the aperture. See Jackson for a description of the particular case of circular apertures.

32 5 Scattering of Radiation

So far, we have discussed the radiation produced by a harmonically moving source without discussing the origin of the source’s motion. In this section we will address the elds which set the source in motion. Consider the case where the motion is excited by some incident radiation. The incident radiation is absorbed by the source, which then begins to oscillate coherently, and thus generates new radiation. This process is generally called scattering.

Incident Radiation Radiation Scattered

Source

5.1 Scattering of Polarized Light from an Electron

For simplicity, let’s consider a plane electromagnetic wave incident upon a single electron of charge e. We will assume that the incident eld has the form i(k x ωt) Ein(x, t) = 1E0e (117)

Since the equations which govern the motion of the electron are linear, we expect the electron to respond by oscillating along 1 at the same frequency (assuming that the magnetic force on the electron is small as long as the electron’s motion is nonrelativistic). This will produce a time varying electric dipole moment. We write the electron position as

x(t) = x0 + x(t) (118)

33 incident waves k0 or particles K λ ≈ | a 1 | or | a 2 | θ k0 k k

K = k - k0

a1 θ d a2 d sin(θ)

Figure 1: Scattering of waves or particles with wavelength of roughly the same size as the lattice repeat distance allows us to learn about the lattice structure. We will assume that each electron acts as a dipole scatterer.

If E0 is small enough, then x will be small compared to the wavelength of the incident radiation. λ

δx e-

This approximation is also consistent with the non-relativistic assumption. When this is case, we can write

i(k x ωt) i(k x0 ωt) e e (119) so that the eld of the incident radiation does not change over the distance traveled by the oscillating electron. If we assume a harmonic form for the electronic motion

34 iωt 10 x(t) = xe , insert this into Newton’s law and solve, we get

ik x0 eE0e x = 1 (120) mω2 !

iωt The dipole moment of the oscillating electron will then be p(t) = pe , with

2 ik x0 e E0e p = ex = 1 (121) mω2 !

Now we will solve for the scattered radiation. Let the geometry of the scattering be as shown below.

scattered wave incident wave ε n' k k' =k k = k n e -

From our electric dipole formulas, the angular distribution of the scattered power Eq. (45) is

4 d scat 2 ck 2 2 P = r < S > n0 = [ p n0 p ] (122) d 8 | | | | or 4 2 d scat e E0 2 P = [1 (n0 ) ] (123) d 8m2c3 1

Note that gives zero radiated power for n0 along 1. I.e. there is no power radiated in the direction along which the dipole oscillates.

10 ik x 2 F = ma eE = e E e 0 = mx(t) = mω x → 1 0

35 We may determine the dierential cross section for scattering by dividing the radiated power per solid angle obtained above by the incident power per unit area (the incident Poynting vector).

c c 2 S = (E B) = E (124) in 8 8 0 thus 4 d 1 d scat e 2 = P = 1 (n0 ) (125) d S d m2c4 1 | in| or d 2 2 = r 1 (n0 ) (126) d 0 1 where r = e2/mc2 = 2.8 1013 cm is the classical scattering radius of an electron. 0 This is the formula for Thomson scattering of incident light polarized along 1.

5.2 Scattering of Unpolarized Light from an Electron

Now suppose that the incident light is unpolarized. Then the cross section is the average of the cross sections for the two possible polarizations 1 and 2 of the incident wave (why?).

2 d 1 2 2 = r0 1 (n0 i) (127) d! 2 unpol Xi=1 2 2 1 2 2 = r r (n0 ) . (128) 0 2 0 i Xi=1

From the fact that (1, 2, and n) form an orthonormal triad of unit vectors, one may show that d 1 2 2 = r0 1 + (n0 n) (129) d! 2 unpol This may be rewritten as

d 1 2 2 = r0 1 + cos () (130) d! 2 unpol

36 where is the scattering angle of the unpolarized radiation. From this we can see that the scattering in predominantly forward ( = 0), and backscattering ( = ). The total cross sections may be obtained by integrating the dierential cross sections. Thus 8 = r2 (131) pol 3 0 and 1 = ( ( ) + ( )) = (132) unpol 2 pol 1 pol 2 pol

5.3 Elastic Scattering From a Molecule

Let us now consider the elastic scattering from a molecule11 in which many electrons are generally present. In calculating the eld produced by the molecular electrons when light is incident on them, we will again assume that we are making our observation very far away from the molecule, so that

d λ x

x >> d source x >> λ

We will not assume however, that d, since we want to use the scattering to learn something about the structure of the molecule. This is only possible if our resolution (limited by ) is smaller than d. Thus we expect our results to include interference eects from dierent charges in the molecule

11Here we shall use molecule to identify any small collection of charges

37 Cu O R O

Figure 2: Rays scattered from dierent elements of the basis, and from dierent places on the atom, interfere giving the scattered intensity additional structure described by the form factor S and the atomic form factor f, respectively.

From page 9, the vector potential produced by the i’th electron in the molecule is eikr 3 ik(n x0) Ai(x) = d x0 Ji(x0)e . (133) cr Z This expression includes just the current of the i’th electron, and assumes x and x d, as well as a harmonic time dependence. To nd the current J (x0) for the i’th i electron, we note that the classical current density of an electron is

J (x0, t) = ev (t)(x0 x ) (134) i i i where (xi, vi) are the location and velocity of the i’th electron. If the electron is exposed to an incident electromagnetic plane wave, we nd that

xi(t) = xi,0 + xi(t) (135)

eE0 ik x iωt i xi(t) = 1 2 e e (136) mω From this we can nd the electron velocity, and hence the current

2 ie E0 ik xi Ji(x0) = 1 e (x0 xi) . (137) mω !

38 In this we have taken out the harmonic time term. Thus the vector potential of the scattered eld generated by the i’th electron in the molecule is

2 ikr ie E0 e i(k k ) x A (x) = e 0 i (138) i 1 mω cr where k0 = kn0 is the scattered wave vector and k = kn is the incident wave vector.

We dene the vector q = k k0 which is (1/h times) the momentum transfer from the photon to the electron, then the total vector potential of the scattered eld is

2 ikr ie E0 e iq x A(x) = A (x) = e i (139) i 1 mω cr Xi Xi This diers from the vector potential due to a single scattering electron only

iq xi through the factor i e . Since this factor is independent of the observation point x, the resulting B Pand E elds are likewise those of a single scattering multiplied

iq xi by a factor of i e . Thus we may immediately write down the dierential cross sections. For pPolarized light

2 d 2 2 iq xi = r0 1 (n0 1) e (140) d!   pol h i i X   where the rst term in brackets is the single-electron result, and the second term is the structure factor. For unpolarized light

2 d 1 2 2 iq xi = r0 1 + (n0 n) e (141) d! 2   unpol i X   We see that from a measurement of the dierential cross section, we may learn about the structure of the object from which we are scattering light. Let’s examine the structure factor in more detail.

iq xi 3 iq xi e = d x e (x xi) (142) V Xi Z Xi 3 iq x = d x e (x xi) (143) V Z Xi

39 However, (x x ) = n (x) (144) i el Xi which is the electron number density at position x, thus

iq xi 3 iq x e = d x e nel(x) = nel( q) (145) V Xi Z where n ( q) is the Fourier transform of the electron density. Thus, our dierential el cross sections are, for polarized light:

d 2 2 2 = r0 1 (n0 1) nel( q) , (146) d! | | pol h i h i and for unpolarized light

d 1 2 2 2 = r0 1 + (n0 n) nel( q) (147) d! 2 | | unpol h i

Since q = k k0 depends upon the scattering angle and the wave number of the incident photon, a scan over and/or k gives you information about the electron density on the target.

5.3.1 Example: Scattering O a Hard Sphere

Single spherical molecule with R a sharp cutoff in its electron distribution z electrons

40 Consider scattering from a single molecule with a sharp cuto in its electron distribution. 3z n (x) = (R x ) (148) el 4R3 | | where R is the radius of the sphere and z is the number of electrons, then

3 iq x 3z nel( q) = d x e nel(x) = 3 3 [sin(qR) qR cos(qR)] , (149) ZV q R where j (x) = (sin x/x cos x) is a spherical Bessel function. Note that this has 1 zeroes at qR = tan(qR) (150) so that we expect zeroes in the scattering cross section. These are clearly related to the size of the molecule. For forward scattering (q = 2k sin(0) = 0), we have

n ( q = 0) = z (151) el This is the maximum value, i.e. all the scattering electrons give constructive interference for forward scattering. A sketch of the unpolarized cross section

d 1 2 2 2 = r0 1 + cos () nel( q) d! 2 | | unpol looks like 100

10-2

10-4 unpol

Ω -6 /d 10 σ d 10-8

10-10 -1.0 -0.5 0.0 0.5 1.0 cos(θ)

If we measured this, we could learn the spatial distribution of electrons by inverting the transform. It is important to note that we could not learn the details of the

41 electron wavefunctions since all the phase information is lost. I.e., the cross section just measures the modulus squared of nel, and so phase information is lost.

5.3.2 Example: A Collection of Molecules

A collection of molecules, each with an electronic distribution n ( ) 0 x

Rj O

In a macroscopic sample, the total electron density is made up of two parts: the electron distribution within the molecules, and the distribution of the molecules within the sample. For simplicity, let’s assume that the sample contains just one type of molecule, each of which has an electron density n0(x). The total electron number density is then n (x) = n (x R ) (152) el 0 j Xj where Rj is the position of the j’th molecule. The Fourier transform of this is

3 iq x iq Rj nel( q) = d x e nel(x) = e n0( q) (153) V Z Xj

42 which is a weighted sum of the Fourier transforms for each molecule. We see that then the cross section will include the factor 2 2 iq R n ( q) e j (154) 0 | | j X From above, we know that n ( q = 0) = z, the num ber of electrons in a molecule, 0 so we may write n ( q) = zF (q) (155) 0 where F (q) is called the form factor of the molecule which is normalized to unity for q = 0. Thus the unpolarized dierential cross section becomes 2 d 1 2 2 2 2 iq R = z r (1 + cos ()) F (q) e j (156) 0 d!unpol 2 | | j X In general, we will not know where all the molecules are,nor do we necessarily care. Thus we will look at an average of the term which describes the distribution of the molecules. 2

iq Rj iq (Rj R ) iq (Rj R0) e = e j0 N e NS(q) (157) * j + jj j X X0 D E X D E where N is the num ber of molecules in the sample, R0 is the position of the origin, and S(q) is the structure factor of the sample. The approximation is justied by the fact that (neglecting nite-size eects) each link between sites occurs about N times.

This link occurrs about N times.

43 iq (R R0) iq (R R0) S(q) = e j = 1 + e j (158) j j=0 X D E X6 D E Note that this goes to 1 as q . Then, using the same procedure detailed before, | | → ∞ this becomes 3 iq x NS(q) = V d x e n(x)n(0) (159) ZV h i where V is the sample volume, and n(x) is the number density of molecules in the sample. Here, n(x)n(0) tell us about correlations between molecular positions. h i n(x)n(0) = (n(0))2 g(x) (160) h i where g(x) is just the probability of nding a molecule at x if there is one at the origin. This is the normalized two-body correlation function. We see that if the molecular form factor is known, then a measurement of the dierential cross section tells us about the Fourier transform of g(x). This is the key to using x-ray (or neutron, etc) scattering to determine the internal structure of materials.

d 1 2 2 2 2 2 = z r0(1 + cos ()) F (q) V (n(0)) g(q) d!unpol 2 | |

Liquid Averages to 1 Crystal at large r g(x) g(x) lattice spacing

1st peak at mean intermolecular spacing

r r

44 6 Diraction

In this section we will discuss diraction which is related to scattering. The proto- typical setup is shown below in which radiation from a source is diracted through an opaque screen with one or more holes in it.

screen

source observer

Clearly, diraction is the study of the propagation of light or radiation, or rather the deviation of light from rectilinear propagation. As undergraduates we all learned that the propagation of light was governed by Huygen’s Principle that every point on a primary wavefront serves a the source of spherical secondary wavelets such that the primary wavefront at some later time is the envelope of these wavelets. Moreover, the wavelets advance with a speed and frequency equal to that of the primary wave at each point in space.12. This serves as the paradigm for our study of geometric optics. Diraction is the rst crisis in this paradigm. To see this consider wave propagation in a ripple tank through an aperture.

12Hecht-Zajac, page 60

45 a) b)

wave λ d flow

In the gure, wavefronts are propagating toward the aperture from the bottom of each box. In case a) the wavelength is much smaller than the size of the aperture d. In this case, the diracted waves interfere destructively except immediately in front of the aperture. In case b), d, and no such interference is is observed. Huygen’s principle cannot explain the dierence between cases a) and b), since it is independent of any wavelength considerations, and thus would predict the same wavefront in each case. The diculty was resolved by Fresnel. The corresponding Huygens-Fresnel principle states that every unobstructed point of a wavefront,at a given instant of in time, serves as a source of spherical secondary wavelets of the same frequency as the source. The amplitude of the diracted wave is the sum of the wavelets considering their amplitudes and relative phases13. Applying these ideas claries case a). Here what is happening is that the wavelets from the right and left sides of the aperture interfere constructively in front of the aperture (since they travel the same distance and hence remain in phase), whereas these wavelets interfere destructively to the sides of the aperture (since they travel two paths with length dierences of order /2). In case b), we approach the limit of a single point source

13Hecht-Zajac, page 330

46 of spherical waves. Beside being rather hypothetical, the Huygens-Fresnel principle involves some approximations which we will discuss later; however, Kircho showed that the Huygen’s-Fresnel principle is a direct consequence of the wave equation.

6.1 Scalar Diraction Theory: Kircho Approximation

This problem could be solved using the techniques we just developed to treat scattering. I.e. by considering the dynamics of the charged particles in the screen, and then calculating the scattered radiation generated by these particles. However, diraction is conventionally treated as a boundary value problem in which the presence of the screen is taken into account with boundary conditions on the wave. Several excellent references for this problem are worth noting:

1. B.B. Baker and E.T. Copson, The Mathematical Theory of Huygens Prin- ciple, (Clarendon Press, Oxford, 1950).

2. Landau and Lifshitz, The Classical Theory of Fields.

3. L. Eyges, The Classical Electromagnetic Field, (Addison-Wesley, Reading, 1972).

4. Hecht-Zajac, Optics, (Addison-Wesley, Reading, 1979).

The Baker reference, especially, has a good discussion of the limits of the Kircho approximation, of course, Landau and Lifshitz have an excellent discussion of the physics, but Hecht-Zajac’s discussion is perhaps the most elementary, and will often be quoted here. The typical question we ask is, given a strictly monochromatic point source S, what resulting radiation is observed at the point O on the opposite side of the screen. At rst this approach may seem to limit us just to point sources. The case of a real extended source which emits non-monochromatic light does not, however, require special treatment. This is because of the linearity of our equations and the complete

47 independence (incoherence) of the light emitted by dierent points of the source. The interference terms average to zero. Thus the total diraction pattern is simply the sum of the intensity distributions obtained from the diraction of the independent components of the light.

screen

iωt e monochromatic observer point source

To treat the theory of diraction, a number of approximations will be necessary.

First, we assume that we can neglect the vector nature of the electromagnetic elds, and work instead with a scalar complex function (a component of E or B, or the single polarization observed in the ripple tank discussed above). In principle, is any of the three components of either E or B. In practice, however, the polarization of the radiation is usually ignored and the intensity of the radiation at a point is usually taken as (x, t) 2. This rst assumption | | limits the number of geometries we can treat.

Second, we will generally assume that /d 1, where d is the linear dimension of the aperture or obstacle.

The third assumption is that we will only look for the rst correction to geometric optics due to diraction. This is often called the Kircho approximation, which will be discussed a bit later. (This set of approximations are sometimes also called the Kircho approximation scheme.)

48 We then impose the boundary condition14

(x, t) = 0 everywhere on the screen (161)

We will assume that obeys the wave equation, and the source is harmonic so that it emits radiation of frequency ω, hence

iωt (x, t) = (x)e . (162)

Thus the spatial wave function obeys the Helmholtz equation

22 = ( 2 + k2)(x) = 0 , with k = ω/c (163) ∇ if we assume that the waves propagate in a homogeneous medium and restrict our observations to points away from the source S. As always, the physical quantities iωt will be the real amplitude ((x)e ), and the modulus squared which denotes the < time-averaged intensity. Let’s consider just the observation points in the volume V below which is bounded by the screen (z = 0 plane), and a hemisphere at innity.

surface S which includes the plane at z=0.

Volume V S

(z < 0) (z > 0) z=0

14Note that we only impose one boundary condition on the screen. This is to avoid the diculties when both boundary conditions (Neumann and Dirichlet) are imposed, as discussed in Jackson on page 429

49 Then, everywhere within V , the Helmholtz equation 22 = 0 is obeyed since the source lies outside of V . To nd within V , we use Green’s theorem

3 2 2 2 d x0 (x0) 0 (x0) (x0) 0 (x0) = d x0 n0 ((x0) 0(x0) (x0) 0(x0)) , ZV ∇ ∇ ZS ∇ ∇ (164) 2 or, adding and subtracting k (x0)(x0) from the rst integrand, we get

3 2 2 2 2 2 d x0 (x0)( 0 + k )(x0) (x0)( 0 + k )(x0) = d x0 n0 ((x0) 0(x0) (x0) 0(x0)) ZV ∇ ∇ ZS ∇ ∇ (165)

This works for any two functions and . We will take (x0) to be the wave amplitude, and take (x0) = G(x, x0), where the Dirichlet Green’s function satises

2 2 ( + k )G(x, x0) = 4(x x0) in V (166) ∇

ik x x e | 0| i.e., it is the response to a unit point source so that in free space G(x, x0) = x x . | 0| However, need to solve for G with boundary conditions

G(x, x0) = 0 for x0 on S . (167) since we will be using Dirichlet boundary conditions on : We will specify the value of (x) (as opposed to its derivative) on the boundary S. Now using the facts that

2 2 ( + k )G(x, x0) = 4(x x0) in V (168) ∇ ( 2 + k2)(x) = 0 in V (169) ∇ and that

G(x, x0) = 0 for x0 on S (170) this Green’s theorem becomes (the Kircho Integral)

1 2 (x) = d x0 n0 ((x0) 0G(x, x0)) (171) 4 ZS ∇

To proceed further we must determine the form of G(x, x0) and we must now specify (x0) on the surface S. For an innite planar screen the Green’s function is

50 given by the method of images (analogously to the electrostatic case)

ik x x ik x y e | 0| e | 0| G(x, x0) = (172) x x x y | 0| | 0|

y' x' image

where x0 = (x0, y0, z0) is in V , but y0 = (x0, y0, z0) (the image point) is not. This 2 2 satises ( +k )G(x, x0) = 4(x x0) in V , and vanishes both on the plane z0 = 0 ∇ 2 and on the hemisphere at innity. Note that it vanishes as 1/r0 as r0 . From → ∞ this it is clear that the integral

1 2 (x) = d x0 n0 ((x0) 0G(x, x0)) (173) 4 ZS ∇ has a vanishing contribution from the hemisphere at innity, since (x0) vanishes at least as fast as 1/r0 as r0 (since it is a solution of the wave equation for a nite → ∞ source). Furthermore, from the boundary condition

(x0) = 0 on the screen (174) we can see that the integral only gets a nonzero contribution from the opening

1 2 (x) = d x0 n0 ((x0) 0G(x, x0)) . (175) 4 Zopening ∇ So far we have made an exact evaluation of our scalar theory. However, to proceed we must make an approximation. We will assume that the value of (x0) in the opening will be the same as if the screen was not there at all. This means that the wavelength we are considering must be small compared to to characteristic size of the problem (i.e. the size of the opening). As a result, our formalism will yield just the the lowest-order correction due to diraction to the results of geometrical or ray optics. This approximation is called the Kircho approximation.

51 Now to implement this approximation we need two things. First, we need the

eld strength at the opening. For a source of unit strength at position x0, the eld

at position x0 in the opening is taken to be

which is just a spherical wave. Second, we need the component of 0G(x, x0) in the ∇ direction of n0 in the opening. Since n0 is the outward normal direction from the volume V , this means we need

ik x x ik x y d d e | 0| e | 0| G(x, x0) = . (176) dz dz x x x y ! 0 0 | 0| | 0|

Since we are using a short wavelength approximation, it follows that k x x0 1 | | and k x y0 1. Consequently, in taking the derivative of G(x, x0), the derivatives | | coming from the denominators are negligible compared to those from the exponentials. Thus ik x x ik x y d ik(z z0) e | 0| ik(z + z0) e | 0| G(x, x0) + . (177) dz x x x x x y x y 0 | 0| | 0| | 0| | 0| We only need to evaluate this in the opening (z0 = 0)

52 observed at x

1 2 (x) = d x0 n0 ((x0) G(x, x0)) (179) 4 Zopening ∇ ik x x0 ik x0 x0 1 2 2ikz e | | e | | = d x0 . (180) 4 opening x x x x x x Z | 0| | 0| | 0 0| We may also write z = cos(0) , (181) x x | 0| where 0 is the angle from the normal to the opening at integration point x0 to the observation point x.

O x'-x x-x' 0 θ' S z

Note that 0 is dierent for each integration point x0. Thus our expression becomes

ik x0 x0 ik x x0 ik 2 e | | e | | (x) = d x0 cos(0) (182) 2 opening x x x x Z | 0 0| | 0| The rst term is the eld from the source located at x0 and received at point x0 in the

opening. The second term is the eld from the “source” at x0 in the opening received at observation point x. The third term is the inclination factor. Thus, this integral is just an expression of the Huygens-Fresnel principle.

6.2 Babinet’s Principle

Let’s explore the consequences of this formula. First, consider two dierent screens which are complimentary.

53 σ σ

One has an opening, labeled , while the other is just a disk. For the second screen the ”opening” is the entire plane z = 0 except for the disk. This opening is the compliment of , so we will call it . We notice something interesting about the amplitudes received at the point O in the two complimentary cases, we have

1 2 (x) = d x0 n0 ((x0) G(x, x0)) (183) 4 Z ∇ 1 2 (x) = d x0 n0 ((x0) G(x, x0)) (184) 4 Z ∇ The sum of these two amplitudes is then

1 2 (x) + (x) = d x0 n0 ((x0) G(x, x0)) (185) 4 Z+ ∇ However, + represents the entire plane z = 0. Thus (x) + (x) represents the amplitude detected at x if there was no screen at all. In other words (x) + (x) includes no diraction. To interpret this, note that (x) may be written as

eik x x0 (x) = f | | + (x) (186) x x diff | 0| where f is a ray optic amplitude. If f = 0, then there is no line-of-sight from the observer to the source, while if f = 1, then there is. The second term is the amplitude due to diraction of the wave. Returning to our amplitudes (x) and (x), we see

54 that one will have f = 0 and the other f = 1, so that the sum will include a term

ik x x0 e | |/ x x . In fact, this is all it will include, since + includes no diraction. | 0| Thus, we see that the diraction amplitudes for complimentary screens cancel. This is Babinet’s Principle. It says that solving one diraction problem is tantamount to solving its compliment. However, Babinet’s principle does not say that the intensities will cancel in the two cases! We will see some consequences of Babinet’s principle in what follows

6.3 Fresnel and Fraunhofer Limits

Lets return to our expression for (x) in the Kircho approximation. Since we assume that the wavelength is small, it follows that

ka 1 kr 1 kr 1 (187) 0 where a is the typical aperture size, r0 is the distance from the screen to the source, and r is the distance to the observer. In general these approximations should be applied to the expression for (x) only after the integral over the opening has been carried out. This can be done for some simple geometries (as in the homework). What we will do here is to insert the above limits into the Kircho approximate expressions for (x). In what follows, we will restrict our attention to apertures, not discs. By doing this it follows that r0 will have an upper limit a, so kr0 1. In practice, the restriction to apertures is not a real limitation since Babinet’s principle allows us to calculate the the diraction amplitude for a disk from the diraction amplitude for the complimentary screen. Once again, our expression for (x) is

ik x0 x0 ik x x0 ik 2 e | | e | | (x) = d x0 cos(0) . (188) 2 opening x x x x Z | 0 0| | 0| Now if a/r0 and/or a/r are not small, then we are limited in what further approximations are possible since r0 is not necessarily small compared to either r or r0.

55 This limit is called Fresnel Diraction, in which the source and/or the observation points are close enough to the aperture that we must worry about the diraction of spherical waves. This presents a dicult problem, which we will not treat (but please see the homework and Landau and Lifshitz, Classical Theory of Fields, Sec. 60). In the opposite limit, which we will treat, is called Fraunhofer Diraction. In this limit ka > 1 kr 1 kr 1 a r and a r (189) 0 0 Note that we specify ka > 1 rather than ka 1 since we want to solve for the rst nite corrections to the latter inequality, so that ka2 a ka2 = ka 1 and 1 (190) r0 r0 r0 Thus we are looking at plane waves instead of spherical waves in this limit. It is possible to simplify our expression for (x) even further in this limit. Then

x0 x0 x0 x0 r0 = r0 + n0 x0 (191) | | r0 where n0 is a unit vector from the source to the origin (taken to be the center of the aperture), so that x = n r . Similarly, 0 0 0

x x0 r n x0 (192) | | where n is a unit vector from the origin to the observer so that x = nr.

n 0 O S n

56 Thus, ik x0 x0 ik x x0 ik(r+r0) e | | e | | e ik(n0 n) x e 0 (193) x x x x rr | 0 0| | 0| 0 where we have dropped second-order terms coming from the denominators. Further- more, since r is large compared to the aperture size, it is appropriate to take

z cos(0) = 1 (194) x x | 0| inside the integral. Thus in the Fraunhofer limit

ik(r+r0) ik e 2 ik(n0 n) x (x) = d x0 e 0 (195) 2 rr0 Zopening But the incident wave vector was kn, so

k(n n) = k k q (196) 0 in diff so we nally have ik(r+r0) ik e 2 iq x (x) = d x0 e 0 (197) 2 rr0 Zopening

7 Example Problems

Let’s consider some examples.

7.1 Example: Diraction from a Rectangular Aperture

Here the opening is given by x0 < a and y0 < b, so the integral above is | | | | a b 2 iq x iqxx iqyy sin(qxa) sin(qyb) d x0 e 0 = dx0 e 0 dy0 e 0 = 2 2 (198) opening a b qx qy Z Z Z The amplitude at x is

2ikab eik(r+r0) sin(q a) sin(q b) (x) = x y (199) rr0 aqx ! bqy !

57 thus, the intensity of the diracted wave is 2 2 2 sin(qxa) sin(qyb) I(x) = (x) = I0 (200) | | aqx ! bqy ! with 4k2a2b2 I0 = 2 2 2 (201) r0r We see that all of the angular dependence of in this is embodied in the dependence on q, the momentum transfer. Since we a looking at small-angle scattering, we are taking q to lie essentially in the x-y plane.

k diff q k in

We see that the nodes of I(x) occur for m n q = , m = 0, and/or q = , n = 0 (202) x a 6 y b 6

The global maximum (I(x) = I0) occurs for qx = qy = 0, i.e. for forward scattering. Thus the intensity of the diracted light looks something like

central maximum

lines of nodes

58 We can recover the formula for single-slit diraction by taking b a, so that our rectangular aperture becomes a slit of width 2a. Then

θ θ in diff x

y 2a sin(θ ) θ in 2a sin( diff )

q = (k ) (k ) = k sin( ) k sin( ) (203) x x in x diff in diff and the condition that there is a node is

m q = k (sin( ) sin( )) = m = 0 . (204) x in diff a 6 Since k = 2/, this may be written

2a (sin( ) sin( )) = m m = 0 (205) in diff 6 which is the usual expression for a node in single-slit diraction.

7.2 Example: Diraction from a Circular Aperture

Here the opening is given by r0 < a. Our Fraunhofer expression for the amplitude at x is then ik(r+r0) a 2 ik e iqr cos( ) (x) = dr0 r0 d0e 0 0 (206) 2 rr0 Z0 Z0

59 where we have taken q to lie entirely in the plane of the aperture, as is appropriate in small-angle diraction. Now

2 iqr0 cos(0) d0e = 2J0(qr0) (207) Z0 where J0 is the zeroth-order Bessel function. Our expression for the diraction amplitude is then

ik(r+r0) a ik(r+r0) ik e 2 e J1(qa) (x) = dr0 r0 2J0(qr0) = ika (208) 2 rr0 Z0 rr0 qa ! where we have used J1(0) = 0. The diracted intensity is then

2 2 4 J1(qa) k a I(x) = I0 , I0 = 2 2 (209) qa ! r r0 this reaches a maximum at q = 0, it then has a minima of zero and maxima which decrease as q increases. For incident light with k normal to the opening,

q = kdiff sin(diff ) = k sin(diff ) (210) since the magnitude of the wavevector does not change. Thus,

2 J1 (ka sin(diff )) I(diff ) = I0 (211) ka sin(diff ) !

The rst minima of J1(x) is at x = 3.83, hence the rst minima of diraction will occur when ka sin(diff ) = 3.83 , or sin(diff ) = 1.22 . (212) 2a! Since, by assumption, is small compared to a, sin( ) , and so the angle diff diff of the rst node is roughly diff = 1.22 . (213) 2a!

60 2a

I (θ ) diff

We notice from these examples of Fraunhofer diraction that all of the light falling on the aperture is deected. Granted, the intensity for forward scattering (q = 0) is nonzero, but the total amount of light with q = 0 is

(q) 2 q2 dq d 0 (214) | | q=0 → Thus, it is legitimate to say that all of the light is deected. Let us apply this observation to a disc which is totally absorbing (i.e. a perfect black-body). If plane waves are incident on this disc, then all the light that falls directly upon it is absorbed. If there was no diraction, then there would be a geometrically perfect shadow cast by the object, and the total cross section of the disk would be its area A. But of course there is diraction, so in addition to absorbing light, some will also be deected by the disk. By Babinet’s principle, as much light will be diracted by the disc as by an aperture of the same shape as the disc but in a screen. But we have said that in Fraunhofer diraction all of the light falling on an aperture is deected. Thus for our disc, the cross section for deection of light will be the disc area A as well. Thus the total cross section, including both absorption (inelastic cross section) and diraction (elastic cross section) is

total = inelastic + elastic = A + A = 2A (215)

61 This is twice the area of the disc. Recall that in this argument we assume that ka 1, where a is the characteristic size of the disc. In this limit in quantum mechanics one nds that the cross section for scattering from a sphere of radius a is 2a2. The factor of two has the same origins here as it does in quantum mechanics.

7.3 Diraction from a Cross

1. Given a normally incident wave of frequency ω calculate the diraction pattern produced by an opaque screen with an aperture in the shape symmetric cross of inner dimension a and outer dimension b in the Fraunhofer limit of the scalar Kircho approximation. Very roughly tell what the diraction pattern will look like.

2. The correct answer to the rst part of this problem also yields the solution to the diraction pattern involving a quite dierent set of apertures. What is this other set of apertures?

a b

Solution. In the Fraunhoer limit, we can use Eq. (197) of the notes.

ik(r+r0) ik e 2 iq x (x) = d x0 e 0 2 rr0 Zopening where r0 is the distance from the aperture to the source, r is the distance from the aperture to the observer, and q is the dierence between the incident and scattered

62 wavevector. We will take q to lie entirely in the plane of the aperture, as is appropriate in the small angle limit. For the geometry in the gure, the integrals may be performed trivially

ik(r+r0) a a a a b a a b ik e iqxx iqyy (x) = dx0 dy0 + dx0 dy0 + dy0 + dy0 dx0 + dx0 e 0 e 0 2 rr0 a a a b a ! a b a !! Z Z Z Z Z Z Z Z so that 2ik eik(r+r0) (x) = [sin(aqx) sin(bqy) + sin(aqy) sin(bqx) sin(aqx) sin(aqy)] qxqy rr0 The corresponding intensity (x) 2 will have a central maximum with I (2ab a2)2, | | ∝ and the whole pattern will be symmetric to rotations modulo /2. By Babinet’s principle we know that the diraction from a cross-shaped screen has an amplitude, that when added to (x), is the same as if there was no diraction at all.

7.4 Radiation from a Reciprocating Disk

A disc of radius a lies in the z = 0 plane and is centered at the origin. It is uniformly charged with surface charge density , and it rotates around the z-axis with an angular velocity cos(ωt) where and ω are constants. Assuming that the motion is nonrelativistic, nd the elds in the radiation zone, the angular distribution of radiated power, and the total radiated power. Solution. Here we will employ the methods developed to treat the radiation of harmonic current sources. The current of the rotation disc is (in cylindrical coordinates) J(x, t) = (z) cos(ωt)ˆ or, in complex notation

iωt iωt J(x, t) = (z)e ˆ = J(x)e

Then,

a 2 1 3 ikn x ikn x f(, ) = d x0 J(x0)e 0 = 0d0 d00ˆ0e 0 c ZV c Z0 Z0 63 We can use the azimuthal symmetry of the problem, and assume that n is in the xz plane. Then n = cos zˆ + sin xˆ, x0 = 0 (cos(0)xˆ + sin(0)yˆ), and ˆ0 = cos(0)yˆ sin(0)xˆ. For non-relativistic motion a/c 1, this allows us to keep only the rst nonvanishing term in the exponential, then

ka4 f(, ) = i sin ˆ 4c Hence, in the radiation zone,

eikr k2a4 eikr B = ik (n f) = sin ˆ r 4c r and k2a4 eikr E = B n = sin ˆ 4c r The power distribution is given by

dP ck2 ck2 2k2a822 = f 2 n f 2 = sin2() d 8 | | | | 8 16c2 h i The total power is obtained by integrating over all solid angles

2k422a8 P = 48c