Lecture 8 February 18, 2010

Transcription

1 Sources of Eectromagnetic Fieds Lecture 8 February 18, 2010 We now start to discuss radiation in free space. We wi reorder the materia of Chapter 9, bringing sections 6 7 up front. We wi aso cover some of 6.4, on Green functions, which I underst you skipped ast fa in 503. Assume a set of charges with given motion. What fieds do they generate? ( wave-guide ast time) We are ignoring back-reaction, the changes in their motion due to the fieds. Therefore inearity. So we can fourier-transform in time.

2 The Basic Equations Consider one fourier component, with a fieds having a time dependence e iωt. They are generated by the fourier components ρ( x, t) = ρ( x)e iωt, J( x, t) = J( x)e iωt. The fieds E( x, t) H( x, t) generated by these charges currents may be described by the eectrostatic potentia Φ( x, t) the vector potentia A( x, t). But as discussed in 6.2, Φ A have a gauge-invariance. To determine them, need Maxwe s equations a gauge condition. Choose Lorenz Gauge: A + 1 Φ c 2 t A Φ are determined. = 0. Then the fieds

3 Then from 6.2, 2 Φ 1 c 2 2 Φ t 2 = ρ/ɛ 0, 2 A 1 c 2 2 A t 2 = µ 0 J. For our fourier mode, / t iω, so each component of A Φ satisfy ( 2 + k 2) Ψ( x) = s( x), (1) with k = ω/c, a with the prescribed source s( x). Soution by Green function, from 6.4. Let us review this.

4 Soution by Green function ( 2 + k 2) Ψ( x) = s( x) is (inhomogeneousy) inear in Ψ, so soution is sum of soutions of the homogeneous Hemhotz equation specific soutions for each piece of the source. The equation is an eiptic partia differentia equation, having a unique soution once boundary conditions are specified. Think of the source as a sum of pieces at each point, s( x) = d x s( x )δ 3 ( x x), sove for a deta function source with the Green function ( 2 x + k 2) G( x, x ) = δ 3 ( x x ). (2) Then the soution for Ψ is Ψ( x) = d 3 x s( x )G( x, x ).

5 Finding G( x, x ) A point charge at origin: V ( x) = q 4πɛ 0 x E = V x = q 4πɛ 0 x 3, E = 2 V = qδ 3 ( x). So we see φ( x) = 1 x has φ = x x 3, 2 φ = 4πδ 3 ( x). On the other h, W := e ±ik x satisfies W = ±ik x x W 2 W = = [ ( ) x ±ik x x 1 x [ ( 3 ±ik x x 2 x 3 ] W k 2 x 2 x 2 W ) k 2 ] W = ± 2ik x W k2 W. Thus ( 2 x + k 2) W φ = ( 2 xw ) φ + 2( W ) ( φ) + W ( 2 xφ ) +k 2 W φ

6 ( 2 x + k 2) W φ = ( 2 xw ) φ + 2( W ) ( φ) + W ( 2 xφ ) +k 2 W φ = ± 2ik x W φ 2ik x x W x x 3 4πW δ3 ( x) = 4πδ 3 ( x) as W ( x)δ 3 ( x) = W ( 0)δ 3 ( x) = δ 3 ( x). So for x = 0, G( x, 0) = W φ 4π = e±ik x 4π x. But the operator ( 2 x + k 2) is transation-invariant, so we may transate: G( x, x ) = e±ik x x 4π x x. (3) We ve ignored boundary conditions. Want outgoing waves ony. Choose upper sign.

7 In Spherica Coordinates We are often interested in sources confined to a compact area how it radiates out to arge distances. Spherica coordinates are most suitabe. In spherica coordinates 2 = 2 δ 3 ( x x ) = r 1 r 2 L2 r r 1 r 2 sin θ δ(r r )δ(θ θ )δ(φ φ ) due to the metric factors h i, From the competeness reation (J3.56) for the spherica harmonics, the anguar part of the deta function can be written as a sum, we have ( 2 r r r 1 ) r 2 L2 + k 2 G( x, x ) = δ(r r ) r 2 Ym (θ, φ )Y m (θ, φ), =0 m=

8 Writing G( x, x ) = R m (r, x )Y m (θ, φ), we have m ( 2 r ) ( + 1) r r r 2 + k 2 R m (r, x )Y m (θ, φ) m = 1 r 2 δ(r r ) =0 m= Y m (θ, φ )Y m (θ, φ), so we see that R(r, x ) = g (r, r )Ym (θ, φ ) where ( 2 r ) ( + 1) r r r 2 + k 2 g (r, r ) = 1 r 2 δ(r r ). For r r this is just the spherica Besse equation, so the soutions are combinations of j (kr) n (kr), or better of j (kr) h (1) := j (kr) + in (kr) r ( i) +1 e ikr /kr. For r < r we need the soution to be reguar at r = 0, so there are no n or h contributions, ony j, g (r, r ) = a (r )j (kr) for r < r,

9 whie for r > r we want ony outgoing waves, with e +ikr, so the soution is pure h (1) with no h (2) (or j ) g (r, r ) = b (r )h (1) (kr) for r > r. But from (3) we see that the Green s function is symmetric under x x, so a (r) = a h (1) write more generay G( x, x )= e±ik x x 4π x x (3) (kr) b (r) = a j (kr), we may g (r, r ) = a j (kr < )h (1) (kr > ), where r < is the smaer of r r r > is the greater.

10 The deta function source for the spherica Besse equation on g (r, r ) means the first derivative is discontinuous, g (r=r +ɛ) g (r=r ɛ) = 1 r 2 = a kj (kr)h (1) (kr) a kh (1) (kr)j (kr). This is ka times the Wronskian of h (1) j, which shoud be r 2. This agrees with the genera statement that the Wronskian satisfies dw/dr = P (r)w, where P (r) is the coefficient of the first order term, here 2/r. Thus we can determine a at any point r, as Jackson tes us for sma x, π π ( x ) j (x) = 2x J +1/2(x), 2Γ( + 3/2) 2 ( ) π 2 +1 n (x) = 2x N 1 +1/2(x) Γ( + 1/2) x 2 π, h (1) = j + in in, so j (r)h (1) (kr) h (1) (r)j (kr) i/(kr)2, a = ik.

11 Green function in spherica coordinates So a together G( x x ) = eik x x 4π x x = ik m j (kr < )h (1) (kr > )Ym (θ, φ )Y m (θ, φ). We are now ready to examine the soutions to the Hemhotz equation, A( x) = µ 0 d 3 x J( x ) eik x x 4π x x =iµ 0 k d 3 x j (kr < )h (1) (kr > )Ym (θ, φ )Y m (θ, φ) J( x ). m

12 If the sources are restricted to some region x < d, we are asking about positions further from the origin, r > d, then r < = r r > = r, A( x) = iµ 0 k h (1) (kr)y m (θ, φ) d 3 x j (kr )Ym (θ, φ ) J( x ). m We see that A has an expansion in specified modes (, m) with the sources ony determining the coefficients of these modes. If the source region is sma compared to the waveength, d λ = 2π/k = 2πc/ω, we have kr 1 whereever J( x ) 0, so we may use the expansion j (x) x /(2 + 1)!!, appropriate for x 1. We see that the owest vaue which contributes wi dominate.

13 (Jackson 9.1) We have derived a quite genera expression for the fieds, but we can amost aways find ourseves in a zone for which things simpify, depending on the reative sizes of d, λ, r. If d r are both much smaer than λ, we are in the near zone, we may set k = 0 whie setting kj (kr < )h (1) (kr >) to i 2+1 r< r> +1. The fieds are essentiay instantaneousy generated by the currents charges. If, in addition we assume d r, the owest vaue wi dominate. If r λ r > d, the fieds osciate rapidy with (kr) ( i) +1 e ikr /r, faing off ony as 1/r, typica of radiation fieds, this is caed the far or radiation zone. If we aso have d λ, kr < is sma whereever J doesn t vanish, the owest mode wi dominate. h (1)

14 We have not bothered to find Φ( x) because the Lorenz gauge iωφ/c 2 = A gives it in terms of A, except for ω = 0, for which Φ( x) is given by the static Couomb expression integrated over a the charges, the eectric fied is given by Couombs aw, there is no magnetic fied arising from Φ.

15 Now et us consider the zone d λ r, which shoud be dominated by the owest. If the = 0 term does not vanish, we may write A( x) iµ 0 kh (1) 0 (kr)y 00 d 3 x Y00 J( x ) = µ 0 e ikr d 3 x J( x ), 4π r because h (1) 0 (x) = ieix /x. We are assuming a sources have an e iωt time dependence, so the continuity equation tes us J = ρ/ t = iωρ, we may write 1 d 3 x J( x ) = iω d 3 x x ρ( x ). The integra is just the eectric dipoe moment, so A( x) iµ 0ω 4π eikr p r. 1 See the ecture notes for some agebra justifying this the messy expression for E beow.

16 That expression, A( x) iµ 0 ω p e ikr /4πr, is accurate for a r > d to owest order in d/λ, provided the dipoe moment isn t zero. Quite generay, H = 1 µ 0 A, whie outside the region with sources, D t = iωɛ 0 E = H = E = iz 0 k H, with Z 0 = µ 0 /ɛ 0. The cur of pf(r) is ˆr p f/ r, so for our eectric dipoe source, we have 2 E( x) = H = ck2 4π ˆr p ( 1 1 ) e ikr ikr r, { eikr k 2ˆr ( 1 (ˆr p) + [3ˆr(ˆr p) p] 4πɛ 0 r r 2 ik )}. r 2 See ecture notes for agebra detais.

17 The eectric fied in the near zone is just what we woud have from a static dipoe of the present vaue at each moment, the E fied dominates the H fied in this zone. Note the first term in E is perpendicuar to x, but the second is not. However this ongitudina term fas off as r 2, so may be negected in the radiation zone r λ, where we can write H = ck2 4π E = k2 e ikr ˆr p eikr r 4πɛ 0 r ˆr (ˆr p) = Z 0ˆr H in the radiation zone In the near zone, that is when d < r λ, we have E = H = iω ˆr p 4πr2 1 4πɛ 0 r (3ˆr(ˆr p) p) 3 in the near zone

18 Power radiated at arge distances: average power per unit soid ange is P dω = r2 2 Re ˆr ( E H ) Z 0c 2 k 4 ˆr (ˆr p) 2 2(4π) 2 = Z 0c 2 k 4 32π 2 p 2 (1 cos 2 θ) = Z 0c 2 k 4 32π 2 p 2 sin 2 θ, where θ is the ange between p x. The tota power radiated is π P = 2π dθ sin θ P 0 dω = Z 0c 2 k 4 π 16π p2 dθ sin 3 θ 0 = Z 0c 2 k 4 12π p2.

19 If = 0 vanishes, need term in the expansion, A (1) = iµ 0 kh (1) With 1 (kr) m=1 h (1) 1 (x) = eix x m=1 m= 1 m= 1 Y 1m (θ, φ) d 3 x j 1 (kr )Y 1m(θ, φ ) J( x ). ( 1 + i ), j 1 (x) = x ( 1 + Ox 2 ), x 3 Y 1m (θ, φ)y 1m(θ, φ ) = 3 4π ˆr ˆr, we see that 3 A (1) = iµ 0 k 3 1 e ikr ( 1 + i ) 4π 3 r kr 3 I disagree with Jackson 9.30 by an overa sign d 3 x ˆr x J( x ).

20 The mutipoe moments invoved here are tensors x J( x ). The antisymmetric part is the integra of the magnetization M( x ) = 1 2 x J( x ), with m = d 3 x M( x ) the magnetic dipoe moment. ˆr m = 1 [ d 3 x (ˆr 2 J( x ) x (ˆr x ) J( x ] ). The symmetric piece is reated to the eectric quadripoe moment Q ij := d 3 x ( 3x ix j x 2 ) δ ij ρ( x ) = d 3 x ( 3x ix j x 2 ) i δ ij ω J = i d 3 x J k ( x ) ( k 3x ω i x j x 2 ) δ ij = i d 3 x J k ( x ) ( 3δ ik x j + 3δ jk x i 2x k ω δ ) ij

21 Again, Q ij = i d 3 x ( 3x ω jj i + 3x ij j 2( x ) J)δ ij. So ˆr Q = ˆr i Q ij ê j = i ( d 3 x 3ˆr ω J( x ) x + 3ˆr x J( x ) 2 x J( x ) ) ˆr. For competeness we need to consider a eectric monopoe term M E = d 3 x x 2 ρ( x ) = 2i d 3 x x ω J. So our compete vector potentia is A (1) = i µ 0k e ikr ( 1 + i ) (6ˆr m + iωˆr Q + iωm E ˆr). 24π r kr Let us evauate H E ony to eading order in 1/r, so we need ony consider the derivative acting on e ikr, needn t worry about ˆr. We can aso drop the i/kr term.

22 Magnetic Dipoe Then H (1) = 1 µ 0 A (1) = k2 e ikr 24π r ˆr (6ˆr m + iωˆr Q + iωm E ˆr). Notice that the eectric monopoe vanishes due to ˆr ˆr = 0. The magnetic dipoe contributes H MD = k2 4π e ikr r ˆr (ˆr m), E MD = iz 0 k H MD = k2 Z 0 4π = k2 Z 0 4π e ikr r ˆr m. e ikr r ˆr (ˆr (ˆr m)) These are of the same form as for the eectric dipoe, but with E H interchanged. The radiation pattern is the same, but the poarization has E m here, whie E ies in the pane incuding ˆr p for the eectric dipoe.

23 Eectric Quadripoe One might be tempted to think the eectric quadripoe aso vanishes, as it invoves ˆr (ˆr Q), Q is symmetric. But that is incorrect: in ˆr (ˆr Q) = ɛ ijkˆr i Q jˆr ê k, the summs are symmetric under i antisymmetric under i j, but that does not make things vanish. Jackson defines the vector Q( n) := Q ij n j ê i, then we have ˆr Q(ˆr). Then again keeping ony 1/r terms, H EQ = ick3 e ikr 24π E EQ = iz 0ck 3 24π r ˆr Q(ˆr) e ikr r ˆr ( ˆr Q(ˆr) ).

24 Power radiated Probaby the most interesting thing one might ask is how much power is radiated, in which directions, as we did for the eectric dipoe. For an eectric quadripoe, Q is a symmetric rea traceess tensor, so we coud rotate the coordinate system so that it wi be diagona. If we take an axiay symmetric case, with Q zz = 2Q xx > 0. The average power per unit soid ange is P dω = r2 2 Re ˆr ( E H ) ˆr (ˆr Q(ˆr)) 2 as shown.