Generation of EM waves Susan Lea Spring 015 1 The Green s funtion In Lorentz gauge, we obtained the wave equation: A 4π J 1 The orresponding Green s funtion for the problem satisfies the simpler differential equation x, x D x δ 4 x and the boundaries are at infinity. Thus D an depend only on the vetor z x x. We find D by taking the 4-dimensional Fourier transform: z D 1 D π k e i k z d 4 k Note that k z ω t t k x x ω t t k x x. This explains the usual onvention of transforming the time variable with the opposite sign from that for the spae variable. Also reall the delta-funtion integral: δ 4 z 1 π 4 d 4 ke ik z The transformed differential equation reads: k k k D 1 π 1
and transforming bak, we get: z D 1 e i k z π 4 d 4 k k k 1 π 4 d 3 ke i k x x dω e iωt t ω / k We ll do the frequeny integral first. For t < t, we lose the ontour with a big semi-irle in the upper half plane, while for t > t we lose downward. The integrand has two poles, at ω ±k, both on the real axis. Sine the result must be zero for t < t event preedes its soure we must evelaute the integral along the real axis by deforming the path to go above the two poles. Contour for inverting the transform For t > t, we use the lower ontour and we traverse it lokwise. Eah of the poles is simple, and the residues are lim ω k e iωt t e it t k ω + k ω / k k eit t k k
and lim ω k Then we have: z D 1 π 4 Θ t t i π 3 Θ t t e iωt t e it t k ω k ω / k k e it t k k d 3 ke i k x x k k πi e it t eit t k k d 3 ke i k x x 1 e it t k e it t k k The first exponential represents an outgoing wave, while the seond represents an inoming wave. Now do the remaining integrals, we hoose our polar axis for k along the vetor z x x. Then: z D i π 3 Θ t t i π Θ t t i π Θ t t π z Θ t t π z Θ t t π z Θ t t +1 π 0 1 +1 0 1 kdk eikzµ +1 0 ikz 1 0 0 0 k dkdµdφe ikzµ 1 k kdkdµe ikzµ e it t k e it t k e it t k e it t k e it t k e it t k dk e ikz e ikz e it t k e it t k dk exp ik z t t exp ik z + t t exp [ ik z + t t ] + exp ik z t t dk exp ik z + t t + exp ik z t t 4πz Θ t t δ z t t δ z + t t The seond term an never ontribute sine both z and t t are positive. Thus we finally have: D z 4πz Θ t t δ z t t 3
and thus the solution to equation 1 is: x A α 4π 4πz Θ t t δ z t t J α x d 4 x 1 z Θ t t δ z t t J α x d 4 x Note: Jakson prefers to write this using the fully ovariant expression [ x ] δ x δ {t x x } δ {t R t + R} 1 R {δ t R + δ t + R} where we wrote t t t, R x x, and used the result Thus δ f x zeros δ x x i f x z D 1 [ x π Θ t t ] δ x This is a ovariant expression in spite of the expliit appearane of t and t, sine the step funtion serves to loate the result on the forward light one from the soure event, and this is an invariant statement. The expression for the potential beomes: x A α 4π [ 1 x π Θ t t ] x δ x J α d 4 x [ x Θ t t ] x δ x J α d 4 x 3 Charge density and urrent for a point harge. The harge and urrent densities are given by ρ x,t qδ x r t 4
and J x,t q vδ x r t and so the 4-vetor is J α ρ, v q, v δ x r t q γ uα δ x r t 4 We an write this ovariantly as J α t 0, x q x u α δ x0 τ 5 where x α 0 t 0, r t 0 is the harge s position vetor at time t 0. To see why this works, note that x δ x0 τ δ t t 0 τ δ x r τ and Thus δ t t 0 τ δ τ t τ t 0 / J α t 0, x q δ τ t τ t 0 γ u α δ τ t τ t 0 δ x r τ γ q γ uα δ x r t 0 whih agrees with equation 4. 3 Radiation from a moving harge- the potential We insert the urrent vetor 5 into the expression 3 for the potential: x A α [ x Θ t t ] x δ x q u α δ x0 τ d 4 x [ x q Θ t t ] x δ x u α δ x0 τ d 4 x 5
Using the delta funtion, the integration over x is easy: x [ x ] A α q Θ t t 0 τ δ x0 τ u α Now again we need to evaluate the delta funtion of a funtion- here the funtion f τ x x0 τ x α x α 0 x α x 0α. The derivative f τ dxα 0 x α x 0α v α x α x 0α and the zeros are those values of τ for whih x α x α 0 x α x 0α 0. Expanding this out, we get where t t 0 x r t 0 0 R x r t 0 t t 0 R/ Only the positive value ontributes to the integral beause of the fator Θ t t 0 τ. The event is on the positive light one from the soure. Thus: or the retarded time. Then the denominator is v α x α x 0α γ, γ v t t 0 + R/ or, using equation 6, γ R/ γ v R t 0 t R/ t ret 6 t t 0, R γ t t 0 γ v R γr 1 ˆR 6
Then A α x q [ x ] Θ t t 0 τ δ x0 τ u α τ qγ, v γr 1 ˆR q 1, R 1 ˆR t ret t ret They may also be written o- These are the Lienard-Wiehert potentials. variantly as: using 6, and with u α A α q u x 7 γ, v q γ, v t t 0, x x 0 1, q t t 0 R q 1, R 1 ˆR all evaluated at the retarded time. x x x 0 4 Fields due to a moving harge To ompute the fields from the potentials or 7, we need to take derivatives. That is, we need to ompare the potentials at two nearby events. The potential A α hanges as x α hanges both beause of its expliit dependene on x α but also beause of the impliit dependene on τ 0 x α that is beause of the need to look bak along the light one to find the soure event. da µ Aµ x ν dxν + daµ 0 0 8 7
Differentiating the light one onstraint x α r α τ 0 x α r α τ 0 0, we get: x α r α τ 0 dx α dr α τ 0 0 9 and from the definition of the partile s veloity So dr α τ 0 u α 0 x α r α τ 0 dx α u α 0 0 where I defined x α x α r α τ 0 Thus from equation 8 and 7, we find qu da µ µ ν dx ν + d qu µ u x 0 u x { } { qu µ u ν x [u γ x γ ] dx ν + q Note that ν x x ν x r τ 0 0 xα r α τ 0 dx α x r τ 0 u xα dx α x u 10 0 a µ u µ a x + u u u x [u γ x γ ] x ν g α x α r α τ 0 g α δ α ν g ν Now we insert the expression 10 for 0, and fator a bit: } da µ qdx ν { u µ [u γ x γ ] u g ν + x ν a µ u µ a x u u u ε x ε } qdx ν { u µ [u γ x γ ] u ν + x ν a µ + u µ a x u ε x ε } 0 So } A µ x q { u µ ν [u γ x γ ] u ν + x ν a µ + u µ a x u ε x ε 8
Now we ompute the field tensor omponents F µν Aν x µ Aµ x ν The symmetri term u µ u ν will anel in the subtration, leaving { } F µν q [u γ x γ ] x µ a ν x ν a µ + a x u ν x µ u µ x ν u ε x ε 11 Now, using the notation in the diagram above, x R 1, ñ u γ 1, and and d u a γ d u γ dγ dγ, + γ d dγ γ3 d 9
so a γ γ d, γ d d + Then we an ompute the dot produts u γ x γ γr 1 1 and d a x γ γ d, γ γ R γ d γ γ R γ d d + R 1, n d ˆn d 1 d ˆn Then from 11, and writing K q [ γr 1 ] 13 we have { } F µν x µ a ν x ν a µ + K u ν x µ u µ x ν γr1 ˆn γ R γ d 1 d K xµ a ν x ν a µ + u ν x µ u µ x ν + γ R d ˆn γr 1 γ 3 d 14 10
The top row is F 0i K x0 a i x i a 0 + u i x 0 u 0 x i + γ R d ˆn γr 1 γ 3 d K R a i n i γ 4 d + γr i n i + γ R d ˆn γr 1 γ 3 d KR γ γ d i + di n i γ 4 d + i n i + γ R d ˆn R 1 γ 4 d where we used 1 for a. Then, inserting 13, the eletri field is given by E i F 0i γ KR n i i γ d + + γ R d ˆn γ R 1 γ d di q R 1 n i i + γ R d γ R 1 di 15 We may divide this result into two parts: Eoulomb i q γ R 1 3 n i i This field has the usual 1/R dependene and is in fat the same as Jakson s equation 11.154. See Jakson page 664 for the details. The other term is the radiation field: it depends on the partile s aeleration. Erad i q R 1 n i i d 1 di 16 It dereases as 1/R rather than 1/R, and so dominates the Coulomb field at large distanes. We an simplify the expression by noting that: [ ] ˆn ˆn d ˆn ˆn d d 1 11
Thus E rad q R 1 [ 3 ˆn ˆn ] d 17 Remember: everything is evaluated at t retarded. Also notie that E rad is perpendiular to ˆn, and is proportional to the aeleration d /. The magneti field is given by: For example: B F 3, F 13, F 1 B x F 3 K x a 3 x 3 a + u 3 x u x 3 + γ R d ˆn γr 1 γ 3 d K n a 3 n 3 a + u 3 n u n 3 + γ R d ˆn γr 1 γ 3 d q γ 3 d n 3 n 3 + γ n d3 n 3 d γr 1 + 3 n n 3 +γ R d ˆn γr1 ˆn γ3 d The first and last terms anel. Thus: q + γ B R d γr 1 ˆn γr 1 ˆn + γˆn d Again we may divide the result into two parts: q B B-S γ R 1 3 ˆn This term, whih goes as 1/R, is the usual Biot -Savart law result, with relativisti orretions. Compare J eqn 11.15 and our expression for E oulomb 1
The radiation term is: B rad q R 1 ˆn ˆn E rad q R 1 d ˆn 1 ˆn + ˆn d d ˆn d + 1 where we used equation 16 for the eletri field. 5 Radiated power The Poynting vetor for the radiation field is S E 4π rad B rad E 4π rad ˆn E rad 4π 4π E radˆn 4πR q R 1 q 1 [ 3 ˆn ˆn [ 6 ˆn ˆn Thus the power radiated per unit solid angle is: ] d ˆn d ] ˆn dp dω R S ˆn q 4π 1 [ 6 ˆn ˆn d ] 18 13
For non-relativisti motion, 1, we retrieve the Larmor formula 704 wavemks notes eqn 35 with a units hange: dp dω q 4π [ ˆn q 4π 3 a sin θ ˆn d where a dv/ is the usual 3-aeleration and θ is the angle between a and ˆn. The total power radiated in the non-relativisti ase is: P dp dω dω q 3 a +1 1 µ µ3 3 π dµ +1 1 0 ] dφ q 4π 3 a 1 µ q 3 3 a 19 When is not small, there are important hanges. The denominator of equation 18 gets small when is lose to 1, that is for diretion of propagation lose to the partile s veloity. Thus the radiation is strongly beamed along the diretion of the partile s motion. 5.1 Covariant generalization From problem 1.15 we know that Θ 0α dv transforms as a vetor. Thus Θ 00 dv transforms in the same way as x 0 t. Thus the ratio, whih is the power, is a Lorentz invariant. But in the instantaneous rest frame of the partile, a α a α a a a So we an write eqn 19 as P q 3 3 aα a α q γ 3 γ + γ d 14
where dot means d/. The term is parentheses simplifies as follows: But So γ γ + γ d γ γ + γ + γ γ d γ 4 d γ + γ 1 γ γ γ γ γ d γ γ 3 d γ γ 4 d d Giving P q 3 γ + γ d whih is positive, as expeted. Now let s look at the two speial ases: 1. parallel to d / : P q 3 γ 1 + 1 where d/ γd/ γa/. Thus. perpendiular to d / : P q 3 q 3 γ4 P q 3 3 γ6 a 0 γ 3 15 q 3 γ4 a 1
Thus we get more radiated power per unit aeleration if a is parallel to linear motion. However, if we relate the power to the fore ating on the partile, see exam problem,we get a different interpretation. 1. F parallel to. In this ase a F/mγ 3 and so P q F 3 3 m. F perpendiular to. In this ase a F/mγ and so P q F 3 3 γ m So the power radiated per unit fore is greater when F is perpendiular to irular motion. 16