Feb. 28, 2011 Larmor Formula: radiation from nonrelativistic particles Dipole Approximation Thomson Scattering The E, B field at point r and time t depends on the retarded position r(ret) and retarded time t(ret) of the charge. Let ( u r ) 0t ret u r ( t ) 0 ret u c velocity of charged particle accelerati on 1 n r r r r r B(r,t) n E (r,t) r r2 r r rÝ r r (n ) (1 ) q n r r E ( r ,t) q 3 ( n ) 3 2 R c1 4 4R 442 4 4 4 43 1 4 44 2 4 4 43 "VELOCITY FIELD" 1 2 Coulomb Law R Field of particle w/ constant velocity "RADIATION FIELD" 1 R Transverse field due to acceleration Qualitative Picture: transverse “radiation” field propagates at velocity c Radiation from Non-Relativistic Particles For now, we consider non-relativistic particles, so u 1 c Then E the RADIATION FIELD is r r r r rÝ q n E rad 3 ( n ) c R simplifies to q r r rÝ 2 n ( n u) Rc and r r r Brad n E rad In u ,n plane B into of paper ra dplane Poynting vector in n directio Magnitudes of E(rad) and B(rad): q u E B sin ra d ra d2 Rc Poynting vector is in n direction with magnitude 22 c 2 S E rad 4 2 cq u S 24sin 4 R c ASIDE: r q r r rÝ E rad 2 n ( n u) Rc If r quÝ E rad 2 sin Rc Show that r r r rr r r r r r r Need two identities: A B C B n n A 1 C C A B r r2 r2 r2 r r r r A B A B 2 A C A B cos So… r r rÝ r r rÝ rÝ r r n n u n n u un n Now r rÝ r rÝ n u n u cos r r n n 1 r r rÝ r r rÝ rÝ r r n n u n n u un n rÝ rÝ rÝ = u cos n u rÝ rÝ = uÝcosn u r r rÝ 2 rÝ rÝ2 n n u = uÝcos n u = uÝ2 cos2 uÝ2 2 uÝ2 cos2 = uÝ2 1 cos2 2 2 Ý = u sin r quÝ E rad 2 sin Rc n Energy flows out along direction with energy dω emitted per time per solid angle dΩ dW SdA dt ergs/s/cm2 cm2 2 2 q uÝ 2 2 sin R d 1 2 3 2 3 4R c dA 22 so 2 dW q u 3sin dtd 4 c Integrate over all dΩ to get total power 2 2 dW q uÝ 2 P sin d 3 dt 4c 2 2 1 q uÝ 2 3 1 d 2c 1 2 2 2q u P 3 3 c LARMOR’S FORMULA emission from a single accelerated charge q 2 2 2q u P 3 3 c NOTES 1. Power ~ q2 2. Power ~ acceleration 2 3. Dipole pattern: No radiation emitted along the direction of acceleration. dW 2 sin dtd Maximum radiation is emitted perpendicular to acceleration. 4. The direction of E is determined by ra d u If the particle is accelerated along a line, then the radiation is 100% linearly polarized in the plane of uand n The Dipole Approximation Generally, we will want to derive with Era d for a collection of particles positions r i velocities u i charges q i 1 ,2 ,... N i You could just add the Era d‘s given by the formulae derived previously, but then you would have to keep track of all the tretard(i) and Rretard(i) One can treat, however, a system of size L with “time scale for changes” tau where L c so differences between tret(i) within the system are negligible Note: since frequency of radiation If L c c then L 1 or L This will be true whenever the size of the system is small compared to the wavelength of the radiation. Can we use our non-relativistic expressions for Let l = characteristic scale of particle orbit u = typical velocity tau ~ l/u tau >> L/c u/c << l/L since l<L, u<c non-relativistic Era d ? yes. Using the non-relativistic expression for E(rad): ) q n ( n u i i E ra d 2 R i c i If Ro = distance from field to system, then we can write n (n d) E 2 rad cR o where d qiri i Dipole Moment L Ro n dW Recall Poy Vec A n dt Emitted Power 2 dP d 2 3sin d 4 c power per solid angle 2 2d P 3 3c power DIPOLE APPROXIMATION FOR NON-RELATIVISTIC PARTICLES What is the spectrum for this Erad(t)? n (n d) E 2 rad cR o d qiri i Simplify by assuming the dipole moment is always in same direction, let E (t)E then d(t)d sin E (t) d (t) 2 cR o Let fourier transform of ˆ d ( t )be d () ˆ fourier ansfo of E(t) be E ( ) then i t ˆ d ( t) e d ) d ( 2 i t ˆ d ( t ) d ( ) ed Then sin ˆ Ý(t) E ( ) 2 fourier transformdÝ c Ro sin 2 2 dˆ ( ) c Ro 2 sin ˆ 2 d ( ) c Ro Recall from the discussion of the Poynting vector: energy dW c2 E ( t ) time area dtdA 4 Integrate over time: dW c 2 E ( t) dt dA 4 (1) Parseval’s Theorem for Fourier Transforms 2 ˆ E ( t ) dt 2 E )d ( 2 (2) Since E(t) is real FT of E 1 ˆ E ( ) E (t)eitdt 2 ˆ( E ) so Thus (2) (See Lecture notes for Feb. 16) ˆ ˆ E ( ) E ( ) 2 2 2 0 ˆ E ( t ) dt 4 E )d ( 2 Substituting into (1) dW ˆ 2 c E ( )d dA0 Thus, the energy per area per frequency substituting 2 sin ˆ ˆ E ( ) 2 d ( ) c R o dW ˆ 2 c E ( ) dAd and dA R d 2 0 2 dW 1 4 2 ˆ d ( ) sin 3 d d c integrate over solid angle 2 dW 8 4ˆ 3 d ( ) d 3 c 2 dW 8 4ˆ 3 d ( ) d 3 c NOTE: 1. Spectrum ~ frequencies of oscillation of d (dipole moment) 2. This is for non-relativistic particles only. Thomson Scattering Rybicki & Lightman, Section 3.4 Thomson Scattering EM wave scatters off a free charge. Assume non-relativistic: v<<c. n E field e = charge electron Incoming E field in direction Incoming wave: assume linearly polarized. Makes charge oscillate. Wave exerts force: F eEo s inot ma mr r = position of charge Dipole moment: so Integrate twice wrt time, t d er d er 2 e d Eo sinot m 2 e E o 2 d sin t o m o So the wave induces an oscillating dipole moment with amplitude e2E o do 2 m o What is the power? Recall time averaged power / solid angle 2 dP d 2 3sin d 4 c 4 2 2 eE 2 o sin 2 sin ot 3 m 4 c 1 t 2 1 sin t t s 2 in t (see next slide) o t 1 4 2 o 2 o So 1 2 42 o 23 eE 2 dP sin d 8 m c Aside: Time Averages The time average of the signal is denoted by angle brackets , i.e., 1 T x(t) lim x(t)dt T 2T T If x(t) is periodic with period To, then 1 x(t) To 2sin2 x 1 cos2x 1 sin x 1/2 cos2x 2 2 To sin 2 x t2 1 1 x sin2x 2 4 t1 t o To to x(t)dt The total power is obtained by integrating over all solid angle: 2 2d P 3 3c or e4Eo2 P 2 3 3mc What is the Thomson Cross-section? Incident flux is given by the time-averaged Poynting Vector c2 ergs 1 S E o 2 8 sec cm Define differential cross-section: dσ scattering into solid angle dΩ dP d S d d Power per solid angle erg /sec /ster Time averaged Poynting Vector erg/sec /cm2 cross-section per solid angle cm2 /ster Thus so since dP c 2 d Eo d 8 d d dP 8 1 d d c Eo2 e 4 Eo2 dP 2 sin 2 3 d 8m c we get 4 d e 2 sin 24 d m c d r2sin 2 o d classical electron radius 2 e ro 2 mc (polarized incident light) Thomson cross section 13 r 2 . 82 10 cm o Integrate over dΩ to get TOTAL cross-section for Thomson scattering d T d d 2 ro2 sin d 8 2 T ro 3 Thomson Cross-section 0 . 6 1 6 cm 0 5 24 2 T NOTES: 1. Thomson cross-section is independent of frequency. Breaks down when hν >> mc2, can no longer ignore relativistic effects. 2. Scattered wave is linearly polarized in ε-n plane Electron Scattering for un-polarized radiation Unpolarized beam = superposition of 2 linearly polarized beams with perpendicular axes , : direc of line pol on 1 2 n d of s irecti catte wav We chose 1 to be in n kplane k direction of inciden ave then is perpendicu lar n to 2 angle b1 etween and n and hence the angle between 2and n angle between n and k . Also, is 2 2 Differential Cross-section d 1 d( ) d unpol 2 d pol 1 2 ro 1 sin 2 2 Thomson cross-section for unpolarized light d( 2 ) d pol Average for 2 components d 1 2 2 ro 1 cos d 2 angl betwee inci eand d scattered waves