Feb. 28, 2011 Larmor Formula: radia7on from non- rela7vis7c par7cles Dipole Approxima7on Thomson ScaEering The E, B field at point r and 7me t depends on the retarded posi7on r(ret) and retarded 7me t(ret) of the charge. Let B ( r,t) = n E ( r,t) [ ] E ( r,t) = q ( n β )(1 β 2 ) + q n κ 3 R 2 c κ 3 R ( n β ) β "VELOCITY FIELD" 1 Coulomb Law R 2 "RADIATION FIELD" 1 R Field of par7cle w/ constant velocity Transverse field due to accelera7on 1
Qualita7ve Picture: transverse radia7on field propagates at velocity c Radia7on from Non- Rela7vis7c Par7cles For now, we consider non- rela7vis7c par7cles, so Then E the RADIATION FIELD is E rad = q c simplifies to n κ 3 R ( n β ) β { } = q n ( n u ) Rc 2 and B rad = n [ E rad ] 2
Magnitudes of E(rad) and B(rad): Poyn7ng vector is in n direc7on with magnitude 3
ASIDE: Need two iden77es: So If Show that n n u Now E rad = E rad q n ( n u ) Rc 2 = q u Rc sinθ 2 A ( B C ) = B ( A C ) C ( A B n n ) =1 A B 2 = A 2 + B 2 2 A C A B cosθ ( ) = n ( n u ) ( ) u n n n u = n u cosθ n n =1 n n u ( ) = n ( n u ) n n u ( ) 2 = E rad ( ) u n n = u cosθ n u = u cosθn u u cosθ n u 2 = u 2 cos 2 θ + u 2 2u 2 cos 2 θ ( ) = u 2 1 cos 2 θ = u 2 sin 2 θ = q u Rc 2 sinθ 4
Energy flows out along direc7on with energy dω emieed per 7me per solid angle dω ergs/s/cm 2 cm2 = q2 u 2 4πR 2 c 3 sin2 θ R 2 dω da so Integrate over all dω to get total power P = dw = q2 u 2 sin 2 θdω dt 4πc 3 = q2 u 2 1 ( 1 µ 2 )dµ 2c 3 1 LARMOR S FORMULA emission from a single accelerated charge q 5
NOTES 1. Power ~ q 2 2. Power ~ accelera7on 2 3. Dipole paeern: No radia7on emieed along the direc7on of accelera7on. Maximum radia7on is emieed perpendicular to accelera7on. 4. The direc7on of is determined by If the par7cle is accelerated along a line, then the radia7on is 100% linearly polarized in the plane of The Dipole Approxima7on Generally, we will want to derive for a collec)on of par7cles with You could just add the s given by the formulae derived previously, but then you would have to keep track of all the t retard (i) and R retard (i) 6
One can treat, however, a system of size L with 7me scale for changes tau where so differences between t ret (i) within the system are negligible Note: since frequency of radia7on If then or This will be true whenever the size of the system is small compared to the wavelength of the radia7on. Can we use our non- rela7vis7c expressions for? yes. Let l = characteris7c scale of par7cle orbit u = typical velocity tau ~ l/u tau >> L/c u/c << l/l since l<l, u<c non- rela7vis7c 7
Using the non- rela7vis7c expression for E(rad): If R o = distance from field to system, then we can write L where Dipole Moment EmiEed Power power per solid angle power DIPOLE APPROXIMATION FOR NON- RELATIVISTIC PARTICLES 8
What is the spectrum for this E rad (t)? Simplify by assuming the dipole moment is always in same direc7on, let then Let fourier transform of then 9
Then ( ) E ˆ (ω) = sinθ fourier transform d (t) c 2 R o = sinθ ω 2 d ˆ (ω) c 2 R o ( ) = ω 2 sinθ c 2 R o d ˆ (ω) Recall from the discussion of the Poyn7ng vector: Integrate over 7me: (1) Parseval s Theorem for Fourier Transforms (2) Since E(t) is real FT of E (See Lecture notes for Feb. 16) so Thus (2) 10
Subs7tu7ng into (1) Thus, the energy per area per frequency subs7tu7ng and integrate over solid angle NOTE: 1. Spectrum ~ frequencies of oscilla7on of d (dipole moment) 2. This is for non- rela7vis7c par7cles only. 11
Thomson ScaEering Rybicki & Lightman, Sec7on 3.4 Thomson ScaEering EM wave scaeers off a free charge. Assume non- rela7vis7c: v<<c. E field electron e = charge Incoming E field in direction Incoming wave: assume linearly polarized. Makes charge oscillate. Wave exerts force: r = posi7on of charge 12
Dipole moment: so Integrate twice wrt 7me, t So the wave induces an oscilla7ng dipole moment with amplitude What is the power? Recall 7me averaged power / solid angle (see next slide) So 13
Aside: Time Averages The 7me average of the signal is denoted by angle brackets, i.e., If x(t) is periodic with period To, then x(t) = lim 2T T 1 +T T x(t)dt x(t) = 1 T o t o +T o t o x(t)dt 2sin 2 x =1 cos2x sin 2 x =1/2 1 2 cos2x T o sin 2 x = 1 2 x 1 4 sin2x t 1 t 2 The total power is obtained by integra7ng over all solid angle: or 14
What is the Thomson Cross- sec7on? Incident flux is given by the 7me- averaged Poyn7ng Vector Define differen7al cross- sec7on: dσ scaeering into solid angle dω Power per solid angle erg /sec /ster Time averaged Poyn7ng Vector erg/sec /cm 2 cross- sec7on per solid angle cm 2 /ster Thus so since we get (polarized incident light) Thomson cross sec7on classical electron radius 15
Integrate over dω to get TOTAL cross- sec7on for Thomson scaeering Thomson Cross- sec7on NOTES: 1. Thomson cross- sec7on is independent of frequency. Breaks down when hν >> mc 2, can no longer ignore rela7vis7c effects. 2. ScaEered wave is linearly polarized in ε- n plane Electron ScaEering for un- polarized radia7on Unpolarized beam = superposi7on of 2 linearly polarized beams with perpendicular axes 16
Differen7al Cross- sec7on dσ dω unpol = 1 dσ(φ) 2 dω = 1 2 ro2 ( 1+ sin 2 φ) pol + dσ(π 2) dω pol Average for 2 components Thomson cross- sec7on for unpolarized light dσ dω = 1 2 r 2 o ( 1+ cos 2 θ) 17