Scattering in Electromagnetism: A Detailed Analysis - Prof. Phillippe Piot, Study notes of Physics

Download Scattering in Electromagnetism: A Detailed Analysis - Prof. Phillippe Piot and more Study notes Physics in PDF only on Docsity! Northern Illinois University, PHY 571, Fall 2006 Part V: Scattering Last updated on November 16, 2006 (report errors to piot@fnal.gov) 1 Two types of scattering: • e- (q = −e, me = 9.1× 10−31 kg) → high energy loss, small deflection, • nuclei (q = Ze, mn À me) → low energy loss, large deflection There are more e- than nuclei (factor Z) so Z more time e- scat- tering... q, E q, E’<E matter 2 So we see that e- are much more efficient that nuclei at extracting energy from incident particles. But when is the IA valid? Let’s check the assumptions: 1- incident particle travels on straight path: θ = ∆pe γMv = 2qe γMv2b = 2 γ qe/b Mv2 = 2 E electrostat. incident Energy = 2 V E . (6) So θ ¿ 1 ⇒ V ¿ E. 2- target remains stationary ⇒ recoil distance during collision is d ¿ b. The interation time is τ ∼ bγv, and corresponding recoil distance is d ∼ ∆pem τ so d ¿ b ⇒ ∆pe m b γv ¿ 1 ⇒ 2qe/b γmv2 ¿ 1. (7) this is a stronger condition than the one coming from θ ¿ 1 by a factor M/m. 5 So let’s keep the stronger condition (and rewrite it by making the classical radius of e- appearing): 2 γ q e e2/(mc2) b c2 v2 ¿ 1 (8) So IA is valid when: 2 βγ q e re b ¿ 1 (9) The NR limit implies: ∆pe mc ¿ 1 ⇒ 2qe mvcb ¿ 1 ⇒ 2q e e2/(mc2) βb ¿ 1 ⇒ 2 β q e re b ¿ 1. (10) Same as condition for IA to be valid but with γ → 1. 6 Now consider the case when the charge q passes through a bulk material (many e-), let ne be the electron density. We have to add all their respective energy gain to deduce their influence on q’s slowing down. We do not need to consider the nuclei to a good approximation. The total number of e- in a cylindrical shell of radius b and thickness db is: Ne = ne(vdt)(2πbdb) (11) z vb db 7 Note that d∆Tq/dt = dE/dt (E is total energy of q), and (1/v)d/dt = d/dz so we can write: dE dz = −4πne(qe) 2 mv2 ln γ2mv3 qeω0 [JDJ Eq (13.9)]. (19) This equation has been derived under the IA. Compare to Bohr’s results (1915) more carefully derived: dE dz = −4πne(qe) 2 mv2 [ ln 1.123γ2mv2 qe〈ω〉 − 1 2 v2 c2 ] , (20) where 〈ω〉 represents the average angular frequency of the bound electron in taget. The agreement between Bohr’s and the equation we derived is no bad: the IA seems to contain the essential physics. 10 Influence of Dielectric Screening For particles not too relativistic the observed energy loss is accu- rately given by the Bohr’s formula for all kinds of particles in all types of media. For ultra-relativistic particles observed energy loss less than what predicted with Bohr formula ⇒ reduction of energy loss is due to “density” effects. In dense media, dielectric polarization alter the particle’s field com- pared to free-space Problem of finding field in the medium can be solved using the Fourier transform. 11 Consider a dielectric medium, ² = ²(ω), µ = 1. In Gaussian units we have: ¤Aα = 4π c Jα (21) where ¤ ≡ ∂α∂α = ²c2 − ∂2t ∇2. Aα = (Φ, −→ A) and Jα = (ρc/², −→ J ). Define F (−→x , t) = 1√ 2π ∫ +∞ −∞ dωF (−→x , ω)e−iωt F (−→x , ω) = 1√ 2π ∫ +∞ −∞ dωF (−→x , t)e+iωt F (−→x , ω) = 1 (2π)3/2 ∫ +∞ −∞ d −→ k F ( −→ k , ω)e −→ k .−→x F ( −→ k , ω) = 1 (2π)3/2 ∫ +∞ −∞ d −→ k F (−→x , ω)e− −→ k .−→x two first eqns: Fourier transform in time, two last eqns: Fourier transform in space. 12 The electric field is then given by−→ E = −−→∇Φ− 1c ∂A∂t or −→ E ( −→ k , ω) = i( ω c2 ²(ω)−→v −−→k )Φ and −→ B = −→∇ ×−→A → i−→k ×−→A = i²(ω)c −→ k ×−→v Φ. Hence ( −→ E ( −→ k , ω)−→ B ( −→ k , ω) ) = i   ω²(ω) c2 −→v −−→k ²(ω) c −→ k ×−→v   Φ(−→k , ω) (25) We want to find the flow of energy away from the incident particle’s trajectory ⇒ find the Poynting flux ⇒ find −→E (−→x , ω) and −→B (−→x , ω): −→ E (−→x , ω) = i (2π)3/2 ∫ +∞ −∞ d −→ k [ ω c ²(ω) −→v c −−→k ]Φ(−→k , ω)e+i −→ k .−→x (26) Let’s specify the problem: consider −→x = bx̂ and take −→v = vẑ. 15 −→ E (−→x , ω) = i (2π)3/2 2q ² ∫ ∫ ∫ dkxdkydkz [ ω c ²(ω) −→v c −−→k ] δ(ω − kzv) k2 − ²ω2 c2 eikxb = i (2π)3/2 2q ² ∫ ∫ ∫ dkxdkydkz [ −kxx̂− kyŷ + ( ω c v c ²(ω)− kz ) ẑ ] ×δ(ω − kzv) k2 − ²ω2 c2 eikxb. the term in kyŷ has no contribution to the integral of ky. Let’s integrate over dkz: −→ E (−→x , ω) = i (2π)3/2 2q ² ∫ ∫ dkxdky [ −kxx̂ + ω v ( v2 c2 ²− 1 ) ẑ ] × e ikxb k2x + k 2 y + ( ω v )2 (1− ²v2 c2 ) (27) 16 Let λ ≡ ωv √ 1− ²v2 c2 and I ≡ ∫ ∫ dk2 eikxb k2x+k 2 y+λ 2. Then the E-field takes the form: −→ E (−→x , ω) = 1 (2π)3/2 2q ² [ −dI db x̂ + iω v (²β2 − 1)Iẑ ] (28) The integration over dky gives: ∫ +∞ −∞ dky 1 k2x + k 2 y + λ 2 = arctan ( ky√ k2x+λ 2 ) √ k2x + λ 2 ∣∣∣∣∣ +∞ −∞ = π√ k2x + λ 2 (29) So I = π ∫ ∞ −∞ dkx eikxb√ k2x + λ 2 = π ∫ ∞ 0 dkx eikxb + e−ikxb√ k2x + λ 2 = 2π ∫ +∞ 0 dkx cos(kxb)√ k2x + λ 2 = 2πK0(bλ). (30) and dIdb = −2πλK1(bλ). 17 So dEf dz = − b 2 ∫ ∞ −∞ EzBydt = − b 4π ∫ ∞ −∞ dt [∫ ∞ −∞ dωEz(ω)e −iωt ] [∫ ∞ −∞ dω′By(ω′)e−iω ′t ] = − b 2 ∫ +∞ −∞ Ez(ω)By(−ω)dω = − b 2 ∫ +∞ −∞ Ez(ω)B ∗ y(ω)dω = Re ( −b ∫ +∞ 0 Ez(ω)B ∗ y(ω)dω ) (35) Expliciting Ez and By we have: dEf dz = −bRe    ∫ +∞ 0 dω  −i √ 2 π q v ω v (1/²− β2)K0(λb)   ×   √ 2 π q c λ∗K1(λ∗b)      (36) dEf dz = Re {∫ +∞ 0 dω 2 π q2 v2 [iω(1/²− β2)λ∗b]K0(λb)K1(λ∗b) } = 2 π q2 v2 Re {∫ +∞ 0 dω(iωλ∗b)(1/²− β2)K0(λb)K1(λ∗b) } (37) The equation was first derived by Fermi. Note that λ or ² need to be complex to have dEf dz 6= 0. To proceed with our calculation we now need to introduce a model for ²(ω). Use the same model as the one used to study Thom- son Scattering: we model the bound target electron as a damped harmonic oscillator: −→x (ω) = − e m −→ E (ω) ω20 − ω2 − iωΓ (38) 20 The dipole moment is just −e−→x and the polarization is defined as the dipole moment density that is −nee−→x : −→ P (ω) = nee2 m −→ E (ω) ω20 − ω2 − iωΓ = ²(ω)− 1 4π −→ E (ω). (39) So we can write ²(ω) = 1 + ω2p ω20 − ω2 − iωΓ (40) wherein ωp ≡ √ 4πnee2/m is the plasma frequency. Now we just plug this into dEf dz and perform the integral. 21 Two sources of poles: −ω20 + ω2 − iωΓ = 0 from the ln(...), and ω2p + ω 2 0 − ω2 − iωΓ = 0 from denominator of 1−²² . All the poles are in the lower part of the complex plane. Consider the integral along C. This gives: I1 + I2 + I3 = 0 , so I = iI1 = i(−I2 − I3). The i comes from the fact we drop the i when evaluating the integrals In. I1 I3 I2 0 24 Let’s evaluate the integrals: I3 = ∫ 0 +i∞ dωω(...) ln(...) (46) Let ω ≡ iΩ with Ω ∈ R then: I3 = − ∫ 0 ∞ dΩΩ(...) ln(...) = ∫ ∞ 0 dΩΩ ω20 + Ω 2 + ΩΓ ω2p + ω 2 0 + Ω 2 + ΩΓ × ( ln 1.123c bωp − ln iΩ + 1 2 ln(−Ω2 − ω20 −ΩΓ) ) (47) the bracket simplifies: (...) = ln 1.123c bωp − ln i− lnΩ + 1 2 ln−1 + 1 2 ln(Ω2 + ω20 + ΩΓ) (48) 25 the ln i and 1/2 ln(−1) cancel each other. So I3 becomes: I3 = ∫ ∞ 0 dΩΩ ω20 + Ω 2 + ΩΓ ω2p + ω 2 0 + Ω 2 + ΩΓ × ( ln 1.123c bωp − lnΩ + 1 2 ln(Ω2 + ω20 + ΩΓ) ) (49) So I3 is real, so iI3 is pure imaginary and therefore its contribution to ReI is zero. Now consider I2, let ω ≡ Reiθ, then I2 = lim R→∞ ∫ π/2 0 idθReiθReiθ ω2p ω2p + ω 2 0 −R2e2iθ − iReiθΓ × ( ln 1.123c bωp − lnReiθ + 1 2 ln(−ω20 + R2e2iθ + iReiθΓ) ) (50) Cerenkov radiation We now consider density effect in the extreme limit bλ À 1 and look at the energy deposited in the target. The large argument approximation for the modified bessel function gives: K0(bλ) = K1(bλ) = √ π 2 e−bλ√ bλ . So the fields are: −→ E (−→x , ω) = q v e−bλ√ bλ ( λ ² x̂− iω v ( 1 ² − β2)ẑ ) , (54) −→ B (−→x , ω) = q v e−bλ√ bλ ŷ. (55) to get radiation λ or ² ∈ C. Let’s take ² ∈ R (no dielectric screening). Then λ = ω v √ 1− ²(ω)β2 (56) To have λ ∈ I, 1− ²β2 < 0 ⇒ ²β2 > 1, this is the Cerenkov condition. 28 Now replace the field in the expression for dEf dz : dEf dz = q2 v2 Re (∫ ∞ 0 dω(iωλ∗b)(1 ² − β2)K0(bλ)K1(bλ∗) ) (57) = q2 v2 Re   ∫ ∞ 0 dω(iωλ∗b)(1 ² − β2)e −b(λ+λ∗) b √ λλ∗   (58) = q2 v2 Re   ∫ ∞ 0 iω √ λ∗ λ ( 1 ² − β2)   (59) but radiation only when λ ∈ I that is for a frequency band ω ∈ [ωl, ω0]. 29 dEf dz = q2 v2 ∫ ω0 ωL dωω ( 1− 1 ²β2 ) (60) This is Frank-Tamm (1937) equation. 1 e(w) w1 w2 1/b2 qc q vt ct/e1/2 vt ct/e1/2 30 So t− t′ = R(t′)cm = |−→ζ +−→v (t−t′)| cm . ⇒ (t− t′)2 = 1 c2m [ζ2+2 −→ ζ .−→v (t− t′)+v2(t− t′)2]. ⇒ (v2−c2m)(t− t′)2+ 2 −→ ζ .−→v (t− t′) + ζ2 = 0; solve to get (t− t′)± = −−→ζ .−→v ± √ ( −→ ζ .−→v )2 − (v2 − c2m)ζ2 v2 − c2m . (64) For cherenkov radiation v > cm to obtain t − t′ > 0 ∈ R we need:−→ ζ .−→v < 0 and (−→ζ .−→v )2 > (v2 − c2m)ζ2, which means ζv cos theta < 0 or theta > π/2, and cos2 θ > 1− c2m/v2. So θ > arccos(− √ 1− (cm/v)2), (65) which lies in [π/2, π]. 33 So potential and fields exist at time t only within a cone which the apex lies at ζ = −→x −−→t (i.e. the present position of incident charge) and for which the apex angle is π − arccos(− √ 1− (cm/v)2). The 4-potential is Aα = Aα− + Aα+ where the ± corresponds to (t− t′)±. Now, [κR]ret = |(1− 1/cm−→v .n̂) −→ R | = |−→R − n̂ cm −→v .[−→ζ +−→v (t− t′)]| = |−→R − n̂ cm −→ ζ .−→v − n̂ cm v2(t− t′)| = |n̂[cm(t− t′)− −→ ζ .−→v cm − v 2 cm (t− t′)]| = 1 cm |(c2m − v2)(t− t′)− −→ ζ .−→v |. (66) 34 Expliciting (t− t′) in the latter equation (using 64), we get: [κR]ret = ζ cm √ c2m − v2 sin2 θ = ζ √√√√1− v 2 c2m sin2 θ. (67) both for (t− t′)±. So the potentials are given by: ( √ ²Φ(−→x , t)−→ A(−→x , t) ) = 2q√ ² 1 ζ √ 1− v2 c2m sin2 θ ( 1−→v cm ) (68) The potentials have a singularity (a hock front) at sin2 θ = (cm/v)2, which corresponds to the earlier results cos2 θ = 1− (cm/v)2. Note that when the frequency-dependence of ² is introduced the shock wave-front is smeared. 35 There are Z times more e- than nuclei, so the net effect is that nuclei scattering is Z times stronger than e- scattering. Average deflection angle in a material: To get the mean-square deflection angle, evaluate: 〈θ2〉 = ∫ dΩθ2dσ/dΩ∫ dΩdσ/dΩ ' ∫ dθθ31/θ4∫ dθθ1/θ4 (74) = ∫ θmax θmin dθ1/θ ∫ dθ1/θ3 = ln θmaxθmin 1 2(1/θ 2 min − 1/θ2max) (75) So 〈θ2〉 ' 2θ2min ln θmaxθmin for a single scattering event. This is just few times θ2min which is a small number. 38 Let’s estimate θmin from physical arguments: bmax ' a, the atomic radius because atomic electrons almost completely screen the nu- cleus if b > a. So θmin = 2qZe γbmaxMv2 ' 2qZe γaMv2 (76) ∼ e 2 ampc2 ∼ e 2/(mec2) ampc2 ∼ re 1836a ¿ 1 (77) So to achieve a sizeable deflection angle, the incident charge needs either to undergo manu small-angle scattering of a few large-angle scattering. • Case of many small-angle scattering: Net effect: charge q random-walk through the target 〈Θ2〉 = N〈θ2〉 and: ⇒ d〈Θ 2〉 dz = nσ〈θ2〉 ' 2nσθ2min ln(θmax/θmin) (78) 39 The distribution of angle after many small-scattering event (random- walk) is given by: PRW (θp) ∝ e −θ2p 2〈Θ2p 〉. • Case of few large-angle scattering: Consider the distribution of scattering angle for a single scattering event: dσ dΩ dΩ = ( 2qZe γMv2 ) 1 θ4 dφθdθ (79) In terms of projected angle θp = θ sinφ, this takes the form: dσ dΩ dΩ = ( 2qZe γMv2 ) 1 θ3p dθp sin 2 φdφ (80) Upon integration over φ we find that the distribution scales as: P1(θp)dθp ∝ dθp θ3p ↔ P1(θp) ∝ 1 θ3p 40