Explore the generation of X-rays in a Coolidge tube, quantum effects from releasing core electrons, and radiation during collisions of particles. Understand the non-relativistic and relativistic limits in radiative emissions during collisions with different polarizations and frequencies.
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Presentation Transcript
PHY 712 Electrodynamics 9-9:50 AM MWF Olin 103 Plan for Lecture 32: Start reading Chap. 15 Radiation from collisions of charged particles 1. Overview 2. X-ray tube 3. Radiation from Rutherford scattering 4. Other collision models 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 1
Generation of X-rays in a Coolidge tube https://www.orau.org/ptp/collection/xraytubescoolidge/coolidgeinformation.htm 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 4
Radiation during collisions r (t) (t+ t) Intensity: 2 ( ) r r r 2 2 d I d d q ( r d dt ( ) ( ) t c r R / i t = dt e q 2 4 1 c ) ( = ) ( ) + r Note that For a collision of duration emitting radiation with polarization and frequency 0: 2 ( ) ( ) t r + t 2 2 d I d d q = ( ) ( ) t + 2 r 4 1 1 c t 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 6
Radiation during collisions -- continued For a collision of duration emitting radiation with polarization and frequency 0: 2 ( ) ( ) t r + t 2 2 d I d d q = ( ) ( ) t + 2 r 4 1 1 c t Non-relativistic limit: 2 2 d I d d Relativistic collision with small q ( ) ( ) ( ) ) 2 = + t t 2 4 c ( ( ) t + : t 2 ( ) + r 2 2 r d I d d q = ( ) 2 2 4 c 1 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 7
Radiation during collisions -- continued Relativistic collision with small : 2 ( ) + r 2 2 r d I d d q r = ( ) 2 2 4 c 1 Also a ssume is perpendicular t o plane r Expressions (averaging over ) for || or polarization: ( 1 ) 2 2 polarization in r and plane d I d d cos cos 2 q 2 || = ( ) 4 2 8 c polarization perpendicular to r and plane 2 2 1 cos d I d d q 2 = ( ) 2 2 8 c 1 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 8
Some details: sin = = z = + r x z cos + y x z = cos = + r cos sin ( ) + x y sin = ( ) ( ) ( ) + r r 1 ( ) ( ) ( ( ) + = r sin 1 cos ( ) ( ) ) + = r c os c s o 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 9
Some details: r = = = + = + cos x z = sin cos y z r + z sin x x ( ) y cos sin 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 10
z Radiation during collisions -- continued Intensity expressions: ( 1 ) 2 2 d I d d cos cos 2 q 2 || = r ( ) 4 2 8 c y 2 2 1 cos d I d d q 2 = ( ) 2 2 8 c 1 x Relativistic collision at low and with small and perpendicular to plane of and , as a function of where Integrating over solid angle: = + r = cos ; r 2 d I d d 2 2 2 d I d d dI d q c 2 || = 2 d 3 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 11
Estimation of q q (t) (t+ t) Momentum transfer: + p ( ) ( ) t c 2 p Qc t Mc mass of particle having charge q 2 2 2 2 dI q q 2 = 2 2 Q 2 3 3 3 d c M c 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 12
Estimation of -- for the case of Rutherford scattering q q (t) (t+ t) Ze Assume that target nucleus (charge Rutherford scattering cross-section: ) has mass M; Ze 2 2 1 d d Zeq pv = p ( ) ( ) 4 2sin '/ 2 Q Assuming elastic scattering: ( ) ( ) ( ) p 2 = = 1 cos ' 2 2 2 sin p '/ 2 2 Q p 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 13
Case of Rutherford scattering -- continued Rutherford scattering cross - section : q 2 2 1 d Zeq = Q ( ( ) ) 4 Ze d pv 2 sin / ' 2 d d d = ' d dQ d dQ ( ) ( ) ( ) 2 = = 1 cos ' 2 2 2 sin p '/ 2 2 Q p 2 1 d dQ Zeq = 8 3 c Q 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 14
Case of Rutherford scattering -- continued q Q Ze Differenti radiation al cross section : 2 2 2 2 1 d dI d q Zeq = = 2 8 Q 2 3 3 3 d dQ d dQ M c c Q ( ) 2 2 2 16 1 1 Ze q = 2 2 3 c Mc Q 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 15
Differenti radiation al cross section continued - - Integratin momentum over g transfer ( ) 2 Q 2 2 2 16 1 Q d d Ze q max = = max ln dQ 2 2 3 d d dQ c Mc Q min Q min Comment on frequency dependence -- expression Original radiation for intensity : ( ) 2 1 2 2 r r d I q d ( ) ( ) t i r R / t c = dt e q that d 2 r 4 d c dt In the t previous derivation c assumed have we s, ( ) ( ) t r R / . 1 q t ( ) ( ) ( ) t ( ) t 0 = r r r R / ' ' 1 t c t dt q PHY 712 Spring 2017 -- Lecture 32 04/10/2017 16
Differenti radiation al cross section continued - - Radiation cross section in terms of momentum transfer ( ) 2 Q 2 2 2 16 1 Q d d Ze q max = = max ln dQ 2 2 3 d d dQ c Mc Q min Q min ( ) = = 2 2 Note that: In general, 1 cos ' is determined by the collision time 2 1 Q v 2 2 Q Q p Q p max min Zeq condition min 2 Radiation cross section for classical non - relativist ic process ( ) 2 2 d Zeq 2 3 16 1 Ze q Mv = fudge factor of order unity = ln d 2 2 3 c Mc 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 17
Electromagnetic effects in energy loss processes (see Chap. 13 of Jackson) Again consider Rutherford scattering now of a nucleus (or alpha particle ze incident on an electron e in rest frame of electron: ze Rutherford scattering cross-section: 2 Q 2 1 d d ze -e = ( ) ( ) 4 2 pv sin '/ 2 d dQ 2 = d d d dQ ) ( = ' d 2 2 ( ) ( ) = 1 cos ' 2 2 2 sin p '/ 2 2 Q p 2 2 d dQ ze cQ = 4 2 2 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 18
Energy loss continued Let represent energy loss due to electron of m / 2 2 dT mc ass : T m = 2 T Q m 2 4 e T d z = 2 2 2 Estimate of energy loss per unit distance in the presence of NZ electrons per unit volume max T dE dx d dT NZ dTT minimum energy transfer + 2 4 e 2 2 2 2 z mc = 2 ln (quantum effects) NZmc 2 2 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 19
Energy loss continued Refining this result, Bethe and Fermi noticed that the analysis lacked consideration of the effects of electromagnetic fields. Representing the colliding electrons in terms of a dielectric function ( ) and the energetic particle of charge ze in terms of the charge and current density: In Fourier space: ze = J k v k 2 4 ( = ( , ) k 2 ( , ) k ( ) k 2 ) c 2 4 = 2 ( , ) A k ( , ) J k ( ) k 2 c c = v k ( , ) k ( ) 2 ( , ) ( , ) 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 20
v k 2 ( ) ze Energy loss continued = ( , ) k ( ) 2 ( ) 2 k 2 c v = ( ) ( , ) A k ( , ) k c The energy loss will be calculated from the work on the electron by the field: = = ( ) * v E r E ( ) 2 ( ) E e dt t e d i 0 The resultant loss estimate is 2 2 e 2 P 2 2 2 4 dE d z mc NZ m e 2 P ln where 2 2 2 P 2 x c 04/10/2017 PHY 712 Spring 2017 -- Lecture 32 21
