Evaluation of Kramers-Kronig Relation and Dielectric Function Properties

phy 712 electrodynamics 9 9 50 am mwf olin 103 n.w
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Explore the practical evaluation and analytic properties of the dielectric function through the Kramers-Kronig transform in electrodynamics. Review equations, examples, and singular integral evaluations for better understanding.

  • Kramers-Kronig
  • Dielectric Function
  • Electrodynamics
  • Analytic Properties
  • Singular Integral
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  1. PHY 712 Electrodynamics 9-9:50 AM MWF Olin 103 Plan for Lecture 36: Review Part II: Further comment of Kramers-Kronig transform Some equations for top of your head Example problems Course evaluation forms 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 1

  2. 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 2

  3. 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 3

  4. Review topic analytic properties of dielectric function Material from Chapter 7 in Jackson The displacement field D is related to the electric field E = = + D E E P 0 = + Dielectric f unction ( ) ( ) ( ) i R I can be shown to be analyic for f or ( ) 0 z z Kramers-Kronig transform for dielectric function: ( ) 1 ( ) 0 1 ' 1 = 1 ' R I P d - ' 0 ( ) with ( ) ; ' 1 = d ' 1 I R P - = ' 0 0 ( ) ( ) ( ) ( ) = R R I I 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 4

  5. Practical evaluation of Kramers-Kronig relation ( ) 1 ( ) 0 1 ' 1 = 1 ' R I P d - ' 0 ( ) with ( ) ; ' 1 = d ' 1 I R P - = ' 0 0 ( ) ( ) ( ) ( ) = R R I I ( ) ( ) ( ) = ( ) = Let R I 1 2 0 0 ( ') ( ') 1 2 ( ) 1 = = 2 2 P d P d 1 2 2 0 2 ( ') 1 ( ') 1 1 ( ) = = 1 1 P d P d 2 2 2 0 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 5

  6. Practical evaluation of Kramers-Kronig relation ( ') 1 ( ) 1 = 2 P d 1 0 ( ') ( ') 1 + = 2 2 P d d 0 ( ') ( ') + 1 + = 2 2 P d d 0 0 Singular integral can be evaluated numerically: 0 0 W ( ') ( ) ( ') ( ') W = + ( )ln + 2 2 2 2 P d P d d 2 W 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 6

  7. Evaluation of singular integral numerically: W ( ') ( ') ( ) ( ') W = + ( )ln + 2 2 2 2 P d P d d 2 0 0 W 2( ) ( ) ( ') 2 2 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 7

  8. Evaluation of Kramers Kronig transform using Mathematica (with help from Professor Cook) 2( ) 1( ) 1 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 8

  9. Another example 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 9

  10. Some equations worth remembering -- 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 10

  11. = = P M SI units; Microscopic or vacuum form ( Coulomb's law: 0; 0): = E / B 0 E 1 c = B J Ampere-Maxwell's law: 0 2 t 0 + = E Faraday's law: 0 t = B No magnetic monopoles: 1 = 2 c 0 0 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 11

  12. 1 = + = D E P 0; = H B M SI units; Macroscopic form ( ): 0 0 = 0 = D Coulomb's law: free D = H J Ampere-Maxwell's law: 0 free t B + = E Faraday's law: 0 t B No magnetic monopol es: 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 12

  13. = + = D E P 0; = H B M Gaussian units; Macroscopic form ( Coulomb's law: 4 4 ): = D D 4 free 1 c 4 = H J Ampere-Maxwell's law: free t B c 1 c 0 + = E Faraday's law: 0 t = B No magn etic monopoles: 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 13

  14. Energy and power (SI units) 1 2 ( ) u + E D H B Electromagnetic energy density: S E H Poynting vector: Equations for time harmonic fields : ( ) ( ) 1 ~ ~ ~ = ( E r + * i t i t i t E r E r E r E r ( ,t) ( , ( )e ( , )e ( , )e 2 * ( , ) D r ) 1 4 ) ( ) = ( , ) + ( , ) B r ( , ) r * r H u ,t avg t ( ) 1 2 ( ) ( ) = ( , ) E r ( , ) H r * S r ,t avg t 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 14

  15. Solution of Maxwell's equations: E 1 c = = E B J / 0 0 2 t B + = = E B 0 0 t Introduction of vector and scalar potentials: 0 = + = + = = B B A A B + = E E 0 0 t t A A = E E or t t 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 15

  16. Scalar and vector potentials continued: / : = = E 0 ( ) A = 2 / 0 t E 1 c B J 0 2 t ( ) 2 A 1 c ( ) + + = A J 0 2 2 t t 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 16

  17. Analysis of the t scalar vector and potential equations : ( ) A = 2 / 0 ( ) t 2 A 2 t 1 c ( ) + + = A J 0 2 1 c + = A Lorentz gauge form - - require 0 L L 2 t t 2 1 c + = 2 / L 0 L 2 2 2 A t 1 c + = 2 A J L 0 L 2 2 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 17

  18. Solution methods for scalar and vector potentials and their electrostatic and magnetostatic analogs: 1 / L c t + = In your bag of tricks: Direct (analytic or numerical) solution of differential equations Solution by expanding in appropriate orthogonal functions Green s function techniques 2 + = 2 L 0 2 2 2 A 1 c 2 A J L 0 L 2 2 t 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 18

  19. How to choose most effective solution method -- In general, Green s functions methods work well when source is contained in a finite region of space Con = sider the electrostatic problem: / L G r 2 0 r = 2 3 r r Define: ( , ) ' 4 ( ) ' 1 = ( ) ( , ) G r + 3 ( ) r r r d r L 4 V 0 1 2 r ( , ) r r r ( ) r ( , ) r r ( ) . d r G G 4 S 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 19

  20. r For electrostat ic problems where ( ) is contained in a small 1 , ( , ) G = r r r r region of space a nd S ' l 1 4 r ( ) ( ) lm * = l ' ' Y Y , , lm lm + + 1 r r ' 2 1 l r 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 20

  21. Electromagnetic waves from time harmonic sources ( ( continuity that the Note r J t t ( ( ) ) ) ) ( ( ) ) ~ i = t r r Charge density : , , t e ~ J i = t J r r Current density : , , t e condition : ( ) r , t ~ J ( ) ( ) ( ) ~ + = i + = r r , 0 , , 0 )and ( r J r For dynamic problems wh e re ( , , ) are , contained in a small region of space and S r r ' ic e r r = ( , ', ) G r r ' 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 21

  22. Electromagnetic waves from time harmonic sources continued: c For scalar potential (Lorentz gauge, ) k r r ' ik 1 e ~ ~ ( ) ( ) ( ) ~ = + 3 r r , ' r , , ' d r 0 r r 4 ' 0 For vector potential (Lorentz c gauge, ) k r r ' ik ~ A ~ A e ~ J ( ) ( ) ( ) = + 3 r r , ' r 0 , , ' d r 0 r r 4 ' 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 22

  23. Electromagnetic waves from time harmonic sources continued: Useful expansion : r r ' ik e ( ) ( h ) ( ) r ( ) ' r = * ik j kr kr Y Y lm l l lm r r 4 ' lm ( ) ( ) kr h l Spherical Bessel function : j kr l ( ) kr ( ) kr = + Spherical Hankel function : j in l l , r ~ 0 ) ~ ~ ( ) ( ( ) ( ) lm Y lm = + r r , , r lm ~ ik ( ) ( ) ( j ) ( h ) ( ) ' r ~ = 3 * , ' r , ' r d r kr kr Y lm lm l l 0 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 23

  24. Radiation from a moving charged particle Variables (notation) : ( ) r t R d ( ) r t R q v q dt z r ( ) r t ( ) r t R r R R q v. r-Rq(tr)=R q r Rq(t) 2 dP q 2 = 2 v sin 3 4 d c y x 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 24

  25. Linard-Wiechert potentials (Gaussian units) ( ) q r q r r dt t t R r R R R d t ( ) t R v ( ) r ( ) r q 2 v v v q v R R c v c R c = + E ( , ) r R R R 1 t 3 2 2 c R c R v v R 2 v R R v / q c v c c = + B ( , ) r 1 t 3 2 2 2 c v R R R c c R E r ( , ) R t = B ( , ) r . t 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 25

  26. Example: Problem 15.2 in Jackson: A nonrelativistic particle of charge e and mass m collides with a fixed, smooth, hard sphere of radius R. Assuming that the collision is elastic, show that in the dipole approximation (neglecting retardation effects) the classical differential cross section for the emission of photons per unit solid angle per unit energy interval is: 2 2 2 2 1 d d R e v c ( ) = + 2 2 3sin ( ) 12 d c where is measured relative to the incident direction. 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 26

  27. z r v x v Suppose that ' sin cos = = = = = v v r z v ( ) + + x y z z sin cos + x sin sin a cos v a b b a cos y 1 + x z sin 2 2 v v Cross section depends on ( ) i 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 27

  28. Low frequency radiation from charged particle during a collision as analyzed by Eq. 15.2 : 2 2 I d c d r r I d d d d d 2 2 d e e ( ) 2 = v v ' 2 2 3 c 4 1 1 4 2 2 2 2 d d d e R ( ) 2 = = v v ' ( ) 2 3 c 4 4 ( ) = = + + v z v x y z For: ' sin cos a sin sin a cos v v b b a ( ) ( ) = + + v v x y z ' sin cos a sin sin a cos 1 v b b a 4 ( ) 2 2 = = = 2 2 v v y ( ) cos sin sin a v d b da b v 1 1 3 = + x z cos sin 2 4 ( ) ( ) 2 2 = cos si + (cos = + 2 2 2 2 v v ( ) cos n cos a sin 1) c s o 4si n v d b da b a v 2 3 2 2 2 e v R 2 31 cos + d ( ) = + 2 2 4sin ( ) d d 12 c 04/24/2015 PHY 712 Spring 2015 -- Lecture 36 28

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