classical electrodynamics i - dept. of physics &...

Seoul National University Classical Electrodynamics I

Department of Physics & Astronomy Dai-Sik Kim

John David Jackson

Dai-Sik Kim

Nano Optics Lab.

School of Physics and Astronomy

Seoul National University

Classical Electrodynamics I



above resonance condition

below resonance condition

Mechanical Spring

7. Plane Electromagnetic Waves and Wave Propagation Day 20



tieeExxxm 0

2

0 )( 0

e

tieEE 0

)( 22

0

0

im

eEx

)( 22

0

0

2

im

Eeexp

0022

0

0

2

)()(

Eim

ENeNexP

i

p

22

0

2

0 1m

Nep

0

22


where



22222

0

2

022222

0

22

0

2

022

0

2

0)()(

)(11

pppi

i

22222

0

22

0

2

0)(

)(1

p

r

22222

0

2

0)(

p

i




2

2

0 1

p

For 0

If we consider the bound charge effect,

2

2

002

2

6.13)(

pp

b

Why?

nn

ppp

iii

n

22

0

2

2

22

02

2

1

22

01

2

0211

If we consider the near resonance frequency, all other contributions are

small except n0

All other contributions are included in the 13.6 factor

2

2

0 6.13

p

i

p

22

0

2

0 6.13High frequency limit


여기서 high frequency란진정한 의미의 high frequency,즉 X, gamma-ray가 아니라Omega_0보다 크고 다른 주요공명보다 작은; ‘비겁한’ 의미의High frequency.



00 In the case of but no free electron, and therefore no J.

i

p

b

2

2

0

EDEJ b

;

EiiEiEH bb

t

DJH

ii

p

bb

2

2

0

i

p

2

0


t

E

t

DH



),()(),( xExD

dexDtxD ti),(2

1),(

Fourier transformation

')',(

2

1),( 'dtetxDxD ti

dexEtxD ti),()(2

1),(

,')',(2

1),( '

dtetxExE ti

with

')',(')(2

1),( titi etxEdtedtxD


5.9 Causality in the connection between D and E

The substitution of gives,),( xD



),()(),(),( 0 txEGdtxEtxD

iedG 1

)(

2

1)(

0

where

Application

i

p

22

0

2

0

1

i

ededG

ipi

22

0

2

0 21

)(

2

1)(

With the help of the residue theorem, you can calculate

022

0 i24

22

0

i




)(4

sin

4

)(2

2

02

2

0

2/2

eG

p

: Heaviside step function)(

)(




* Kramers – Kronig Relation

'

'

1/)'(

2

11

)( 0

0

d

zi

z

C

Cauchy’s theorem gives,

'

'

1/)'(

2

11

)( 0

0

d

zi

z

No contribution from the great semi circle

'

'

1/)'(

2

11

)( 0

0

d

ii)'(

'

1

'

1

iP

i

By using the principal part,




'

'

1/)'(11

)( 0

0

dP

i

0 22

00

0

''

/)'(Im'21'

'

/)'(Im11

)(Re

dPdP

0 22

00

0

''

1/)'(Re2'

'

1/)'(Re1)(Im

dPdP


감동적인 결과 하나

)'(2

)'(Im 0

0

K

22

0

0)(Re

K



Classical Electrodynamics 1

10. (25 points) This problem concerns the Kramers-Kronig relationship and causality.

(a) (10 points) Derive the Kramers-Kronig relations.

(b) (5 points) For a strong absorption line with Im ε ω′ =πK

2ω0𝛿 𝜔′ − 𝜔0 , show that

Re 휀 𝜔′ = 휀0 +𝐾

𝜔02 − 𝜔′2

(c) (5 points) Now, instead of the Kramers-Kronig relationship, suppose we wrongly assumed that Im ε ω′ =

πK

2ω0𝛿 𝜔′ − 𝜔0 , and Re 휀 𝜔′ = 휀0; ε 𝜔′ = 휀0 + 𝑖

πK

2ω0𝛿 𝜔′ − 𝜔0 . Show that, for a pulsed excitation

E t = 𝐸0𝜏𝛿(𝑡)𝑒−𝑖𝜔𝑡 , this dielectric constant gives the following unphysical result for the polarization:

P t =𝐾

2ω0𝑠𝑖𝑛(𝜔0𝑡). In other words, we have polarization at t<0, even before the pulse arrives at t=0.

Obviously, you need the real part of the dielectric constant to have the right form to cancel the polarization at

t<0.

(d) (5 points) From causality, can you guess that 𝑑𝜔𝑒−𝑖𝜔𝑡

𝜔02−𝜔2 ∝ θ t − θ −t 𝑠𝑖𝑛(𝜔0𝑡)? Can you prove it by

the contour integral-Cauchy theorem?



Day 21

End of Chapter 7

classical electrodynamics i - dept. of physics &...

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