Classical Electrodynamics I - SNUphysics.snu.ac.kr/php/subject_list/Notice/data/1399441208.pdfSeoul National University Classical Electrodynamics I Department of Physics & Astronomy

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



Perfect conductor case

xtkzEE ˆ)(cos0

ytkzBB ˆ)(cos0

After normal incidence reflection:

tkzxEtkztkzxEE sinsinˆ2)](cos)([cosˆ00

tkzyBB coscosˆ2 0

22 k

Perfect conductor

x

y

z

E

B

k

0

8. Waveguides, Resonant Cavities, and Optical Fibers Day 21

못들어옴



Boundary condition for the perfect conductor

0// E KHn

ˆ

K

: Surface current

cf) previous lecture: Quasi-Static Magnetic Fields in Conductors (section 5.18),

z

etz

yHH

)(cosˆ0

z

etz

yHE

)4

(cos2

2ˆ

0

phase difference.

For the perfect conductor case,

0)(ˆ cEEn

0)(ˆ cBBn

Where cc BE

, are the fields inside the conductor.




BiE

EiB

0 B

0 E 022

B

E

Because of the cylindrical geometry it is useful to single out the spatial variation of

the fields in the z direction and to assume

tiikzeyxEtzyxE ),(),,,(

tiikzeyxBtzyxB ),(),,,(

In general,22 k

8.2 Cylindrical Cavities and Waveguides




Assuming propagation in the positive z direction and the nonvanishing of at

least one of Ez and Bz , the transverse fields are

ztztt BzEkk

iE

ˆ

)( 22

ztztt EzBkk

iB

ˆ

)( 22

pf) Suppose that the z dependence is given by

0)( 222

B

Ekt

2

222

zt

tiikzeyxEtzyxE ),(),,,(





It is useful to separate the fields into components parallel to and transverse

to the z axis:

tz EEE

where

zz EzE ˆ

zEzEtˆ)ˆ(

The Maxwell equation can be written out in terms of transverse and parallel

components as

BiE

EiB

0 B

0 E

zttt EBzi

z

E

ˆ

zttt BEzi

z

B

ˆ

ztt BiEz

ˆ

ztt EiBz

ˆ

z

EE z

tt

z

BB z

tt




BiE

양변 x z tBziBziEz

ˆˆ)(ˆ

?)(ˆ Ez

)()()()()( abbaabbaba

)(ˆ,ˆ Ezz

EEEbzaLet z

tt

ztz Bziz

EE

z

EEEzTherefore

ˆ)(ˆ,



It is evident from the previous equations that if Ez and Bz are known the

transverse components of E and B are determined.

zttt EBzi

z

E

ˆtiikzeyxEtzyxE ),(),,,(

zttt EBziEik

ˆ

zttt BEziz

B

ˆzttt BEziBik

ˆ

zttztt EEziBik

ziEik

ˆ1

ˆ

tztt EziBik

B

ˆ1


- ①

- ②

From ②

With ①,




The result, after some simplifications, is

ztztt BzEkk

iE

ˆ

)( 22

ztztt EzBkk

iB

ˆ

)( 22

ztztt Bzk

E

kiik

E

ˆ1

2

Likewise,

ztztt Bzk

Ek

iikE

ˆ

2




x

y

z

E

B

)(cosˆ0 tkzxEE

)(cosˆ0 tkzyBB

22 k

Ex) TEM mode

02 tE

02 tB

This means that ETEM is a solution of an electrostatic problem in two dimensions.

It is necessary to have two or more cylindrical surfaces to support the TEM mode.

An important property of the TEM mode is the absence of a cutoff frequency.




Transverse Magnetic (TM) waves

0zB everywhere; boundary condition, 0SzE

Transverse Electric (TE) waves

0zE everywhere; boundary condition: 0

S

z

n

Bn: surface normal

)( 22 k

EikE zt

t

)(

ˆ22 k

EziB zt

t

)(

ˆ22 k

BziE zt

t

)( 22 k

BikB zt

t


0

S

z

x

BCan be deduced from Ey =0 on y=0

surface plus Bn =0, in--------------------- zttt BEzi

z

B

ˆ



ztztt BzEkk

iE

ˆ

)( 22

ztztt EzBkk

iB

ˆ

)( 22

It is found from

that the transverse magnetic and electric fields for both TM and TE waves are

related by

tt EzZ

H ˆ1


8.3 Waveguides



TM waves

)( 22 k

EikE zt

t

)( 22 k

BikH zt

t

TE waves

)(

)(

0

0

TEk

k

k

TMk

kk

Z

Where Z is called the wave impedance and is given by




0)( 22

2

2

2

2

zHk

yx

TE mode

0)( 222

B

Ekt

0

S

z

n

Hat x=0, a and y=0, b

b

yn

a

xmHyxH z

coscos~),( 0


8.4 Modes in a Rectangular Waveguide



If a>b, the lowest cutoff frequency, that of the dominant TE mode, occurs for

m=1, n=0:

a

0,1

22

22

2

mnmnb

n

a

mk

Cutoff frequency

2/122

b

n

a

mmn




This corresponds to half of a free-space wavelength across the guide. The

explicit fields for this mode are

tiikz

z ea

xHH

cos0

tiikz

x ea

xH

ikaH

sin0

tiikz

y ea

xH

aiE

sin0


22

22

2

mnmnb

n

a

mk

ztztt BzEkk

iE

ˆ

)( 22

ztztt EzBkk

iB

ˆ

)( 22



We note that, for mn ,the wave number is real: waves of the mode can propagate in the guide.

For frequencies less than the cutoff frequency , the wave number is imaginary;

such modes cannot propagate and are called cutoff modes or evanescent modes.

n

c

kv

mnmn

p

1

1

1

2

2

Note that

Vp : phase velocity n: refractive index




pv

gvv

gpvvv 2

dkkd )2()2(

n

cvk

dk

dv mnmn

mn

g

11

12

2

vg : group velocity




Hollow cylinder case

TM

0zE 0)( 222 zt Ek

0)(1 22

2

2

2

2

k

d

d

d

dm

Solution: Bessel function

0)(1 22

2

2

2

2

k

d

d

d

dm

im

z eE )(~),(

Rz




222 k

0)(1

2

22

2

2

m

d

d

d

d

Assume m=0,

tiikz

z eJEE )(00

Boundary Condition 0)( rEz

...)654.8

(),520.5

(),405.2

(,0 000 R

JR

JR

Jm

.....654.8520.5405.2

030201RRR




tiikz

z eR

JEE )405.2

(00

tiikzeR

JHH )405.2

(ˆ10

Take the lowest zero,

1405.21,0

R

Cut off frequency

2

22 405.2

Rk

Dispersion relation




Because of reflections at the end surfaces, the z dependence of the fields is that

appropriate to standing waves:

zkBzkA cossin

end surfaces

z

0 d

The boundary conditions can be satisfied at each surface only if

...),2,1,0( pd

pk


8.7 Resonance Cavities



For TM fields the vanishing of Et at z=0 and z=d requires

...),2,1,0(cos),(

p

d

zpyxEz


From the cylinder waveguide result,

...),2,1,0(cos)(0

p

d

zp

R

xJeEE mn

m

im

z

mnxWhere are the n-th roots of 0)( xJm

2

22

2

21

d

p

R

xmnmnp



The lowest TM mode has m=0, n=1, p=0

R

405.2010

ti

z eR

JEE )405.2

(00

tieR

JEiH

)405.2

(10




earth

ionosphere

요사이의공명은아마도빛이지구를7바퀴반도는것과관련!?;

7.5 Hz?

http://sedonanomalies.weebly.com/schumann-resonance.html





이건조은데...; air index of refraction과관계된문제를내봐?

n_air=1.00029; cf) n_water=1.33



Eikonal equation

nn ||

Assume

t

BE

t

DH

)(x

0)();()(

ExDxx

EE

0))()(

()(2

2

22 x

x

EExn

cE

Slowly varying




0)(2

2

22 Exn

cE

0)(2

2

22

xn

c

If n is constant, xkn

ci

e

ˆ

For slowly varying n, )(xS

ci

e

Substitution gives,

0)( 22

2

2

Sc

iSSxnc

SSxn

)(2

)(ˆ)()( xkxnxS

)(ˆ xk

is a unit vector in the direction of )(xS




The form of is that of a plane wave with wave vector, ).(ˆ xk

)()( xSds

rdrn

ds

rdk

ˆ

ds

dSxS

ds

d

ds

rdrn

ds

d

)()(

But ,ˆ

kds

dso that )()(ˆˆ rnrnkk

ds

dS

Finally, we obtain

r

Cnrn

ds

rdrn

ds

d

4);()( 원운동
































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Classical Electrodynamics I - SNUphysics.snu.ac.kr/php/subject_list/Notice/data/1399441208.pdfSeoul National University Classical Electrodynamics I Department of Physics & Astronomy