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Classical Electrodynamics

Chapter 4Multipoles, Electrostatics of

Macroscopic Media, Dielectrics

A First Look at Quantum Physics

2011 Classical Electrodynamics Prof. Y. F. Chen



Contents

§4.1 Multipole Expansion §4.2 Multipole Expansion of the Energy of a Charge Distribution

in an External Field§4.3 Elementary Treatment of Electrostatics with Ponderable Media§4.4 Boundary-Value Problems with Dielectrics §4.5 Molecular Polarizability and Electric Susceptibility§4.6 Models for Electric Polarizability§4.7 Electrostatic Energy in Dielectric Media



§4.1 Multipole Expansion

It is just the expansion result for under spherical coordinates.

Localized charge distribution

( ')x

'x

x

O

( ')x x

The potential given by a localized charge distribution ( ') is x

3

0

1 ( ') '( )4 | ' |

x d xxx x

Similarly, we can use Taylor expansion to expand under Cartesian coordinates. 2

22

1( ) ( ) | ( ) | ( ) ......2!x a x a

df d ff x f a x a x adx dx

Let | | , ' | ' |r x r x

According to addition theory for spherical harmonics,

Consider r>r’, we know that

12 2 2 1/ 2

1 1 1 ' (cos )' '| ' | ' 2 'cos [1 ( ) 2( ) cos ]

l

lll

r Pr r rx x r r rr rr r

*4(cos ) ( ', ') ( , )2 1

l

l lm lmm l

P Y Yl

*1

0

1 4 'Thus, ( ', ') ( , )2 1| ' |

ll

lm lmll m l

r Y Yl rx x

1/ | ' |x x

1/ | ' |x x

※Taylor expansion for single variable function:

(expand around x=a)


Consider vector field :2

0 0,

1( ) ( ) | | ......2! i jx x

i i ji i j

f ff x a f a x x xx x x

If we expand 1/ | ' | at ' 0x x x

2

' 0 ' 0,

1 1 1 1 1 1( ) | ' ( ) | ' ' ......' 2! ' '| ' | | ' | | | | ' | | ' |i i jx x

i i ji i j

x x xx x xx x x x x x x x x

2 2 2

1 1( ) ( )' '| ' | ( ') ( ') ( ')i ii ix xx x x x y y z z

1 22 2 2 3/ 2 2 2 2 3/ 2 2 2 2 3/ 2

( ') ( ') ( ')ˆ ˆ[( ') ( ') ( ') ] [( ') ( ') ( ') ] [( ') ( ') ( ') ]

x x y y z za ax x y y z z x x y y z z x x y y z z

2 2 2 2 2

11 12 135 5 5,

1 3( ') [( ') ( ') ( ') ] 3( ')( ') 3( ')( ')( ) | | |' ' | ' | | ' | | ' | | ' |i j i j

x x x x y y z z x x y y x x z zx x x x x x x x x x

2 2

21 22 235 5 5

3( ')( ') 3( ') | ' | 3( ')( ')| | || ' | | ' | | ' |

y y x x y y x x y y z zx x x x x x

2 2

31 32 335 5 5

3( ')( ') 3( ')( ') 3( ') | ' || | || ' | | ' | | ' |

z z x x z z y y z z x xx x x x x x

22,

3 5' 0 ' 0, ,

31 1From above discussion, we know that ( ) | ; ( ) |' ' '| ' | | ' |

i j i jix x

i i i j i ji i j

r r rrx r x x rx x x x

2,

3 5,

31 1 1Therefore, ' ' ' ......2!| ' |

i j i jii i j

i i j

r r rr x x xr r rx x

(The result under Cartesian coordinates)


For a charge distribution , the scalar potential observed at can be regarded as the contributions from different parts of multipole expansion.

( ')x

x

23, 3

3 5,0 0

31 ( ') ' 1 1 1( ) ( '){ ' ' ' ...} '4 4 2| ' |

i j i jii i j

i i j

r r rrx d xx x x x x d xr r rx x

2,3 3

2 5,0

31 1 1ˆ{ ( ') ' ' ( ') ' ' ' ...}4 2

i j i ji i i j

i i j

r r rq r x x d x x x x d xr r r

3 3 3Define: ( ') ' (monopole) ; ( ') ' ' (dipole) ; 3 ' ' ( ') ' (quadrupole) i i ij i jq x d x p x x d x Q x x x d x

2,

,2 5,0

(3 )1 1 1ˆTherefore, ( ) { ... ..}4 6

i j i ji i i j

i i j

r r rqx p r Qr r r

Furthermore, from the symmetry of Qi,j we know that ij jiQ Q

11 12 13

21 22 23

31 32 33

Only 6 independent componentsij

Q Q QQ Q Q Q

Q Q Q

It is a reducible tensor under Cartesian basis representation.

We can use proper transformation and constraint to require and define 11 22 33 0Q Q Q

an irreducibly traceless quadrupole tensor : 2 3,(3 ' ' ' ) ( ') 'ij i j i jQ x x x x d x

2 2 2 311 22 33 1 2 3,which satisfy [3( ' ' ' ) 3 ' ] ( ') ' 0Q Q Q x x x x x d x

If we expand the scalar potential under spherical coordinates:3

* 31

00 0

1 ( ') ' 1 4 '( ) ( ')( ( ', ') ( , )) '4 4 2 1| ' |

ll

lm lmll m l

x d x rx x Y Y d xl rx x


* 3Define: ' ( ') ( ', ') 'llm lmq r x Y d x

* 31 1

0 00 0

( , )1 4 1 1 4( ) ( ) ( ' ( ') ( ', ') ') ( , ) ( )4 2 1 4 2 1

l ll lm

lm lm lml ll m l l m l

Yx r x Y d x Y ql r l r

2 1 ( )!We know that ( , ) (cos )4 ( )!

m imlm l

l l mY P el m

2 / 2 2 / 2 21( ) (1 ) ( ) (1 ) ( 1)2 !,where

( )!( ) ( 1) ( )( )!

m l mm m m l

l lm l l m

m m ml l

d dP x x P x x xdx l dx

l mP x P xl m

0000

1 101 2 1/ 2

1 11

0 2 22 20

1 2 1/ 22 212 2

2

(cos ) 1/ 4( ) 1

( ) (cos ) 3 / 4 cos( ) (1 ) (cos ) 3 / 8 sin (cos sin )

3 1 3 1( ) (cos ) 5 / 4 ( cos )2 2 2 2

( ) 3 (1 ) (cos ) 5 / 4 3cos sin( ) 3(1 )

YP x

P x x YP x x Y i

P x x Y

P x x x YP x x

2 2

22

(cos sin )

(cos ) 5 / 96 3sin (cos sin )

i

Y i

3

00 ( ') 1/ 4 ' 1/ 4q x d x q

3 310 ( ') ' 3 / 4 cos ' ' 3 / 4 ( ') ' ' 3 / 4 zq x r d x x z d x p

3 311 ( ') ' 3 / 8 sin '(cos ' sin ') ' 3 / 8 ( ')( ' ') ' ' 3 / 8 ( )x yq x r i d x x x iy d x p ip

2 2 3 2 2 320 33

3 1( ') ' 5 / 4 ( cos ' ) ' 5 /16 ( ')(3 ' ' ) ' 5 /162 2

q x r d x x z r d x Q

2 3 321 5 / 24 ( ') ' 3cos 'sin '(cos ' sin ') ' 5 / 24 ( ') '(3 ' ') 'q x r i d x x z x iy d x

13 235 / 24 ( )Q iQ 2 2 2 3

22 11 22 125 / 96 ( ')3 ' sin '(cos ' sin ' 2 sin 'cos ') ' 5 / 96 ( 2 )q x r i d x Q Q iQ


The discussion above help us to find the relation between multipole moment under spherical coordinates qlmand multipole moment under Cartesian coordinates.

Consider electric fields expressed by multipole expansion with given l,m :

20

20

20

( , )1(2 1)

1 1 1Under spherical coordinates : ( , )(2 1)

1 1 1 ( , )sin (2 1) sin

lmr lm l

lm lml

lm lml

YlE qr l r

E q Yr l r

imE q Yr l r

If there is a dipole at , then the electric field observed at p

0x

( )E x

x

030 0

ˆ ˆ3 ( ) ˆ( ) , where is the unit vector for ( ).4 | |

n p n pE x n x xx x

The result can also be derived from the scalar potential contributed by the dipole moment.

32

0

ˆ1( ) ; ' ( ') '4

idipole i i i

i

rx p p x x d xr

If we shift the coordinates and let 0x x r

03

00

( )1( )4 | |

idipole i

i

x xx px x

0 0 ,5 3

,0 0 0

3 ( )( )1( ) ( ) ( ) { }4 | | | |

i i i j j i i jdipole dipole dipole

j i jj

p x x x x pE x x x

x x x x x

21 1 01 1 1 01 2 02 1 1 01 3 03 1

5 5 5 30 0 0 0 0

3 ( ) 3 ( )( ) 3 ( )( )1 {4 | | | | | | | |

p x x p x x x x p x x x x px x x x x x x x

2 22 2 02 2 2 02 2 2 02 3 03 2

5 5 5 30 0 0 0

3 ( ) 3 ( ) 3 ( )( )| | | | | | | |

p x x p x x p x x x x px x x x x x x x

23 3 03 1 01 3 3 03 2 02 3 3 03 3

5 5 5 30 0 0 0

3 ( )( ) 3 ( )( ) 3 ( ) }| | | | | | | |

p x x x x p x x x x p x x px x x x x x x x


0 05 3 3

0 0 0 0 0

ˆ ˆ[3 ( )]( )1 3 ( )( ) ( ) { }4 | | | | 4 | |

dipole dipolep x x x x p n p n pE x x

x x x x x x

0 5 30

1 3( )If 0, ( ) { }4

dipolep r r px E xr r

0 5 3 30 0

ˆ3 cos1 3 cos 1ˆAnd if , 0 ( ) { }4 4

rdipole

p a ppr r pp pz x E xr r r

When the observing point is very close to the electric dipole, then the result must be modified.

Assume an effective charge distribution to model the electric dipole is in the following form:

a

0( ) cos

Calculate the dipole moment contribute by this charge distribution :2

0ˆ & 2 cos ; 2 sin cosd p z dq z z a dq a d

/ 2 / 22 3 2 30 0 00 0

4ˆ ˆ ˆ2 cos cos 2 sin 4 cos sin3

p a a d z a d z a z

2 2/ 20 0 02 2 20

0 0 0 0

2 cos sin 2 cos sin1 1 2ˆˆ ˆ ˆAnd cos cos 4 4 4 3z

a d a dddE z r E z za r a

Thus, the electric field contribute from the dipole moment at the origin (center of dipole) is 3

0

4( ) /( )3 3

pE a

§4.2 Multipole Expansion of the Energy of a Charge Distribution in an External Field

Therefore, the total contribution from the electric dipole moment is


030 0

ˆ ˆ1 3 ( ) 4( ) [ ( )]4 3| |

dipolen p n pE x p x x

x x

( ')x

( )x

The interaction energy between a localized charge distribution and an external scalar potential can be expressed as

3( ) ( )W x x d x

(※)

If the scalar potential is slowly varying near the charge distribution , then the total interaction energy can be expressed by multipole expansion:

( )x

23 3

0 01 , 1

1Expand the external scalar potential around a proper origin ( ) (0) | | ......2 i jx x

i i ji i j

x x x xx x x

3

, 1

1Use ( ) (0) (0) (0) .....2

ji

i ji i

EE E x x E

x x

2

For external potential (field) 0 0i ii i i

Ex x x

32

,, 1

1( ) (0) (0) (3 ) (0) .....6

ji j i j

i j i

Ex x E x x r

x

Substitute back into (※)3

3 3 2 3,

, 1

1( ) (0) ( ) (0) ( )(3 ) (0) ......6

ji j i j

i j i

EW x d x x x d x E x x x r d x

x

dipole with field3

,, 1monopole with potential

quadrupole with the field gradient

1(0) (0) (0) ......6

ji j

i j i

Eq p E Q

x


§4.3 Elementary Treatment of Electrostatics with Ponderable Media

extE

indE

Matter

( )tot in matter ext indE E E

If the charge distribution in the matter won’t induce higher multipole moment density other than dipole moment under the external field, then the potential in a small region in the matter would be:

v

30

1 ( ) ( ') ( ')( , ') [ ' '] , (dipole moment per unit volume)4 | ' | | ' | i

i

x P x x x Px x v v P N pvx x x x

3 33

0

contribution from free charge contribution from the bound charge in matter

1 ( ) ( ') ( ')( , ') [ ' ' ]4 | ' | | ' |

x P x x xx x d x d xx x x x

3 33

0

1 1 ( ') 1 ( ) 1We know that ( ) '( ) , then ( ) [ ' ( ') '( ) ']4| ' | | ' | | ' | | ' | | ' |

x x xx d x P x d xx x x x x x x x x x

Use ( ) ( ) ( ) ( ) ( ) ( )f A f A f A f A f A f A

3 31 1 ( ') ˆ' ( ( ') ) ' ( ' ( ')) '| ' | | ' | | ' |

P xP x d x P x d x ndax x x x x x

31 ( ' ( ')) '

| ' |P x d x

x x

3 3 3

0 0

1 ( ) ' ( ') 1 ( ) ' ( ')Therefore, ( ) [ ' '] [ '4 4| ' | | ' | | ' |

x P x x P xx d x d x d xx x x x x x

We can regard the total charge in the matter at this time to be , where tot f b b P


0 0

1According to Gauss's law: [ ]totfE P

0 0Then, ( ) , where is the electric displacement.f fE P D D E P

For isotropic and linear dielectric, 0eP E

0 0 0Thus, (1 ) , where (1 ).e r r eD E P E E E

※ The boundary condition for dielectric:

Ad

D

21

0d

d

E

0d

21

According to Maxwell's equation: fD

BEt

211 2 1 20

lim ( ) ( )f fdD nds D D nA Q D D n

0 0 0

lim ( ) lim limCd d d

E nds E dl B ndst

2 1 2 1 21ˆ0 0 ( ) 0t tE E E E n

1 2211 2 1 2

2 1 21 1 2

( ) | | , if 0

ˆ( ) 0 | |

f x s x sf

x s x s

D D nn n

E E n

<Example 1>

d source chargeq

(0,0, )d

q12

' image chargeq

(0,0, )d

0z

(Azimuthal symmetry)

Q. For a point charge q in dielectric 1, evaluate the scalar potential under this case.

We can use image charge method to solve this problem.Assume the induced charge at their image position and

enforce the boundary condition to get their values. '&q q

(1) Region 1: 0, the effective charge seen in this region are , 'z q q

1 2 2 2 21

1 '( , , ) [ ]4 ( ) ( )

q qzz d z d


(2) Region 2: 0, the effective charge seen in this region are , z q q

2 2 22

1 "Let '' ( , , )4 ( )

qq q q zz d

Consider the boundary condition :

1 2 1 2(1) | | ( 0) ( 0)x s x s z z 2 2 2 2

1 2 1 2

1 ( ') 1 " 1 1( ') " (a)4 4

q q q q q qd d

1 2 1 21 2 1 0 2 0(2) | | | |x s x s z zn n z z

1 2

0 02 2 3/ 2 2 2 3/ 2 2 2 3/ 21 2

( ) ( ) ' "( )[ ] | | ( ') " (b)4 [( ) ] [( ) ] 4 [( ) ]z z

z d q z d q q z d q q qz d z d z d

1

2 2 1 2 1 21 2

1 1 2 2 1 2

2

2 ' ( 1) "1 1( ') " 2Solve (a) & (b) " ( ) , ' ( ) " ( )22 ( 1) "( - ') "

q qq q q

q q q q qq qq q q

1 2 1 22 2 2 2

1

22 2

2 1 2

( ) /( )1[ ] , 04 ( ) ( )

Therefore, ( , ,z)21[ ( )] , z<0

4 ( )

q zz d z d

qz d

2 12 2 120 2 10.2 2 10.02


We can use to calculate the induced surface chargebP

dA

12

1 2 21 2 1 21ˆ ˆ( ) ( )b bP P n A Q P P n

0 1 1 0 1 2 2 0 2From ( ) ; ( )D E E P P E P E

1 2 0 2 1 21ˆAnd on the surface ( - )bE E E E n

1 2 1 1 22 2 3/ 2 2 2 3/ 2

1

1 2 1 1 22 2 3/ 2 2 2 3/ 2

1

( ) /( )1 ˆ( , , ) [ ]4 [( ) ] [( ) ]

(For 0)( )( ) /( ) ˆ( , , ) [ ]

4 [( ) ] [( ) ]z z

qE z az d z d

zz dq z dE z a

z z d z d

2 22 2 3/ 2

2 1 2

2 22 2 3/ 2

2 1 2

21 ˆ( , , ) [ ( )]4 [( ) ]

(For 0)2 ˆ( , , ) [ ( )]

4 [( ) ]z z

qE z az d

zq z dE z a

z z d

2 2 1 22 1 2 2 3/ 2 2 2 3/ 2 2 2 3/ 2

2 1 2 1 1 2 1 2 1

2 21 1 1 1(0) (0) ( ) ( ) ( )4 ( ) 4 ( ) 2 ( )z z

qd qd qdE Ed d d

0 2 12 2 3/ 2

1 2 1

( )2 ( )bq d

d

2 12 2 120 2 10.2 2 10.02


§4.4 Boundary-Value Problems with Dielectrics

Consider a sphere with dielectric constant ε2 locates in an uniform dielectric matter ε1 and be affected by an external electric field

0 ˆzE E a

0 ˆzE E a

1

2

a

P

zUnder this condition, the scalar potential at r

0 2( ) cos (bound charge polarization due to sphere)r E r

Since the case satisfies azimuthal symmetry, its general solution in spherical coordinates is

1

1( , ) [ ] (cos )ll l ll

lr A r B P

r (require solution converge)

0(For ), ( , ) (cos )l

in l ll

r a r A r P

( 1)

0(For ), ( , ) [ ] (cos )l l

out l l ll

r a r B r C r P

2 0When , the contribution from sphere can be neglected: ( ) cosr r E r ( 1)

00

( ) lim [ ] (cos ) cosl lout l l lr l

r B r C r P E r

According to the orthogonality of Legendre polynomial, as only 1 contributes.r l

( 1)1 1 0

0

0

( , ) (cos ) &

( , ) (cos )

lout l l

l

lin l l

l

r B r C r P B E

r A r P


Consider the boundary condition:

1 22 1 21

1 21 22 1 21

| |ˆ( ) , if =0

| |ˆ( ) 0

s s

s s

D D n

E E nn n

( | | )in r a out r a 2 3

1 0 1 1 0 1( 1) (2 1)

1, cos | [ ]cos | (i)

1, (cos ) | (cos ) | (ii)r a r a

l l ll l r a l l r a l l

l A r E r C r A E C a

l A r P C r P A C a

2 1( | | )in outr a r ar r

3 32 1 1 0 1 2 1 1 0 1

1 ( 2) (2 1)2 1 2 1

1, cos [ ( 2) cos | ] [ 2 ] (iii)

1, (cos ) | ( 1) (cos ) | ( 1) (iv)r a

l l ll l r a l l r a l l

l A E C r A E C a

l lA r P l C r P l A l C a

(2 1) (2 1) (2 1) (2 1)2 2 2 1 2 1From (ii) & (iv) ( 1) [ ( 1) ] 0l l l l

l l l l l l lA C a lA l C a l A l C a C a l l

It must be valid for any 1 0, 0 (for 1)l ll C A l

31 0 1 3

2 1 0 2 1 232 1 1 0 1

From (i),(iii)[ 2 ]

A E C aA E C a

A E C a

3 2 11 0

2 130 2 1 1 2 1

2 1 11 0 0 0

2 1 2 1

( )2

( ) ( 2 )3( )

2 2

C E aE C a

A E E E

2 11 0 0

0 2 12 1 3

3 2 12 1 0 2

0 02 2 1 2 1

3 ( , ) cos ( ) cos( , ) ( ) cos 22Therefore, we have

( , ) cos ( )( , ) cos ( ) cos 22

inin

outout

due to the spplied field

r E r E rr r E

aa r E rr E r E rr

0

arg

cos

due to the bound ch e polarization

E


3 5,0

1 ( ') 1Compare with multipole expansion, ( ) [ ...]4 2| ' | | ' | | ' |

i jij

i j

x xq p x xx Qx x x x x x

3

2 102

2 1

We can see ( ) cos as the contribution from electric dipole moment2

a Er

332 1 2 1

0 0 03 20 2 1 2 1

1 cos ˆThus, ( ) cos 4 ( )4 2 2 z

pr a E p a E ar r

1 1 0 12 1 21 1 1 2 2

2 2 0 2

( )ˆAnd, ( ) , & on the surface

( )pol

P EP P n E E

P E

2 2 0 2 1 1 0 1 21 0 1 2 21ˆ ˆ( ) ( )pol E E E E n E E n

We can also calculate the electric field:

2 1 10 0 0

2 1 2 1

2 1 10 0 0

2 1 2 1

3ˆ ˆ[ cos cos ( )] ( ) cos2 2

31 ˆ ˆ[ sin ( )sin ] ( ) sin2 2

in inr r r

in in

E a E E E ar

E a E E E ar

32 1

0 032 1

32 1

0 032 1

2ˆ [ cos ( ) cos ]2

1 ˆ [ sin ( ) sin ]2

out outr r

out out

aE a E Er r

aE a E Er r

Substitute | and | into ,in outr r a r r a polE E

2 1 2 1 2 10 0 0 0 0

2 1 2 2 2 2

2[2( ) cos ( ) cos ] 3 ( ) cos2 2 2pol E E E

a0 cos 0

00

ˆRecall the internal field for a sphere with surface charge distribution cos3 zE a

2 1 2 10 0 0 0 0 03

2 1 2 1

ˆ ˆThen in this case 3 ( ) and 3 ( )2 (4 / 3) 2 z z

pE P E a aa


2 10.1

2 11

2 110


§4.5 Molecular Polarizability and Electric Susceptibility

0 ˆzE E a

dense media

az

P

0We know that in linear dielectric, eP E

If the medium is composed by high density molecule, then the total contribution to the electric field in this region is .

0 (internal field)near pE E E

We can calculate the internal electric field of the spherical region in an applied field and surrounded by polarization :

0 0 ˆzE E a

0//P E

0

| | (a)According to boundary condition :

(b)

in r a out r a

in out r rD D D Pr

12

0 1

( , ) cosFrom the example in section 4.4 :

( , ) cos cosin

out

r A r

r E r C r

3 31 0 1 0 1 0 0 0 1(B.C. (a)) 2 2 2A E C a A E C a

30 1 0 0 0 1 0 1 0 0

3 31 0 1 1 0

0 0

(B.C. (b)) 2 3 3

( ) & ( )3 3

A E C a P A E PP PA E C A E a a

3

0 0 20 0

Therefore, ( , ) ( ) cos & ( , ) cos ( ) cos3 3in outP P ar E r r E r

r

0 00 0

ˆ ˆ( ) cos ( )3 3

near

p

inin r z

EE

P PE E a E ar

P.S. In amorphous & simple cubic matter, 0nearE

0 0 00

Substitute into ( ) (1 )3 3

ein e e e

PE P E P E P E

Consider the total polarization, molP N p

0 0And ( ) ( ) (averaged polarization per molecule),3mol i molmolPp E E E

molwhere is microscopic molecule polarizability


0 0 0 01Thus, ( ) ( ) (1 ) ( )13 3 3 1

3

molmol e mol mol

mol

NP PP N E E P N N E P EN

0

0 0

( / ) 13 3We have, ( ) , ( ) [ ] (we have used 1)1 3 ( / ) 213

mol ee mol e

emol

NN NN

§4.6 Model for the Molecular Polarizability

Effective spring model:

k e

m

assume atom fix

20 2

0

The equation of motion: eEF k x m x eE xm

2

20

mole EP exm

2

0 20 0

And mol mol moleP E

m

If the molecule is composed by ion with different mass and valence, then there are different spring constant ω0 .

※Calculate the thermal average for Pmol (The weighting can be decided from the Boltzmann factor)

The interaction energy between electric dipole and the external electric field is cosW p E pE

cos( cos ) exp( )

cosexp( )B

mol

B

pEp dk TP pE d

k T

-sin

Let cos

d dyy

1 1 2 1 21 11

exp( ) [( ) | ( ) | ] [( )( ) ( ) ( )]B B B B B B

pEy pEy pE pE pE pEk T k T k T k T k T k TB B B B

B

k T k T k T k TpEypy dy p e y e p e e e ek T pE pE pE pE

1

1exp( ) ( )( )B B

pE pEk T k TB

B

k TpEy dy e ek T pE


2[( )( ) ( ) ( )]Therefore, {coth( ) } cos

( )( )

B B B B

B B

pE pE pE pEk T k T k T k TB B

B Bmol pE pE

k T k TB

k T k Tp e e e ek T k TpE pEP P PpE pEk T e e

pE

It is called Langevin-Debye formula.2 2 2

22 3 2 3

1 (1 / 2) (1 / 2) 1 1 1 1If is very small ( - ) ( )(1 )(1 / 2 / 3!) (1 / 2 / 3!) 2 6 2 6 12

x x

x x

e e x x x x x x x xx xe e x x x x x x x x x x

13

neglect

x

201Thus, when is very small,

3molB B

PpE P Ek T k T

Usually, the polarization in molecule may be originated from (1) permanent polarization (2) induced polarization

2

0

0

13mol i

Bpermanentinduced

Pk T

classical electrodynamicsocw.nctu.edu.tw/course/classical_electrodynamics/electrodynamics... ·...

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