Diel Lecture2
Transcript of Diel Lecture2
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Lecture 2
Lecture 2
1.1. Dipole moments and Dipole moments and
electrostatic problemselectrostatic problems
2.2. Polarizability Polarizability α
3.3. Polarization and energy Polarization and energy
4. 4. Internal field. LangevenInternal field. Langeven
functionfunction
5.5. Non-polar dielectrics.Non-polar dielectrics.
6.6. Lorentz's field.Lorentz's field.
7.7. Clausius-Massotti formulaClausius-Massotti formula
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electrostaticelectrostatic
problemsproblems
electrostaticelectrostatic
problemsproblems
ε
ε
1
2
Z E o
a
Let us put a dielectric sphere of
radius aa and dielectric constant ε2, in
a dielectric extending to infinity
(continuum), with dielectric constant
ε1, to which an external electric fieldis applied. Outside the sphere the
potential satisfies Laplace's equation
∆φφ=0=0 , since no charges are present
except the charges at a great
distance required to maintain theexternal field. On the surface of the
sphere Laplace's equation is not
valid, since there is an apparent
surface charge.
Figure1
Inside the sphere, however, Laplace's equation can be used
again. Therefore, for the description of φ , we use twodifferent functions, φ 1 and φ 2, outside and inside the sphere,
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Let us consider the center of the sphere as the origin of thecoordinate system, we choose z-axis in the direction of theuniform field. Following relation in the terms of Legendre polynomial represents the general solution of Laplace’s
equation:
)(cos
)(cos
012
011
θ φ
θ φ
n
nn
nn
n
n
nn
nn
n
P r
Dr C
P r
Br A
∑
∑∞
=+
∞
=+
+=
+=
The boundary conditionsare:( ) θ−=−=φ ∞→ cosr E z E
r 001(2.1)
( ) ( )ar ar == φ=φ 21 Since φ is continuous across a
boundary
(2.2)
1.
2.
3.ar ar dr
d
dr
d
==
φ
ε=
φ
ε 22
11 (2.3)
since the normal component of D must be continuous at the surfaceof the sphere
At the center of the sphere ( r=0 ) φ2 must not have asingularity.
4.
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On account of the first boundary condition and the fact thatthe Legendre functions are linearly independent, allcoefficients An are zero except A1, which has the value A1 = -
Eo. On account of the fourth boundary condition, all
coefficients Dn are zero. Thus, one can write:
cosr E )(cos P r
Bn
nn
n θ−θ=φ ∑∞
=+ 0
011
(2.4)
)(cos P r C n
n
n
n θ=φ ∑
∞
=02
(2.5)
Applying the second and third boundary condition to (2.4) and(2.5), we have for any of n except n=1:
B
a C an
n n
n
+ =1
(2.6)
− = −ε ε1 2
1( +1)nB
an
n+2nC a
n
n (2.7)
and
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From these equations it follows that Bn=0 and Cn=0 for all
values of n except n=1. When n=1, it can be written:
1203
1
1
102
1
2C E
a
B
,aC a E a
B
ε−=
+ε
=−
B a E 1
2 1
2 1
3
0
2
=−
+
ε ε
ε ε0
21
11
2
3 E C
ε ε
ε
+=
Substitution of these values in (2.4) and (2.5) gives:
Hence:
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z E r
a03
3
12
121 1
2
−
ε+εε−ε
=φ (2.8)
(2.9) z E 021
1
2 2
3
ε+ε
ε
−=φSince the potential due to the external charges is given by
φ =-Eo z , it follows from (2.8) and (2.9) that the contributions φ '1
and φ '2 due to the apparent surface charges are given by:
φ ε εε ε
' 1
2 1
2 1
3
3 02
= −+
ar
z E (2.10)
(2.11)
z 0
12
122
2' E
ε ε
ε ε φ
+−
=
The expression (2.10) is identical to that for the potential dueto an ideal dipole at the center of the sphere, surrounded by adielectric continuum, the dipole vector being directed alongthe z-axis and given by:
m E =−
+ε ε
ε ε2 1
2 1
02
a3
(2.12
)
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The total field E2 inside the sphere is according to (2.9),
given by:
0
21
12
2
3 E E
ε ε
ε
+= (2.13
)
A spherical cavity in A spherical cavity in
dielectricdielectric In the special case of a spherical cavity in dielectric ( ( ε ε 11==ε ε ;;
ε ε 22=1=1),), equation (2.13) is reduced to:
This field is called the "cavity field". The lines of
dielectric displacement given by Dc=3Do /( 2ε+1 ) are
more dens in the surrounding dielectric, since D is
larger in the dielectric than in the cavity
ε ε 22=1=1
ε ε 11==ε ε
012
3 E E
+
=
ε
ε
C (2.14)
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A dielectric sphere in vacuum A dielectric sphere in vacuum
For a dielectric sphere in a vacuum (ε1=1; ε2=ε), the equation
(2.13) is reduced to:
012
3 E E
+=
ε (2.15)
where E is the field inside thesphere. The density of the lines of dielectric displacement Ds is
higher in the sphere than in the surrounding vacuum, sinceinside the sphere Ds=3ε Eo /( ε +2). Consequently, it is larger
than Eo.
ε ε 11=1=1
ε ε 22==ε ε
According to (2.12) , the fieldoutside the sphere due to theapparent surface charges is the
same as the field that would becaused by a dipole m at the centerof the sphere, surrounded by avacuum, and given by:
m E =−
+
ε
ε
1
20
a3 (2.16)
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This is the electric moment of the dielectric sphere.electric moment of the dielectric sphere.
Therefore (2.16) can be checked in another way. The uniform
field Es in the dielectric sphere, given by (2.15), causes a
homogeneous polarization of the sphere. The induced dipole
moment per cm3 is according to the definition of P , given by:
0
4
3
2
1
4
1 E E P
π ε
ε
π
ε
+
−=
−= s
(2.17)
Hence the total moment of the sphere is:
m P = E =−+
4
3
1
2
3
0πεεa a
3
(2.18)
which is in accordance with (2.16).
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PolarizabilityPolarizability
α PolarizabilityPolarizability
α When a body is placed in a uniform electric field Eo inin
vacuumvacuum, caused by a fixed charge distribution, its dipole
moment will in general changed.
In most cases polarizable bodies are polarized linearly, that is,
the induced moment mm is proportional to EEoo. In this case one
have:
0 E m α= (2.19)
where α α is called the (scalar) polarizability of the body.(scalar) polarizability of the body.
The difference between the dipole moments before and after
the application of the field Eo is called the induced dipole
moment m. If a body shows an induced dipole moment
differing from zero upon application of a uniform field Eo, it is
said to be polarizable. polarizable.
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From the dimensions of the dipole moment, [e][l], and the
field intensity, [e][l]-2, it follows that the polarizability has the
dimension of a volume. Using the above definition of the
polarizability, we conclude from equation (2.12) that a
dielectric sphere of dielectric constant ε and with radius aa has
a polarizability:m E =
−+
ε εε ε
2 1
2 1
02
a3
(2.12)
3a2
1
+−
=ε
ε α (2.20)
For a conducting sphere in the case ε→∞ε→∞ from the relation
(2.20) one can obtain that a conducting sphere with radius aa
has a polarizability:
3a=α (2.21)
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There is the fundamental difference between the two types of
polarization. In the case of dielectric sphere every volume
element is polarized, whereas in the case of a conducting
sphere the induced dipole moment arises from true surface
charges. In some simple cases of spherical molecules, it is possible with
α α to describe the induced polarization. In general, it is not true
and we have to use a polarizability tensor α provided the
effects remain linear. It leads to:
0 E m α= (2.22)
In such a case the induced dipole moment need to have thethe
same directionsame direction as the applied fieldapplied field. In general this direction
will depend on the position of the body relative to the
polarizing field.
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Polarization and EnergyPolarization and Energy
Very often it is useful to collect some of the elementarycharges into a group forming an atom, a molecule, a unite cellof a crystal, or some larger unit. Let the jth unit of this typecontain the s elementary charges e j1, e j2,....., e jk ,.....,e js, and let
) ,... ,...., ,( x j j j j j sk 21 r r r r = (2.23)
m r ( ) x ei jk jk
k
s
==∑
1
(2.24)
be an abbreviation for the set of all their displacementsr j1,....., r js.
is the electric moment of this jth group of charges,and
Then
M( ) ( ) X x j
j
= ∑ m (2.25)
where the sum extends over all therou s.
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The vector sum of their individual moments mm(( x x j j)) thus forms
the total moment MM( ( X X ). ). The main aim of this part is to find
the average displacements, and hence the average electricaverage electric
moment moment under the influence of an external electric field.
In order to obtain a preliminary idea about the average
contributions of certain displacements to the electric moment
we shall consider two cases, each of which has its
characteristic type of displacement:1) The displaced charge is bound elasticity to an equilibriumposition; (induced dipole moment)(induced dipole moment)
In the first case the displacement of the charge e, carried out
by a particle of mass m on a distance r , restoring force
proportional to -r acts on the particle in a direction opposite to
the displacement (hence the - sign). Thus if the constant
2) A charge has several equilibrium positions, each of which
it occupies with a probability which depends on the
strength of an external field.(.(Permanent dipole moment)Permanent dipole moment)
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where ω o /2π denotes the frequency of oscillation, and -mω o2r
is the restoring force. Equation (2.26) can be written
E r r
m
e
dt
d 2
+2
2
0ω −= (2.26)
)()(2
r' -r r' -r 2
0ω −=
2
dt
d (2.27)
E r' 2
0
=ω m
e(2.28)
where
i.e. dr' /dt=0. The charge e, therefore, carries out harmonic
oscillations about the position r' which thus represents the
time average of its displacement, i.e. if C and δ are constant
)t Ccos( ' δ+ω+= 0r r
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The average electric moment (induced) is, therefore,
(2.29)µα ω=
er E ' =e
m 02
For an example of the second mechanism of polarization let
us consider a particle with the charge ee that can be in two
different equilibrium position A A and BB, separated by adistance bb. In the absence of an electric field the particle has
the same energy in each position.
Thus, it may be assumed to move in a potential field of the
type shown in Fig.2.2. In equilibrium with its surroundings itwill oscillate with an energy of order kT about either of the
equilibrium positions, say about A.A.
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r'
B
A
r'
e bE
Fig.2.2Fig.2.2
Occasionally, however,
through a fluctuation, it will
be acquire sufficient energy
to jump over the potentialwall separating it from BB.
On the time of average,
therefore, it will stay in A A as
long as in BB, i.e. the
probability of finding it in
either A A or BB is 0.5.0.5. The presence of a field E will affect this in two ways. Firstly, as
in case (1), the equilibrium position will be shifted by amount r'
which for simplicity will be assumed to be the same in A A and BB.Secondly, the potential energies V A, V B of the particle in the
two equilibrium positions will be altered because its interaction
energy with the external field differs by e( bE ), i.e.
)(bE eV V B A
=− (2.30)
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Therefore, in average, the particle should spend more time
near BB than near A A. Actually, since according to statistical
mechanics, the probability of finding a particle with energy V
is proportional to e-V/kT ,
are the probabilities for positions A A and BB respectively. Theyhave been normalized in such a way as to make
in agreement with the physical condition that the particlemust be in one of the two positions. Thus from (2.31) and
(2.30)
1 p p B A =+ (2.32)
pe
e e p
e
e e A
V kT
V kT V kT B
V kT
V kT V kT
A
A B
B
A B
=+
=+
−
− −
−
− −
/
/ /
/
/ / , (2.31)
.bE
bE
01
1/)(
/)(
>+−=−
kT e
kT e
A Be
e p p (2.33
)
It follows from the definition of the probabilities p pAA and p pBB that
if the condition of the system over a long time t t 11 is considered,
the particle will spend a time (use 2.32)
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[ ] 121
21
1 )( t p pt p A B A −−=
[ ] 12
1
2
11 )( t p pt p A B B −−=
in position A A, and
in position BB. It has thus been displaced by the distance b from A A to BB during the fraction 1/2( pB- pA) of the time t t 11. The
average moment induced by the field is thus
(2.34
)
)(21
A B p pe −b
Hence, if θ θ is the angle between b and E, the projection of theinduced moment into the field direction is, using (2.34) and(2.33) given by
.0
1
1cos
2
1/
/
>
+
−kT oscbE e
kT oscbE e
e
eeb
θ
θ
θ (2.35)
In most cases it is permissible toassume
kT E e <<b (2.36)
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for putting ee=4.8⋅10-10 e.s.u.,EE=300 v/cm.=1 e.s.u., bb=10-8
cm.≈ distance between neighboring atoms in a molecule, and
T=300oC (room temperature) one finds
ebE
kT ≅
× × ×× ×
≅− −
−
−4 8 10 10 1
1 38 10 30010
10 8
16
4.
.
Developing (2.35) in terms of ebE/kT , the average inducedmoment in the field direction is found to be
(1
2)2eb
kT
cos'
2 θ E er + (2.37
)
where eer'r' is a term similar to those considered in case (1)
which has been added to account for the elastic displacement.Often two charges +e+e and -e-e are strongly bound, forming anelectric dipole µ=e=eaa, where aa is the distance between them. The above case (2) then leads to the same as that of a dipole
µ having two equilibrium positions with opposite dipole
direction, but with equal energy in the absence of a field.
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In a field E the energy of interaction between field and dipoleis given by
)( - μE (2.38)
so that 22 µ µ EcosEcosθ θ is the energy between the two positions if ∠θ ∠θ is the angle µ and EE. This equivalent to equation (2.30) if
Actually putting an immobile charge -e-e halfway between A A
and BB turns case (2) into the present case. Clearly theinduced moment must be the same for both cases becausethe charge -e-e is immobile, and its distance from A A and BB is 1/21/2
bb, leading to a dipole µ. Introducing (2.39) into (2.37) yieldsfor the induced moment in the field direction
b μ e21= (2.39
)
'cos22
er E kT +
θ µ (2.40)
In contrast to case (1) the electric moment (orientationpolarization) now depends on temperature. A matterconsisting of a great number of such dipoles will have a
temperature-dependent dielectric permittivity in contrast to a
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This means that in the dipolar case the entropy of thedipolar case the entropy of the
substance is decreased by the field substance is decreased by the field.
In the case (1), the field exerts a force on elastically bound
charge, thus shifting its equilibrium position. In case (2) this
force of the field on the charge again leads to contribution of
type (1) denoted by er ' in equations (2.37) and (2.40). It
would be wrong, however, to assume that the field by thisforce turns a dipole from one equilibrium position into
another. This is affected in a more indirect way because the
field slightly alters the probabilities of a jump of a dipole from
one equilibrium position to another.
The difference between the action of the field in theThe difference between the action of the field in the
two cases (1) and (2) is essential for the wholetwo cases (1) and (2) is essential for the whole
theory of dielectric permittivity.theory of dielectric permittivity.
The difference between the action of the field in theThe difference between the action of the field in the
two cases (1) and (2) is essential for the wholetwo cases (1) and (2) is essential for the whole
theory of dielectric permittivity.theory of dielectric permittivity.
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Internal field; Langeven functionInternal field; Langeven function
In dielectric two essentially different types of interactiontwo essentially different types of interactionforces should be distinguished.forces should be distinguished. Forces due to chemical bonds,van der Waals attraction, repulsion forces, and others have allsuch short ranges that usually interaction between nearest neighbors only need be considered. Compared with these
forces dipolar interaction forces have a very long range.dipolar interaction forces have a very long range.As it was shown above a polarized dielectric can bea polarized dielectric can be
considered as composed of small regions each having aconsidered as composed of small regions each having a
certain dipole moment, and the total dipole moment of thecertain dipole moment, and the total dipole moment of the
body is the vector sum of the moments of these regions.body is the vector sum of the moments of these regions. It is
also known from macroscopic theory that the energy per unit energy per unit volume of a macroscopic specimen depends on its shapevolume of a macroscopic specimen depends on its shape.
This implies that interaction between dipoles must be takeninto account even at macroscopic distances.Due to theDue to the long rangelong range of the dipolar forces an accurateof the dipolar forces an accurate
calculation of the interaction of a particular dipole with allcalculation of the interaction of a particular dipole with all
other dipoles of a specimen would be very complicated.other dipoles of a specimen would be very complicated.
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A very good approximation can be made by considering thatA very good approximation can be made by considering that
the dipoles beyond a certain distance, saythe dipoles beyond a certain distance, say aamm can be replacedcan be replaced
by aby a continuous mediumcontinuous medium, having the macroscopic dielectric, having the macroscopic dielectric
properties of the specimen.properties of the specimen.
Thus the dipole whose interaction with the rest of the Thus the dipole whose interaction with the rest of thespecimen we are calculating may be considered asspecimen we are calculating may be considered as
surrounded by a sphere of radiussurrounded by a sphere of radius aamm containing a discretecontaining a discrete
number of particles, beyond which there is a continuousnumber of particles, beyond which there is a continuous
mediummedium To make this a good approximation To make this a good approximation the dielectricthe dielectric
properties of the whole region within the sphereproperties of the whole region within the sphere should beshould beequal to those of a macroscopic specimen,equal to those of a macroscopic specimen, i.e. it shouldi.e. it should
contain a sufficient number of molecules to make fluctuationscontain a sufficient number of molecules to make fluctuations
very small.very small. Lorentz's method Lorentz's method for the treatment of dipolar interaction:for the treatment of dipolar interaction:
from a macroscopic specimen select a microscopic sphericalfrom a macroscopic specimen select a microscopic spherical
region, which is sufficiently large to have the same dielectricregion, which is sufficiently large to have the same dielectric properties as a macroscopic specimen. properties as a macroscopic specimen. The interactionThe interaction
between the dipoles inside the spherical region will then bebetween the dipoles inside the spherical region will then be
calculated in an exact way, but for the calculation of their calculated in an exact way, but for the calculation of their
interaction with the rest of the specimen the letter isinteraction with the rest of the specimen the letter is
considered as a continuous medium.considered as a continuous medium.
Lorentz's method Lorentz's method for the treatment of dipolar interaction:for the treatment of dipolar interaction:
from a macroscopic specimen select a microscopic sphericalfrom a macroscopic specimen select a microscopic spherical
region, which is sufficiently large to have the same dielectricregion, which is sufficiently large to have the same dielectric properties as a macroscopic specimen. properties as a macroscopic specimen. The interactionThe interaction
between the dipoles inside the spherical region will then bebetween the dipoles inside the spherical region will then be
calculated in an exact way, but for the calculation of their calculated in an exact way, but for the calculation of their
interaction with the rest of the specimen the letter isinteraction with the rest of the specimen the letter is
considered as a continuous medium.considered as a continuous medium.
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We shall now investigate the dependence of the polarizationon the electric fields working on a single molecule. For theinduced polarization P
α we write:
P E α α= ∑ N k k i k
k
( ) (2.41
)
where NN is the number of particles per cm3, α α the scalarscalar
polarizabilitypolarizability of a particle and EEii the average field strength
acting upon that particle. The index k k refers to the k -th kind of
particle.The field The field EEi i is called theis called the internal field.internal field. It is defined as theIt is defined as thetotal electric field at the position of the particle minus the fieldtotal electric field at the position of the particle minus the field
due to the particle itself.due to the particle itself. The calculation of EEii is one of the
important problems associated with the theory of dielectrics.
This calculation can be executed both in the continuum
approach for the environment of the molecule and with the
help of statistical mechanics. The orientation polarizationorientation polarization can be written as:
P µ = ∑ N k k
k
µ (2.42)
where µ k is the value of the permanent dipole vector permanent dipole vector avera edavera ed over all orientations.
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The value µk k can be computed from the energy of the
permanent dipole in the electric field. This energy isdependent on the part of the electric field tending to directthe permanent dipoles. This part of the field is called the
directing field Edirecting field Ed d .. The dependence of µk k onon EEdd is computed from the energy of
dipole in electric field as we did above:
θµ−=⋅= cos E -W d d E μ (2.43)
where θ θ is the angle between the directions of EEdd and µ. SinceSince
W W is the only part of the energy, whichis the only part of the energy, which depends on thedepends on the
orientation of the dipoleorientation of the dipole, the relative probabilities of the, the relative probabilities of the
various orientations of the dipole depend on this energy various orientations of the dipole depend on this energy W W
according toaccording to Boltsmann’s distribution lawBoltsmann’s distribution law..
From Boltzsmann's law
Let us consider the average value of <cos<cosθ θ >.>. For a random
distribution of the dipoles in matter we have <cos<cosθ θ >=0>=0,
whereas <cos<cosθ θ >=1>=1 if all the dipoles have the same direction
as EEdd.
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),(1
cot1
][
][1
1
sin2
1
sin2
1cos
cos
0
cos
0
cos
a La
anhaaee
ee
e
e xe
a
dxe
xdxe
ad e
d e
aa
aa
a
a
x
a
a
x x
a
a
x
a
a
x
kT
E
kT
E
d
d
=−=−−+
=−
=
===
−
−
+−
+−
−
−
∫
∫
∫
∫ π θ µ
π θ µ
θ θ
θ θ θ
θ
(2.44)
x
kT
d E
=θ µ coswhere
and a
kT
d E
= µ
L(a)L(a) is called Langeven functionLangeven function
In Fig.2.3 the Langevenfunction L(a)L(a) is plottedagainst aa. L(a)L(a) has a limitingvalue 11, which was to beexpected since this is the
maximum of coscosθ θ . For smallvalues of aa, <cos<cosθ θ >> is linearin EEdd::
10 if 33
1cos <<≤== a
kT d
E a
µ θ (2.45
)
y=1/3 a
0 1 2 3 4 5 6 7 8 0.0
0.2
0.4
0.6
0.8
1.0 y=
y=L( a )
a
y
Fig.2.3
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The approximation of equation (2.45) may be used as long as
a E
d kT
E d
kT = < <
µ
µ0 1
0 1.
..or
At room temperature (T=300o K) this gives for a dipole of 44DD:
µ
kT d
E 1.0< = 3 105 v/cm
For a value of µ smaller than the large value of 4D, the valuecalculated for EE
ddis even larger. In usual dielectric
measurements, EEdd is much smaller than 105 v/cm and the use
of (2.45) is allowed.From equation (2.45) it follows that:
d kT
cos E μ3
2µ=θµ= (2.46
)
Substituting this into (2.42) we get:
P µ = ∑ N k
k
µ
3κΤ
2
( )d
Ek
(2.47)
From the first lecture wehad
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We now substitute (2.41) and (2.47) into (2.48) and find:
µ α π
ε P P = E +
−4
1 (2.48)
This is the fundamental equationfundamental equation is the starting point for
expressing EEi i and EEd d as functions of the Maxwell field E and
the dielectric constant ε.
+
−
∑k d
k
k ik k
k kT N )(3)(4
1 2
E E = E
µ
α π
ε
(2.49)
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Non-polar dielctrics. Lorentz's field. Non-polar dielctrics. Lorentz's field.
Clausius-Massotti formula.Clausius-Massotti formula. Non-polar dielctrics. Lorentz's field. Non-polar dielctrics. Lorentz's field.
Clausius-Massotti formula.Clausius-Massotti formula.For a non-polar systemnon-polar system the fundamental equation for thedielectric permittivity (2.49) is simplified to:
επ
α−
∑1
4 E = E N
k k
k i k ( ) (2.50
)
In this case, only the relation between the internal field andthe Maxwell field has to be determined.
Therefore the polarization in the environment of a real cavityis not homogeneous, whereas the polarization in the
environment of a virtual cavity remains homogeneous.
The field in such a cavity differs from the field in a real cavity,given by (2.14), because in the latter case the polarizationadjusts itself to the presence of the cavity.
Let us use the Lorentz approachLorentz approach in this case. He calculatedthe internal field in homogeneously polarized matter as theas the
field in a virtual spherical cavity.field in a virtual spherical cavity.
0
12
3 E E
+
=
ε
ε
C (2.14
)
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The field in a virtual spherical cavity, which we call the Lorentzfield EELL, is the sum of:
1.1. the Maxwell fieldthe Maxwell field EE caused by the external charges andcaused by the external charges and
by the apparent charges on the outer surface of theby the apparent charges on the outer surface of the
dielectric, anddielectric, and2.2. the fieldthe field EEsphsph induced by the apparent charges on theinduced by the apparent charges on the
boundary of the cavity (see fig.2.4).boundary of the cavity (see fig.2.4).
θ
θ d
+
+
E
+
_
Fig.2.4Fig.2.4
The field EEsphsph is calculated
by subdividing theboundary in
infinitesimally small ringsperpendicular to the fielddirection. The apparentsurface charge density onthe rings is -Pcos-Pcosθ θ ,, their
surface is 22π π r2sinr2sinθ θ ddθ θ sothat the total charge oneach ring amounts to:θθθπ= cos P d sinr -2de 2 (2.51
)According to Coulomb's lawCoulomb's law, a charge element dede to the field
component in the direction of the external field , given by:
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dE de
r =
2cosθ (2.52
)
Combining (2.51) and (2.52), we find for the component of
EEsphsph in the direction of the external field :
P P E 3
4cossin2
0
2 π θ θ θ π
π
== ∫ d sph
(2.53)
For the reasons of symmetry, the other components of EEsphsph
are zero, and we have with EELL==EE++EEsphsph:
This is Lorentz's equation for the internal field.This is Lorentz's equation for the internal field.
Substituting (2.54) into (2.50), wefind:
E E L
=+ε 2
3(2.54)
∑=+−
k
k k N α π
ε
ε
3
4
2
1 (2.55)
This relation is generally called the Clausius-Massotti equation.Clausius-Massotti equation.
d i i
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For a pure compound it isreduced to:
απ=+ε−ε
N 3
4
2
1 (2.56)
From the Clausius-Mossotti equationClausius-Mossotti equation for a pure compound, itfollows that it is useful to define a molar polarization [molar polarization [PP]:]:
[ ] P M
d =
−+
εε
1
2(2.57)
where dd is the density and MM the molecular weight. When theClausius-Mossotti equationClausius-Mossotti equation is valid [P][P] is a constant for agiven substance:
[ ] P N A
=4
3
πα (2.58)
The Clausius-Mossotti equationClausius-Mossotti equation can also be used for polarsystems in high-frequency alternating fields:
εε
πα∞
∞
−+
= ∑1
2
4
3 N
k k k
(2.59)
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Often this equation is used for still higher frequencies, where
the atomic polarization too cannot follow the changes of the
field. If according to Maxwell relation for the dielectric
constant and the refractive index ε ε ∞ ∞ =n=n22, it is possible to write:
in which ε ε ∞ ∞ is the dielectric constant at a frequency at which
the permanent dipoles (i.e. the orientation polarization) can
no longer follow the changes of the field but where the atomic
and electronic polarization are still the same as in static
fields.
n
n
N k k
e
k
2
2
1
2
4
3
−
+= ∑
πα (2.60)