1. The Debye theory . Gases and polar molecules in non-polar solvent.
2. The reaction field of a non-polarizable point dipole 3. The internal and the direction fields for polar dielectrics.
4. The Onsager equation . The dielectric properties of polar non-associative liquids.
1

In Debye theory the internal field one, given by (2.54):
E
i and directing field
E
d were not distinguished from each other. Debye used for both fields the Lorentz's
E
L
   2
E
3 If we will substitute this expression into (2.49)   1 4 
E =

k N k
  
k
(
E
i
)
k

k
2  3
kT
(
E
d
)
k
  for both
E
i
and
E
d ,
then we can find:    1  2
= 4

3

k N k
 
k
 
k
2 3
kT
 (3.1) 2

This is generally called the Debye equation. It was the first relationship that made the connection between the
of the substance being tested and the
parameter that can be experimentally measured.
Using the molar polarization
[ P ]
 , defined by (2.57)     1 2
M d
we can write the
Debye equation
for a pure compound as:  4  3
N A
    2 3
kT
 (3.2) Thus, according to the
, the molar polarization of a compound at a given temperature
. It must be independent of the pressure and have the same value in gaseous and the liquid state. 3

But
in many cases, however, the Debye equation is in considerable
with the experiment. It works very nice for
. In this case one has 
and equation (3.1) can be written as:   1
=
4  
k N k
  
k
 
k
2 3
kT
  (3.3) In this form, the
is still used. For gases at normal pressures the difference between the fields
E , E
i ,
E
d , and
E
L can be neglected. For pure compounds equation (3.2) changes into:   1
=
4 
N
     2 3
kT
  (3.4) 4

When a molecule with other particles, the
inhomogeneous field of the permanent dipole polarizes its environment .
In the surrounding particles moments proportional to the polarizability are induced, and if these particles have a permanent dipole moment their orientation is influenced. To calculate this effect one can use a simple model:  is surrounded by
The field of the dipole in such a cavity polarizes the surrounding matter, and the resulting in homogeneous polarization of the environment will give rise to a field at the dipole, which is called the reaction field
For reasons of symmetry vector 
R
will have the same direction as the dipole , and it is obvious that no saturation effects occur.
R.
will be proportional to  as long as Thus:
R

f
μ
(3.5) The factor
f
is called the factor of reaction field . If the cavity is not spherical, this factor must be replaced by a reaction field tensor .
5

The model consists of a spherical cavity with radius dielectric of dielectric permittivity surroundings, the value chosen for 
in a continuous . The point dipole with moment  is situated in the center of the cavity (fig.3.1). In this very simplified model of the interaction between a permanent dipole moment and its influences the results.
z Let us assume that only one kind of molecule is presented:  
a
4 3 
Na
3  1 (3.6) Fig.3.1
It implies that the volume of the cavity is equal to the volume available to each molecule.
To calculate the reaction field we have to know the potential in the cavity due both to the dipole itself and to the interaction of the dipole with the surrounding dielectric. 6

Let the center of the dipole to be the origin of a coordinate system and choose the direction of the z-axis along the dipole vector. It will be the symmetry about the z-axis. As it was shown in the previous lecture the general solution of following relation in the terms of Legendre polynomial represents Laplaces’s equation :  1 
n
   0  
A r n n

r B n n
 1  
P n
 2   
n
 0  
n

r D n n
 1  
P n
where  1 is the potential outside the sphere and inside the sphere.
 2 is the potential 1.
The boundary conditions in this case are:   1
r
   0 2.
  1
r

a
   2
r

a
(3.7) (3.8) 3.
 
dr
1
r

a

dr
2
r

a
(3.9) 7

The only source of the field lines within the cavity is the permanent dipole by: 
.
The potential due to an ideal dipole along the z-axis is given   
r
2 cos  thus all coefficients
n are zero except
1 , which has the value
1
 Using the bound conditions and the same procedure that we used in previous lecture, we can get .
 2  1   3
r
2    cos 1 2 
r
 2  cos  2 (    1 ) 
a
3
r
cos  (3.10) (3.11) According to (3.11), the field in the cavity is a superposition of the dipole field in vacuum and a uniform field
R
, given by:
R

a
1 2 3 2 (    1 )  1  (3.12)   1
=
4 
N
reaction field is given by:     3  2
kT
  (3.4) 8

f

a
1 2 3 2 (    1 )  1 (3.13) Formally, the field of dielectric can be described as the field of a virtual dipole  c at the center of the cavity, given by: 
c
 2  3   1  (3.14) The presented model involves a number of simplifications, since the
dipole
is assumed to be
ideal
and
located at the center of the molecule ,
which is supposed to be
continuous dielectric. spherical and surrounded by a
9

Let us now consider the case of a polarizable permanent dipole, having an average dipole 
R
polarizability  .
In this case the reaction field and therefore satisfies the equation:
R
induces a
R

f(
μ
 
R
)
where  is the permanent dipole moment. Therefore: (3.15)
R
 1 
f f
  Eliminating
f
by using (3.13) we have:
R
 2 2 (   1  1  1 )  
a
3 1
a
3 
f

a
1 2 3 2 (    1 )  1 (3.16) (3.17) 10

To give an approximate calculation of the numerical values of we use approximation (3.6) and the relation:
R
in the case pure nonassociated polar liquids containing one kind of particle 
a
3 
n
2
n
2  1  2 (3.18) where
is the index of refraction. The substitution of (3.6) and (3.18) into (3.17) leads to:
R
 4  3
N
2 (  2    1 )
n
2
n
2 3  2  (3.19) The number of particles per
cm 3
,
, can be computed from:
N

d M N A
(3.20) where
being the molecular weight of the substance, Avagadro's number. For usual values of  ( magnitude of the reaction field strength is 5-50 10 6
density and 0.5-5 D ) the order of
A 11

Under the influence of the reaction field
; the increased moment is:
μ
*

μ
 
R
(3.21) The combination of (3.21) and (3.16), and taking into account (3.13) gives:    1  
f
  1  
a
3  2 2 (     1 ) 1 (3.22) Using (3.18) we can calculate the increase of the dipole moment:     2  2    1
n
2
n
2  2 3 (3.23) 12

In the case of polar dielectrics, the molecules have a permanent dipole moment  , and both parts of the fundamental equation (equation 2.49 from previous lecture) must be taken into account.  4  1 
E =

k N k
  
k
(
E
i
)
k
 
k
2 3
kT
(
E
d
)
k
  In the case of non-polar liquids the internal field can be considered as the sum of two parts; one being the cavity field and the other the reaction field of the dipole induced in the molecule
E
i
=E
c
+R.
For polar molecules the internal field can also built up from the cavity field and the reaction field, taking into account now the reaction field of the total dipole moment of the molecule .
13

The angle between the dipole moment and the
of the permanent part of the
will be constant during the movements of the molecule.
It means that
Therefore, this reaction field moment of the molecule under consideration, and does not contribute to the directing field
R E
d does not influence the direction of the dipole .
On the other hand, the reaction field does contribute to the internal field
E
i , because it polarizes the molecule. As a result, we find a difference between the internal field
E
i and the directing field
E
d .
Since the reaction field
R
belongs to one particular orientation of the dipole moment, the difference between
E
i and
E
d will give by the value of the reaction field averaged over all orientations of the polar molecule:
E
i
E
d

R
(3.24) 14

The direction field
E
d can be obtained by the following procedure:  )
polarizability; of a molecule without changing its  ) let the surrounding dielectric adapt itself to the new situation;  ) then fix the charge distribution of the surroundings and central molecule.
the The average field in the cavity so obtained is equal to the value of
E
d that is to be calculated, since we have eliminated the contribution of
R
to
E
i by removing the permanent dipole of the molecule .
With the aid of this procedure it is possible to calculate
E d
in a simple way. If in the above procedure we had also to removed the polarizability of the molecule before fixing the charge distribution of the surroundings, we would formed a
, i.e. a cavity where the surroundings are allowed to adapt themselves to the new situation.
15

For a spherical cavity, the homogeneous cavity field (2.14) from the previous lecture:
E
c
E
C
 2  3   1
E
is given by eqn To calculate
E
d , the polarizability of the molecule must now be taken into account. The field
E
d causes a dipole 
E
d with reaction field
f

E
d where
f
is the reaction field factor given by (3.13). Thus,
E
d will be given by the equation:
E
d

E
c

f

E
d
(3.25) or
E
d
 1  1
f

E
c
(3.26) 16

Combining (2.14) and (3.26) we have:
E
d
 1  1
f
 2  3   1
E
(3.27) If the dielectric consists of different kinds of molecules, we must calculate the directing field for each kind of molecule separately. We then have for the k-th kind of molecule: (3.28) (
E
d
) k  1  1
f k

k
2  3   1
E
with
f k

a
1 2 ( 3
k
2    1 )  1 (3.29) where  is the dielectric constant of the mixture and
a
k the cavity belong to the particle of the
kind.
the radius of 17

The internal field is now found from eqn.(3.24):
E
i

E
d

R
(3.30) where
R
is the total reaction field connected with the permanent part of the dipole moment. This will be given by the expression for the reaction field of the polarizable point dipole, (equation 3.16):
R
 1 Therefore,
is given by: 
f f
 
R
 1 
f f

μ
(3.31) in which

is the value of  averaged over all orientations. given by equation (2.46) from the previous lecture:

μ
  2 
E
d
Hence: 18 is

E
i
=
E
d
 1 
f f
  2     1  1 
f f
  2    1  1
E
d

f
 2  3   1
E
(3.32) For a mixture of different kinds of molecules we find for the internal field at
kind of molecule:  
i
k
=
  1  1 
f k f k

k
 2
k
   1  1
f k

k
3  2   1
E
(3.33) It must be stressed that all these relations are hold only
It is not necessary that the dipole be situated in the center of the sphere, however, if the polarizability of these particles is considered to be
in the center of the molecule or homogeneously distributed over the whole sphere. 19

We can now substitute eqns. (3.28) and (3.33) for the directing and the internal field, respectively, into the fundamental equation, eqn. (2.49) from the previous lecture: We then find:   1 4 
E =

k N k
  
k
(
E
i
)
k
  2
k
3
kT
(
E
d
)
k
  (  1)(2  12  + 1)
=

k N k
1  1
f k

k
  
k
 1 3
kT
1  
f
 
k

k
  (3.34) in which
f k
is given by (3.29):
f k

a
1 2 ( 3
k
2    1 )  1 20

Taking for the radius of the cavity and the polarizability the values of these quantities as found for each pure compound (eqn. (3.6) and (2.59) from the pervious lecture)
4

Na k
3
3
 1      1  2  4  3 
k N k

k
we obtain for the ratio 
k a
3
k

k a k
3  ( (     ) )
k k
 1  2 (3.35) 21

From this we find: 1  1
f k

k
   
k
    
k
  1  (3.36) Hence, (3.34) can be written in the form: (  1) 4 
=

k N k
  2  
k
    2
k
   
k
   3  2
k
   2   2   
k
  1     3
kT
   (3.37) For pure dipole liquids it is useful to substitute  from the equation of Clausiuss-Mossotti eqn.(2.59)  1 4 
=
3  4   2      1        2  2     2    2  1  
N
  9
kT
(3.38) Hence: 22

 2  9
kT
4 
N
(     (    )( 2    2 ) 2   ) (3.39)
This equation is generally called the Onsager equation . It makes possible the computation of the permanent dipole moment of a molecule from the dielectric constant of the pure dipole liquid if the density and
 
are known.
23