Lecture 5.

5 Dielectric Polarizability

1) Microscopic Approach

Earlier we saw, polarization in a system with N dipoles per unit volume can be expressed as

P = N. μ i.e.

P=Nqδ (17)

This gives
P Nqδ
χ = εr − 1 = = (18)
εo E εo E

Hence, greater is δ, greater is χ and hence larger in εr i.e. more polarizable a medium is, more
is its dielectric constant.

Further, polarizability of an ion/atoms in a dipole of type, i, is defined as

𝜇𝑖 P𝑖
α𝑖 = = (19)
Eloc N𝑖 Eloc

where Eloc is the local electric field experienced by an atom or ion or molecule which can be
different than the applied field. However, magnified of local field can be modified quite
significantly the polarization of surrounding medium.

As a result P can expressed as

P = N𝑖 α𝑖 Eloc (20)

Note that equation (20) relates a macroscopic parameter, P, to the microscopic parameters N,
 and Eloc.

As a result, in general, susceptibility is

Nα
χ = εr − 1 = (21)
εo

where  is sum of all types of polarizabilities due to different mechanisms

(electronic+ionic+dipolar+interface), as we will see later.

2) Analytical treatment of Polarizability

Here we discuss the simple analytical solutions for determining the polarizability and
polarization for each of the above polarization mechanisms. The analysis will shed some light
on the dependence of polarizability on the material parameters as well as any external
parameters.

Electronic Polarization
To achieve this, let us first imagine an atom as a perfect sphere, having +Ze charge at the center of the
nucleus and an equivalent –Ze charge of electrons around it. Here R is the radius of the atom.
2R
d

-q +q
E
Figure 1 Schematic of electronic polarization

Under the application of a electric field, E, the force F1, on the charges is given as

F = Ze.E (29)

As a result, positive and negatives charges experience forces in different direction and their center
move away from each other by a distance, d.

However d has to be finite values because this force is balanced by Coulomb force of attraction
between opposite charges which can be calculated by simple electrostatistics and is given as

 q(nucleus ) * q(electrons inside the distance d) 

F2   
 4 o .d 2 
  (4 / 3) d 3  
 Ze * Ze  (4 / 3) R 3    2 2  (30)
    Z e d
 4 o .d 2
  4 o .R 3 
 
 

Equating two forces, gives the equilibrium separation distance do, i.e.
4 o .R3E
do  (31)
Ze

Induced dipole moment, , would be

  Ze.d o or

  4 o .R 3 E (32)

If dipole density was N, the polarization is given as

 P  4 o . N .R 3.E
(33)

Using equation (15), we can get electronic susceptibility, e, and electronic polarizability, e,
as

 e  4 NR 3 and  e  4 o R 3 (34)

This is a very important result as it allows us to work out the electronic contribution to the
dielectric constant. This also shows the linear relationship between polarization and electric
field for electronic polarization mechanism. Another point to note is that polarizability of
atoms with s- and p- atoms i.e. alkali and rate-earth is more than transition elements i.e. d-
atoms due to higher shielding of nucleus by electrons in d-atoms.

The equation (34) also shows that larger the atom is, large is the polarization! Moreover
anions are more polarizable than cations as electrons in anions outer shells are more loosely
bound.

Values for atomic polarizability of some atoms are given below:

e (×1024 cm-3)

Halogens Alkalis

F 2 Li 0.03

Cl 3 Na 0.3

Br 4.5 K 0.9

I 7 Rb 7

Example:

Assume a typical value for dipole density, N, and atomic radius, R, for a material made up of
atoms with spherical symmetry of orbitals, i.e.

N = 4*1019 cm–3 and R ≈ 7*10– 9 cm, yielding

e = 7*10-4

which is a very small value and the relative dielectric constant, r, does not increase by much.
What it says that electronic polarization effect is generally very weak and contributes very
little to the overall polarization. Having said that, it must be borne in mind that some
materials like Si which are covalently bonded and orbital do not have spherical symmetry and
as a result the dielectric constants can be higher (for example Si has about r ~ 12). For
electronic polarization, r is also equal to n2 and hence, if one knows the refractive index, this
works as a check for the values calculated.

Values for some materials where electronic polarization is the principle mechanism of
polarization are given below (source: “Solid State Physics” by N.W. Ashcroft and N.D.
Mermin):

Carbon (diamond) 5.7 Si 12.0

Ge 16.0 SiC 6.7

GaP 8.4 GaAs 10.9

Ionic Polarization
Figure below shows that in an ionic solid, in the absence of external electric field, all the
dipoles (formed by each Na+ and Cl- pair with an equilibrium separation distance as do)
cancel each other due to crystal symmetry and hence net dipole moment is equal to zero.
Remember, in these solids, no dipole rotation is allowed.

However, when a finite field, E, is applied, the force experienced by the ions leads them to
move away from their equilibrium positions , as shown in the figure, giving rise to unequal
dipole moments in different directions and as a result, material will have net dipole moment.

- +
X M


d d0

E=0 E
Figure 2 Schematic of the ionic polarization
As you can now see, that the distance between the ions decreases by ‘d’ in one direction and
increase by ‘d’ in opposite direction.

Now we calculate the magnitude of ‘d’.

The force, F1, which increases the distance between the ions of charge, q, can be expressed as

F1 = q . E (35)

However, there is a force, F2, in other direction trying to restore the equilibrium between the
ions which is expressed as

F2 = ki . d (36)

Where ki can be considered as spring constant of the bond between ions assuming ionic
dipoles behave like springs.
The spring constant can be expressed in terms of elastic modulus, Y, and can be expressed as

ki = Y . do (37)

At equilibrium, F1 = F2 and by combining equation (36), (.437) and (38), we get an estimate
of d i.e.
𝑞.𝐸
𝑑= (38)
𝑌.𝑑𝑜

Hence, induced ‘extra’ dipole moment, , will be

𝑞2 .𝐸
𝜇 = 𝑞. 𝑑 = = 𝛼𝑖 𝐸 (39)
𝑌.𝑑𝑜

where i is ionic polarizability. Thus the polarization, P, can be written as

N .q 2 .E
P  N . 
Y .d o
(40)

where N is the dipole density per unit volume.

Note: Here we considered electric field, E, parallel to the main crystallographic axis. If this is
not the case, one needs to take the component of dipole moment in the direction of the field
before adding them together.

This is a very rough guide to calculation of ionic polarization and can be more complex in
case of many ionic solids, especially when ions do not have similar charges. For example, in
calcium fluoride, CaF2, a material used for making lenses for lithography machine, the
dielectric constant is approximately equal to n2. This enable the lenses to made of
dimensions of about 0.1 m.

Above expressions also tell us that stronger bonds lead to smaller polarization which seems
obvious because then you wouldn’t be able to stretch the atoms too far!

Following are values of dielectric constant of some materials where ionic polarization
contributes (in addition to electronic polarization of course) (source : “Solid State Physics”
by N.W. Ashcroft and N.D. Mermin):

ZnO 4.6 ZnS 5.1

ZnSe 5.8 CdS 5.2

CdSe 7.0 BeO 3.0

MgO 3.
Dipolar or Orientation Polarization
In case of dipolar polarization, we have materials where dipole are present independent of
each other i.e. they don’t interact and they can be rotated freely by an applied field unlike in
case of ionic polarization.

One such example is water where each H2O molecule has dipole moment and these dipole
moments are free to rotate and can have any orientation with respect to neighbouring
molecules. Due to ability of molecules to move around randomly, liquids like water have
very limited dipolar polarization contribution despite having a permanent dipole moment for
each molecule.

A two dimensional image of water molecules is shown below which should remain almost
similar at any instant. Arrow depict the direction of dipole moment associate with each
molecule.

=0

Randomly oriented dipole moments cancel

each other in the absence of any electric field.

Figure 3 Dipoles in a polar material

It is quite obvious by looking at the figure that in the absence of applied field, all the
moments are randomly distributed (huge number for even a small amount of water, say 10
ml) and net dipole moment would be zero. Now when we apply an external electric field say
parallel to the paper along +X-axis as shown below, the dipoles would tend to align with the
applied field to lower their electrostatic energy and lead to minimum dipole energy (basically
positive end of dipole would like to join the negative end of the applied field).

For instance, water does have a pretty large dielectric constant of ~80 which means that there
is obviously some orientation along the field.

Excercise:

You can show it to yourself that this dielectric constant is several orders of magnitude
smaller than for fully oriented dipole moments at some normal field strengths (see
here for solution).
In reality, tendency of dipoles to orient along into the field direction will be counteracted by
random collisions between molecules, a process driven by the thermal energy ‘kT’ at a
temperature ‘T’. Moreover, the dipole containing molecules move around at all times due to
thermal energy and associated entropy i.e. disorder. So, basically when we apply external
field, there is a competition between field induced order and thermal driven disorder. Latter
prevails in liquids more than in solids. Hence, perfect alignment of dipoles would be possible
only at very low temperature and possibly in solid crystals.

As a result of intermolecular collisions and thermally induced random movement, the dipoles
do not completely align themselves along the field unless the field is extremely high. So
basically, the configuration of dipoles looks something like shown below.
E
+
Without field
+
With field

Figure 4 Effect of electric field on the dipole orientation

This basically highlights that the orientation of all the dipoles is just a little bit shifted
towards the field direction leading to an average non-zero dipole moment in the direction of
the applied electric field. Thermodynamically speaking, we will minimize the free energy, G
which is equal to H – TS where H is the internal energy and S is the entropy. So for this, now
we take the following configuration where the charge dipole of an molecule makes an angle,
, to the applied field.

+
 
+
E

Figure 5 2-D schematic diagram of a charge dipole aligned w.r.t. applied field
 The internal energy, U, of a dipole depends on its orientation with respect to the
applied field, i.e. it is function of ‘’. U is minimum when the dipole is completely aligned
with the field i.e.  = 0 and maximum if  = 180. The energy U() of a dipole with dipole
moment, μ, under a, applied field, E, can be written as

U(δ) = -μ.E = - |μ| |E| cos  (41)

What it means in 3-D world is that we would have a cone of dipoles making angle  around
the applied field E which points along the axis of the cone as shown below. The total internal
energy of the whole material would be sub of internal energies of all these cones.





Dipole cone made by

all dipoles at one angle
d to the applied field
Figure 6 Schematic representation of dipoles around the electric field
Now we need to calculate the number of dipoles at an angle , N() which can then
be multiplied with the energy for that  and then be integrated between  ranging from 0 to
180 to give us the total internal energy. However, to calculate the free energy, we also need
to calculate the entropy, S, which is not very straightforward.

This is where Boltzman statistics (from classical thermodynamics) rescues us which

allows us to work out the minimum of the free energy using a distribution function. We treat
our system, to a good approximation, as a classical system containing a number of
independent entities (the dipoles), occupying various energies as determined by their angle
w.r.t. applied field. The distribution of these dipoles at various energies can be expressed by a
distribution function in such a manner that it minimizes the free energy of the system.

Now, at a given temperature, T, number of dipoles with energy U can expressed as

 U   
N (U ( ))  A exp   
 kBT  (42)

where A is an constant and kB is Boltzmann constant. Equation (42) provides us the number
of dipoles in a cone as shown above. Now, we should calculate the component of the dipole
moment parallel to the applied field using the solid angle dΩ for an unit sphere in the
segment between  and  + d.

dW


Dipole cone

Figure 7 Spherical representation of dipoles

This implies that the number of dipoles between  and  + d is equal to N(U()) · dΩ.

The total dipole moment is nothing but the sum of the components μE of the dipole moments
in direction of applied field i.e.

μE = (N dΩ) × (μ cos ) (43)

The average dipole moment, <μE>, is calculated by adding up the contributions from all the
solid angles i.e.


E 

0
N (U ( ))  cos  d W

0
N (U ( )) d W
(44)

The incremental solid can be expressed

dΩ = 2π sin d (45)

which together with (42) yields

   E cos  
0
sin  cos  exp 
 k BT 
 d
E 
   E cos  
0 sin  exp  kBT  d
(46)

To solve this equation, we make substitutions as = (μ E/ kBT) and x = cos  in the above
equation which yields
 1
 . x.exp   x  dx
E  1
1
 exp   x  dx
1 (47)

or rather

E  .L   
(48)

where L(β) = Langevin function, named after Paul Langevin, and is defined as

L(β) = coth (β) –(1/β) (49)

The function coth(x) which is nothing but (ex+e–x)/(ex–e–x). We won’t go into details of L(x)
except that it is not an easy function to deal with and L(x) values are between 0 and 1 for our
purposes. Langevin function is plotted as shown below.

L(  ) Asymptotic values

0 
1 3 10
Figure 8 Schematic plot of Langevin function

As we can see, for large values of β (i.e. large electric fields or very low temperatures, close
to 0 K),

L(β) ≈ 1 (50)

while for small values of β (β < 1) i.e. nominal fields and higher temperatures, the slope is 1/3
for β → 0 and hence the Langevin function is approximated by

L     13 
(51)

In general, β is always much smaller than one i.e. β << As a result, the induced dipole
moment, often called as Langevin - Debye equation is given as

2E
E   d E
3k BT
(52)
Where d is dipolar polarizability. Now, the average polarization is

N  2E
P
3k BT
(53)

These equations hold pretty well for small values of μ and E and/or large enough T which we
can presume as normal conditions. In fact when fields are very high and temperature is very
low (i.e. minimizing the randomization), all the dipoles would be parallel to the applied field
and then the average dipole moment would be equal to the theoretical dipole moment.

Equation (52) also shows the temperature dependence of dipolar polarizability as  = .E
unlike in the case of electronic and ionic polarization.

3) Summary

Here, we calculated the electronic and ionic polarizability and the polarizations of the
dielectric using electronic charge dipoles and ionic lattice having charges coupled as springs.
In case of electronic polarizability, we see that large the ion is, higher the polarizability is. On
the other hand, ionic polarizability is inversely proportional to the elastic modulus which is
related to the bond strength. Hence, higher the bond strength is, lower is the ionic
polarization. Finally we calculated the dipolar polarization using random dipole model and
Boltzman statistics. Here, we saw that polarizability is inversely related to the temperature
which explains the drop in the polarization of a polar solid upon increasing the temperature.
Next, we will investigate the behavior of dielectric under alternating fields.