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Unformatted text preview: 11 Inside Dielectrics 11—1 Molecular dipoles In this chapter we are going to discuss why it is that materials are dielectric. ll—l Molecular dipoles
We said in the last chapter that we could understand the properties of electrical
systems with dielectrics once we appreciated that when an electric ﬁeld is applied
to a dielectric it induces a dipole moment in the atoms. Speciﬁcally, if the electric 11—3 Polar molecules; orientation ﬁeld E induces an average dipole moment per unit volume P, then K, the dielectric polarization
constant, is given by 11—2 Electronic polarization 11—4 Electric ﬁelds in cavities of a K — ] == (H 1) dielectric
e E '
0 11—5 The dielectric constant of
We have already discussed how this equation is applied; now we have to dis “quills; the ClauSius'MOSSOtﬁ cuss the mechanism by which polarization arises when there is an electric ﬁeld equatlon
inside a material. We begin with the simplestposhsible example—the polarization 11_6 Solid dielectrics
of gases. But even gases already have complications: there are two types. The _ I I
molecules of some gases, like oxygen, which has a symmetric pair of atoms in each 11"7 F 91' 1' oelecmcny; BaTIOa molecule, have no inherent dipole moment. But the molecules of others, like water
vapor (which has a nonsymmetric arrangement of hydrogen and oxygen atoms)
carry a permanent electric dipole moment. As we pointed out in Chapters 6 and 7,
there is in the water vapor molecule an average plus charge on the hydrogen
atoms and a negative charge on the oxygen. Since the center of gravity of the nega ReVieW Chaplet 31, VOL I, The Origin tive charge and the center of gravity of the positive charge do not coincide, the 0f the Ref’aC‘il’e Index ‘
total charge distribution of the molecule has a dipole moment. Such a molecule is Chapter 40, _Vf’1 1’ The rm'
called a polar molecule. In oxygen, because of the symmetry of the molecule, the 01.0793 of 51011311“, MeC/wm“ centers of gravity of the positive and negative charges are the same, so it is a
nonpolar molecule. It does, however, become a dipole when placed in an electric
ﬁeld. The forms of the two types of molecules are sketched in Fig. 11—1. 11—2 Electronic polarization We will ﬁrst discuss the polarization of nonpolar molecules. We can start with
the simplest case of a monatomic gas (for instance, helium). When an atom of
such a gas is in an electric ﬁeld, the electrons are pulled one way by the ﬁeld while
the nucleus is pulled the other way, as shown in Fig. 10—4. Although the atoms are (a)
very stiff with respect to the electrical forces we can apply experimentally, there is a
slight net displacement of the centers of charge, and a dipole moment is induced. For small ﬁelds, the amount of displacement, and so also the dipole moment, is a CENTER OF
+ AND ‘ CHARGE proportional to the electric ﬁeld. The displacement of the electron distribution
which produces this kind of induced dipole moment is called electronic polarization. We have already discussed the inﬂuence of an electric ﬁeld on an atom in
Chapter 31 of Vol. I, when we were dealing with the theory of the index of refrac CENTER OF
tion. If you think about it for a moment, you will see that what we must do now is ' CHARGE e
exactly the same as we did then. But now we need worry only about ﬁelds that do CENTER OF . . . . . . . + CHARGE not vary With time, while the index of refraction depended on timevarying ﬁelds. In Chapter 31 of Vol. I we supposed that when an atom is placed in an oscilla
ting electric ﬁeld the center of charge of the electrons obeys the equation Fig, 1 11, (a) An oxygen molecule
with zero dipole moment. (b) The water
molecule has a permanent dipole moment
Po (b) d2x 2
m W + mwox = qu. (11.2) 11—1 The ﬁrst term is the electron mass times its acceleration and the second is a restoring
force, while the righthand side is the force from the outside electric ﬁeld. If the
electric ﬁeld varies with the frequency w, Eq. (11.2) has the solution x — ——‘E— (11.3) ‘ m(wg — (.12) ’ which has a resonance at w = we. When we previously found this solution, we
interpreted it as saying that we was the frequency at Which light (in the optical
region or in the ultraviolet, depending on the atom) was absorbed. For our
purposes, however, we are interested only in the case of constant ﬁelds, i.e., for
w = 0, so we can disregard the acceleration term in (11.2), and we ﬁnd that the displacement is
qu x = mwz (11.4)
0
From this we see that the dipole moment p of a single atom is
2
_ _ qu _
p — qex — mwg (11.5) In this theory the dipole moment p is indeed proportional to the electric ﬁeld. People usually write
p = aeoE. (11.6) (Again the 60 is put in for historical reasons.) The constant a is called the polariz
ability of the atom, and has the dimensions L3. It is a measure of how easy it is to
induce a moment in an atom with an electric ﬁeld. Comparing (11.5) and (11.6),
our simple theory says that " 2 2
eomwo mwo 2 2
‘16 47” . (11.7) If there are N atoms in a unit volume, the polarization P—the dipole moment
per unit volume—is given by P = Np = NaeoE. (11.8) Puttmg (11.1) and (11.8) together, we get P
K—1=;0—E=Na or, using (11.7),
47rNe2
K — l — mwz  (11.10) 0 From Eq. (11.9) we would predict that the dielectric constant K of different
gases should depend on the density of the gas and on the frequency we of its optical
absorption. Our formula is, of course, only a very rough approximation, because in Eq.
(11.2) we have taken a model which ignores the complications of quantum me
chanics. For example, we have assumed that an atom has only one resonant
frequency, when it really has many. To calculate properly the polarizability a of
atoms we must use the complete quantummechanical theory, but the classical
ideas above give us a reasonable estimate. Let’s see if we can get the right order of magnitude for the dielectric constant
of some substance. Suppose we try hydrogen. We have once estimated (Chapter
38, Vol. I) that the energy needed to ionize the hydrogen atom should be approxi mately 4
E z (11.11) 11—2 For an estimate of the natural frequency (.00, we can set this energy equal to hwo—
the energy of an atomic oscillator whose natural frequency is coo. We get
w ~ 1 me4 _
0 ~ 2 h3 If we now use this value of we in Eq. (11.7), we ﬁnd for the electronic polarizability (11.12) The quantity (hZ/mez) is the radius of the groundstate orbit of a Bohr atom (see
Chapter 38, Vol. I) and equals 0.528 angstroms. In a gas at standard pressure and
temperature (1 atmosphere, 0°C) there are 2.69 X 1019 atoms/cm3, so Eq. (11.9)
gives us K = l + (2.69 X 1019)l61r(0.528 X 10—8)3 = 1.00020. (11.13) The dielectric constant for hydrogen gas is measured to be
Kexp = 1.00026. We see that our theory is about right. We should not expect any better, because
the measurements were, of course, made with normal hydrogen gas, which has
diatomic molecules, not single atoms. We should not be surprised if the polariza
tion of the atoms in a molecule is not quite the same as that of the separate atoms.
The molecular effect, however, is not really that large. An exact quantum
mechanical calculation of a for hydrogen atoms gives a result about 12% higher
than (11.12) (the l67ris changed to 181r), and therefore predicts a dielectric constant
somewhat closer to the observed one. In any case, it is clear that our model of a
dielectric is fairly good. Another check on our theory is to try Eq. (11.12) on atoms which have a
higher frequency of excitation. For instance, it takes about 24.5 volts to pull the
electron oﬂ‘ helium, compared with the 13.5 volts required to ionize hydrogen.
We would, therefore, expect that the absorption frequency 000 for helium would be
about twice as big as for hydrogen and that a would be onequarter as large. We
expect that Khelmm z 1.000050.
Experimentally,
Khellum = so you see that our rough estimates are coming out on the right track. So we have
understood the dielectric constant of nonpolar gas, but only qualitatively, because
we have not yet used a correct atomic theory of the motions of the atomic electrons. 11—3 Polar molecules; orientation polarization Next we will consider a molecule which carries a permanent dipole moment
po—such as a water molecule. With no electric ﬁeld, the individual dipoles point
in random directions, so the net moment per unit volume is zero. But when an
electric ﬁeld is applied, two things happen: First, there is an extra dipole moment
induced because of the forces on the electrons; this part gives just the same kind of
electronic polarizability we found for a nonpolar molecule. For very accurate
work. this effect should, of course, be included, but we will neglect it for the
moment. (It can always be added in at the end.) Second, the electric field tends to
line up the individual dipoles to produce a net moment per unit volume. If all the
dipoles in a gas were to line up, there would be a very large polarization, but that
does not happen. At ordinary temperatures and electric ﬁelds the collisions of the
molecules in their thermal motion keep them from lining up very much. But there
is some net alignment, and so some polarization (see Fig. 11—2). The polarization
that does occur can be computed by the methods of statistical mechanics we
described in Chapter 40 of Vol. 1. 11—3 (b) Fig. 11—2. (0) In a gas of polar
molecules, the individual moments are
oriented at random, the average moment
in a small volume is zero. (b) When there
is an electric field, there is same average
alignment of the molecules. Fig. 11—3. The energy of a dipole
pa in the ﬁeld E is —po  E. To use this method we need to know the energy of a dipole in an electric ﬁeld.
Consider a dipole of moment pa in an electric ﬁeld, as shown in Fig. 11—3. The
energy of the positive charge is q¢(1), and the energy of the negative charge is
—q¢(2). Thus the energy of the dipole is U = q¢(1) — q¢(2) = 44" Vela,
or U = —p0E = —p0Ecos 0, (11.14) where 6 is the angle between 170 and E. As we would expect, the energy is lower
when the dipoles are lined up with the ﬁeld. We now ﬁnd out how much lining up occurs by using the methods of statis
tical mechanics. We found in Chapter 40 of Vol. I that in a state of thermal equili
brium, the relative number of molecules with the potential energy U is proportional
to e—U’”, (11.15)
where U(x, y, z) is the potential energy as a function of position. The same argu
ments would say that using Eq. (11.14) for the potential energy as a function of
angle, the number of molecules at 0 per unit solid angle is proportional to e‘”””. Letting n(0) be the number of molecules per unit solid angle at 0, we have n(6) = n0e+poE°°°9”°T. (11.16)
For normal temperatures and ﬁelds, the exponent is small, so we can approximate
by expanding the exponential: oEcos 0). (11.17) = 710+ LT We can ﬁnd no if we integrate (11.17) over all angles; the result should be just
N, the total number of molecules per unit volume. The average value of cos 0 over
all angles is zero, so the integral is just no times the total solid angle 41r. We get __1K.
—41r (11.18) "0 We see from (11.17) that there will be more molecules oriented along the ﬁeld
(cos 0 = 1) than against the ﬁeld (cos 0 = — 1). So in any small volume contain
ing many molecules there will be a net dipole moment per unit volume—that is,
a polarization P. To calculate P, we want the vector sum of all the molecular
moments in a unit volume. Since we know that the result is going to be in the
direction of E, we will just sum the components in that direction (the components
at right angles to E will sum to zero): We can evaluate the sum by integrating over the angular distribution. The
solid angle at 0 is 21r sin 0 d0, so P = / n(0)po cos 0271' sin 0 d9. (11.19)
0
Substituting for n(0) from (11.17), we have
N " E
__ _ _ ELL
P — 2 0 (1 + M. cos 0)po cos 0 d(cos 0),
which is easily integrated to give
_ NpﬁE
P — 3kT  (11.20) 114 The polarization is proportional to the ﬁeld E, so there will be normal dielectric
behavior. Also, as we expect, the polarization depends inversely on the tempera
ture, because at higher temperatures there is more disalignment by collisions. This
l/T dependence is called Curie’s law. The permanent moment p0 appears squared
for the following reason: In a given electric ﬁeld, the aligning force depends upon
p0, and the mean moment that is produced by the lining up is again proportional
to p0. The average induced moment is proportional to pg. We should now try to see how well Eq. (11.20) agrees with experiment. Let’s
look at the case of steam. Since we don’t know what p0 is, we cannot compute P
directly, but Eq. (11.20) does predict that x — 1 should vary inversely as the tem
perature, and this we should check. From (11.20) we get
P N173 = RTE = 360”, (11.21) Kl so K —— 1 should vary in direct proportion to the density N, and inversely as the
absolute temperature. The dielectric constant has been measured at several
different pressures and temperatures, chosen such that the number of molecules in
a unit volume remained ﬁxed.* [Notice that if the measurements had all been
taken at constant pressure, the number of molecules per unit volume would
decrease linearly with increasing temperature and K — 1 would vary as T‘2
instead of as T‘1.] In Fig. 11—4 we plot the experimental observations for K — l
as a function of UT. The dependence predicted by (11.21) is followed quite well. There is another characteristic of the dielectric constant of polar molecules——
its variation with the frequency of the applied ﬁeld. Due to the moment of inertia
of the molecules, it takes a certain amount of time for the heavy molecules to turn
toward the direction of the ﬁeld. So if we apply frequencies in the high microwave
region or above, the polar contribution to the dielectric constant begins to fall
away because the molecules cannot follow. In contrast to this, the electronic
polarizability still remains the same up to optical frequencies, because of the
smaller inertia in the electrons. 11—4 Electric ﬁelds in cavities of a dielectric We now turn to an interesting but complicated questionthe problem of the
dielectric constant in dense materials. Suppose that we take liquid helium or
liquid argon or some other nonpolar material. We still expect electronic polari
zation. But in a dense material, P can be large, so the ﬁeld on an individual atom
will be inﬂuenced by the polarization of the atoms in its close neighborhood. The
question is, what electric ﬁeld acts on the individual atom? Imagine that the liquid is put between the plates of a condenser. If the plates
are charged they will produce an electric ﬁeld in the liquid. But there are also
charges in the individual atoms, and the total ﬁeld E is the sum of both of these
effects. This true electric ﬁeld varies very, very rapidly from point to point in the
liquid. It is very high inside the atoms—~particularly right next to the nucleus——and
relatively small between the atoms. The potential difference between the plates is
the line integral of this total ﬁeld. If we ignore all the ﬁnegrained variations, we
can think of an average electric ﬁeld E, which is just V/d. (This is the ﬁeld we were
using in the last chapter.) We should think of this ﬁeld as the average over a space
containing many atoms. Now you might think that an “average” atom in an “average” location would
feel this average ﬁeld. But it is not that simple, as we can show by considering what
happens if we imagine different—shaped holes in a dielectric. For instance, suppose
that we cut a slot in a polarized dielectric, with the slot oriented parallel to the
ﬁeld, as shown in part (a) of Fig. 115. Since we know that V X E = 0, the line
integral of E around the curve, 1‘, which goes as shown in (b) of the ﬁgure, should * Stinger, Steiger, and Gachter, Helvetica Physica Acta 5, 200 (1932).
11—5 K1 0004 0.003 0.002 0 0 0.001 0.002 0.003
1/T PK") Fig. 114. Experimental measure ments of the dielectric constant of water
vapor at various temperatures. Fig. l l—5. The ﬁeld in a slot cut in a
dielectric depends on the shape and
orientation of the slot. DIPOLE FIELD 1. 1. Fig. 11—7. The electric ﬁeld of a
uniformly polarized sphere. be zero. The ﬁeld inside the slot must give a contribution which just cancels the
part from the ﬁeld outside. Therefore the ﬁeld E0 actually found in the center of
a long thin slot is equal to E, the average electric ﬁeld found in the dielectric. Now consider another slot whose large sides are perpendicular to E, as shown
in part (c) of Fig. 11—5. In this case, the ﬁeld E, in the slot is not the same as E
because polarization charges appear on the surfaces. If we apply Gauss’ law to
a surface S drawn as in (d) of the ﬁgure, we ﬁnd that the ﬁeld E0 in the slot is
given by E0 = E + 5. (11.22) 60
where E is again the electric ﬁeld in the dielectric. (The gaussian surface contains
the surface polarization charge 03,01 = P.) We mentioned in Chapter 10 that
60E + P is often called D, so eoEo = Do is equal to D in the dielectric. Earlier in the history of physics, when it was supposed to be very important
to deﬁne every quantity by direct experiment, people were delighted to discover
that they could deﬁne what they meant by E and D in a dielectric without having
to crawl around between the atoms. The average ﬁeld E is numerically equal to
the ﬁeld E 0 that would be measured in a slot cut parallel to the ﬁeld. And the ﬁeld
D could be measured by ﬁnding E 0 in a slot cut normal to the ﬁeld. But nobody
ever measures them that way anyway, so it was just one of those philosophical
things. Fig. 116. The ﬁeld at any point A
in a dielectric can be considered as the
sum of the ﬁeld in a spherical hole plus
the ﬁeld due to a spherical plug. For most liquids which are not too complicated in structure, we could expect
that an atom ﬁnds itself, on the average, surrounded by the other atoms in what
would be a good approximation to a spherical hole. And so we should ask: “What
would be the ﬁeld in a spherical hole?” We can ﬁnd out by noticing that if we
imagine carving out a spherical hole in a uniformly polarized material, we are just
removing a sphere of polarized material. (We must imagine that the polarization
is “frozen in” before we cut out the hole.) By superposition, however, the ﬁelds
inside the dielectric, before the sphere was removed, is the sum of the ﬁelds from
all charges outside the spherical volume plus the ﬁelds from the charges within the
polarized sphere. That is, if we call E the ﬁeld in the uniform dielectric, we can
write E = Ehole + Eplugs where E1101e is the ﬁeld in the hole and Eplug is the ﬁeld inside a sphere which is
uniformly polarized (see Fig. 11—6). The ﬁelds due to a uniformly polarized sphere
are shown in Fig. 11—7. The electric ﬁeld inside the sphere is uniform, and its
value is P
Eplug = —§3 (11.24)
Using (11.23), we get
Ehole = E + £3 (11.25) The ﬁeld in a spherical cavity is greater than the average ﬁeld by the amount
P/3eo. (The spherical hole gives a ﬁeld 1/3 of the way between a slot parallel to
the ﬁeld and a slot perpendicular to the ﬁeld.) 11—5 The dielectric constant of liquids; the ClausiusMossotti equation In a liquid we expect that the ﬁeld which will polarize an individual atom is
more like Ehole than just E. If we use the Ehole of (11.25) for the polarizing ﬁeld in 11—6 Eq. (11.6), then Eq. (11.8) becomes P = Naeo(E + (11.26)
60
01'
Na P — m3 60E. Remembering that K — 1 is just P/eoE, we have
Na
K— 1 —1—:ij3 which gives us the dielectric constant of a liquid in terms of a, the atomic polar
izability. This is called the ClausiuSMossotti equation. Whenever Na is very small, as it is for a gas (because the density N is small),
then the term Nat/3 can be neglected compared with l, and we get our old result,
Eq. (11.9), that x — 1 = Na. (11.29) Let’s compare Eq. (11.28) with some experimental results. It is ﬁrst necessary
to look at gases for which, using the measurement of K, we can ﬁnd a from Eq.
(11.29). For instance, for carbon disulﬁde at zero degrees centigrade the dielectric
constant is 1.0029, so Na is 0.0029. Now the density of the gas is easily worked out
and the density of the liquid can be found in handbooks. At 20°C, the density of
liquid CS2 is 381 times higher than the density of the gas at 0°C. This means that
N is 381 times higher in the liquid than it is in the gas so, that—if we make the
approximation that the basic atomic polarizability of the carbon disulﬁde doesn’t
change when it is condensed into a liquid—Na in the liquid is equal to 381 times
0.0029, or 1.11. Notice that the Not/3 term amounts to almost 0.4, so it is quite
signiﬁcant. With these numbers we predict a dielectric constant of 2.76, which
agrees reasonably well with the observed value of 2.64. In Table 11—1 we give some experimental data on various materials (taken
from the Handbook of Chemistry and Physics), together with the dielectric constants
calculated from Eq. (11.28) in the way just described. The agreement between
observation and theory is even better for argon and oxygen than for CS 2—and
not so good for carbon tetrachloride. On the whole, the results show that Eq.
(11.28) works very well. Table 11—1 Computation of the dielectric constants of liquids from the dielectric constant of the gas. Gas Liquid
Substance 1: (exp) Na Density Density Ratio“ Na K (predict) 1: (exp)
CS2 1.0029 0.0029 0.00339 1.293 381 1.11 2.76 2.64
02 1.000523 0.000523 0.00143 1.19 832 0.435 1.509 1.507
CC14 1.0030 0.0030 0.00489 1.59 325 0.977 2.45 2.24
A 1.000545 0.000545 0.00178 1.44 810 0.441 1.517 1.54
‘ Ratio = density of liquid/density of gas. Our derivation of Eq. (11.28) is valid only for electronic polarization in liquids.
It is not right for a polar molecule like H20. If we go through the same calcu
lations for water, we get 13.2 for Na, which means that the dielectric constant for
the liquid is negative, while the observed value of K is 80. The problem has to do
with the correct treatment of the permanent dipoles, and Onsager has pointed out
the right way to go. We do not have the time to treat the case now, but if you are
interested it is discussed in Kittel’s book, Introduction to Solid State Physics. 11—7 Fig. ll—8. A complex crystal lattice
can have a permanent intrinsic polariza
tion P. @ 02 .Ti‘" 0 Bot? Fig. 1 1—9. The unit cell of BaTiO3.
The atoms really fill up most of the space;
for clarity, only the positions of their
centers are shown. 11—6 Solid dielectrics Now we turn to the solids. The ﬁrst interesting fact about solids is that there
can be a permanent polarization built in—which exists even without applying an
electric ﬁeld. An example occurs with a material like wax, which contains long
molecules having a permanent dipole moment. If you melt some wax and put a
strong electric ﬁeld on it when it is a liquid, so that the dipole moments get partly
lined up, they will stay that way when the liquid freezes. The solid material will
have a permanent polarization which remains when the ﬁeld is removed. Such a
solid is called an electret. An electret has permanent polarization charges on its surface. It is the electrical
analog of a magnet. It is not as useful, though, because free charges from the air
are attracted to its surfaces, eventually cancelling the polarization charges. The
electret is “discharged” and there are no visible external ﬁelds. ‘ A permanent internal polarization P is also found occurring naturally in some
crystalline substances. In such crystals, each unit cell of the lattice has an identical
permanent dipole moment, as drawn in Fig. 11—8. All the dipoles point in the same
direction, even with no applied electric ﬁeld. Many complicated crystals have, in
fact, such a polarization; we do not normally notice it because the external ﬁelds
are discharged, just as for the electrets. If these internal dipole moments of a crystal are changed, however, external
ﬁelds appear because there is not time for stray charges to gather and cancel the
polarization charges. If the dielectric is in a condenser, free charges will be induced
on the electrodes. For example, the moments can change when a dielectric is
heated, because of thermal expansion. The eﬂ'ect is called pyroelectrz'cz'ty. Similarly,
if we change the stresses in a crystal—for instance, if we bend it—again the mo
ment may change a little bit, and a small electrical effect, called piezoelectricity,
can be detected. For crystals that do not have a permanent moment, one can work out a theory
of the dielectric constant that involves the electronic polarizability of the atoms.
It goes much the same as for liquids. Some crystals also have rotatable dipoles
inside, and the rotation of these dipoles will also contribute to K. In ionic crystals
such as NaCl there is also ionic polarizability. The crystal consists of a checkerboard
of positive and negative ions, and in an electric ﬁeld the positive ions are pulled
one way and the negatives the other; there is a net relative motion of the plus and
minus charges, and so a volume polarization. We could estimate the magnitude
of the ionic polarizability from our knowledge of the stiffness of salt crystals, but
we will not go into that subject here. 117 Ferroelectricity; BaTi03 We want to describe now one special class of crystals which have, just by
accident almost, a builtin permanent moment. The situation is so marginal that
if we increase the temperature a little bit they lose the permanent moment com
pletely. On the other hand, if they are nearly cubic crystals, so that their moments
can be turned in different directions, we can detect a large change in the moment
when an applied electric ﬁeld is changed. All the moments ﬂip over and we get a
large effect. Substances which have this kind of permanent moment are called
ferroelectric, after the corresponding ferromagnetic effects which were ﬁrst dis
covered in iron. We would like to explain how ferroelectricity works by describing a particular
example of a ferroelectric material. There are several ways in which the ferro
electric property can originate; but we will take up only one mysterious case—that
of barium titanate, BaTiO3. This material has a crystal lattice whose basic cell is
sketched in Fig. ll—9. It turns out that above a certain temperature, speciﬁcally
118°C, barium titanate is an ordinary dielectric with an enormous dielectric con
stant. Below this temperature, however, it suddenly takes on a permanent moment. In working out the polarization of solid material, we must ﬁrst ﬁnd what are
the local ﬁelds in each unit cell. We must include the ﬁelds from the polarization ll—8 itself, just as we did for the case of a liquid. But a crystal is not a homogeneous
liquid, so we cannot use for the local ﬁeld what we would get in a spherical hole.
If you work it out for a crystal, you ﬁnd that the factor 1/3 in Eq. (11.24) becomes
slightly different, but not far from 1/3. (For a simple cubic crystal, it is just 1/3.)
We will, therefore, assume for our preliminary discussion that the factor is 1/3
for BaTiOa. Now when we wrote Eq. (11.28) you may have wondered what would happen
if Na became greater than 3. It appears as though it would become negative. But
that surely cannot be right. Let’s see what should happen if we were gradually to
increase a in a particular crystal. As a gets larger, the polarization gets bigger,
making a bigger local ﬁeld. But a bigger local ﬁeld will polarize each atom more,
raising the local ﬁelds still more. If the “give” of the atoms is enough, the process
keeps going; there is a kind of feedback that causes the polarization to increase
without limit—assuming that the polarization of each atom increases in proportion
to the ﬁeld. The “runaway” condition occurs when Na = 3. The polarization
does not become inﬁnite, of course, because the proportionality between the in
duced moment and the electric ﬁeld breaks d0wn at high ﬁelds, so that our formulas
are no longer correct. What happens is that the lattice gets “locked in” with a high,
selfgenerated, internal polarization. In the case of BaTi03, there is, in addition to an electronic polarization, also
a rather large ionic polarization, presumed to be due to titanium ions which can
move a little within the cubic lattice. The lattice resists large motions, so after the
titanium has gone a little way, it jams up and stops. But the crystal cell is then left
with a permanent dipole moment. In most crystals, this is really the situation for all temperatures that can be
reached. The very interesting thing about barium titanate is that there is such a
delicate condition that if Na is decreased just a little bit it comes unstuck. Since
N decreases with increasing temperature—because of thermal expansion—we can ’vary Na by varying the temperature. Below the critical temperature it is just
barely stuck, so it is easy—by applying an external ﬁeld—to shift the polarization
and have it lock in a different direction. Let’s see if we can analyze what happens in more detail. We call T, the critical
temperature at which Na is exactly 3. As the temperature increases, N goes down a
little bit because of the expansion of the lattice. Since the expansion is small, we
can. say that near the critical temperature Na = 3 — [3(T — Tc), (11.30) where 13 is a small constant, of the same order of magnitude as the thermal expansion
coefﬁcient, or about 10‘5 to 10‘6 per degree C. Now if we substitute this relation
into Eq. (11.28), we get that ﬁ(T—Tc)/3 Since we have assumed that B(T — T c) is small compared with one, we can ap
proximate this formula by 9 K —— 1
This relation is right, of course, only for T > T c. We see that just above the
critical temperature K is enormous. Because Na is so close to 3, there is a tremen
dous magniﬁcation effect, and the dielectric constant can easily be as high as 50,000
to 100,000. It is also very sensitive to temperature. For increases in temperature,
the dielectric constant goes down inversely as the temperature, but, unlike the case
of a dipolar gas, for which K — 1 goes inversely as the absolute temperature, for
ferroelectrics it varies inversely as the difference between the absolute temperature
and the critical temperature (this law is called the CurieWeiss law).
When we lower the temperature to the critical temperature, what happens?
If we imagine a lattice of unit cells like that in Fig. 11—9, we see that it is possible 1 1—9 (0) —e+—e>e>e>—e>
+ t (b) Fig. 11—10. Models of a ferroelec
tric: (0) corresponds to an antiferro
electric, and (b) to a normal ferroelectric. to pick out chains of ions along vertical lines. One of them consists of alternating
oxygen and titanium ions. There are other lines made up of either barium or
oxygen ions, but the spacing along these lines is greater. We make a simple model
to imitate this situation by imagining, as shown in Fig. 11—10(a), a series of chains
of ions. Along what we call the main chain, the separation of the ions is a, which
is half the lattice constant; the lateral distance between identical chains is 2a.
There are lessdense chains in between which we will ignore for the moment. To
make the analysis a little easier, we will also suppose that all the ions on the main
chain are identical. (It is not a serious simpliﬁcation because all the important
effects will still appear. This is one of the tricks of theoretical physics. One does
a diﬂerent problem because it is easier to ﬁgure out the ﬁrst time—then when one
understands how the thing works, it is time to put in all the complications.) Now let’s try to ﬁnd out what would happen with our model. We suppose that
the dipole moment of each atom is p and we wish to calculate the ﬁeld at one of
the atoms of the chain. We must ﬁnd the sum of theﬁelds from all the other atoms.
We will ﬁrst calculate the ﬁeld from the dipoles in only one vertical chain; we will
talk about the other chains later. The ﬁeld at the distance r from a dipole in a
direction along its axis is given by = _.. (11.32) At any given atom, the dipoles at equal distances above and below it give ﬁelds in
the same direction, so for the whole chain we get p 3.( 2 3 l ...)__I:0383.
2+§+27+64+ _eo a3 Echain — (11.33) 41150 a3
It is not too hard to show that if our model were like a completely cubic crystal—
that is, if the next identical lines were only the distance a away—the number 0.383
would be changed to 1/3. In other words, if the next lines were at the distance a
they would contribute only —0.050 unit to our sum. However, the next main
chain we are considering is at the distance 20 and, as you remember from Chapter 7,
the ﬁeld from a periodic structure dies oﬁ exponentially with distance. Therefore
these lines contribute much less than —0.050 and we can just ignore all the other
chains. It is necessary now to ﬁnd out what polarizability a is needed to make the
runaway process work. Suppose that the induced moment p of each atom of the
chain is proportional to the ﬁeld on it, as in Eq. (11.6). We get the polarizing ﬁeld
on the atom from Echm, using Eq. (11.32). So we have the two equations 1’ = “EOEchain
and with E and [7 both ﬁnite. Thus if a is as large as a3/0.383, a permanent polarization
sustained by its own ﬁeld will set in. This critical equality must be reached for
barium titanate at just the temperature Tc. (Notice that if a were larger than the
critical value for small ﬁelds, it would decrease at larger ﬁelds and at equilibrium
the same equality we have found would hold.) For BaTiOa, the spacing a is 2 X 10—8 cm, so we must expect that a =
21.8 X 10—24 cm3. We can compare this with the known polarizabilities of the
individual atoms. For oxygen, or = 30.2 X 10—24 cm3; we’re on the right track!
But for titanium, a = 2.4 X 10—24 cm3 ; rather small. To use our model we should
probably take the average. (We could work out the chain again for alternating 11—10 atoms, but the result would be about the same.) So a(average) = 16.3 X 10‘“,
which is not high enough to give a permanent polarization. But wait a moment! We have so far only added up the electronic polariz
abilities. There is also some ionic polarization due to the motion of the titanium
ion. All we need is an ionic polarizability of 9.2 X 10—24 cm3. (A more precise
computation using alternating atoms shows that actually 11.9 X 10"24 is needed.)
To understand the properties of BaTiO3, we have to assume that such an ionic
polarizability exists. Why the titanium ion in barium titanate should have that much ionic polar
izability is not known. Furthermore, why, at a lower temperature, it polarizes along
the cube diagonal and the face diagonal equally well is not clear. If we ﬁgure out
the actual size of the spheres in Fig. 11—9, and ask whether the titanium is a little
bit loose in the box formed by its neighboring oxygen atoms—which is what you
would hope, so that it could be easily shifted—you ﬁnd quite the contrary. It ﬁts
very tightly. The barium atoms are slightly loose, but if you let them be the ones
that move, it doesn’t work out. So you see that the subject is really not onehundred
percent clear; there are still mysteries we would like to understand. Returning to our simple model of Fig. 11—10(a), we see that the ﬁeld from one
chain would tend to polarize the neighboring chain in the opposite direction, which
means that although each chain would be locked, there would be no net permanent
moment per unit volume! (Although there would be no external electric effects,
there are still certain thermodynamic effects one could observe.) Such systems exist,
and are called antiferroelectric. So what we have explained is really an anti
ferroelectric. Barium titanate, however, is really like the arrangement in Fig.
ll—10(b). The oxygentitanium chains are all polarized in the same direction
because there are intermediate chains of atoms in between. Although the atoms
in these chains are not very polarizable, or very dense, they will be somewhat
polarized, in the direction antiparallel to the oxygentitanium chains. The small
ﬁelds produced at the next oxygentitanium chain will get it started parallel to the
ﬁrst. So BaTi03 is really ferroelectric, and it is because of the atoms in between.
You may be wondering: “But what about the direct effect between the two OTi
chains?” Remember, though, the direct effect dies off exponentially with the
separation; the eﬂ‘ect of the chain of strong dipoles at 2a can be less than the effect
of a chain of weak ones at the distance a. This completes our rather detailed report on our present understanding of the
dielectric constants of gases, of liquids, and of solids. ll—ll ...
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 Spring '09
 LeeKinohara
 Physics

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