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In the case of
orientation polarization we have
a material with built-in dipoles that
are independent of each other, i.e. they
can rotate freely - in sharp contrast to
ionic polarization. |
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The prime example is liquid water, where every water molecule is a
little dipole that can have any orientation with respect to the other
molecules. Moreover, the orientation changes all the
time because the molecules
moves! Orientation polarization for dielectric dipoles thus is
pretty much limited to liquids - but we will encounter it in a major way again
for magnetic dipoles. |
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A two-dimensional "piece of
water" may - very graphically - look somewhat like the picture below that
captures one particular moment in time. It
is like a snapshot with a very, very short exposure time. A few nanoseconds
later the same piece of water may look totally different in detail, but pretty
much the same in general. |
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In a three-dimensional piece of water
the blue and red circles would not have to be in the same plane; but that is
easy to imagine and difficult to draw. |
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Shown is a bunch of water molecules
that form natural dipoles because the negatively charged oxygen atom and the
two positively charged H - atoms have different centers of charge. Each
molecule carries a dipole moment which can be drawn as a vector of constant
length. If we only draw a vector denoting the dipole moment, we get - in two
dimensions - a picture like this: |
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Again, remember that both pictures
are "snap shots" that only appear
unblurred for very small exposure times, say picoseconds, because the dipoles
wiggle, rotate, and move around rather fast, and that in three dimensions the vectors would also point out of
the drawing plane. |
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The total dipole
moment is the vector sum of the individual dipole moments.
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For dipoles
oriented at random, at any given moment
this looks like the picture below if we draw all vectors from a common origin:
The sum of all dipole moments will be zero, if the dipoles are randomly
oriented. |
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We can see this most easily if we have all
dipoles start at the same origin. The picture, of course, is two-dimensional
and crossly simplified. There would be a lot more (like 10
20) dipoles for any appreciable amount of water - you really will
average them to zero pretty well. |
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In reality, the orientation into the
field direction will be counteracted by random
collisions with other dipoles, and this process is energized by the
thermal energy "kT"
contained in the water. |
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Again, the dipoles are not sitting
still, but moving around and rotating all the time - because they contain
thermal energy and thus also some
entropy. |
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Whenever two molecules collide, their
new orientation is random - all memory of
an orientation that they might have had in the electrical field is lost.
This is
analogous to what happens to electrons carrying an electrical current in an
electrical field. |
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The electrical field only induces a
little bit of average orientation in field
direction - most of the time an individual dipole points in all kinds of
directions. This is the simple truth even so some (undergraduate) text books
show pictures to the
contrary. The "real" picture (in the sense of a snapshot with a
very short exposure time) looks like this: |
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| Without field |
With field |
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The orientation of all dipoles is
just a little bit shifted so that an average orientation in field direction
results. In the picture, the effect is even exaggerated! |
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In fact, the state of being liquid by necessity implies quite a bit of
entropy, and entropy
means disorder. |
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Perfectly aligned dipoles would be in
perfect order without any entropy - this is
only possible at extremely low temperatures (and even there quantum theory
would not allow it) where we will not have liquids any more, or more generally,
dipoles that are able to rotate freely. |
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In other words, we must look for the
minimum of the free enthalpy G
and not for the minimum of the internal
energy U. At finite temperatures the minimum of the free
enthalpy requires some entropy
S, i.e.
randomness in the dipole orientation, so we should not expect perfect
orientation. |
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If you are not
familiar with the basics of thermodynamics, you have a problem at this point.
If you do know your thermodynamics, but are a bit insecure, turn to the basic
module "Thermodynamics" (in
the "Defects"
Hyperscript) to refresh your memory. |
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We obviously need to calculate the
free enthalpy G =
U – TS to see what
kind of average orientation will result in a given field. Note that we use
U, the common symbol for the (internal) energy instead of
H, the common symbol for the enthalpy, because U
and H are practically identical for solids and liquids anyway.
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Moreover, a mix up
with the magnetic field strength usually designated by H, too,
would be unavoidable otherwise. (The possible mix-up between internal energy
U and voltage U is not quite so dangerous in this
context). |
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The internal energy od a dipole is
clearly a function of its orientation with respect to the field. It must be
minimal, when the dipole is aligned with the field and thedipole moment has the same direction
as the electrical field, and maximal if the direction is reversed. |
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This is the easy
part: The energy U(d) of a dipole with
dipole moment m in a field E as
a function of the angle d ("delta")
between the dipole moment direction and the field direction. |
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From basic electrostatics we have
have |
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| U(d) = |
– m ·
E = |
– |m| · |E|
· cos d |
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The minimum
energy U thus would occur for d = 0o, i.e. for perfect alignment in
proper field direction (note the minus
sign!); the maximum energy for d =
180o, i.e. for alignment the wrong way around. |
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That was for two dimensions - now we must look at
this in three dimensions. |
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In 3D we see that all dipoles
with the same angle d between their axis and
the field still have the same energy - and this means now all dipoles on a
cone with opening angle 2d around the field axis if we consider possible
orientations out of the plane of drawing. |
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In order to obtain the total internal energy
Utotal of a bunch
of dipoles having all kinds of angles
d with the field axis, we will have to sum up
all cones. |
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This means we take the number of
dipoles N(d) having a particular
orientation d times the energy belonging to
that d, and integrate the resulting function
over d from 0o to
180o. This is something that we could do - if we would know
N(d). |
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However, just calcuating
Utotal will not be of much use. We also must
consider the entropy term – TS, because we do not want to
calculate the total internal energy
Utotal, but the total free
enthalpy G = Utotal –
TS. |
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We need to consider that term as a
function of all possible angle distributions and then see for which
distribution we can minimize G. |
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But what is
the entropy S(N(d))
of an ensemble of dipoles containing N(d) members at the angle d as a function of the many possible distribution
N(d)? Not an easy question to answer
from just looking at the dipoles. |
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Fortunately, we do not have to calculate
S explicitly!
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We know a formula for the distribution of (classical)
particles on available energy levels that automatically gives the minimum of the free
enthalpy! |
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We have a classical system where a number of independent
particles (the dipoles) can occupy a number of energy levels (between
Umin and Umax) as defined by
d = 0o or d = 180o, respectively. |
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Basic thermodynamics asserts that
in equilibrium, the distribution of the
particles on the available energy levels is given by the proper distribution function which is defined in such a way
that it always gives the minimum of the
free enthalpy. |
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Since we deal with classical
particles in this approach, we have to use the Boltzmann
distribution. We obtain for
N(U) = number of dipoles with the energy U |
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With a constant A that
has yet to be determined. |
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This Boltzmann distribution equation
gives us the number of dipoles with a certain angle relative to the field
direction, i.e. the number of dipoles that have their tips on a circle with an
opening angle 2d relative to the field
directions as shown below. |
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We are, however, only interested in
the component of the dipole moment parallel to the
field. For this we look at the
solid
angle increment dW defined on
the unit sphere as the segment between d and
d + dd. |
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The number of dipoles lying in
the cone angle increment defined by d and
d+ Dd is the same
as the number of dipoles with tips ending on the surface of the unit sphere in
the incremental angle dW. It is given by
N(U(d)) · dW.
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Note that dW is
a measure of an incremental area; a kind of
ribbon once around the unit sphere. |
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The sum of the components mF of the dipole moments in field direction is then
. |
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| mF = (N ·
dW) · (m ·
cos d) |
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If you are not familiar with
spherical coordinates, this (and what we will do with it), looks a bit like
magic. Since we do not want to learn Math in this lecture, the
essentials to spherical coordinates are
explained in detail in a basic module. |
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The average
dipole moment, which is what we want to calculate, will now be
obtained by summing up the contributions from all the dWs |
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| <mF> =
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p
ó
õ
0 |
N(U(d)) · m · cosd · dW |
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And the integrals have to be taken
from the "top" of the sphere to the "bottom" , i.e. from
0 to p. |
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dW and d are
of course closely
related, we simply have |
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Putting everything together, we
obtain a pretty horrifying integral for mF
that runs from 0 to p |
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| <mF> =
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| m · |
p
ó
õ
0 |
sind ·
cosd · exp |
m · E
· cosd
kT |
· dd |
p
ó
õ
0 |
sind ·
exp |
m · E
· cosd
kT |
· dd |
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One advantage is that we got rid of
the undetermined constant A. The integral, being a determined
integral, is now simply a number depending
on the parameters of the system, i.e. the temperature T, the
dipole moment m and the field strength
E. |
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The problem
has been reduced to a mathematical exercise in solving
integrals. |
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Since we are not interested at doing
math, we just show the general direction toward a solution: |
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Use the substitutions |
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The integral reduces to
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| <mF> = |
| m · |
–1
ó
õ
+1 |
x · exp (b · x) · dx |
–1
ó
õ
+1 |
exp (b · x)
· dx |
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The final result
after quite a bit of fiddling around is |
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With L(b) = Langevin function, named after
Paul Langevin, and defined
as
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y = coth x |
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The "coth" is the hyperbolic cotangent, defined as coth x =
(ex + e–x)/(ex
– e–x) = 1/tanh x. |
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L(b) is a tricky function, because the coth
x part looks pretty much like a hyperbola, from which the real
hyperbola 1/x is subtracted. What's left is almost nothing -
L(x) values are between 0 and 1 |
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The polarization (always on average,
too) is accordingly
. |
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This is a definite result, but it
does not help much. We need to discuss the mathematical construct
"Langevin function L(b)" to get some idea of what we obtained. We look
at the graph in general units and in units of the dipole moment and electrical
field (in red). |
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Since b is proportional to the field strength
E, we see that the dipole moment and the polarization increases
monotonically with E, eventually saturating and giving <µF> =
µ which is what we must expect. |
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The question is, what range of b values
is accessible for real materials. i.e. how close to the saturation limit can we
get? |
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For that we look at some simple
approximations. |
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If we develop L(b) into a series (consult a math textbook), we get
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| L(b) =
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b
3 |
– |
b3
45 |
+ |
2 b5
945
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– ..... |
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For large values of
b we have L(b) » 1,
while for small values of b (b < 1), the Langevin function can
be approximated by . |
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The slope thus is
1/3 for b ®
0. |
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For
"normal" circumstances, we always have b << 1 (see below), and we obtain as final
result for the induced dipole moment
the Langevin - Debye
equation |
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| <mF>
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= |
m2 · E
3kT |
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| <P> |
= |
N · m2 ·E
3kT |
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These equations will be rather good
approximation for small values of m and
E and/or large values of T. For very large fields and very
small temperatures the average dipole moment would be equal to the built in
dipole moment, i.e. all dipoles would be strictly parallel to the field. This
is, however, not observed in "normal" ranges of fields and
temperatures. |
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Lets see that in an example. We
take |
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E = 108
V/cm which is about the highest field strength imaginable before we have
electrical breakdown,
m = 10–2 9 Asm, which is a
large dipole moment for a strongly polarized molecule, e.g. for HCl, and
T = 300 K. |
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This gives us
b = 0,24 - the approximation is still valid.
You may want to consult exercise 3.2-1 again (or for the first time)
at this point and look at the same question from a different angle. |
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At T = 30 K, however,
we have b = 2,4 and now we must think twice:
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1. The approximation would no
longer be good. But |
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2. We no longer would have
liquid HCl (or
H2O, or liquid whatever with a dipole moment), but solid
HCl (or whatever) , and we now look at
ionic polarization and no longer at
orientation polarization! |
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You may now feel that this was a
rather useless exercise - after all, who is interested in the DK of
liquids? But consider: This treatment is not restricted to electric dipoles. It is valid for
all kinds of dipoles that can rotate freely, in particular for the magnetic dipoles in paramagnetic materials
responding to a magnetic field. |
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Again, you may react with stating
"Who is interested in paramagnets? Not an electrical engineer!" Right
- but the path to ferromagnets, which
definitely are of interest, starts exactly where orientation polarization ends;
you cannot avoid it. |
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It is important to be aware of the
basic condition that we made at the
beginning: there is no interaction between the
dipoles! This will not be true in general. |
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Two water molecules coming in close
contact will of course "feel" each other and they may have preferred
orientations of their dipole moments relative to each other. In this case we
will have to modify the calculations; the above equations may no longer be a
good approximation. |
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On the other hand, if there is a
strong interaction, we automatically have
some bonding and obtain a solid - ice in the case of water. The dipoles most
likely cannot orientate themselves freely; we have a different situation
(usually ionic polarization). There are, however, some solids where dipoles
exist that can rotate to some extent - we will get very special effects, e.g.
"ferroelectricity". |
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© H. Föll (Advanced Materials B, part 1 - script)
