In the last couple of lectures, we have been
talking about the various quantities
associated with electromagnetic field. For
instance, we found out that electromagnetic
field carries energy; it carries momentum.
And today we will try to talk about another
application that is electromagnetic field
also carries angular momentum. And in fact
we
will see some very interesting consequences
that might arise, because of conservation
of
angular momentum.
We had introduced the, what we called as a
Maxwell’s stress tensor and in terms of
which we had tried to explain the conservation
of linear momentum. And let us so in
today the first part of the lecture, I will
be talking about angular momentum and its
conservation with respect to electromagnetic
field. And later, we will go over to an
application of time dependent phenomena, namely
propagation of electromagnetic
waves.
.
.As always we have these work hours before
us; the 4 equations known as the Maxwell’s
equation. The del dot of E is rho by epsilon
0, del dot B equal to 0, then Faraday’s
law,
del cross E equal to minus d B by d t, del
cross H, the Ampere Maxwell’s law given
by J
free plus d D by d t. In addition, we have
constitutive relations of this type: D equal
to
epsilon 0 E plus P; H equal to B by mu 0 minus
M. In addition, we have things like, for
instance, the continuity equations; or, for
instance, if you are talking about a metal,
may
be ohms law is valid, and things like that.
So, we will we will be talking about this
as
you go on.
.
This was our expression for the Maxwell stress
tensor, in terms of which we had defined
the momentum density of the electromagnetic
waves. Basically, it has 2 terms- one is for,
from electric field, the other is from the
magnetic field; and the expression is fairly
simple.
..
And, so let us look at, how does one get,
define the angular momentum. The angular
momentum, of course, if electromagnetic field
has a momentum then it stands to reason
that about an origin, let us take any origin,
I should be able to define the angular
momentum by the standard relationship, namely
l is equal to r cross P.
.
So, since, I am talking about the densities,
I am using small letters. So, this is angular
momentum density, which is equal to r cross
the momentum density of the
electromagnetic field. And, so what we will
do is this, that rate of change of angular
.momentum, this is all in terms of the density,
this is equal to r cross; remember that
angular momentum’s change is related to
the torque that acts on it. So, therefore,
r cross
the force, as usual this is rho E that is
the force exerted by the electric field; and
of
course, J cross B, which is the usual Lorentz
force equation, J cross B.
So, therefore, if we, I had already shown
how this expression is simplified by going
over
to Maxwell’s stress tensor; and so as a
result, since I have r cross everywhere, I
will be
able to borrow this expression literally.
So, therefore, what I would get is, d by d
t of the
angular momentum density, plus 1 over c square
r cross s, if you recall that I had 1 over
c
square S, S is the pointing vector, so I have
1 over c square r cross s; and that was equal
to the part which ultimately translated as
the Maxwell’s stress tensor. So, I had epsilon
0
r cross E del dot E, I am not going to repeat
that algebra, minus E cross del cross E, and
a
corresponding term from the magnetic field,
which is r cross B del dot B, minus B cross
del cross E.
Now, we had already simplified this term before.
So, therefore, this would lead to an
equation of this type, d by d t of l, this
I will now put in mech, because this is mechanical
angular momentum, plus the electromagnetic
angular momentum; these are the total
change in the angular momentum of the sources
and the electromagnetic fields. So, that
is equal to r cross, if you recall, this term
was divergence of the second rank tensor,
namely Maxwell’s stress tensor. So, this
of course, is S by c square, this is my
electromagnetic angular momentum, and this
is r cross the divergence of the stress
tensor. So, this is what we had seen. So,
if you now integrate it over the complete
volume, what you would get, would be that
statement of conservation of angular
momentum.
..
And, now that, it is being integrated; the
only dependence is through time. So, therefore,
this is d by d t. L mechanical, plus integral
over the volume of l electromagnetic that
is
the momentum density d cube x. And, this change,
this change results in a, if you like a
flux of torque through the surface, because
this is, we know that rate of change of
angular momentum is related to torque, and
this is of course, the total angular
momentum. So, therefore, this would be a surface
integral, the derivation is exactly the
same as before; and this is r cross Maxwell’s
stress tensor, dotted with d S.
The interpretation, as I said, the left hand
side means, what is the total rate of change
of
the momentum on of the combined system, namely
this angular momentum of the source
and the angular momentum of the electromagnetic
field; and that can change only if
there is a flow through the surface or if
you like it is a flow of torque.
..
There is a, an interesting problem which I
will be doing. And, this is known as the
Feynman paradox; Feynman had talked about
this. We are going to be using a very
similar problem. So, the problem consists
of the following. I have a line charge, infinite
line charge, which has a charge density minus
lambda; surrounding this, surrounding this
is a cylindrical surface, and this cylindrical
surface contains a charge density, surface
charge density which is lambda, which is the
same as the magnitude of this one, divided
by 2 pi, namely the circumference. So, of
course, this then has the dimension of charge
density.
Now, the question is this, that in this situation
there is electric field only within, say if
this radius is a, then only between 0 and
a there is electric field; that is because
we have
adjusted this charge density in such a way,
that is equal and opposite to the charge
density of this line charge. So, as a result,
outside, if you use your standard Gauss’s
law,
the electric field is 0.
So, the question is this; that when, supposing
in, through this, I have a current is flowing,
and I decrease the current to 0; this is,
so if I decrease the current to 0, then you
will find
this cylinder will start rotating. The, how
does it happen that the cylinder start rotating.
So, this is, this as a charge density lambda,
and there is a current density j. So, let
us look
at, why does it rotate? The Feynman paradox
was because the since the system was
initially at rest, it should continue to be
at rest, because otherwise it will violate
angular
.momentum conservation. The reason is that,
when you talk about the conservation of
angular momentum, you have to also take into
account the angular momentum of the
electromagnetic field, and that is what we
going to be talking today.
.
So, let us look at what we said; we said that
angular momentum L was, this is I am
talking about the electromagnetic field angular
momentum only; in my initial angular
momentum of the mechanical system was 0, because
everything was at rest. So, this is r
cross, the pointing vector S by c square;
and, since it is total, I get 2 pi r d r and
integrated overall. The, I know, what is the
electric field. So, electric field is minus
lambda divided by 2 pi epsilon 0 r. And, I
know and of course, it is along the radial
direction; and this is valid for r less than
a only; there is no electric field outside,
because
the charge densities have been adjusted that
way.
So, let us look at, what does it give me for
the momentum charge density. So,
momentum density is 1 over mu 0 c square.
What I am doing is, to write down an
expression for S. S, if you recall is E cross
H. So, H is written as B by mu 0. So,
therefore, this is E cross B, this is; and
that is equal to 1 over mu 0 c square, the
minus
lambda divided by 2 pi epsilon 0 r, r, and
the external magnetic field which we have
put
in; let us say there is an external magnetic
field along the z direction, which is B 0
z.
Now, what I can do is, since I have a cylindrical
geometry, so I have got r phi z. So,
therefore, this quantity gives me mu 0, well
c square; and recall mu 0 times epsilon 0
is
.equal to 1 over c square. So, that will cancel
this c square there, I will left with the
B 0
lambda divided by 2 pi r, and it is azimuthal,
because there is a r cross z with a minus
sign there.
So, if this is my momentum density, the angular
momentum L becomes, let me pull out
the constants first, B 0 lambda divided by
2 pi, integral from 0 to a, r cross phi divided
by r, then 2 pi r d r. So, r cross phi, of
course, is in the z direction. So, this is,
I have got
B 0 lambda divided by 2 pi; 2 pi also actually
goes because of this, there is an another
factor there; and this as the magnitude r,
so r and r goes; I am left with r square by
2. So,
when integrated gives me, a square by 2 and
the direction z. So, this is the situation
that I
have got an angular momentum the field, which
I have calculated, is given by this
expression. Now, what happens, when we switch
off the magnetic field?
.
Now, when I switch off the magnetic field,
in the presence of the magnetic field, there
was a flux through the, this surface here;
the, any of the cross sectional surface there
was
a flux. And, when this magnetic field is,
let say slowly reduced to 0, there is a change
in
the flux; now this change in the flux, because
you are reducing the magnetic field in the
z
direction from sum value is 0 to 0 results
in an azimuthal current; and this azimuthal
current I can calculate.
So, this is, this is the expression for my
angular momentum of the electromagnetic field,
which actually happens to be the total angular
momentum, because my system was,
.mechanical system was at rest. So, this is,
this is the initial angular, total angular
momentum. Now, let us then see what is happening
at the end.
.
So, what I have done is, I have gradually
reduced the magnetic field to 0, which has
resulted in a azimuthal current; and I can
calculate that azimuthal current density,
J phi,
which is Q by t. And, this is, there is a
charge which is being, supposing the velocity
that
has come up, the angular velocity that has
come up is omega. So, therefore, this is 2
pi a
sigma, that is the amount of charge that I
had, into omega by 2 pi, and, that is lambda
omega by 2 pi. So, when it rotates, this is
what we get.
Now, these results in a magnetic, a equivalent
magnetic field; let me call it B final that
is
mu 0 J z. So, which is mu 0 lambda omega over
2 pi, along the unit vector z. Now, since
the angular momentum is in the same direction,
I can borrow this expression. So, if I
know that the induced final magnetic field
happens to be B f, then I should be able to
write down what is the final angular momentum
of the field, which is B final lambda a
square by 2, along the z direction. And, if
you plug the value of the B final, which we
just now have calculated, then this works,
what I have is mu 0 by 4 pi lambda square
a
square omega. So, this is my final angular
momentum of the electromagnetic field.
You notice that the initial angular momentum
is not the same as this. So, in order to
compensate for this change in the angular
momentum, my system, let us assume that the
moment of inertia of that system is I, then
I will have I omega, plus this quantity there,
.mu 0 by 4 pi lambda square a square omega;
this quantity must be my l I which is B 0
lambda square a square by 2. So, this is,
this is what I get B, B 0 lambda a square
by 2.
And, this allows us to determine what should
be the value of the, what should be the
value of the angular velocity with which the
disc will rotate.
The summary of this is, that in considering
conservation of angular momentum of a
system which has sources, and sources of electromagnetic
field, one should take into
account that total contribution, both the
contribution to angular momentum from the
sources as well as due to the electromagnetic
field.
The next application that I talk about is
that, we have seen that the electromagnetic
field
carries momentum. So, as a result, supposing
I put electromagnetic field in an enclose,
a
cavity, then since this carries momentum,
the walls of that cavity will experience
pressure in the presence of electromagnetic
field; and this actually is something which
you can, sort of verify experimentally, when
electromagnetic waves falls on, let us say
a
mirror. So, the way we will calculate this
is this, that the force, supposing I have
a cavity.
.
So, this is my cavity; inside there is electromagnetic
field; and let suppose I am talking
about this wall, which has this has the direction
of its normal. So, the electromagnetic
waves or electromagnetic field will exert
a pressure on this right hand wall. And, I
know
that this force, supposing I look at a small
area d s on this wall, then the force on that
wall will be given by the Maxwell’s stress
tensor, dotted with d s.
.Now, since my force is in the x direction,
the direction of d s is in the x direction;
all that
I require is just the x x component of the
Maxwell stress tensor, and that we have see
is
epsilon 0, E x square, minus half E square,
plus 1 over mu 0 B x square minus half B
square.
Now, if my radiation is isotropic that it
is exerting identical forces in all directions,
then
it makes sense to say that E x square is actually
a third of E square, and likewise B x
square is one third of B square. So, we plug
this in, what you get is d f, of course, only
the x component, is one sixth epsilon 0 E
square plus one sixth B square over mu 0.
And,
if you look at this expression, you notice
this is one third of half epsilon 0 E square,
plus
half B square over mu 0, which is nothing
but the energy, this is nothing but the energy
density of electromagnetic field. So, this
is energy density divided by 3.
We will talk about some applications of this;
incidentally, this is the starting point for
derivation of something like a Stephen Boltzmann
law. With this I have completed the
discussion of various conservation laws, which
arise or which I have to be taken into
account when we discuss electromagnetic fields.
What I am going to do now, is to go over to
the set of equations that we wrote down
earlier, and obtain what I known as plane
wave solutions to the electromagnetic field
equations. The electromagnetic field equations
are a set of equations which have some
special solutions, and one of those special
solutions is the solution of the plane wave
solution, we will be talking about that.
..
To begin with, I will be talking about a linear
isotropic medium. Linear, implying that
relationship between D and E is linear; the
constant of proportionality is the dielectric
permittivity; and the relationship between
B and the H field are also linear. So, B is
equal
to mu H. I will also talk about regions which
are source free that is the solutions that
I
am going to look for, are in regions where
rho is equal to 0 and J is equal to 0.
.
And, that tells me that in the source free
region, del dot of E must be equal to 0 because
there is no rho; and of course, del dot of
B is always equal to 0. Now, del cross E,
which
.is the Faraday’s law is equal to minus
d B by d t; and del cross B, I had a mu 0
J, which I
will drop because my J is equal to zero, so
this is equal to mu epsilon, let me, let me
take
mu epsilon, rather than mu 0 epsilon 0. If
I had mu 0 epsilon 0, it would mean I am
taking the solution in vacuum, but this is
mu epsilon d E by d t. So, these are the basic
equations with which we will work.
Now, notice, take any one of these equations.
For example, take del cross E. Now, if you
take del cross of this equation, del cross,
del cross E, I get minus d by d t of del cross
B;
and of course, we have seen several times,
this gives me del of del dot of E, minus del
square E, that is equal to minus d by d t
of, for del cross B, I will replace from
Maxwell’s, Ampere’s law, mu epsilon d
E by d t. We had seen that in the source free
region, dell dot E is equal to 0. So, I get
a del square E is mu epsilon, d square E by
d t
square.
.
So, let me write that down. Identically, you
could have converted this into equation for
the B field; and all that you need to do then
is to take the del cross B equation; take
del
cross of this, replace for the del cross of
E from this equation. So, I would get very
similar equation, del square of B is mu epsilon
d square B by d t square.
Now, this, these are equations, which, sort
of, look very simple, but there is a problem.
See, when you take del square of E, it does
not mean the, there is actually 6 equations,
because E and B are vectors, so I have to
take component vice. But del square of E,
if
.you take x component, is not in general equal
to del square of E x; this, however, is true,
if I work in Cartesian coordinate system.
The, if you were to solve this in, for instance,
the spherical polar, then of course, it will
mix up the various components; we had seen
that del square mixes up. But the Cartesian
coordinate system, you of course, have del
square E x, del square of E is d square of
d x square E x, E x plus d square about d
y
square E y, plus d square about d z square
E z. So, as a result solving these in Cartesian
coordinate system is fairly strict simple,
and that is what we are going to do for the
momentum.
Now, what is it that we get now? Look, firstly,
what we have seen is this; we have got
del dot E is equal to 0, and del dot of B
is equal to 0; this will let, I will, I will
return
back to this picture, in a second. So, I am
going to be looking for some special solutions
of this equation. And these are the, what
are known as the plane wave solution. What
are
plane waves? The plane waves are those for
which the surfaces of constant phase, which
are also known as the wave fronts, there planes.
So, for example, a spherical wave would
be, that if you are looking at, finding the
locus of all points which have the same phase,
then in a spherical waves those surfaces are
spheres, but in plane wave the surfaces of
constant phases are planes.
So, the solution, I am looking for is this;
I am looking at E vector as E 0, e to the
power i,
k dot r minus omega t. Now, this is very useful,
because what we can always do is to; of
course, our waves are real. So, what we could
do is to take an exponential form and
ultimately take for example, either real part
or imaginary part. If I take a real part,
I can
for example, if E 0 is taken as a real, then
I will get cosine k dot r minus omega t. But
otherwise, this is the way the algebra becomes
very simple, if you work with this. Now,
and similarly, B is equal to B 0 e to the
power i, k dot r minus omega t.
So, let us, let us look at what is meant by
then surfaces of constant phase. So, surfaces
of
constant phase will be those surfaces for
which k dot r minus omega t is constant. So,
I
am looking at, incidentally I should point
out that, I can also have a plus sign there,
the
difference is in the farmer case, when k dot
r minus omega t is there, it is what is known
as a forward moving wave; and if k dot r plus
omega t is there it is a backward moving
wave. So, the surfaces of the constant phase
will be k dot r, well, that let me take a
minus
r omega t, that is equal to constant.
.So, let us, let us assume that, at any given
time, omega t is constant. So, this implies
that
k dot r is constant. Now, let us say, the
component of r vector along the direction
of k, k
is known as a propagation vector, is zeta
let us say. So, therefore, this zeta, which
is
defined as k dot r, this is constant. So,
therefore, what I have is, minus omega t,
plus k
zeta, that is equal to constant. And, if you
differentiate it now, with respect to time,
what
you get is, at a, sorry, this is, this is
the constant, and so I have got k dot r. So,
that is k
zeta; and this quantity is constant. So, therefore,
I get minus omega; if I differentiate it
with respect to time, I get k d zeta by d
t 
that is equal to omega, which means d zeta
by d
t is omega by k, plus or minus if your chosen
the other sign, and which is equal to plus
or
the velocity.
So, omega by k is the, is what is known as
the phase velocity. We will see, why it is
not
just the velocity, but the phase velocity
later. So, these are the harmonic solutions,
the
plane wave solutions of this equation.
.
So, let us look at the plane wave solution
little more carefully. So, we had already
seen
that del dot of E is equal to 0.
..
Now, since, I have got E is e to the power
i k dot r minus omega t, if you took a
divergence, this will lead to k dot of E is
equal to 0. Parallelly, since del dot of B
is equal
to 0, I will get k dot B is equal 0; which
tells me, remember that we are in a non
conducting medium, which tells me that electric
and the magnetic field vectors are
transverse to the propagating vector. Thus,
that is not all; that is something more that
happens.
Let us take the, one of these equations. Let
say del cross B equation. I know that del
cross B is mu epsilon d E by d t. Now, this
will result in i k cross B; operation of del
is
equivalent to multiplying with i k because
of the exponential form that we have taken.
So, that is equal to the operation. So, therefore,
let me write this, the del operator is same
as multiplication by i k. And, d by d t operator
is same as multiplication by minus i
omega, because I am looking only at forward
moving wave. So, this is minus i omega
mu epsilon E.
Now, what you do is this; i and i will go
away; multiply the, take the cross product
of
both sides, with k for instance. Let us find
out what is k cross k cross B, and that is
equal
to, well, there is a minus sign; I have minus
omega mu epsilon, and then k cross E; this
is
k cross k cross B, and I know, which can be
written as k times k dot B; this left hand
side
is k times k dot B, minus B times k dot k
which is k square, that is equal to minus
omega
.mu epsilon k cross E; k dot B is equal to
0, because the magnetic field is transfers
to k.
So, this term will go away; minus will take
care of that.
So, therefore, the magnetic field B is omega
mu epsilon by k, k cross E. So, you could
rewrite this as omega mu epsilon, unit vector
k cross E. Let us examine this, mu times
epsilon; mu times epsilon, we have seen is
1 over velocity square. This is seen from
this
equation directly, because this is then a
wave equation whose velocity is 1 over mu
epsilon. So, therefore, this quantity is written
as; this is omega; I should have add a k
square there, so therefore, there is a 1 over
k still outstanding there. So, I have a omega
by k, and mu epsilon is 1 over v square k
cross E, and omega by k gives me 1 velocity.
So, therefore, this is 1 over v k cross E.
And, if I am doing this thing in vacuum, then
this
velocity is of course, just the c, namely
the velocity of light. Let us see what I have
got
so far.
.
What I have got is, k dot E is equal to 0,
k dot B is equal to 0, B is equal to 1 over
velocity times k cross E. What is the import
of these equations? These tells me that E
and
B are perpendicular to k; and this tells me
that E and B themselves are perpendicular
to
each other. So, it means that the electric
field, the magnetic field and the propagation
vector k, they form a triad, a right angle
triad; they are mutually orthogonal and they
form a right handed triad.
..
I will take you back to this picture that
I had shown you earlier. So, what it means
is this;
the, since the electric field magnetic field
and the direction of propagation r
perpendicular to each other, the, suppose
I take the direction of propagation as the
z
direction, then electric and magnetic field
will lie in x y plane; and in that plane the
electric field and the magnetic field will
be perpendicular to each other. So, what it
means is this, that if I have a propagation
like this; suppose I have a electric field
this
way, this is magnetic field and this is my
propagation vector.
And, as the wave progress, this one moves
along k, so that, E and B always remains in
one plane. And, what you see there, it is
just a pictorial representation; since, E
is
varying with time sinusoidally, this is the
green picture, perpendicular to that the B
is
also varying with time; one section of this
is mixing, but that is what the whole thing
is
about.
..
The, let me also then talk about a few other
quantities associated with the
electromagnetic field, electromagnetic waves.
For example, we had shown that the
energy density is given by an expression like
this. Half epsilon 0 E square, plus B square
by 2 mu 0. And, we have we have already seen
that B is equal to 1 over v k cross E.
.
So, I can simply rewrite this; that supposing
I want to write down, what is energy
density. So, I have got half, well, there
is epsilon 0 E square, but I am in a medium.
So,
let me just generalize. Half epsilon E square
plus B square by 2 mu 0, 2 mu. And, if I
.take half epsilon out, for instance; I will
get E square plus B square by 2 mu, B square
by
mu epsilon; mu epsilon is, 1 over mu epsilon
is v square. So, 1 over 2 epsilon E square
plus v square B square. So, this is nothing
but E square itself. So, therefore, this is
epsilon times E square. So, this for example,
is one of the ways in which you can write
down the total energy density.
The pointing vector S, or let us take it magnitude
is E cross H, I would digress a little bit
and, sort of, you must have noticed that,
whenever we write pointing vector, we prefer
to
write it as a E cross H, and not as a E cross
B, though occasionally that would be true
if
B and H are linear, but a general expression
is E cross H; the reason is that if you recall,
the H field is the field due to real sources;
and the difference between the B field and
H
filed for the instance, could arise because
of the magnetization; and the magnetisation
currents being bound currents, they cannot
transport energy. So, therefore, E cross H
is
the pointing vector.
And, this is nothing but well, I am talking
about linear medium. So, it is E B by mu 0.
You could also write it as, for example, c
epsilon 0 E square. If you look at the pointing
vector itself, this will be then c epsilon
0, E 0 square; supposing I am looking at the
real
part of that exponential, then cos square
k z minus omega t along the direction k.
The intensity, which is defined as the time
average of the pointing vector; at then take
an
average of this over a time period, which
works out to half. So, this is equal to half
c
epsilon 0, E 0 square k. So, this is the intensity.
..
Now, before I proceed further, I would like
to make a comment on what is known as the
state of polarization; this is something which
is already familiar to you, but let me make
some comments in any case. We have said, let
me for the moment talk only about
electric vector, because the electric vector
B being transverse to the magnetic vector.
So,
if I talk about electric vector, you can make
similar conclusion about the magnetic vector
as well. So, what we have said is that, the
electric vector lies in a plane, perpendicular
to
the direction of propagation. Now, if that
is true. So, this is my direction of propagation,
and let us say that this is a plane which
is perpendicular to it, the electric vector
can be in
any direction there.
Now, so we had written that since electric
vector is perpendicular to the z direction,
I
can express a general electric vector E which
depends upon z and t, by let us say the real
part. Now, since this is in the x y plane,
I can write it as, for example, E 0 x times
i, plus
E 0 y times J, and of course, the exponential
factor e to the power i omega t minus, well,
minus i omega t, plus i k z. Now, in general,
these quantities E 0 x and E 0 y, they are
complex.
..
Now, let us take some special relationship.
So, suppose my E 0 x, supposing is equal to
E
0 x magnitude, times e to the power i phi.
Now, let me also say that E 0 y is also equal
to
its magnitude, and suppose I have the same
phase, suppose let it have the same phase,
then my electric field will be written as,
well, real part of; so E 0 x magnitude i,
plus E 0
y magnitude J, times cosine of k z, minus
omega t, plus phi. Notice, this that, in this
case,
the magnitude of the vector changes from 0,
because cosine can take 0 value, to, of
course, E 0 x square plus E 0 y square. But
its direction remains constant.
So, the electric field vector, as it propagates,
points in the same direction; such a thing
is
what we are calling as linearly polarized
wave. Now, what we could do is, to take
different values; for example, supposing there
is a phase difference between the x
component and the y component, supposing this
is E 0 y e to the power i phi; now this
will then be, in general, an elliptically
polarized wave. If these 2 amplitudes are
equal
and the phase difference happens to be pi
by 2 phi is equal to pi by 2; and E 0 x is
equal
to E 0 y, I get what is known as a circularly
polarized wave.
Next lecture, we will be talking about, how
to obtain from electromagnetic theory
standard relations, like for example the reflection
and refraction of electromagnetic
waves.
.
