Good morning. Last 2 lectures, we have been
working with the wave equation. And today
I am going to continue that go little bit
more in detail, let me remind what you have
done last time.
We have taken a special problem where you
assume the electric field this pointing along
x directions the magnetic field is pointing
along y directions. And we assume that the
electric field along x direction was some
function of z and t that is it did not vary
if moved in x, it did not vary moved in y,
but it did vary if you moved in z. Similarly,
Hy was some other function of z and t. And
last time we worked out and we found that
if you solve this problem you find that the
electric field Ex actually a function of a
very specific form. It is the sum of 1 arbitrary
function depending on z minus ct and another
arbitrary function z plus ct. This f and g
are completely up to us they can be anything
at all where C was equal to 1 over square
root of mu naught epsilon naught this is in
vacuum.
I did not prove it I gave it to as an assignment,
but you can show where the equation satisfied
by Ex namely del squared Ex del z square minus
1 over C squared. I should say 1 over C squared
minus mu epsilon del squared Ex del t square
is 0. This equation we have derived exactly
the same equation can be derived for Hy I
did not prove it I have left it to you work
out. So, it a same equation in both cases
therefore, which is in solution and you will
also end up with Hy is equal to some other
function f delta of z minus ct plus g delta
of z plus ct. So, we have 2 solutions of course,
2 solutions in terms of 4 arbitrary functions
which means that the we have tremendous amount
of freedom means specifying what you want
to specify. Now, in addition what I also did
last time was I said supposing this is x;
this is y; this is z.
Supposing the electric field as a certain
shape, that is it points along the x axis.
There is a triangular profile plus there is
a rectangular profile you imagine both of
these are pointing along x now drawn it only
on z axis. But really when I draw this arrow
it mean that all x and y is the same electric
field because the electric field depends only
on z which uniform in x and uniform in y.
So, now, this take this electric field and
suppose this part of the electric field was
dependent on z minus ct and this part depended
on z plus ct. I think I have inverted the
case the rectangle was actually here now what
we saw was this was z plus ct that was z minus
ct. This triangle moves in the positive z
directions as you vary time and this rectangle
move in a negative z direction. They meet
when they meet super position means that you
end up with complicated shapes. But after
sufficient amount of time this triangle would
have reached of here and this rectangle would
have reach that down here..
So, you would have 
this triangle moving up and this rectangle
moving down, and this is why called waves
they move these are disturbances that are
not like coulombs law at all it is an electric
field. So, you are used to electric fields
which are due to charges and therefore, this
stay put you put a charge somewhere a electric
around the charge is fixed to the charge.
But this electric field is not fixed to a
charge it is just moving there is no charge
in the derivation when we derived this we
assume this there is no charge density. There
is no current density just had a disturbance
and the disturbance self consistently just
keep moving. So, it is a very different object
from what we have seen earlier coulomb law
is a fixed static charges. These kinds of
moving structures are coming out of Maxwell’s
equations the generalized amperes law. Now,
what I am going to do is to relate Ex and
Hy last time I just said where Ex and Hy have
same shape that’s reasonable, because Ex
is connected to Hy through paradise and amperes
law.
So, if there is an triangular Ex there is
something you do not know what does something
at the same z for that t. Because otherwise
you could have curl of E being related to
del b del t of the same point you did not
have curl of H related to del D del t. So,
at the same Z where there is a electric field
there must also be magnetic field and if this
electric field is moving in time like this.
So, 2 must be the magnetic field and if this
electric field is moving backward in Z. So,
2 must be it is magnetic field. So, we can
see that there is a correspondence between
whatever is the forward moving disturbance
due to E and the forward moving disturbance
due to h, because paradise law connects them
up, amperes law connects them up. Similarly,
there is a connection between the backward
moving disturbance g and the backward moving
disturbance c delta. Let us put all of them
little more solidly.
So, what I am going to say is that let z minus
ct be equal to u z plus ct is equal to v alright
and I am going to assume that Ex is some function
of u only there is no v in it. So, I am going
to try and find out what is the kind of H
that comes out of such an electric field.
So, that corresponds to considering only the
triangle part of the disturbance. I am not
doing anything terrible by assuming this.
Because I can always solve this problem then
I can solve a problem where Ex is equal to
some g of v. And then I add up the 2 solutions
that it the wave equation is linear it means
that if I have solve this problem and this
problem and solve the full problem alright.
This is very much like a pendulum problem
x double dot plus kx equal to 0 well I can
study x I put k squared Cos k t I can study
x equals Sin kt. I can study the properties
and then I can write the general form which
is A Cos kt plus B sin kt.
So, just in the same way I can look, at just
E is equal to f u, and I can looks separately
E is equal to g of v. And then I can add up
the linear combination of u there is no need
to add up linear combinations because each
of this is a arbitrary function. So, it includes
a scaling factor as well. So, let us see what
we get I am going to use this form Ex is equal
to f of u, what is Faradays law? Faradays
law tells us curl of E is equal to mu naught
del H del t I know that this is the y direction.
So, I need y components of this. So, the y
component of curl of E is going to be del
del Z Ex is equal to mu del H y del t, so
the minus sign, because it is Faradays law.
What is amperes law curl of H is equal to
plus epsilon naught del E del t now E is in
the X direction, so del E del t is in X direction.
So, I need x component of curl of H which
is minus del del z Hy is equal to epsilon
naught del E del Ex del t I have derived this
several times. So, you should be quite comfortable
with it, now I have assumed Ex is f of u which
is f of z minus ct. So, I can substitute for
Ex portion, so what do I get. I get that minus
mu naught del Hy del t is equal to del del
z of Ex which is df du, because du dz is 1.
What do I do? What do I get here? I get minus
del Hy del z is equal to epsilon naught times
del Ex del t.
So, when I take del u del t when I get a minus
C times df du now as I said H is really f
delta of u plus g delta of v. But you look
at these expressions you find that del H del
t del H and del z of functions of u only there
are not functions at v at all. You can see
the right hand sides have constants multiplying
functions of u which means Hy cannot depend
on v. This is not saying anything surprising
it is simply saying that if that the electric
field disturbance is moving in this way. The
magnetic field disturbance must also move
in the same directions otherwise supposing
they were magnetic disturbance moving other
way. Then what would happen? After certain
time this triangular disturbance would have
reached here. But the magnetic field disturbance
whatever it shape was moving in the opposite
direction which would have reached some way
here. But you cannot have Faradays law connecting
curl of E here to del reach del t here curl
of E is related to del H del t in this point.
So, whatever directions the electric field
is moving magnetic field must move in the
same directions. So, it means there is no
g of v. So, I now know that H y is f delta
of u.
So, I can substitute for these derivative
as well, so what do I get I get minus mu naught
del f delta del t. So, when I take del del
t I get a minus c d f delta du is equal to
df du this 1 says minus del del z does not
give me a factor df delta du is equal to minus
C epsilon naught df du. Actually both of these
expressions are saying the same thing because
C is equal to 1 over square root of mu naught
epsilon not. So, we look at this side this
says mu naught C, which is equal to square
root of mu naught over epsilon not, because
C is 1 over square root of mu naught epsilon
naught mu naught over square root of mu naught
epsilon naught is going to give me a square
root of mu naught the numerator. And this
epsilon naught gives me epsilon naught in
the denominator. This term C epsilon naught
is going to be epsilon naught over square
root of mu naught epsilon naught this is going
to give me square root of epsilon naught over
mu naught.
Now, this is on the left hand side that is
on the right hand side. So, I have bring everything
to the right hand side I get 1 equation out
of this says df tidle du is equal to square
root of epsilon naught over mu naught df du.
This is just a constant this is the u derivative
of the electric field this is the u derivative
of the magnetic field may be integrate both
sides with you, what will you get? You will
get f tilde is equal to this factor square
root epsilon naught over mu naught times f
plus constant the constant does not matter
we are interested in moving objects. So, a
constant is coulombs law. So, we would not
worry about constants, we are only interested
in disturbances of the type which give us
waves. So, it tells us that whatever shape
you take for f you can look like a man you
can look like a triangle you can look like
a square you can look like anything at all.
The same shape is there for the magnetic field
it is scaled by a factor epsilon naught over
mu naught square root. But otherwise magnetic
field and the electric field are the same
function moving in the same direction and
not only that they are even the same function.
So, we write properly the magnetic field points
in the y direction 
and it points that way. So, this electric
field points in this direction magnetic field
point in that direction and the both move
together. It is a very important thing to
understand that in coulombs law where the
charges where the source of electric field.
So, since the charges were stationary electric
field was stationary in this case we look
at this equations actually what they are saying.
Let us look at this kind of equations what
they are saying is that the source of the
electric field is the magnetic field the source
of the magnetic field is the electric field.
So, it means that the electric field causes
the magnetic field which in term causes the
electric field. But there is a space shift
that going on, because of the space shift
they are force to move in order to keep in
step and that motion is nothing more than
the speed of light. This is 3 into 10 to the
8 meters per second I think it is very important
to understand this point. That I have not
meet any assumption about what this shape
was this is arbitrary can be any shape at
all whatever shape I assume magnetic field
of the existing machine. Now, everything as
said which was for f of u can be repeated
for g of v if I do that what I will get is
that this df du will comes dg dv dg dv and
I will be then forced throughout f and keep
g g delta and I write now the equations connecting
g delta and g. I will get once again the same
set of 2 equations which are identical and
they again will say the same thing. And I
will get g delta is square root of epsilon
naught over mu naught g. Once again for an
arbitrary shape in electric field I get the
same shape in the magnetic field scale by
a fact. So, ultimately my picture looks like
this is my z axis y axis x axis.
I have 
my electric field 
this part of the electric field is moving
in the positive z direction this part of the
electric field is moving in z negative direction.
Corresponding to this electric field I have
the magnetic field along the y direction and
corresponding to this also I have an magnetic
field pointing the y direction. And this magnetic
fields Hy Z plus ct in this magnetic field
Hy of z minus ct. So, both these together
moving the positive z direction, both these
together moving in a negative z direction
and if they happen to cross each other which
is very possible as in time they move in opposite
direction. They will go through each other
all that will happen is at every point the
local electric field is the sum of this 2
fields. So, they will distort the shape will
look funny for a while the go right through
each other 1 step. They come through each
other once again you get back this shape down
here further down and this shape appear moving
further up. Now, that we are worked out 1
1 dimensional problem let us do more ordinary
job of the mathematics.
We had the vector relation curl of E equal
to minus del B del t and we had curl of H
is equal to an ignoring currents and charges
at I do not have a j del D del t. Now, I want
to do exactly what I did earlier what I did
was I took this equation and notice this added
time derivatives, but I needed a space derivatives.
So, I took a space derivative of this equation
in order to use this to eliminate B that is
what I did for the 1 d case. So, I am going
to do the same thing, I am going to take a
space derivative of this equation. So, that
I can eliminate this, but which space derivatives
do I need I need curl of H. So, I need to
take curl of this which means I will have
to take curl of the whole equation curl of
curl of E is equal to curl of minus del B
del t. I have just taken this equation and
taken it is curve this is acting on x y z
this is acting on time. So, they are independent
operations, so I can commute them. So, I get
minus del del t of curl of B, but b I am going
to write as mu H minus pull the mu out and
H.
So, I have written B as mu naught times H,
but I know what the curl of H is. So, I can
write this as minus mu naught del del t of
epsilon naught del E del t. So, I am following
exactly the same procedure I followed to get
the solution in the 1 d case, how to get the
equation in a 1 d case? So, finally, I can
write down an equation the equation says curl
of curl of E is equal to a there is a minus
here yes some minus mu naught epsilon naught
del square E del t square. This is the 3 d
equivalent of the 1 d equation except that
there is very nasty business going on here.
This does not look like del square del z squared
of E x which is what we had in the 1 d case
this was there, but this 1 does not at all
look like what we had earlier. So, we need
to do some more work we know at least in the
1 d case it simplify we can try and see and
simplify even in 3 d case. Well I have thought
you the tricks of how to simplify.
Such equations curl curl of E can be written
as epsilon i j k unit vector along i derivative
along j of curl of E along k that what a curl
is the cross product of the gradient operation
and the function here taking the curl of,
but this itself a curve. So, I can write it
as episilon i j k unit vector along I derivative
along j of epsilon k l n I am introducing
2 dummy variables l and m del del x l. This
is what the k th components of curl of E you
would normally have a E sub area unit vector
along k, but I have taken the dot product
of that and the and this operation.
So, the unit vector is got observed you just
have the k th component. So, I can write this
out this epsilon i j k epsilon k l n is of
I and I have del square del xj del xl of Em.
If only this j and l where both equal then
the beginning to look bit like del square
del z squared of Ex you have to now work on
this piece. As I told you epsilon i j k only
cares about cyclic or anticyclic. That is
you can write a circle put 1 2 3. If I j k
involves going clockwise it is one if i j
k involves going counter clockwise it is minus
1. So, I can always if I got 1 2 3 I can write
3 1 2 both of them involves going the same
directions. So, I can write this down as epsilon
k I j epsilon k l m of ei del squared del
xj del xl Em.
Now, k i j must each be different k l m must
be different otherwise this expressions are
0 which means either I equals l j is equals
m or I equals m j equals l if I equals l j
equals m. Both of this are clockwise or both
of these are counter clockwise in 1 case it
is 1 square the other case is minus 1 squared.
So, both ways it becomes plus. So, I will
try i equal l and j equals m. So, E sub I
del square by del xj del xi of Ej, because
i is equal to l j is equal to m minus because
it i is equal to m and j is equal to l. So,
one of them is clockwise one of them is counter
clockwise. So, that is 1 into minus 1, so
minus, so now i is now m j is now l ei del
squared by del xj squared of Ei.
Now, we need to make sense of this it is equal
to ei del del xi acting on Ej del xj minus
del squared del xj square acting on ei Ei
I just collected terms. So, del del xi are
combine with Ei del del xj combine with a
capital ej similarly, here the ei and capital
Ei are combined together. But each of this
are vector operation and let me write down
the vector operation this is gradient and
this is divergence at e. So, it is gradient
of divergence of E minus this is del squared
acting on vector E just simple as that. So,
I can reduce the wave equation to a simpler
form the simpler form is let me rewrite the
equation.
Curl curl E is equal to minus mu naught epsilon
naught del squared E del t squared. This curl
curl E became gradient of divergence of E
minus del squared E is equal to minus mu naught
epsilon naught del squared E del t square.
Now, I am assuming we are working in vacuum
and we have already assume divergence of D
is equal to 0 no charge this D is nothing,
but epsilon naught E. So, divergence of E
is also 0 this goes away which leaves as finally,
with the equation that is the generalization
of the 1 d problem del square E minus mu naught
epsilon naught del squared E del t squared
is equal to 0. Supposing the electric field
depended only on z. Then this del squared
this del squared is del squared del x squared
plus del squared del y squared plus del squared
del z square. But if E depended on z only
this term matters this 2 would be 0 and that
would give me an equation that said del squared
E del z squared minus mu naught epsilon naught
del squared E del t square is equal to 0.
And if further said E is nothing but Ex along
x this is the equation we got earlier. So,
this equation now includes this equation as
a simple special case, what are the kinds
of solutions? We get for this equation before
I discuss that let me go back to the 1 d problem
lets graph a little bit more.
I drawn x y and I said let me guess that the
electric field is in the H direction and the
magnetic field in a y direction. And from
that I managed to get the solution. Now, calling
this x and calling this y is bit a arbitrary.
I could just as well as said electric field
is in this direction in which case I got the
magnetic field in this direction after all
if I rotate by 90 degrees it should not make
really a difference. So, I am not restricted
to talking about electric field along x I
can also talk about the electric field along
y. To be honest I can talking about the electric
field along z z could not move in this direction
z anymore. So, I could right an electric field
that was partly in the x direction and partly
in the y direction let me do that. I am going
to assume I have a triangular electric field
in a x direction 
which I know that.
That means, I have a magnetic field that is
in the y direction. Let say that I also have
an electric field does in the y direction
in that case my magnetic field would be in
the minus x direction. So, this is Ex of let
us say z minus ct this is the chalk Hy of
z minus ct this is Ey of z minus ct and this
is Hx in the minus sign minus Hx of z minus
ct. Both of this solutions are valid and the
both will travel in the positive z direction,
because they depends on z minus ct. You can
have 2 more waves which will depends on z
plus ct 1 of which has Ex the other has Ey.
And these different direction of Ex and Ey
are quite arbitrary, because as I said choosing
what is the direction of x and direction of
y is something that we do a scientist or as
the engineers. The wave does not know what
is x and what is y what is z wave arbitrary
said that direction in which the wave is moving
called z.
But the direction x of the electric field
could be anything and in fact, is anything.
In general the electric field need along x
nor along y it will point along some general
direction. But luckily if we had these 2 kinds
of solutions then we can write down the full
solution. We can say the electric field vector
as a function of Z minus ct is going to be
equal to some function f of z minus ct along
the x direction plus some other function.
I am going to call it as g even though last
time I was using g plus z plus ct here running
out of symbols g of z minus ct along the y
direction. Now, this is in general vector
field this is the x component and this is
y component. If as specify the x and y component
I have more or less specified everything I
could specify a z component. But if I specify
a z component as I will show you later the
wave cannot travel in the z direction. So,
i by knowing the electric field along x and
know the electric field along y I have covered
all possible direction of the electric field.
Because the component along x the component
along y if I show in the x y diagram and are
some components along x some component along
y I can vectorially add and I will have component
in the general direction. So, this is the
way of solving the full problem for any direction
of the electric field what so ever. I do not
have to solve the infinite number of problems
I just solve 2 problems the problem along
x and a problem along y. Once I have done
that I got the components therefore, I can
built up any electric field at all. These
components the components along x and y there
are called polarization, that is I can find
the solution where the electric field is along
x magnetic field is along y that will be called
an x polarized way. But I could also find
the wave where the electric field points along
y magnetic field points along negative x direction
that will be called y polarized way.
I could actually also find a waves where the
electric field pointed either along x nor
along y put pointed safe in this direction.
In which case the magnetic field need a point
along y not along minus x will point in some
intermediate direction and it will look like
that. But that wave in fact, can be broken
up into combination of this wave and this
wave, because any general direction of the
electric field can be broken up into component
along x and a component along y. So, that
is a very important idea when we compute this
equations because you see this equation is
more complex equation in the previous equation.
I write the previous equation when you look
at that equation was del square Ex del z squared
minus mu naught epsilon naught del squared
Ex del t square is equal to 0. This part is
the same this del squared now is del squared
del x squared plus del squared del y squared
plus del squared del z squared.
So, it also a little more complicated with
the real increase in complication is here
this is a vector which means it is Ex x hat
plus Ey y hat plus Ez z hat. So, that is why
we did lot of algebra and got rid of these
particular points this is also a wave equation
and this is also a wave equation. But this
particular wave equation looks quite nasty
we do not quite know what to do it. Whereas,
this one looks sufficiently familiar to us
after looking at this equation that we feel
is almost like the same equation. Well there
is one way of looking at this equation that
makes it simpler what we can do is we can
say I will take this equation and write it
as 3 equations it is a vector equation. So,
as the way I will write it as I will say unit
vector along x of del squared Ex minus mu
naught epsilon naught del squared Ex del t
squared plus unit vector along y del squared
Ey minus mu naught epsilon naught del squared
Ey del t square plus unit vector along y del
squared Ey minus mu naught epsilon naught
del square Ey del t squared plus unit vector
along z del squared Ez minus mu naught epsilon
naught del squared Ey del t squared whole
thing is equal to 0. I just written it out
this has an Ex x hat plus Ey y hat plus Ez
z hat.
So, does this, so the unit vector along x
does not is a constant. So, it is not effected
by del del t it is not effected by del squared
I can pull it out. So, I have collected all
terms with x hat in it all terms with y hat
in it all terms with z hat in it. But this
is a vector equation and if a vector at any
x y z is equal to 0, which means that x component
of that vector is 0. The y component of that
vector is 0 and the z component of the vector
is 0 which means each of this brackets is
separately 0. They cannot I mean they cannot
do anything in this bracket that could make
this brackets 0. I cannot do anything here
that would make that 0 each of them must separately
go to 0 because that is what it means to say
the overall thing is 0. So, that allows me
to simplify things at great deal instead of
having a very complex vector equation i now
have 3 scalar equations.
The 3 scalar equations are del squared Ex
minus mu naught epsilon naught del squared
Ex del t squared equal 0, that came from setting
this to 0. Then I have del squared Ey minus
mu naught epsilon naught 
is equal to mu epsilon naught del square Ey
by del t squared that came from setting this
to 0. And finally, del squared Ez minus mu
naught epsilon naught is 
equal to 0. So, I now have 3 equations in
3 scalar fields and in fact, I can see a very
familiar equation here now. If only I get
rid of the x and y dependence I have gone
back to the 1 d problem. And I have this equation
which is now telling me whether y component
of the electric field obeys the same equation
as the x component and in fact, z component
obeys the same equation as well. Now, I am
going to simplify, because the whole point
of this not to do a very general analysis
it was to help us understand very simple.
Supposing the vector field E which is the
function of x y z and t is only a function
of z and t it does not depend on x and it
does not depend on y. Then these equations
are going to simplify I will just write them
in place. So, I get 3 equations the 3 equations
are del squared del z squared of Ex minus
mu naught epsilon naught del squared dx del
t squared similarly, dy similarly, dz. Now,
these 2 equations there are quite familiar
with let me write down the corresponding equation
for also it will be del squared del z squared
of Hx minus mu naught epsilon naught del squared
Hx del t square equal to 0 del squared Hy
del z square minus mu naught epsilon naught
del squared Hy del t squared equal to 0 del
square Hz del z squared minus mu naught epsilon
naught del square Hz del t squared equals
0.
So, I have 6 equations and I would like to
know what kind of solutions I can get out
of this 6 equations clearly from what we have
understood of 1 d problems. We know that Ex
as a preference for Hy and we can guess that
Ey as a preference for Hx which makes as question
mark at what this 2 are doing. But we have
to get this relation out now out of faradys
law. So, we have curl of E is equal to minus
mu naught del H del t. Now, curl of E is x
hat y hat z hat 0 0 del del z, because I do
not have del del z del del y Ex Ey Ez which
is equal to mu naught times Hx x hat plus
Hy y hat plus Hz z hat. Now, if you look at
this curl you can see that there are 2 zeros
here. So, what; that means, is if I take the
x component let me write down what this is
equal to it is x component times del del y
is missing. So, I do not have this term minus
del del z of Ey, the y 
component has del del z of Ex minus del del
x is missing. So, this does not have and the
z components requires del del x or del del
y both of which are missing 0.
So, there is no Hz the curl of E is links
to Hx and Hy and how is it link Hx is connected
to del del z of Ey Hy is connected to del
del z of Ex. Now, we can write the same equation
for amperes law 
except for a change in sign and this epsilon
naught this is the identical equation. So,
it is going to tell us that Ex is related
to del del z of Hy, Ey is related to del del
x del del z of Hx. So, once again we find
that there is no z component. So, we think
of these waves as being sources for each other.
There is no source for the z component even
though we have this equation here we have
no way of connecting up this z components
to anything else. This kinds of waves can
exist they cannot exist in vacuum they can
exist provided we have a media and we may
look at that later on.
And those kinds of waves are what are called
compressive waves and they are waves where
the field is along the direction in which
it is moving. Because if you have variation
in only z it means the solution are going
to be z minus ct and z plus ct. The wave is
moving in z and the fields points along z
those kinds of solutions are very rare they
cannot exist in vacuum. In vacuum what you
can have are Ex related to Hy and you can
have Ey related to Hx you can see it coming
out of these relation directly. So, it is
it just coming out of the vector notation
and vector relationship that you will only
have solutions in this 2 combinations. That
is why when I talked about polarization I
said Ex will cause Hy Ey will cause minus
Hx. I will go little bit more into this next
lecture, but please revise and verify for
yourself with these equations are the way
I put in them.
