So, in this class we will discuss the wave
equations, for electromagnetic waves and their
solutions.
The content of this discussion is like this,
the solution in the Cartesian coordinate system
which will give rise to the plane waves or
speaking otherwise that a plane wave satisfies
the Cartesian system of coordinates. The solution
in spherical coordinates which will give rise
to spherical waves and the solution in cylindrical
coordinates systems, which will give you our
cylindrical waves.
So, these are the main three categories of
electromagnetic waves, which are the direct
outcome of the solutions of the Maxwell's
equation the waves wave equation. You may
recall that if you have a point source, then
immediately around this point source; there
are spherical waves, but as you move away
from the spherical waves the curvature becomes
less and less and you end up with a plane
wave. But this if the source is a line source,
then the waves corresponding waves which will
come out from the source will be cylindrical
waves. So, these three different kinds of
waves we will try to analytical solved starting
from the wave equation and will also look
at the properties .
So, the pair of wave equations in free space
is represented by an electric field equation
and a magnetic field b vector equation. Where
c is the velocity of the electromagnetic waves
in free space .
Now, from this set of two equations as we
have seen earlier that it can be put into
a general and compact form using psi, where
this psi represents e or b fields or any of
the component. So, it could vector wave equation
or it could be a scalar wave equation.
This equation when we solve that then we will
see that if we solve it in the Cartesian coordinate
systems, we will find that it will ah give
a solution which is a plane wave and when
we solve this equation in the spherical polar
coordinates, then the resulting wave will
be a spherical wave; in the same way if we
solve it for cylindrical coordinate system
the waves will be cylindrical wave these are
very basic and very fundamental it is important
to know the behavior of the waves.
Therefore for a plane wave solution we start
with the Cartesian system of coordinates.
As I have mention that ah electromagnetic
wave at a large distance from the source becomes
almost a plane wave .
So, in Cartesian coordinate system as you
know we are very familiar, that your point
is represented by three components x y and
z.
So, in this system the Laplacian is represented
by del square del x square, del square del
y square plus del square del z square. So,
in this Cartesian coordinate system we can
write the wave equation in this form that
the three components equal to 1 by c square
and the time derivative double derivative
of times. So, for such equation it is well
known that the ah the separation of variables
should be applicable to find out the solutions.
So, expressing the psi which is a function
of x, y and z and t for this case. We can
write this psi of x y z and t as psi of psi
x of x y of y z of z and t of t because these
are the four independent variables associated
with this equation.
Now, if I replace substitute psi by this x
y z and t, then the first quantity will give
you del square x by del x square multiplied
by this quantity. In the same way the second
quantity will give you del square capital
y by del y square multiplied by the other
three quantities and so on. For this quantity
this will be del square capital T by del t
square into 1 by c square multiplied by the
three x y z parameters .
And now if you divide that equation, if you
divide the equation where you have already
substituted this value by psi that is by x
y and z, then the equation will take the form
of this that is 1 by x del square x by x square
and so on. You can see that the variables
are now separated out, each variable term
is now independent they are not connected
they are coupled with any of the other. So,
if it is so, each of the terms are independent
then they must be equal to a constant. Putting
that constant for each of them for each of
them, this I put equal to minus k x square
this one equal to minus k y square.
You can see here I can put this minus k x
square k y square k z square and put this
I use minus omega square. Where this then
if you if you put it back minus k x square
minus k y square minus k z square equal to
minus omega square into by c square. So, that
is what appears here. So, omega square by
c square equal to this, but this is equal
to k.
So, the solutions of each of this equation
solution of each of this equation is of this
form x equal to this, y equal to e to the
power plus minus i k y y and similar z equal
to e to the power of plus minus i k z z and
the time part will become t of t equal to
e to the power of plus minus i omega t.
So, the total solution is now you have started
from this equation so, I put them back I get
that A is a constant, and then you multiply
these three quantities which will appear in
this form and this is the plane wave solution.
So, starting with the Cartesian coordinate
system for the Laplacian, we end up with the
solution of psi. Because psi represents the
electromagnetic waves the e vector or any
of the components of psi the v vector or any
of the components of b in general each component
will satisfy this plane wave equation
Now, a plane wave is usually written in this
form and if you write this, this is the compact
form using this r. So, if you decompose this
r, k dot r as k x x plus k y y plus k z z
then this quantity represents the phase of
the of the electromagnetic wave. Now if you
look at the definition of the plane wave,
a wave whose phase is constant over a plane
surface at any instant of time is a plane
wave.
So, to look for that, I have this phase which
is this quantity and now that has to be constant.
So, we look for the locus, where this phase
remains constant. Now because at a given time
we will look for this phase constant phase
plane. So, this omega t is a constant as a
result this quantity is also constant this
is the well-known equation of a plane. If
you write in the form of the direction cosines
then x by l, y by m and z by n equal to constant
where this l m and n are represented by 1
by k x 1 by k y and so on. So, these are the
direction cosines of. So, the reciprocal of
the components of the propagation vector represents
the direction cosine for the plane electromagnetic
waves .
So, electromagnetic waves as you see that
the phase remains constant over a particular
plane, these are the periods over reach the
phase remains constant is a propagation of
pictorially.
A traveling three dimensional plane wave is
represented by this I have seen this now,
if I look at the wave very explicitly, E vector
will be represented by ah assumed i E x j
E y k E z and similarly B vector will be represented
by this. But each of the components E x E
y and E z will satisfy the plane wave solution.
So, I can write this equation in this form
for E y I can write in this form and similarly
for E z I can write this form and for B x
B y B z I also get this expressions .
Now, we will look for the solution of the
three dimensional wave equation, for the a
spherical wave solution in spherical polar
coordinates; waves from a point source is
the spherical wave source .
Look at this spherical polar coordinate system,
you have r theta is the inclination and phi
the azimuth r is the radius. So, any point
p is represented and the spherical system
in this way.
In spherical coordinates the Laplacian can
be written in this from where r is the radius
vector, theta is the inclination and phi is
the azimuth angle. So, look at this this part
is purely the radial part and this part represents
the angular part.
. So, for a wave which is spherically symmetric
that is there is no dependents on the angular
part, we can remove these three terms from
the Laplacian. So, the Laplacian simplifies
to this form, that is del square del r square
plus 2 upon r del del r. We will use this
reduced Laplacian to solve the wave equation
in the spherical geometry .
So, the Laplacian reduced to reduces to this
form, which can be written in this form also
by doing that the manipulation. Therefore,
the wave equation becomes 1 upon r del square
r psi del r square equal to 1 by v square
del square psi del t square, where v is the
velocity of the electromagnetic wave. So,
this equation becomes using this we can write
this equation in this form. So, this is the
reduced wave equation for in the spherical
coordinates for the electromagnetic waves
.
Now, I use this equation for that we choose
a function r into psi equal to u, which is
also a function of r then if I substitute
into this equation, it becomes del square
u by del r square and 1 by u square del square
u by del t square. Look at this equation this
is only the function of r, this is function
of time. So, where r in psi equal to u satisfies
one dimensional equation. So, this represents
a one dimensional wave equation .
The one dimensional wave equation this yields
plane wave solution for the transformed function
this. So, by doing this through this root,
we get a plane wave solution, but if we. So,
the plane wave solution will have a form like
this u the r of this, where r minus v t plus
g of r plus v t.
So, the solution for psi is like this, but
this is now the spherical wave solution .
So, the solution of the above equation is
thus r into psi will become like this, and
hence the solution for psi takes the form
of psi of r t equal to A by r e to the power
of i k r minus omega t. You can see that if
you are far away from the source then the
intensity which is the mode of this function
falls by r square, and hence the amplitude
falls by 1 upon r. So, this is consistent
and this represents the spherical wave.
There is. So, for a spherical wave we write
psi of r t equal to this.
Now, let us check that the phase of the electromagnetic
waves is a constant over a spherical surface.
So, k is now the mod of k and r is the magnitude
of r. A wave having the same phase over a
constant spherical surface at any instant
of time will represent a spherical wave. So,
because it is constant at a given time therefore,
we set omega into t is also a constant. Therefore,
k into under root of x square plus y square
plus z square which is the mod value of this
r vector is equal to constant at a given time.
Therefore, if you square both sides we get
x square plus y square plus z square equal
to constant and this is the equation of a
sphere. So, all this is very consistent .
Alternatively there is an alternative way
of arriving at the spherical wave solution,
the Laplacian reduced form of the Laplacian
you have seen like this, which can be reorganized
into this form and then the wave equation
becomes like this 1 by r square del del r
of r square del psi del r, which is equal
to 1 by v square del square psi del t square
Now, will solve this wave equation to get
the. So, in this case we put .
There is one step that let us put psi of r
equal to u of r by r. So, the equation del
psi del r now becomes 1 by r del u del r minus
u by r square. Just del del r if you do of
this you will get this equation now this equation
will give you when multiplied by r you get
r square del psi del r equal to r del u del
r minus u, hence del del r of r square del
psi del r is equal to this, which will give
you two terms which are identical, but opposite
in sign. So, they will cancel these two terms
will cancel.
So, this left hand side now becomes r del
2 u by del r square; del 2 u by del r square
this quantity and the wave equation now transforms
to this form. So, because 1 by r, 1 by r cancels
from other side.
So, this is again a well-known equation and
we can use the separation of variables because
r is a representation for the radial function
and t is for the time part . So, if I substitute
this u equal to R of r and T of t into this
equation, then we will get this there will
be t multiplied by this and r multiplied by
this. Now if you divide both sides by R and
T I get this equation. Now each term is independent
and they must be constant so, using the same
approach, we can write each of them is equal
to minus k square.
So, the solutions are now this equal to minus
k square this equation is a is again well
known equation whose solution is R of r equal
to e to the power of plus minus i k r, and
t this equal to minus k square will give a
solution, which is e to the power of plus
minus i omega t where I have used this omega
equal to k into v the frequency of the wave.
So, the total solution takes the form of u
of r equal to e to the power of i k r plus
minus omega t.
But we started with psi. So, to get back with
a with an equation of psi we should write
psi of r equal to 1 by r e to the power of.
So, again will end up with the spherical wave
and the general solution of this wave is psi
of r 1 by r e to the power of i k r minus
omega t plus e to the power of oh this should
be plus . So, this is spherically outgoing
waves and if you put a plus here it should
be spherically incoming wave.
A traveling spherical waves you have a source,
if you move radially outwards then they will
form the wave front, which are the surface
of the spheres. So, at a distance r the phases
will be constant, at a distance 2 r the phases
are constant and so on.
Now, we look at the cylindrical wave; cylindrical
wave is realized from a line source if you
have a linear chain of ah radiating dipoles
or linear chain of ah radiating system, then
all around the line in a cylindrical surface
the phases will be constant.
So, look at this cylindrical coordinates if
I if I have my source sitting here, then on
the surface of the cylinder the phases will
be constant. The representation is this is
the azimuth phi, this distance this is rho
and the height is z is a usual notation for
cylindrical coordinate system.
So, the Laplacian in cylindrical coordinate
system comes through this equation for psi.
So, del square psi will be involving del square
del rho square 1 by r rho del rho and the
azimuth part. The Laplacian in the wave equation
this Laplacian gives you the wave equation
in this form.
With angular and azimuthal symmetry, that
is if I assume that it is independent of the
azimuthal symmetry, we can write this equation
only as a function of rho and the wave equation
takes the form of this. Because you can remove
these two terms because of the independent
azimuthal symmetry the solutions of this wave
equation.
The solutions of this equation are well known
Bessel functions and can be represented by
this where this Bessel function has the argument
of this propagation constant, and the radial
coordinate rho times the e to the power of
i omega t. For large distances rho this cylindrical
waves are approximated as psi of rho t is
equal to this.
So, 1 upon under root of rho this is again
very consistent. Travelling spherical waves
look like this we have a line source, there
is a cylindrical surface with constant phase
at some larger distance the phases are constant
over a cylindrical surface and so on and so
forth. So, by doing this we have discussed
different types of waves particularly the
spherical wave, the plannar wave and cylindrical
waves, which we have ah solved from starting
from the Maxwell's equation and the wave equation
.
Thank you.
