In this section we will discuss the Plane
Electromagnetic Waves in Free Space .
The contents are as follows, first we will
consider plane electromagnetic waves equation
in free space which is ah derived from the
basic set of Maxwell's equations. Then we
will look at the electric field, the magnetic
field and the propagation vector, their directions
and how they are oriented in the space and
their relative orientation also. Followed
by this we will discuss the polarisation properties
of electromagnetic waves in the free space.
So, we first consider the basic wave equation
which is in this form that del square of psi
equal to 1 upon c square del square psi del
t square. Here, the psi represents the electric
field or the magnetic field or any of their
components. So, if we consider the electric
field or their comp[onent] or any of their
components then we can write this equation
in this form. Whereas, if we considered the
magnetic field or their components we will
consider this equation; basically both of
them are of the same form that is this equation
.
Then we look for the solution for this equation,
as we know we have already seen that the plane
wave solution is given by this equation in
the Cartesian coordinate system that the total
psi solution is equal to psi 0 and e to the
power of k dot r minus omega t. Which, if
you decompose this k dot r into the Cartesian
coordinates then you can write that psi 0
equal to e to the power of i k x x plus k
y y plus k z z minus omega t.
Your omega t it is the frequency of the electromagnetic
waves, k x k y and k z are the components
of the propagation constants in the three
orthogonal cartesian coordinate directions.
So, this plane wave solution for the electromagnetic
waves in vacuum is the same where, we will
use this psi for a particular electric field
or a magnetic field component .
So, if I considered that the electric field
in the Cartesian space it will be represented
by this quantity that is iE x plus jE y and
kE z. So, these are the three cartesian components
of the electric field E x, E y and E z. And
similarly, for the magnetic field we write
that the components of the magnetic field
along x is equal to B x, along y it is B y
and along z this B z.
Now, each of the Cartesian components that
is E x, E y and E z will follow the plane
wave solution, as we have seen the plane wave
equation will be satisfied by each of the
Cartesian components of the electric field
as well as the magnetic field. So, the electric
field component E x will be equal to some
amplitude of the electric field E x 0 into
e to the power of this space factor. And similarly,
for E y we have an amplitude of the electric
field for the y component which is E y 0 and
the space factor and similarly for E z.
So, this is a complete set of the field components
representing the total electric field, which
is given by this equation. In the same way
for the magnetic field we have this magnetic
field components B x, B y and B z. So, each
of them is related to the amplitude with the
space factor by this equation and for B y
by this equation and so on. So, we have seen
explicitly the electric field components in
terms of its amplitude and the space factors
.
Now, our objective is to look at the orientation,
the relative orientation of the electric field
and the magnetic field and the direction of
propagation of the electromagnetic waves;
when you considered the wave is propagating
in the free space . So, for that first will
bring in the first Maxwell's equation for
free space that is del dot E equal to 0. Having
seen that we can see that this del dot E is
equal to this del operator and this is the
electric field. So, del dot E can be represented
by this del E x del x plus del E y del y plus
del E z del z.
Now, since we know that the electric field
x satisfies this equation, the solution of
electric field E x is equal to E x 0 e to
the power of i k x x k y y k z z minus omega
t. Where omega is the frequency of the electromagnetic
waves, k x, k y and k z as I have mentioned
these are the three Cartesian components of
the propagation vectors in the free space.
So, del E x del x when you operate on E x
del del x when you operate on E x then it
will detect this i k x and will leave the
E x 0 into this part, that is E x as it is.
So, by doing so, we get del E x del x is equal
to i k x 0 i k x into E x and this is for
only the x component del E x del x. If we
considered the other 3 component, other 2
components that is del E y del y and del E
z del z then we will end up with similar expressions
for del E y del x del E y del y and del E
z del z which you can put together in this
form. So, we get that del E x del x plus del
E y del y plus del E z del z is equal to this
quantity i k x E x plus i k y E y plus i k
z E z, which is which can be written as i
k dot E.
Now, from this del dot E we have seen that
i k del dot E is equal to i k dot E, but del
dot E itself is equal to 0. Therefore, you
readily obtain that k dot E is equal to 0.
This tells you that k and E because there
dot product is 0 so, they must be perpendicular
to each other. So, we conclude by saying that
del dot E is equal to k dot E is equal to
0, which means that the wave vector k is perpendicular
to the electric field E when the electromagnetic
wave is propagating in the free space
Now, if we consider instead of del dot E equal
to 0 the first equation, the second equation
that is del dot B equal to 0 and we go back
with this equation del dot B equal to 0, you
can again write this equation del dot B will
be equal to i del del x j del del y plus k
del del z dot i B x plus j B y plus k B z.
As a result we will obtain this del dot B
will be equal to del del x of B x del del
y of B y and del del z of B z.
Now, in the same way if we calculate this
del del x of B x we will obtain i k x B x
and so on. Therefore, this del del x of B
x plus del del x del del y of B y plus del
del z of B z will be equal to i k x B x plus
i k y B y plus i k z B z, which will be equal
to i k dot B in the same way.
Therefore, we obtain in the same way the del
dot B since, del dot B is equal 0 k dot B
is also equal to 0; which will lead to the
conclusion that the wave vector k is perpendicular
to B. So, we have been able to achieve two
relations that is k is perpendicular to E
and and k is also perpendicular to B.
This means that k and k and B are perpendicular,
E and k their also perpendicular. But, that
does not ensure that E and B whether they
are perpendicular or not . So, there is a
there is a more rigorous way of looking at
the relation between k B and E . So, to do
that we consider Maxwell's third equation
the curl equation.
Now, the curl of E if we have to represent
this we write in this way that i del del x
j del del y k del del z E x E y E z. So, this
is the curl operator operating on E which
if we take the x component of this curl then
it will be del del x del del y E z del del
z of E y. So, this quantity is the x component
of the curl which will give you i k x i k
y E z minus i k z E y because, del E z del
y will detach only the y component that is
k y will be out. So, i k y E z 0 and this
factor will remain as it is which is equal
to i k y E z.
So, del cross E the x component of this quantity
will give you i k cross E the x component
of that and if we consider all the other 2
components, namely the del cross E y component
and del cross E z component we will end up
with two more expressions; i cross i into
k cross E y component, i into k cross E z
component. And then, if we put together all
the 3 components we can write that del cross
E is equal to i k cross E.
So, this is again a very interesting relation
now, we will have to evaluate the right hand
side that is del B del t. So, because you
know that B vector can be represented with
the Cartesian forms like i B x j B y and k
B z. And if you take the time derivative of
the Cartesian component then we will see that
del B x del t we will we will take out this
omega and i factor outside. And this equation,
expression for the magnetic field x component
will remain as it is, as a result we can write
del B x del t is equal to i omega B x.
So, this is only for the x component of the
magnetic field when you take the time derivative.
And similarly, if you do the same thing for
the other components that is del del t of
B y del del t of B z then we can write this
equation all put together in this form which
is equal to i omega and B. And this is a very
interesting way of looking at things that
the right hand side just boils down to i omega
B. So, del B del t equal to i omega B. Now,
we have ready expression for del cross E as
well as del B del t .
If we know put together so, del cross E has
given us i k cross E and del cross B has given
as del B del t equal to minus i omega B. So,
we can use these two expressions to arrive
at the conclusion that k cross E k cross E
must be equal to omega B. So, that is a very
beautiful relation which says that the the
B vector is perpendicular to E and B vector
is perpendicular to k. That means, B vector
must be perpendicular to the plane containing
the vector k and the vector E . So, B is perpendicular
to E and B is perpendicular to k.
Now, in the same way if we proceed for the
fourth equation that is del cross B is equal
to is equal to mu naught epsilon naught del
E del t. If we evaluate this equation and
in the same way, if we evaluate this left
hand part of the calculation that is del cross
E del cross B is equal to i k cross B. And
for the right hand side if I do this derivative
time derivative of the electric field, I arrive
at 1 by c square del E del t because, mu naught
epsilon naught is 1 upon c square. So, this
quantity will become i omega by c square and
the vector E. So, I still get the equation
for del cross E in this form del cross B and
del E del t in this from.
Now, if I use this left hand side and right
hand side together I may right that k cross
B must be equal to i ah omega upon c square
into E. So, look at this equation this tells
you that k is perpendicular to E and B is
perpendicular to E; that means, the electric
field E is perpendicular to the plane containing
the vectors k and B. So, this is the complementary
part which we have seen earlier for the electric
field.
But all three of them they are very nicely
related can see that if I remove, if I want
to get rid of this minus sign we can reverse
this position that is B cross k will be equal
to omega by c into E. So, B k and E they form
this equation from here you can conclude that
B is perpendicular to E and k is perpendicular
to E also.
Now, we will look at the consequence of these
two findings B is perpendicular to E, B is
perpendicular to k and also k is perpendicular
to E. This means that B, k, E all of them
are perpendicular to each other, all of them
are or at right angles to each other and they
form a right handed Cartesian coordinate system.
We will see that how they are they are connected
to each other in terms of the direction, in
terms of the coordinate access.
So, so you see k cross E equal to omega B,
k cross B cross k equal to omega upon c square
E. If we consider this example that, let us
suppose that I take the electromagnetic wave
to be propagating along the z direction. Then
I can write k equal to k z an if I if I consider
the electric field is oriented along the x
direction, then k z cross E x should give
me omega mu B y. So, that should yield B y;
that means, if I take k along z and E along
x I should expect that B should be oriented
along y, z x y.
And if I assume that B happens to be along
y axis and then k is along z axis which I
have already considered, then B cross k B
y cross k z equal to omega by c square E x;
that means, y z x. So, they again consistent
then E happens to be along x so; that means,
this is consistent because I have started
with the assumption that k is along z, E is
along x, the result was that B is along y
Now, I have considered that B is along y and
k is along z so, it yields that E is along
x. So, they are consistent and we can we can
write in a compact form that is E B k will
will represent that x y z the axis of the
Cartesian coordinate system and it forms a
right handed triad of vectors. So, this is
how we could see that the electric field,
the magnetic field and the direction of propagation
they are normal to each other. And because,
the direction of propagation is ah is such
that the electric field and magnetic field
they are always perpendicular and there are
also perpendicular to each other than the
wave is a transverse wave.
So, this is how we can also look at the nature
of the electromagnetic waves, the transverse
nature of the electromagnetic waves in the
free space. So, pictorially if we represent
this this is your x axis, this is your y axis
and this is your z axis. So, along the x axis
if you consider the electric fields are oriented
then along the y axis the magnetic fields
will be oriented. And as a matter of fact
that orthogonal triad of vectors will give
you that the propagation direction will be
along the z axis. So E, B, k forms a right
handed Cartesian coordinate system.
Now, we discussed this fact for the case of
electromagnetic waves in free space and you
know not only for free space, but for most
dielectrics this permeability mu is equal
to mu 0. Those this is a nonmagnetic property
and mostly mu you can approximate as mu 0.
Therefore, this B becomes equal to mu naught
into H, where H is the magnetic field, B was
the magnetic induction vector. In that case
as well for the rest part of all formalisms
we can replace that B by H .
So, if you orient the electric field along
the x direction and H along y direction then
again the propagation can vector will be along
z direction. So, in that case E, H and k they
will form the triad of vectors. So, in a in
free space we can see that E, H, k they form
the triad of vectors and also E, B and k that
forms the triad of vectors . Because E and
B, H and k they are related in the same direction
.
