with the wave equations set up in the previous
lecture we are now ready to see how electromagnetic
waves actually come in to b we are going to
keep our discussion very simple and consider
plane waves that means the amplitude is going
to be same over a plane travelling in one
direction in particular i am going to take
a wave travelling in the x direction so i
am going to motivate you by taking e in the
y direction and b in the z direction and show
you that if they vary with time and space
they can sustain each other so let us say
e is at x and as you go little farther out
e changes to e x plus delta x e is in the
y direction similarly b is in the z direction
so let's put z here and as you go little farther
out it changes to b z x plus delta x and it
also is changing with time
now i know from faradays law that curl of
e is equal to minus d b d t in particular
in this case if i take a square like this
or ah rectangle like this traveling or traversing
it counter clockwise as shown these by black
arrows i know that by faradays law i am going
to have integral e dot d l is equal to minus
d by d t of b dot d s where d s is the area
in now since e is in only y direction therefore
the horizontal lines in this do not contribute
to the line integral let the width of this
rectangle be delta y and then i am going to
get by previous whenever we proved this stokes
theorem i am going to get e y at x times delta
y in the negative direction from the left
hand side black line and on the right hand
side i am going to get plus e y at x delta
y plus partial e y partial x delta x delta
y on the right hand side ah side line going
up this is the line integral which is going
to be equal to minus d by d t of b times delta
x delta y now i am going to put a partial
derivative here because now is this thing
is not under the integral sign anymore
lets cancel a few terms this cancels with
this and i end up getting this whole thing
then gives me partial e y partial x is equal
to minus partial b which is in z direction
partial t because these two terms also cancels
so what we get from this is what i am taking
is x direction y direction z direction we
have taken e going this way and b going this
way wave travelling to the right we'll again
see how it travels to the right and i get
partial e y over partial x is equal to minus
partial e b z over partial t that's one equation
let us now apply the other equation which
gives me curl of b is equal to mu zero epsilon
zero d e d t remember this is coming from
the displacement current for this i'll take
a rectangle in the x z plane here and traverse
it counter clockwise again remember this is
my x direction this is my y direction this
is my z direction and this distance is delta
x this is going to be delta z now stokes theorem
gives me b dot d l is integration is equal
to mu zero epsilon zero d by d t of e dot
d s b dot d l like in the previous argument
is going to be b at x and we are going again
in the same direction as z so this is going
to be delta z
the lines parallel to x axis do not contribute
because b is perpendicular to that line minus
b of z at x plus delta x which i can write
as b x plus partial b z over partial x delta
x multiplied by delta z and this is going
to be equal to mu zero epsilon zero d e d
t e is in y direction lets write the magnitude
times delta s which is delta x delta z again
we are going to cancel terms this term will
cancel with this delta x delta z cancels with
this and i end up getting the equation minus
d b z d x is equal to mu zero epsilon zero
d e y d t
so the two equations that we've obtained are
for this e y going in this direction b z we've
obtained are partial e y partial x is equal
to minus partial b z partial t that's one
equation and the other equation is partial
b z over partial x is equal to minus mu zero
epsilon zero partial e y partial t let us
see what we can learn about these fields from
these two equations now from the equation
d e y d x is equal to minus partial b z partial
t we see if the wave is travelling this should
be equal to plus or minus one over v partial
e y partial t where v is the speed of wave
and this immediately tells you that b z should
be equal to plus or minus e y over v i know
electromagnetic waves travel with speed c
so let's write this or we denote it by c e
y over c the plus sign is for the wave travelling
to the left and the minus sign is for the
wave travelling to and therefore if the wave
is traveling to the left minus b z would be
equal to e y over c and therefore if the wave
is travelling to the left b will be pointing
in the negative z direction on the other hand
if i have b z equals e y over c a wave would
be traveling to the right what you can see
is that e cross b gives me the direction of
the wave let us see how we can combine these
two waves these two equations
so the equations that i have is partial e
y over partial x is equal to minus d b z over
d t and partial b z over partial x is equal
to minus mu zero epsilon zero d e y by d t
if i take this first equation differentiate
it once more with respect to x i get d two
e y over d x square sis equal to minus d by
d t of d b z by d x i have switched x and
t which is equal to plus mu zero epsilon zero
d two e y over d t squared remember this is
the wave equation with the speed one over
c square being equal to mu zero epsilon zero
so what we have argued so far is that if i
have two perpendicular fields e and b which
vary in both in time and space then it is
possible for them to propagate and sustain
each other and they propagate with its speed
c which is equal to one over square root of
mu zero epsilon zero
in the next lecture we will be obtaining this
wave equation more rigorously directly from
maxwell's equations
