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PROFESSOR: So welcome back.
And today, I'm going to
do some examples of what
happens when you have more
than one charge radiating.
This is a very
important example,
because whenever you have many
charges radiating, if it's
coherent, you get the
phenomenon of interference,
and it plays a very important
role in many fields of physics,
in particular, in optics.
You may have heard terms
like interference patterns,
diffraction, et cetera.
All that is to do with when
you have more than one source
radiating.
So the example I'm going to
consider is the following.
Suppose you have
two charges, both q.
Each one is oscillating.
It's oscillating with an angular
frequency 2 pi over lambda.
This tells you the
position of the charge
as a function of time.
So it's centered to
distance d from here
and oscillating
within amplitude a
as a cosine 2 pi over lambda t.
Well, lambda is, of
course, the wavelength
of the radiated
electromagnetic wave.
So this one is oscillating
up and down like that,
and there is another one over
here oscillating just for fun
in the other direction.
This one is oscillating
in along the x-axis,
and this one is oscillating
along the y-axis.
And here is the
formula, which tells you
what is the position of this one
at the given instant of time.
They're both given in
terms of the same time,
with the same time.
t equals 0.
Therefore, these are
two coherent sources.
The question is, what will
be the resultant field
anywhere in space?
At some point we want to know
what is the resultant field.
Now, each one of these--
you know what it does.
It radiates a spherical
wave with a certain angular
distribution.
Alright?
And the waves overlap in space.
Now because the
electromagnetic waves
are solutions of the
electromagnetic wave
equation, which is a
linear equation, if we have
a coherent source of
radiation at any point,
we simply add vectorially
the electromagnetic waves.
The electric fields
you add vectorially
all the magnetic field.
Now in principle, we should
go-- to first principles.
Consider Maxwell's equations.
Consider what happens when
a charge is oscillating.
And in fact, consider
both charges,
what they're doing at
any instant of time,
and solve in all of
space Maxwell's equation
to see what the electric
field is everywhere.
We build on our experience,
and what we will do
is we will assume that we've
solved the problem of what
happens when this single
charge is accelerating.
We've done this several times
in the [? PAR sensor ?].
Professor Walter Lewin
has done it in the course.
Whenever a charge is
accelerating, it radiates.
And this is the
formula, which tells
you what is the electric field
from the given charge, which
is accelerating at
some position, r and t.
I'm just reminding
you that what we find
is that the electric field at
any position at some time t
is given by, or related
to, the perpendicular
component of the
acceleration of the charge.
It's perpendicular to the
direction of propagation,
but at a point t in space,
the electric field there
is related to the acceleration
at an earlier time.
At the time, t prime, which
is earlier than the time when
we are considering
by the amount r/c.
All that this does--
it reflects the fact
that if I have some with
an accelerated charge,
at that instant, the
electromagnetic wave starts
to propagate, and it takes a
time r/c to reach a distance r.
That's all this is telling us.
The other thing this is
telling us-- the wave front
is a sphere.
The amplitude of the electric
field drops off like 1/r.
And it is not uniform
in all direction,
as we've seen in the past.
Right?
For example, if I
take this charge
as it's oscillating up and down,
the maximum amplitude radiated
is in this plane
where we are looking,
where the
perpendicular component
of the acceleration of
this is the maximum.
And its minimum upwards is
0, because this acceleration
has no component
of x perpendicular
component to this direction.
And similarly, for this.
The other thing which
I wish to point out
is that the polarization of
the electromagnetic wave,
the electric vector, is
parallel to the perpendicular
component of a.
So from this, the
electric vector,
when it gets to
here or here, will
be in the direction of
x, while from this one,
it'll polarized in
the direction y.
The other thing which
I want to stress,
that we will in the
other examples--
always consider
a situation where
the point where we're looking
at the electromagnetic field
is far away from
where the charges are.
The formula, which I
just showed you here,
applies only in the far field.
Nearby, the electric field
and the magnetic field
are very complicated.
They can be calculated, but they
cannot be summarized in that
simple form.
But if I'm far away from the
accelerating charge then,
it is possible to write
it in this simple form.
So our discussion is only valid
if the wave length radiated
is much smaller
than the distance
from where we're considering.
This distance is also
much smaller than that.
And this is to allow us
to approximate sine theta
by theta, et cetera.
And that's the case, alright?
OK.
With that, we can now
immediately solve the problem.
Simply consider the
acceleration of each charge
from our knowledge of
where the charge is.
We can differentiate
it twice, and we
can get the acceleration.
So we know that the acceleration
of the first charge,
which is oscillating
along the x-axis,
is given by minus
a omega squared
cosine omega in the x-direction.
I want to emphasize,
because this
is in the same direction as the
perpendicular component of a1.
And, I'm sorry.
I misspoke.
The answer-- what
I said is correct
but for the wrong reason.
In this case, by definition, the
acceleration of the first one
is this because the
displacement of the first one
is in the x-direction.
Similarly, the acceleration
on the second one
is in the y-direction,
because the displacement
is in the y-direction.
Alright?
So these are the
two accelerations.
This is just some--
occasionally I'll use omega,
occasionally lambda just
to remind you of that.
OK.
So all I now have to do
for each of the charges
is calculate this
quantity at the position
where I want to calculate
the electric and the magnetic
field.
So, first of all
in this problem,
we're asked to do
it in two places.
We're asked to do
it at position p1
and at p2. p1 is
along the z-axis,
and p2 is off the z-axis.
First, let's do it for p1.
So, the electric field,
due to the first charge
at position p1 at time t,
is given by this formula
if I insert in this for the
distance r, which is L. OK?
And I put in the
acceleration of this one
and the perpendicular
component of the acceleration
perpendicular to the line
drawing my two charges.
And the position
p1 is-- in fact,
that line is along the z-axis.
So the perpendicular
direction is, in fact,
in the x-direction.
OK.
So I don't even have to take
the-- the actual acceleration
is the perpendicular component.
OK.
And I have to calculate
it at a time which
is t minus the distance p1 is
from in my charges, which is L.
So it's minus omega L over c.
I calculate t prime in here.
Notice, I have not
tried to calculate
the distance between the
charge and p1 exactly,
because I told you that
the distance d is very much
smaller than L. And so the
overall distance is negligently
different from being
just L. So that's
the electric field
due to the first one.
Similarly, I can do the
same for the second one.
Everything is the same.
The only difference
in that case is
that now the electric
vector-- the acceleration
is along the y-axis and
therefore, the electric field
is polarized along
the y-direction.
So at the point p1, the two
electric fields are the same.
They're both
oscillating in phase
with the same
frequency, et cetera.
But this one is pointing
in the x-direction.
This in the y-direction.
Now, as I mentioned at the
beginning, if at any point
you have electric field
due to two sources,
they simply add because the
system is a linear system.
So we simply have to
add the electric fields.
Electric fields are
vectors, therefore
we don't add them
algebraic as scalars,
but we have to add
them vectorially.
And so, we take this vector
and that vector and add it.
And that gives us the total
electric field at point p1.
And simply from here,
I get d0 like that,
which I can rewrite in
terms of a unit vector.
Alright?
This is a unit vector,
which is at 45 degrees.
In other words, if that's
the z-direction x and y,
it's at 45 degrees to
both the x and y-axes.
That's the unit vector.
This is the amplitude of it.
And this is d-- tell us what
is the oscillating frequency
and what is the phase.
Notice, although each
one of these sources
radiated there an electric
field of magnitude E0,
when I add them, I did not get
to 2E0, because as I mentioned,
electric fields
are vectors, and we
have to add them vectorially.
And that's why you
get here the root 2.
So this is the first
example, which I've done.
Now I want to take
the same geometry
but calculate what is
the electric field.
What we've just done,
we've calculated here.
Now I want to do
the same situation--
two sources
oscillating coherently.
But I want to look at what is
the electric field at the point
p2.
And I've taken p2 to be
in that the xz-plane,
but in a direction that's
an angle lambda over a to d.
d is this distance.
Lambda is the wavelength of the
radiated electromagnetic wave.
And I've taken this
crazy number simply
to make the arithmetic
come out easier at the end.
And so we now want to calculate
the electric field here.
And we do exactly the same
except we go a little faster.
What I will do is I will
calculate the electric field
here.
Due to this charge
oscillating, I'll
then calculate the
electric field here.
Due to this charge, this one
is oscillating like that.
I'm reminding you this one
is oscillating like that.
And we will vectorially
add the two.
Now, from the point of
view of the amplitude,
the difference in distance
between this and here or this
and here or this and
here is insignificant.
And so I have ignored that
in the previous calculations,
and I'll do the same here.
But when you add
to the two, we have
to worry about
the relative phase
of the alternating
the electric fields.
And the difference of
this distance compared
to that distance is no
longer insignificant
when it comes to the relative
phase of the two radiations.
So, in this formula when I'm
calculating the electric field,
this r-- it doesn't
matter whether I
call this L or L plus
a little bit or less.
This doesn't change very much.
But in calculating the
phase of the radiation,
I cannot ignore that.
In the previous case,
the phase was the same
because this distance
and that distance
are exactly the
same by symmetry.
For this position, that
distance and that distance
are not the same.
And so I what I will do is I'll
use as reference that distance,
and when I'm calculating
this distance,
I will calculate by how much
I have to subtract from here
to end up with this length
and for this one, by how much
I've add to it to get to here.
Pay attention to that when
I'm doing that in a second.
And the rest is very
straightforward and similar
to what I've just done.
So now, as I say, I want to
calculate the electric field
at the point p2, which is this.
This is the x-coordinate,
y, and z-coordinate.
And I pointed it out
to you a second ago.
Alright.
So I don't have to rewrite
all those q's and omegas,
et cetera, by
analogy, what we've
done-- the electric field due to
the first charge at position p2
will be that the
amplitude that we've got
is the same for the
previous case of P1.
So it's-- E0 is the amplitude.
It's pointing in the
x-direction as before.
And we know that the description
of the electric field
will be given by cosine omega
t1 prime where t is-- so far
I've always talked
about t prime,
because I only had one when
I was considering just one
charge.
But here the t1 is
different to t2.
So quickly, this term comes
from the perpendicular component
of a right as before.
And t1 prime is
equal to the time
when I'm looking at the electric
field minus the time it takes
for this signal from the
charge to get to the point
where I'm looking at the
electric field to point p2.
So, it will be-- as I
told you a second ago,
I'll take L as the reference.
And I'm subtracting from it that
distance d sine lambda over 8d.
For a second, let me
go back to that picture
because you can
easily get confused.
So I'm coming back here.
What we're trying to calculate--
the distance from here to here.
So I'm taking this
distance, which is L,
and I'm subtracting from
it this distance, which
is d sine this angle here,
which is the same as this angle.
So that's what I'm
calculating there.
So this is L minus d sine
lambda over 8d divided by c.
Then the electric field is
then immediately followed
from this d0 in the x-direction,
cosine omega t minus this.
And if you assume that
this angle is small,
and I told you that that's
given in the problem,
the sine of an angle
is equal to an angle.
Approximately, they're small.
And if I multiply this out,
I simply get an angle here.
That's pi/4.
So I find that the
electric field at point p2
is pointing, due to the
first charge, is pointing
in the x-direction,
as you'd expect,
because of the direction
which the charge is moving
times E0 times cosine omega
t minus omega L over c
plus a phase like that.
How about from the other charge?
If I take the other
charge, everything
will be the same except now
this charge is oscillating
in the y-direction, so
the electric field will
be in the y-direction.
And this time,
it's omega t2 prime
where t2 will be
very similar to t1
but now the distance is
greater by this amount.
For a second, let's go
back to this picture
so you see what
I'm talking about.
If I calculate
this distance, it's
the same as that distance
plus d sine this angle, which
is this lambda over 8d,
and I'm subtracting.
And I want to
emphasize that what
I'm focusing on here,
what is important,
is the difference
between this and that
and not the absolute value.
So when I'm
approximating-- if you
do this calculation
for yourself,
you'll see I'm doing
a slight approximation
to calculate that.
But this and this is
the same quantity.
But the important thing is
the difference of this phase
here is plus pi/4.
Here is minus pi/4,
because here it's
minus that little
bit of a distance,
and here it's plus a
little bit of the distance.
So we finished.
The total electric
field at p2 and time t
is the sum of the electric field
due to the first charge plus
due to the second challenge.
And as I mentioned a second ago,
electric fields are vectors.
Thus, we have to add
these two vectorially.
This is in the x-direction.
This is in the y-direction.
And so what we
have at that point
is two electric oscillating
electric fields.
They're oscillating
coherently but out of phase.
This is a plus pi/4 phase.
This is a minus pi/4 phase.
So the difference of phase
between those two is pi/2.
It's 90 degrees.
So this is out of phase with,
one with respect to the other,
by 90 degrees.
They're coherent, but
90 degrees out of phase.
The magnitudes are
the same, but this one
is pointing in the x-direction
and this in the y-direction,
and you know what
that corresponds to.
If I have at the location
two electric fields, one
doing that and the
other one doing this,
if they're out of phase by 90
degrees, and I add the two,
what do I get?
I get a constant
size of a radius,
and the thing is rotating.
That is the description
of a rotating
vector of magnitude E0.
In the other case, at P1, there
was not this phase difference,
so the two were-- one was
oscillating like this.
The other one was
oscillating like that.
But they were oscillating
at the same phase.
So this is what they were doing.
And if you add those up, clearly
you get a line diagonally.
And that's why before, the
result was a linearly polarized
electric field there at 45
degrees to the x and y-axis.
In this case, these two are
out of phase by 90 degrees.
So when one is doing
this, the other one
is also but out of phase.
And so when this
one's at the maximum,
this one's at the minimum, and
so that's what's happening.
So they're out of phase.
And then if you
add the two up, you
get something doing
this, which we
called circularly
polarized light.
And that's why I chose
that crazy angle,
because it came out like this.
So that's the end
of that problem.
I will now do another
one, which to emphasize
some technique-- a different
technique of doing it.
OK.
Let's move.
So the next problem
is the following.
We're again dealing with several
charges oscillating coherently.
Their cohering sources.
And we're asking, what
are the electric fields?
Somewhere in space?
And if we found the
electric field, of course,
we could always calculate
the magnetic field
at that location using Maxwell's
equations or our knowledge
of the relation between
electromagnetic field
in a progressive
electromagnetic way.
Now I'll take three charges.
But that's not the hope.
It's more different, but
what this problem adds
is I'll use it to show a
technique that is often used,
which helps in the
solution of such problems.
So the problem now
is the following.
I have three charges.
Each of the same magnitude.
They're located along
the x-axis at position 0
minus d and minus 2d.
So these distances are the
same-- each distance d.
And what the problem
is-- up to t equals 0,
these three charges are
displaced from equilibrium.
I put y is 0.
So, at the height, y0 here.
Then, at t0-- don't ask me how.
Magically, I get these three
charges to start oscillating.
Such that at time
yt, the displacement
is y0 cosine omega t.
And they continue-- this
continues like that forever.
The question is, as a result
of this oscillation of charges,
what will be the electric
field a long way away from here
at position L along the x-axis,
so we call that position p.
So the problem is, calculate
E at position p for all times.
Once we've calculated, we could
calculate the magnetic field,
But to save you time--I mean,
you could do it for yourself.
You know if you calculate
the electric field,
there is a progressive
wave over here.
if E is like this,
then b will be
perpendicular to it
of the amplitude which
is just a dE over c.
So that will be straightforward,
so I'm not asking it.
Now, let's just
think for a second
what goes on in this problem.
Initially, the charges
are stationary,
so there will be a
Coulomb field around.
But we are specifically asked
to ignore static fields.
So in the last problem too,
I ignored static fields.
We were only considering
the time-dependent fields.
You can always have superimposed
on the time-dependent field
some static field that
doesn't add anything.
So what we have here is at
time up to time t equals 0,
the charges are here.
So let's take one
of the charges here.
Then it starts moving.
It will have an
acceleration, which
is perpendicular to the
direction in which I'm
interested the propagation
of the electromagnetic wave.
So it will certainly
radiate in this direction.
So this charge, which was
oscillating, will radiate.
Over here at the point
p, the electric field
will be initially 0.
And it will continue being 0
until the electric field, which
is generated here,
propagates that distance.
So, only after a time, L/c, will
the electric vector get here?
So up to the time L/c, there
will be no electric field here.
How about this charge?
This charge is initially
is displaced at y0.
Then it starts
oscillating the same.
Initially, it's stationary.
It'll produce a Coulomb field,
which is a static field.
We're not interested in it.
But once it starts accelerating,
it starts radiating.
And that radiating progresses.
So that will also produce
a field over here.
But since these two
distances are different,
the radiation from here, first
of all, will get there a little
later.
But also, once it gets here,
it will have a different phase
because the radiation
traveled a different distance.
And same for the last one.
So now, how do we do this?
I almost sound like
a broken record.
As always, I can calculate
the electric field
from each charge, and it'll be
given by this formula, which
we've seen over and over
again where t prime is
the distance from the
charge to the point p.
Furthermore, I know the
perpendicular acceleration
and also the perpendicular
component of it,
because in this problem,
the perpendicular component
is the same as the
actual acceleration.
After time t equals
0, the acceleration
of every one of these charges
a1, a2, a3, is the same.
It's the same direction.
And it's given by
that simply because we
know what is the
displacement y of t.
If I differentiate it twice,
I get the acceleration.
If I plug this into
here for each charge,
I will know what is the electric
fields from each charge.
I can then add the electric
fields from each charge,
and I'll get the
total electric field.
You have to add
them vectorially.
So, once again, for
t less than L/c,
the electric field at
position p will be 0.
There was no earlier
accelerated charge
which radiated an
electric field which
got here before this time.
So that's nice and easy.
How about for a later period,
a period between L/c or just
after L/c.
For t just after
L/c, the acceleration
of the charge at x equals 0
would produce an electric field
which propagated and would
have got to my point p.
So between that time
L/c and the time when
the radiation from the second
charge got to the point p,
I will have an
electric field but only
from the first radiation
due to the first charge.
I can forget the second
and third charge at minus d
and a minus 2d but not
the charge at x equals 0.
So that will be I
use this formula.
I plug it in.
I use the acceleration of a1.
I put it in here, and I know
that t prime is t minus r/c.
r in this case is
L. So this formula
immediately tells me that's
what it is and lets me save--
I don't want to write
this over and over again.
This-- I'll call that E0.
So from the first
charge, this is
what the electric field-- it's
an oscillating electric field--
what it looks like
at position p.
But I'm reminding you, this is
only true for this time window
when the radiation
from the charge at q
equals 0 has got to point p,
but from the other one's not.
How about from a little later?
From the time L/c plus d
over c-- in other words,
now radiation over
a distance L plus d
has had time to
reach my point p.
So now, the second charge,
the one that minused d,
has had enough time
to reach my point p.
But if I limit to the window,
this and that, in other words,
there's still a difference
of d/c in time between those.
I will get at point
p the radiation as
from the first charge.
That's exactly the same as this.
That's obviously got there.
But now from the second
charge, it has got there.
And they're very similar.
They are both-- the
charges are accelerating
along y-direction in phase.
So they have the same
frequency, and the amplitude
is going to be the same,
because in this formula,
this r is the total
distance between the charge
and the point p.
Now, there is a tiny
difference in that distance
for the first and second charge.
But if I take them
here a tiny difference,
this will not change very much.
I'm ignoring that difference.
So I'm calling both of
them E0 and ignoring
that tiny difference.
But I cannot ignore the
difference on the phase.
So the t prime into
two cases is different.
In one case, it's t minus L/c.
And the other case,
it's minus L/c
minus this little extra
time where the radiation
takes from the second
to the first charge.
And the third one,
the radiation could
not to reach the point p yet.
So that is now the total
electric field from the two.
And finally-- and
I have to do this.
I haven't done the addition.
And finally, if we take a time
which is greater than this,
then we are in a situation
where the oscillation
on the last charge
has had enough time
to get to the point p.
Let me go back to this
picture and just repeat this
so that you don't get lost.
So these charges have started
to move all at the same time.
After a time L/c, the
radiation from this one
will have got to here.
If I add to the time
d/c, that's how long
the electromagnetic radiation
takes to get to from here
to here.
So a little bit
later, the radiation
from this plus the radiation
from this is getting here.
And a little bit later,
the radiation from this
also gets there.
So at the end, there
is now radiation
from all three charges
getting to this point.
And from then on,
it'll continue forever
as long as these
are oscillating.
So from then on, you have
the radiation from the three.
And the only
difference between them
is they all have a
slightly different phase.
So you might be satisfied
with just knowing
that these are the
electric fields as sums
of the algebraic
sum between those
all pointing the same direction.
So I don't have to worry
about the vector addition,
but I have to worry about the
addition of the amplitudes.
If we want to find
out what this sum is--
what happens when you add one
or two or three oscillatory
functions like this?
What is the resultant
oscillation?
We have to algebraically,
or trigonometrically, add
these three or these two.
And I do this example
in order to introduce
a mathematical technique, which
in situations of this kind,
makes life much, much easier
than just going and adding
cosines.
And it is by using the
so-called complex amplitudes.
So the issue is, how do we add
these three cosine functions
and where they each have a
slightly different phase?
One way is brute force.
Do it by using trigonometrical
[? formulae ?].
For example, you could
add the first and second
by using the formula cosine
a plus cosine b equals twice
cosine half the
sum of the angles
cosine half the
difference of the angle.
And you could do that for
adding the first to the second,
and then later, once
you've got an answer,
you could add to that
the third, et cetera.
But you can see that if
you have five, six, seven,
eight sources, or many, many
more, this becomes cumbersome.
There's a nice
mathematical trick.
And that is by using
complex numbers.
You know that this is
De Moivre's theorem
that E to the j
theta can be written
as the cosine of an angle
plus j sine of an angle.
I can use this
mathematical trick
to solve the problem
of adding these.
Remember that at this stage,
this is pure mathematics.
As always, we've converted
an experimental situation
into a mathematical problem.
And we've got to solve
this using mathematics.
You don't have to
ask yourself what's
the meaning of j
sine theta something.
It is a mathematical expression.
And we're going
to use mathematics
to solve this problem.
Using this, I could
always write--
that suppose I have the
cosine of some function,
I can always write it
as the real part of E
to the j that angle.
So if cosine, for
example, of omega
t minus kL, and you'll see it
somewhere up there for example.
It can be written as the real
part of E to the j omega t
minus j to the kL, but that is
the same as multiplying these
as E to the j omega t times E
to the minus jkL, where here I'm
just reminding you of k
omega c, et cetera, is.
I can do this for
every-- if I want
to d-- let's say we're
doing this one first.
I want this to that.
I can write this as the real
part of a complex number.
And I can write this as a real
part of the complex number.
What I will then do-- I
will first solve this.
I will do the addition by
adding the two complex numbers,
knowing full well that if I
added the complex numbers,
I will have in the process
added the real parts and also
the imaginary parts.
And since you cannot have a real
number equal to an imaginary
number in any way, if I take
during the process of addition,
I would have continuously kept
separate the real and imaginary
part.
So for example, for this third
part, this addition here,
I can write the first term as
the real part of each of the E
to the j omega t, E to the minus
jkL in the y-direction times
E0.
This is, of course,
nothing other
than E cosine omega t minus Lc.
The second one I
can do the same.
This is the same.
And the only difference
between those two
is that phase minus d/c,
which is kd [? using ?] here.
And so, the real
part of this will
be the answer to
the sum of those.
So I'll do this in
a second, but then
let me immediately go to the
third one so I do both of them
at the same time.
And in this third case,
the answer I'll want
is the real part.
This is the same as before.
And here the three terms.
The first term is just E0,
because all the phases are out
here.
The second one differs from
that by minus d/c, which
is minus kd, so it's
E to the minus jkd.
And the last one is E to
the minus j times 2 kd,
because here we have a 2d.
Now, why did I
bother to do this?
Because this addition is
trivial while the other one
was not trivial.
It needed hard labor.
Why do I say this is trivial?
Because I've converted
this algebraic problem
into a geometry one.
I can represent each
one of these terms
on an Argand diagram.
So for example, here to here.
Let's take the third case.
We're adding these two vectors.
This is only a real
part, so it's E0.
That's a vector of length
E0 along the real axis.
I'm reminding you on
an Argand diagram,
this direction is the real axis,
and this is the imaginary axis.
With that, so the
E0 is only real,
and I'm adding to it E to
the minus jkd, which is what?
Has a magnitude of E0.
And this is the angle,
theta, with respect
to the real axis, which
in this case is minus kd.
So this is the
angle, so it's minus.
So I'm going down.
So it's a vector,
which is length E0,
and it's pointing in
an angle of minus kd.
It's this angle.
I suppose, not to confuse
you, I'll call it minus kd.
This is minus and minus,
just so not to confuse you.
With the distance
between the charges of d,
kd is 2 pi over 3,
which is a 120 degrees.
So this term is a
vector like that.
And we have to add
those two vectors.
Well, this is easy to do.
This angle is 60 degrees.
This is 120.
OK?
And this length
is equal to that,
so the result of
this plus of that
will be this red vector
here, which has magnitude E0.
And at what angle is it?
This angle here 240 degrees.
I'm sorry.
I need that later in the
second for the next part.
I don't need it at this stage.
I'm sorry.
I'm adding this to that.
The result is this vector.
And this vector
has a magnitude E0.
And this angle
here is, of course,
60 degrees, which
is 2 pi over 6.
And so, so adding those two
will give me just the real part
on each of E to the j
omega t to the minus
jkL in the y-direction.
And adding those two
is E0 in the direction
of 60 degrees, which
is minus 2 pi over 6.
Bracket right there.
So this describes this addition.
And this, of course, I
can now take the real part
of this whole thing,
and there is the answer.
E0 in the y-direction
cosine omega
t minus omega over c and
with a phase of minus 2 pi
over 6, which is of
course, 60 degrees.
For the first one, so
this is the answer.
And we didn't have
to add to any cosine.
We just do a vector addition
on the Argand diagram.
Let's do the last case.
In the last case we have
three terms-- one, two, three.
If I write these as the real
part of a complex amplitudes,
by analogy with this, the
first term is the same.
The second term is the same.
And the third one is almost
the same as the second time,
except the d has now become 2kd.
This was d/c.
Here it's 2d over c.
So here we have minus j 2kd.
So now I have to add these
three complex numbers.
And again, this is much easier
to do it geometrically then
trigonometrically.
The first one I can represent
by this on the Argand diagram.
The second one by this.
And the third one, of course,
is at 240 degrees to here.
This was 120.
The next one is 240, which
means that the last one, all
that it is, I'd
remove this vector,
and it's in this direction.
So now we're adding
this to that to this,
and even I can solve
that in my head.
If I add this vector to
that to that, I get 0.
I'm back where I started.
So this quantity is 0.
And so the electric
field will be 0.
And see how
relatively easy it is
to do if you use this
complex amplitude method?
One can do many problems to
do with waves and vibrations
using complex amplitudes.
In general, it makes
the algebra easier.
But for most cases, I have
not done that in order
to-- that you don't have the
double difficulty of trying
to understand the physics and
struggling with the mathematics
that you may not be so familiar.
By the time we come to many
sources of the radiation,
it is so much easier to
do using complex amplitude
that I would urge you to learn
it on simple cases like this,
and then use it in more
complicated situation.
Finally, I just want
to save one word,
and that is the following.
Some of you may be surprised.
How is it that I got--
I have three charges.
Three charges
oscillating, radiating.
How is it that when
you get to here,
you get after a certain
time, you get to [INAUDIBLE].
And the answer is--
pictorially you
can see what happens--
at this point,
the radiation from one of
the charges looks like that.
From the second one,
it's out the phase
by 1/3 of the wavelength,
and the next one by 2/3.
And so you're adding three
waves, three oscillating
motions on top of each other.
And if you add these,
the result is 0.
That is, pictorially,
the same as what
I did here in this diagram.
You simply-- you do have three
waves arriving from the three
charges, but they
add to 0 because each
has a different phase
from the previous one.
And you can see it
here what happens.
Here I'm plotting as
a function of time.
The amplitude at the point p.
