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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.