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PROFESSOR: So welcome back.
And today, we will
solve one type
of problem, a very
important problem,
basically, the following.
Suppose all that you have is a
single charge, but it can move.
Here is the origin
of coordinates.
This is the charge,
Q. Its position is
given by some vector, z of t.
And the charge can move.
It can have a velocity.
It can have an
acceleration, et cetera.
And the problem is,
somewhere in space,
at a point, p, what
is the electric field?
What is the magnetic
field at all times?
That's the problem.
Now the differences between
different particular problems
will be where we are and what
the charge is moving et cetera.
In particular, we
will consider problems
where this charge is oscillating
in a simple harmonic motion.
It's along a line.
And then, in another
problem, we'll
consider it's rotating
with uniform velocity.
But in principle, we
could be doing anything.
So how do you calculate
the electric field?
How do you find out what the
magnetic field is everywhere?
Also, if there is an
electric and magnetic field,
there will be flow of energy and
energy flux or pointing vector.
What will that be
at that location?
And finally, if a charge does
accelerate here and it does
radiate energy, and
the question is,
what is the average total power
radiated by such a charge?
Now this kind of problem
is extremely important.
It's important
just from the point
of view of understanding
how the world works.
There are charges
all over the place.
And they are moving.
And they do produce
electric magnetic fields.
A problem like this
gives you insight
how the phenomena takes place.
But let me tell you
immediately, in general,
this is a horrendous problem.
You know that, if you
have charges in space,
and you have electric
and magnetic fields,
the charges, the currents, the
electric and magnetic fields
have to satisfy Maxwell's
equations, those
four Maxwell's equations.
So you can imagine
how complicated
it must be to find electric and
magnetic fields, in general,
for some arbitrary
motion like this.
And the answer is, you would
never be asked to do that.
And you would never have to do
that, in general, analytically.
If you really had a problem
like this that you had to solve,
in general, you would
do it on a computer.
All right?
But let me continue.
But it turns out that,
under certain conditions,
this problem is not so hard.
And yet, it contains
all the physics
and gives you all the
insights you need.
And that is the
following situation.
If you have the charge moving
in a confined region, which
is much smaller,
whose dimensions
are much smaller than the
distance at which you are
interested to calculate the
electric magnetic fields,
suddenly, this problem does
not become hard, or in fact,
not hard at all,
relatively speaking.
So in some sense-- and this
often happens in physics
when you are given
the problem-- you
have to have an understanding
of the situation
in order to understand what the
person is asking you to solve.
So in a situation like
this, the buzz words,
or what you would find
is, for example, say,
there is a charge.
It's accelerating.
Calculate the electric field at
a distance far from the charge,
or the far field,
as it's called,
rather than the near field.
And the reason for
it is the following,
in this case, that
the hard part of this
is calculating the field
relatively near the charge,
because there, you
have a superposition
of this static field, for
example, the Coulomb field.
If a charge is moving,
it's like a little current.
And so it produces
the magnetic fields
through the Biot-Savart
Law, et cetera.
But these fields drop off,
like 1/r, from the charge.
And so, if you are
far away, then,
in general, in most
situations where you're
interested in understanding
what's going on,
you can ignore those fields,
because the further you
are away, the smaller they are.
And they become negligible.
There is another
contribution to the field,
to the electromagnetic field,
a very important contribution.
And that is-- and
in fact, you know
that, in vacuum, without
any charges there,
you can have electric and
magnetic fields present
that come from propagating
electromagnetic waves.
And it turns out that, if
you have a charge which
is accelerating,
that does produce
an electromagnetic wave.
And that does not drop
off as 1 over r squared,
but as 1/r, much more slowly.
So if you are far
away, you are dominated
by the propagating
electromagnetic wave.
Now it's not that
this is trivial,
but certainly, the propagating
electromagnetic wave
is much easier to calculate
without the presence
of these other fields.
Now Professor Walter Lewin
has done that in his classes.
He has shown how, if you
have somewhere a charge which
is accelerating, how it
generates an electric field far
away from that charge
and a magnetic field.
I won't derive it
for you, but I'll
summarize what you have learned.
Now for a second, I'll
just want to point out
why the problem of
electromagnetic waves
is more difficult than,
say, that of a string.
The analogous problem we
want to recount for a string
is the following.
Suppose you have a string,
you hold it at one end.
Right?
And suppose I move this
one end sinusoidally?
You know that a string
like this is a line.
And in there are
oscillators, infinitely
close to each other,
infinite number of them.
All right?
Each one moves along a single
line, say, up and down.
So for each one,
it's a single number
tells you what the
displacement is.
You know that the string
satisfies the wave equation.
If you put a
boundary at one end,
the oscillator, the first one,
is somehow externally driven.
That excites the
next one, et cetera.
And you know that
the wave propagates.
Now in the case of
electromagnetism
of this kind of a problem,
the situation is very similar,
but you have this
extra complication.
In the string, each oscillator
was on the straight line
and only moved up and
down in one direction.
For a charge, every point
in space you could imagine,
there is an oscillator.
The excitation of
that oscillator
is the electric and
the magnetic field.
That's the analogous
to the displacement.
But notice, already,
two quantities,
electric and magnetic.
Here, for the string,
you had the displacement,
a single number
in one direction.
Here, the electric and
the magnetic fields
are vector quantities.
So not only do you have them
spread in three dimensions,
instead of along the
string, the displacement
is a vector quantity.
So is the directions in
three-dimensional space,
et cetera.
So that, in principle,
is the same,
but it is much more complicated.
But coming back
now, fortunately,
if you analyze what happens,
if you take a charge,
if it's stationary,
it does not radiate.
It has the Coulomb field,
which, as I told you,
drops off like 1 over r squared.
If it's moving with
uniform velocity,
for example, it does
generate magnetic fields.
But again, they drop off.
It doesn't radiate.
But if it accelerates, then
it does produce, far away,
an electron, a field
which propagates.
And as Professor Walter Lewin
has showed in his lectures,
you find the following.
That if I have a charge
which accelerates,
at the time when
it's accelerating,
it produces an
electric field, which
I can summarize
with this formula.
So it's the following.
Suppose, at the origin,
somewhere I have a charge.
And it accelerates, say,
in some direction, up.
And I am interested what
is the electric field that
propagates outwards
in some direction, r.
What one can show
is that this charge
will propagate in all
directions, including
this direction.
But there's only one
component of the acceleration
of the charge which gives
rise to the production
of the field which propagates.
And that is the
component which is
perpendicular to the
direction in which I'm
interested to calculate
the electric field.
So if the charge, for example,
is accelerating upwards,
like this, then it's
only the component
of this acceleration which is
perpendicular to this line that
gives rise to the
electric field.
And one can calculate it.
And its formula is
rather straightforward.
You find the electric
field at some distance, r,
from the charge.
At time t, is
related to, as I say,
this acceleration of the charge,
but only the component, which
I call aperp, the component
of a perpendicular
to this line, that's
this quantity.
And what is interesting is the
field here that is produced
is proportional to this
perpendicular component
of the acceleration.
But there is a difference
in time because,
as this charge
accelerates-- it's
analogous to take the string.
When I took the
end of the string
and I moved it, at
that instant, the kink
got produced which
propagated along this string.
Same happens here.
If I have a charge
which accelerates,
it produces an electric
field which propagates.
And it propagates with
the velocity of light, c.
So the field over here is
given by the acceleration
of the charge at a time,
t prime, which is earlier.
If you're interested in
the field here at time t,
then you have to know what was
the acceleration of the charge
here at the earlier
time, t prime,
which is equal to
t minus r/c, which
is the time the electric field
took to get to that point.
That's the tricky part.
This is the only tricky
part of this equation.
So the electric field at
the position r at time t
is related to the
acceleration of the charge,
the perpendicular
component of it,
at an earlier time times
some constants, the charge
and a constant.
And it drops off like 1/r.
OK?
So this is the crucial equation.
It tells us that, as I
say-- I'm repeating this,
because it's so important.
Whenever I have a charge
which is accelerating,
it, at that instant,
produces an electric field
which is moving outwards
at the speed of light.
Its amplitude is dropping
like 1/r, all right?
And goes on forever.
Now you know from
Maxwell's equations,
if you have an electric field
which is propagating like that,
associated, there will
be a magnetic field.
This is now just the
same as you've always
seen for electromagnetic waves.
I like sort of visualizing.
If an electromagnetic wave
moves in that direction,
and the electric field is this
direction, at that instant,
there will be a magnetic
field perpendicular to it.
And the B is proportional
to E. And they both
move forward like that.
OK?
So if this is E, then
you can get from a B.
They are proportional
to each other.
There's just that constant,
C, because the units.
And they are perpendicular
to each other.
OK.
So if you understand
this formula,
all the problems that you
would be normally asked
can extremely
easily be done, OK?
Once you know E
and B, of course,
you can calculate
at every location.
The Poynting vector is just a
cross product between those.
And I just summarize them on
the formula we may need later.
So I'm now going to
start from scratch
by taking a concrete example.
All right?
So if this refresher didn't
help, maybe the problem
itself will.
So now let's take
a concrete example.
I'll take a single
charge, Q. OK?
So I have a coordinate system.
Here is my origin
of coordinates.
And I'll take a charge
here, charge of magnitude,
Q, moving up and down,
up and down sinusoidally.
So it's given by
the amplitude z0.
It's moving along the
z-axis, cosine omega t.
This is the motion.
Of course, something
must be moving it.
But the assumption here
is it doesn't in any way
affect the electrical
and magnetic fields.
So that's the only
thing that we have
to consider in the universe.
And the question are,
at different places
in space, what is the
electric and magnetic field?
So for example, one place is a
distance, R, along the z-axis
up.
So it's not along the axis.
So the position where we
are interested, where p is,
has coordinates 0,
0, R. In other words,
it's a distance, R,
along the z-axis.
And we are all
asked, at this time,
R divided by C, what is the
electric and magnetic field
at that location?
OK?
That's the first
part of the question.
Well, that, I hope you've
already done in your head.
If the charge is moving up
and down like this-- oh,
and by the way, I
have to emphasize,
this problem, in general--
except this one part--
you are able to do only if
we satisfy what I said here,
that the region within which
the charge is moving or located
is small, compared to the
distance where I'm asking what
the electric and
magnetic fields are.
So here, it says for R
very much greater than Z0.
So whatever I say
from now on, I'm
making the assumption
I'm very far away.
I only have the radiated field.
I can ignore the complicated
fields like Coulomb's field
and Biot-Savart, et cetera.
OK.
So the first part,
even without this,
I can do because
imagine we are asking
what is the electric
field along the z-axis.
And so suppose I am
at that location,
and I'm looking
down at the charge.
What do I see?
I see the charge going
towards me and away from me,
and towards me and away from me.
So it does have
velocity at some stage
and suddenly has acceleration.
But the acceleration is
towards me and away from me.
So what is the
perpendicular component
of the acceleration
in my direction?
0.
At no time do I see the
charge accelerating sideways.
I only see it going
towards and away from me.
So the aperp is 0.
And then I go back
to my equation here.
If aperp is 0, it doesn't
matter what you do with it,
E will be 0.
If E is 0, B is 0, because they
are proportional to each other.
So at all times
this will be 0, OK?
The next part of the question.
And let we quickly tell you, in
the next part of the question
it says, now let's have this
charge not move up and down,
but along the y-- sorry.
No.
No.
I misspoke.
Let's leave this
moving that way.
But instead of looking
along the z-axis,
let's look at the
point along the y-axis.
So at the point 0,R,0,
or in other words,
at distance R from the origin
along the y-axis, again,
at this time.
And then the third one, it asks
let's look at it at 30 degrees.
So now, in this case, we
look at in the direction
where it's in the y,
z plane, but at 30
degrees to the z-axis,
so in some direction
there where the
coordinate is this.
OK?
And the whole purpose
is to just get
you a feel how accelerated
charges radiate
in different directions.
So the first one I've
already done for you.
OK?
But here, let me just
repeat myself or point out.
In each part of
this question, we
are a distance, R,
from the charge.
Now you can say,
well, the charge
is moving a small amount.
But the whole essence of
making this a doable problem
and getting a situation
which you can solve
is where all the
motion of the charges
over distances
which is very much
smaller than where you are.
So because R is very
much greater than z0,
from the distance of
the accelerated charge,
from where you want to
calculate the electric field,
you can assume is
constant in distance R.
So in all parts of
this problem, you
are the same distance away,
but at a different angle.
And in all cases, we are
asked to find the field
at some time, t, when
this is oscillating.
Let's go to the second part.
How about in a direction
which is along the y-axis?
All right?
So in the y direction,
if something
is accelerating in
the z direction,
that is perpendicular
to the y-axis.
And so for the second
part, the acceleration
of the charge at
time t prime, is
simply I have to take this
and differentiate it twice,
all right?
And so we've done it.
This is this.
And this acceleration, that has
a component in the z direction
which is perpendicular
to the y direction.
So in our formula,
this is Eperp.
And so we can plug
it into that formula.
And we get that
the electric field
at position R along the
y-axis is given by this.
And I just took that formula--
let me just remind you--
I have just taken this formula
and plugged in the aperp.
And I end up with this equation.
And so I see that, if a charge
is oscillating up and down
like that, it does radiate
perpendicular to it.
This is the nearest analogy
to that string case.
If the string is
moving like that,
it propagates a wave
along the string.
And here, if a charge is
accelerating like this,
it does radiate
along that direction.
This is the magnitude.
And as a function,
that's what it is.
The actual problem
was not in general
what it is, what will
it be at the time R/C?
Well, at t R/C, this is
R/C minus R/C. This is 0.
So this is cosine
of 0, so this is 1.
And so we get, finally,
that at that time,
the electric field
is equal to this.
It's just that times 1.
This is the electric field.
What is the magnetic field?
Well, as I told you, if you
know the electric field,
B is perpendicular to
it, proportional to it.
And there's just a 1/C. So
they here have it C cubed.
And the magnetic field
will be in the x direction.
The electric field
is always parallel
to the accelerated charge.
The magnetic field is in
the perpendicular direction.
Also, it is important to
remember the minus sign.
I come back to this formula.
This minus sign
is very important.
What it's saying
is, at the instant
when the charge is
accelerating in that direction,
the electric field it produces
is in the opposite direction.
So when this is accelerating,
at the instant this accelerates,
it produces an electric field
which goes in this direction.
And that field
propagates like that.
OK?
And the magnetic field
is perpendicular to it.
And so that is the Part 2.
And now we come to doing the
third one-- it's actually
the same problem--
where we are considering
the radiation at the slightly
more complicated angle.
So far, we've looked at the
charge is moving up and down.
We said there's no
electromagnetic field radiated
upwards, or downwards,
for that matter.
Along the y-axis it is.
And of course, if we
took any other location
in the plane of our x, y,
in the horizontal plane,
in all directions,
it will radiate
in a similar manner
as along the y-axis.
And finally, just
for completion,
let's take one more direction.
And what the problem said, let's
consider going at 30 degrees
to the z-axis.
That's an angle like that.
What is the electric
field that is
generating in that direction?
Again, this is just
to give us practice
at calculating vectors,
components, et cetera.
So I will do that now over here.
So this is the
solving the Part 3.
So what we're
interested in, we have
this charge moving up and down.
And we're interested in the
radiation a long way away,
a distance R away
from this charge,
but at some angle like
that, with just 30 degrees
to the z-axis.
So in order to
use that formula--
it's all the time
the same thing,
I'm just using
the same formula--
I need to calculate the
component of the acceleration
of the charge which
is perpendicular
to that direction, OK?
Now there are many ways.
You can do it in
your head, if you
are good at
manipulating vectors.
Here, I wrote it, in case you
wanted a complete formula.
Here is how the
perpendicular component
is related to the
acceleration, a.
And at the beginning there,
I summarized it for you.
But you can do it
any way you want.
I did it this way.
And so I am taking this charge
and calculating, at some time,
t prime, what is the
component of the acceleration
of the charge in a direction
perpendicular to the direction
in which I am considering
the radiation.
And this is what it comes out.
If you remember, I know what
is the motion of the charge.
You've seen it over there.
So I know what z is,
a function of time.
Therefore, I can calculate this.
And that's what it
comes out as, alright?
Here, I formally
use this equation.
So it's as before.
We will be interested in the
electric field at the position,
R, so at a time which is later.
In other words, t
minus R/C is the time
at which the prime time,
t prime, at which I'm
interested in this acceleration.
And that's what
it looks like, OK?
Now, I can simplify it.
I'm not going to
waste your time.
You can go through this algebra.
Simplify it here.
And then I take
this acceleration
and plug it into that
equation, which tells me,
if I know the perpendicular
component of acceleration
at this earlier time, I
know what the electric field
is at the time t.
And this is what
it comes out at.
So the electric field
in a direction 30
degrees to the z-axis,
or in other words,
at a distance, capital
R, from the origin.
Or in other words,
at the coordinates 0,
R/2, R root 3 over-- and this is
the y-coordinate, x-coordinate,
y, and z-- at this time, I
plug this into that equation.
And I get this
and the direction.
If you do it formally, you
immediately get the direction.
It's not very hard.
It'll be, in fact, it's
60 degrees to the z-axis.
All right.
So that's what the
electric field will be.
Then, knowing the
electric field,
again, I know that E and B are
perpendicular to each other.
They are proportional to each
other, except for the 1/C.
All right?
And so, at the same
location, I can calculate B.
And I get this.
The magnitude here
and here are the same,
except for one power
of C here, C cubed.
The direction is different.
B will be the x direction.
And that, you could
do in your head.
If the charge is oscillating
like this, at all angles,
you'll still have
the magnetic field,
which will perpendicular
in that direction.
All right?
And you can check for yourself.
This half is simply
the sine of 30 degrees.
OK.
So I've gone very
slowly, intentionally,
and taken the simple
situation where
a single charge is
oscillating up and down
and I've calculated for
you the electric field,
the radiated electric
field, up and down, at 0.
In a horizontal plane, I've done
it effectively, all locations.
And I've done it
at an angle, theta,
where theta was 30 degrees.
I could repeat this for every
angle, theta, every location.
So in fact, just taking
more and more cases,
we have calculated the
electric and propagating
electric and magnetic field
in all directions, all right?
And we calculated the magnitude
at a distance from the origin.
OK?
It drops off like 1/R we saw.
And like the other field, we
drop off like 1 over r squared.
And so this charge, which
is oscillating up and down,
radiates in all directions,
but with different magnitudes,
biggest in the
plane perpendicular
to the vector describing
the acceleration.
OK.
Now the next thing that I
will do now is the following.
Let's take the same
situation and try
to calculate the total power
radiated by the charge.
So I've taken this same problem.
There is the charge
oscillating up and down
with amplitude z0
cosine omega t prime.
And what we see-- well, we
know that it radiates energy.
If it radiates energy,
of course, normally,
it would have to finally stop.
But I'm assuming there is
some mechanism which uniformly
maintains the motion of these
charges in a certain motion.
Energy will be radiated.
And I want to calculate what
is the total energy radiated
per second by this charge.
And the way we are going to
do this-- so the question now
is, for this particular charge,
oscillating like we discussed,
what is the total energy
per second-- in other words,
power-- radiated at
the time t prime?
Clearly, if it's
radiating energy outwards,
by conservation of
energy, the amount
of energy coming
out of this will
be the same as the
amount of energy crossing
a sphere of radius,
R, at a later time.
So if I calculate the energy
crossing a sphere at the time
t, where t is this
t prime plus R/C,
at a later time, that
will be the total energy
per second radiated by this
charge at the time, t prime.
So I will now
calculate the energy
per second which is crossing
the surface of this sphere
at the time, t.
Well, how much energy
is crossing the surface?
We know that, at the
surface of this sphere,
there is, at every location,
an electric field which
is parallel to the
surface, because this
is perpendicular to R.
The electric field is
perpendicular to this.
So at this location, for
example, the electric field
will be like this,
crossing it, part
of the surface of this sphere.
There will be a magnetic
field perpendicular to it.
And so, at this
location, there will
be a energy flux, the
so-called Poynting vector,
the energy per second
crossing a unit [INAUDIBLE].
And that's this vector,
s, the Poynting vector,
which, as you know
from Professor Walter
Lewin's lectures, is E cross
B. I forgot my vector sign
on top of the B. These are
vectors, of course, E cross B.
Now we know, for an
electromagnetic wave in vacuum,
that the E and B are
perpendicular and proportional
to each other, just the
fact the C. So this will
be the magnitude of E squared
divided by this mu 0 times
C, because B is E/C.
This is a scalar quant--
these were vectors.
This is a scalar.
So this is the energy flowing
per unit area, per unit time
at the surface of this.
And the E will depend on
the location on this sphere.
So what is the
total energy that's
leaving the charge
at time t prime?
It will be the sum
of all the energies
that are coming
out per unit area
along this surface at time t.
So it's going to be
the integral of SdA.
What I'm doing, basically,
calculating at every location
S multiplied by the area
there-- so that gives me
how much energy goes
through that area--
and adding them
across the surface
everywhere around
the complete sphere.
That will be the total
energy leaving here.
So imagine this is
oscillating up and down,
radiating these
spherical waves outwards.
And they cross this surface.
And at every place, there is
energy flowing, the Poynting
vector.
And here, I'm adding them all.
This will be the
total power radiated
by this charge at that
earlier time, t prime.
OK.
So I now have to just log
through and calculate this.
S, I have to
calculate from this.
And then I have to
calculate the piece of area.
Well, let me tell
you what I will do.
From what we've learned
from the earlier part,
the magnitude of E only
depended on this angle, theta,
and this distance.
So imagine, along the
surface, I take a slice
this surface, like this, where
every point along this surface
is at an angle, theta,
to the z-axis, OK?
And it's a thin slice.
So what will be this distance?
I'll take this to be d theta.
This distance is R, so that
distance is already Rd theta.
If I multiply that
by 2 pi r sine theta,
r sine theta is the
radius of this circle.
So this total area going all
the way around this sphere
is 2 pi r sine theta
multiplied by Rd theta.
OK?
So now along this,
everywhere, S is the same.
So if S is at that
angle, theta, and I've
multiplied by 2 pi R
sine theta Rd theta,
this is the piece of area,
if I multiply by that,
this gives me the total
energy flux or power
that is leaving the sphere
along this part, all
the way behind there,
all the way around.
OK?
Now what is the total
energy per second leaving
this complete sphere
is the addition
of pieces like that for
the complete sphere.
So it would be the
integral of this
for theta going from 0 to pi.
And then I've covered
the compete sphere.
So that will be
this integral, which
is the power radiated
by the charge.
OK, if you've understood
that, then it's
just pure algebra now.
S, I can calculate,
because I know
it's the magnitude of
E squared over C mu 0.
But that, we've already done.
We did it for 30 degrees.
And now do it for theta.
So here, instead of having
sine 30, which was 1/2,
I have sine theta there.
So that squared gives me
this divided by C mu 0 that.
And of course, it depends--
so this is what E is.
But it's oscillating as
cosine omega times this.
And we have to square it,
because this is E squared.
So this is what the
Poynting vector is at,
the angle theta, distance
R away at time t, OK?
And therefore,
the total power is
this multiplied by that
integrated from 0 to pi.
I've just rewritten
it, this quantity,
plugging that into here.
I end up with this.
Everything does not
depend on theta,
except sine theta, which
you have a squared from here
and another sign from there.
So I get sine cubed
theta d theta.
OK, we're home.
So that's the answer.
Because this you can do,
I won't waste your time.
The way to do it is you can
take sine theta d theta to be
d cosine theta, and
then integrating sine
squared theta d cosine theta.
And if you do that, you get 4/3.
So I plug 4/3 in
for that integral.
This is the answer, OK?
So what we see
now, through any--
and it's independent of R, which
I hope did not surprise you.
It's a statement of
conservation of energy,
because you have this
charge oscillating
up and down, radiating.
Through every sphere of radius
R, the same amount of energy
has to go through,
or otherwise, you'
d be gaining or losing energy.
So that's a check.
You can check.
If R came into this formula,
you would have made a mistake.
And the only place
where it does,
is only-- it oscillates,
so at any position, R,
the energy going through
oscillates, because it's
a wave going out, OK?
So this changes the phase,
but not the magnitude.
In the actual question,
it asked for what
is the time average
that's radiated outwards.
So we want to know the
time average of this.
And that's the time
average of that is here.
You know that the time average
of cosine some function of t
is 1/2.
If you take cosine--
sorry, cosine squared,
which was in here, cosine.
So the time average of
cosine squared is 1/2.
So finally, we get that
the average, time average,
power leaving this
oscillating charge
is given by this formula.
It's actually a famous formula.
It's called
[? Lambor ?] formula.
OK?
So that's the total
energy radiated.
Fine.
We have a few minutes
left, and so I'll
do one more quick problem.
OK, just for variety--
so far, we've
considered a charge which
is accelerating up and down.
And be considered all
permutations and combinations
about what happens to the
electromagnetic field going
outwards.
Now let's consider
something else,
a charge which is
going in a circle.
Let's consider that a charge,
Q, which is rotating in the z,
y plane with a
uniform velocity, v,
given by omega z0 where omega
is the angular velocity of this.
So we have a charge going
like this, uniform velocity.
What is the radiated field?
What radiated electric and
magnetic fields are produced?
Well, go back to
where I started.
The charge is moving
with uniform velocity.
Static charges don't radiate,
but accelerated charges do.
Does this charge accelerate?
It's moving with a
uniform speed in a circle,
but does it radiate?
Well, you know that,
if an object is
going with a uniform
speed, V, around a circle,
it's all the time
accelerating, right?
It's accelerating
inwards all the time.
And so this charge is
accelerating inwards.
So at every instant
of time, it's
accelerating like this,
so as this moves more.
So all the time, it has
a constant magnitude
of acceleration, but the
direction changes all the time.
So this charge, at every
instant of time, will radiate.
And it will radiate in all
directions, like before.
For simplicity, so we don't get
overburdened by mathematics,
let me just think for a
second what a-- let's just
talk qualitatively.
I can describe the motion of
a charge going like this by,
at any instant of time,
it has a position.
This is Z0 vector
is the position
of this charge at time t prime.
I could write it as Z0 cosine
theta y plus Z0 sine theta Z,
all right?
That's the y-coordinate.
That's the z-coordinate.
If that is the position, I
can rewrite this like this,
because theta, of course,
is given by omega t prime.
It's rotating, right?
What is the acceleration?
I differentiate this twice.
So here is the acceleration.
So if we consider
the two components,
what I see is that this charge
has a component in the y
direction and in the z
direction, like this.
And so it will, at every
instant of time, radiate.
I can decompose the
radiation, using this formula,
into the two components.
And what we see is that the
two components are out of phase
from each other by 90 degrees.
So for example, if I am
straight ahead looking at it,
I will see a charge which
is accelerating up and down,
like that, and out
of phase, like this.
So it will radiate
straightforward a component
like this, oscillating, and
out of phase by 90, like that.
And what is that?
You know.
That's called circularly
polarized light.
There are many
ways to look at it.
You can either look
at the components,
or you could look
at the rotation
of the electric vector.
At the moment, I
am looking at this
from the point of
view of components.
So if I look at this, straight
at it, what I will see,
an electromagnetic wave
with the electric vector
polarized vertically,
and another one
polarized horizontally,
out of phase by 90 degrees.
And so, at any
location in space,
I'll see an electric vector
which is rotating like this.
Of course, there's a magnetic
vector perpendicular to it.
And that we call,
circularly polarized light.
So this will radiate in that
direction circularly polarized
light.
If I look from
above, what do I see?
The charge is going
like this, so I just
see this component
from above-- you
can look at the two
components-- like that.
So straight up, I'll see
linearly polarized light.
If I look from the side, I
see it's going up and down.
I will see also
linearly polarized,
but in a different direction.
So a charge like
this, as before,
will radiate in all directions.
But the extra complication now
is, in different directions,
it'll be different polarization.
For example, if I'm
looking at this from here,
this crazy angle,
what I will see
is I will see a component from
the acceleration like this,
and from that, but they
have different amplitude.
And so I'll get the
two polarizations which
are different
amplitudes, corresponds
to elliptically polarized light.
So in this direction, I'll
see an elliptically polarized
light.
And using this formula
any place in space,
you can calculate the
electric and magnetic field,
just doing the same as
we did before, but take
the full vector description
of the perpendicular
component of the
acceleration of the charge.
So let me stop there and
just sort of summarize.
In principle, we've
done just very little.
What we showed today
is, in general,
if I have a charge which
moving with some velocity,
some acceleration,
et cetera, it'll
be surrounded by a
very complicated field.
There will be the Coulomb
field, the Biot-Savart field,
the radiated field, et cetera.
If you need to solve the
complete thing properly,
you will need a computer.
You can't do it.
But if you are only interested
in the radiated field which
is far away from the
accelerated charge,
it turns out the situation
is sufficiently simple,
you can do it almost on
the back of an envelope.
All you need to know, if you
are interested in the field
in any location, is
you ask yourself,
at an earlier time, what
was the charge doing?
And by earlier, I mean
at time that light
had time to come from
the charge to me.
So I take that distance, R/C,
and at that earlier time,
I need to know what
the charge was doing.
I ask myself, in which
direction it was accelerating?
I take the magnitude
of acceleration,
which is perpendicular to the
direction in which I'm looking,
and I calculate that and
multiply by some constants
that we've done
over and over again,
and that will tell me what
the electric field is.
From that, I can get
the magnetic field,
and I get the full radiation.
Thank you.
