Prof: I thought I would
go back a little bit to LCR
circuits.
 
I think that's one of the more
useful things you learn in this
course.
 
I want to do it carefully so
everyone's on top of that.
So remember that the problem
you want to solve looks like
this.
 
That's an AC source.
 
That is a resistor.
 
There's an inductor and a
capacitor.
And that's the circuit.
 
And this voltage we take to be
V_0
cosωt.
 
That's the kind of voltage that
will come out of any AC
generator.
 
The omega is controlled by how
the turbine is spinning.
That frequency is the same as
this frequency.
Our job is to find the current
in the circuit.
Now what makes it difficult is
that,
unlike DC circuits where you
take a voltage and divide by
some effective resistance,
the equations here are not
algebraic equations but
differential equations.
So we'll write it down then
I'll tell you how we handle it.
So the equation for this
circuit, you begin as usual,
you go around the whole loop
and you add everything to 0,
then you will find
V_0
cosωt =
R times I L
times dI/dt  1 over
integral of I up to time t
times 1/C.
 
That's the voltage.
 
Everybody with me on that part?
 
That determines the condition,
that determines the current at
every instant in time.
 
So if you can solve this
equation, you solve for the
current, but what's difficult is
that it's not an algebraic
equation.
 
It's not like V = IR,
where you divide by R,
you've got the current.
 
This has got integrals,
it's got derivatives.
But it turns out that if the
voltage is V_0
cosωt,
there's a very clever way to
solve this problem,
which takes it back almost to
the good old days of V =
IR.
No differentials,
no integrals,
nothing.
 
That's the magic.
 
That was invented by an
engineer from GE called
Steinmetz.
 
He put to work some ideas that
may seem esoteric,
because they involve complex
numbers, but it makes life
tolerable for people doing
circuit theory.
And here is the trick:
I ignore this problem,
and I solve the following
problem, purely mathematical
device.
 
The problem I solve has a
voltage V_0
e^(Iωt).
 
We all realize that you cannot
get a source that does that,
but it's pure mathematics.
 
Then the solution to that
problem, that voltage,
I assume, will drive some
current I call I˜.
So I˜ is the
answer to the problem where the
driving voltage is not real,
but this, if you mathematically
write the same equation.
 
But now we realize that
I˜ can be complex.
I˜ is not real.
 
It can be a complex number.
 
Just because the voltage is a
complex number,
it's an equation,
it's got imaginary and real
parts on the left hand side,
it will have imaginary and real
parts on the right hand side,
and so you'll need both to make
it work.
 
But if you have solved this
problem, I claim you have also
solved this problem.
 
Why?
 
Because if you take this
equation,
which is a complex equation,
then the real part on the left
hand side will match the real
part on the right hand side.
Imaginary part will match the
imaginary part.
That's because if you have two
complex numbers,
z_1 and you
say they're equal,
you mean x_1
Iy_1 is
x_2
Iy_2.
And they can be equal only if
these guys are equal,
those guys are equal.
 
The real part has to match,
the imaginary part has to.
You cannot borrow from the real
part and give to the imaginary
part.
 
They're apples and they're
oranges, so you've got to match
them separately.
 
The closest analog is a vector
equation.
If you have a vector equation
and you equate two vectors,
so you say vector
F is m times
vector a,
then F_x will
be m times
a_x and
F_y will be
m times
a_y.
 
So there are two equations in
one single vector equation,
three in 3D and a complex
equation, there are two
equations.
 
So let's take the equivalent of
the real part in the left hand
side and the real part in the
right hand side,
and what do you get?
 
The real part of this is the
most important thing.
The real part of this is
cosωt.
that's because
e^(iωt) is
cosωt 
isinωt.
The real part is just
cosωt.
On the right hand side,
I should take R times the real
part of I˜.
 
Let me just call that guy
I.
Then I should take L
times the derivative of the real
part of I˜,
which I want to call I.
Then 1/C times the
integral of the real part of
I˜,
the real part of
I˜ I want to call
I.
In other words,
let the real part of
I˜ be called
I.
Then we notice that this
I is exactly the I
that we want.
 
It satisfies exactly the
equation we want to solve.
So we have some good news and
some bad news.
The good news is that if you
solve this equation,
you've also solved this
equation.
The bad news is,
if you cannot even solve the
real equation,
what makes you think you can
solve this complex equation,
right?
I cooked up another problem
that looks even more difficult
and I say, "Hey,
if you can do that,
we can do this."
 
Well, it turns out this complex
problem is actually easier to
solve than this guy.
 
I'll try to tell you why it's
easier to solve.
If you come to this equation,
I've told you repeatedly,
any equation involving
derivatives and so on,
you guess the answer and you
put it in.
You try to make a guess.
 
I'm trying to guess,
what kind of function I(t)
has the property that when I
multiply it by R,
I should get a
cosωt,
when I differentiate,
I should get something like
cosωt,
when I integrate,
I should get
cosωt?
That's the only way left and
right will match,
and neither cosine nor sine
will do the trick.
If you pick the sine,
then when you differentiate it,
you will get a cosine that
matches.
This one integrated,
you will get a cosine that
matches this,
but this will contain a sine.
On the other hand here,
since this exponential,
if you make the choice
I˜(t),
it's I˜_0
e^(iωt),
it is going to work.
 
It is going to work because if
you take R times that,
you get R times
I_0
e^(iωt).
 
If you take the derivative of
this guy, that's Iω
times the same current.
 
Integral is 1 over
Iω times the same
thing.
 
That's the beautiful property
of the exponential,
that when you differentiate it,
just like multiplying by
iω.
 
When you integrate it,
it's like dividing by
iω,
and when you multiply by
R, it's like multiplying
by R.
So if you take this assumed
solution and put it into that
equation, let's see what we get.
 
We get e_0
e^(iωt) = (now
allow me to jump one or two
steps, because I did it last
time.
 
Try to do this in your head.)
 
R times I˜
is going to contain R
times I˜_0
e^(iωt).
How about L times
dI˜/dt,
then I come with the
d/dt, this guy is a
constant.
 
This brings me an
Iω,
so I write IωL
times the same thing.
When I integrate,
I get 1 over Iω
times the 1/C that looks
like that.
So the time dependence,
this is a time dependent
problem, but miraculously,
time dependence matches on the
two sides.
 
And now we get this equation,
V_0 = R iωL
1/iωC times
I˜_0.
And our goal was to find
I˜_0,
so I˜_0 =
V_0 divided by
this complex number Z.
 
This complex number Z is
called the impedance,
and it = R iωL
1/iωC,
for this problem.
 
I'm not telling you the answer
for every problem,
but for this problem,
life is very simple.
You find that current as a
voltage divided by a number,
just like dividing by R.
 
The only trick is,
there are two catches now.
First catch is,
this number is not real,
so you've got to get used to
the fact that you will work with
complex numbers.
 
If you're willing to work with
them, you get this.
Second thing is,
this is not the current we were
supposed to find.
 
What's the current we were
supposed to find?
The current that I wanted was
the real part of this
I˜.
 
That means it's the real part
of I˜_0
e^(iωt) divided
by--sorry.
That's correct.
 
That's what I want.
 
And that's going to be then the
real part of V_0
over Z e^(iωt).
 
We always pick V_0
to be a real number.
The amplitude on the voltage
you can take it to be a real
number.
 
Then you write it as a real
part of V_0
divided by Z
e^(i)^(Φ).
So I should tell you what this
is.
I'm saying, imagine the complex
number Z, plotted this way.
This is R, this is iωL
− 1/i −
i/ωC,
and this is the complex number
Z.
 
This is the angle Φ.
 
Then you write the same complex
number in polar form as a
modulus and e^(i)^
(Φ). So now you
find I of t is the
real part of
V_0 over mod
Z e^(iωt -
)^(Φ),
and that we know is
V_0 over mod Z
cos(ωt −
Φ).
 So that's the final
result.
 
That's the answer to the
original question.
Now you know trigonometry well
enough to know that
cos(ωt −
Φ) is
really cosωt
cosΦ
sinωt
sinΦ.
So the answer cannot be just a
cos and cannot be a sine.
It's a suitably chosen
admixture of sines and cosines
that does the trick.
 
So if you want to,
if you say, "I don't want
to deal with complex
numbers"
you can take a guess like this
and put it in the equation,
and after a lot of
manipulation,
you will find there is a
Φ satisfying this
condition.
 
What's the property of
Φ?
Tan of Φ is
ωL −
1/ωC divided by
R.
But the beauty of the complex
numbers is, it just comes out in
one package as the phase of a
complex number.
I also told you to think about
how you could do this with real
numbers.
 
It's not going to be easy to
take a voltage,
which is cosωt,
divided by anything,
to get a current which is
cos(ωt −
Φ).
 
There's nothing you can do to a
cosine which will shift its
phase.
 
But if you're working with
complex numbers,
you can take e^(iωt)
divided by
e^(i)^(Φ) and
turn it into e^(iωt -
)^(Φ).
 
So in the imaginary world,
it's very easy to shift the
phase,
because a complex number,
when it multiplies or divides
another number,
rescales and rotates it.
 
So in the complex plane,
it's very easy to attach the
phi, and at the end of the day,
you take the real part.
So you can imagine doing any
realistic problem.
Suppose I tell you R is
10 ohms, where ω is
some 100Π.
 
L is 3 henries and the
capacitance is 2 microfarads,
you can find Z. Z
for this circuit will be 100--
I'm sorry Z will be 10
ωL,
will be 100Π
times 3 henries −
1/ωC,
is 100Π times
2�10^(-6).
 
That's some complex number.
 
I forgot the i here.
 
That is a complex number.
 
I don't want to calculate it.
 
It's whatever it is.
 
Then it's got a real part which
is 10, an imaginary part,
which is this algebraic sum of
these two.
You can actually plot the real
values you get here in this
problem.
 
I'm just going to show you the
10.
The other guy's whatever it is,
and you can get the phase.
That tells you that in this
problem,
the current will have an
amplitude which is the volts you
apply,
maybe you applied 100 volts,
and you divide by this absolute
value of Z.
So if you want,
I'll complete the last part.
The absolute value of Z,
like for any complex number,
is R^(2) ωL - 1
over ωC squared,
right?
 
Any complex number,
the absolute value is real
squared imaginary squared under
root.
So let's write in all its glory
the current that we want.
I(t) is
V_0/R^(2)
ωL - 1 over
ωC squared cosine
of ωt -
Φ where
tanΦ = ωL
− 1/ωC divided
by R.
 
Here is the picture,
here's R.
ωL minus
1/ωC.
There's Φ.
 
There's a complex number
Z. Okay.
So you can imagine now putting
numbers and getting what you
want.
 
Notice that the current lags
the applied voltage.
For example,
if cosωt is a
maximum of t = 0 and
Φ was 45 degrees,
you'll have to wait till
ωt is 45 degrees
before the current reaches a
maximum,
so it will be lagging the
voltage.
But sometimes Φ
can come out negative.
Can you see how Φ
could come out negative?
This angle Φ need
not be positive,
do you understand that?
 
What would make it negative?
 
Look at the formula there,
ωL −
1/ωC.
 
Yes?
 
Student:  If
1/ωC is bigger than
ωL.
 
Prof: Right.
 
If 1 over ωC is
bigger than ωL,
this will be more negative than
positive.
The complex number may end up
like that.
So I've written it for one
possible sign.
If that flips sign,
the Φ itself will
change sign.
 
It will become ωt
32 degrees.
It can go either way.
 
It depends on who is dominating
it, whether the inductor or the
capacitor is dominating it.
 
So let's look at the answer for
one other interesting feature.
I want you to know two things.
 
This trick works only if the
voltage is a pure oscillatory
function like
cosωt.
But then you can write it as a
real part of
e^(iωt) and do
this.
Secondly, the impedance is not
a constant.
Whereas the resistance is just
10 ohms, impedance varies with
frequency.
 
It's a frequency dependent
number that you divide the
voltage by to get the current.
 
So the consequence of this is
the following.
Suppose you plot here the
magnitude of the current,
let's call it
I_0.
This whole thing,
let me call it
I_0.
 
That's the amplitude of the
current, as a function of
frequency for a given applied
voltage V_0.
So V_0 is
fixed, but when ω
varies, these numbers are
varying.
When ω goes to 0,
you've got a 1 over 0 in the
denominator.
 
That's going to beat everything.
 
And 1 over 0 squared in the
denominator means the whole
thing vanishes.
 
The current starts out as 0.
 
That corresponds to the fact
that if your voltage had been a
DC source instead of an AC
source--
that's what ω = 0
means--
the capacitor will charge to
some point and then stop the
current.
 
That's it.
 
Then it will go up,
then come down,
because at very large
ω,
the ωL is going to
dominate.
You get 1 over ωL
squared under root,
that looks like
1/ωL.
It will fall like
1/ω at very large
frequencies.
 
And it will be the maximum when
these two guys cancel each
other.
 
So you've got R^(2).
 
If you're trying to get the
maximum current,
there's nothing you can do
about R^(2).
It is what it is.
 
But you can play these two guys
against each other and find a
frequency ω
and R so that this
is true and that
ω_0 is
just related to the resonant
frequency of the LC combination.
So at that frequency,
the current amplitude will be
simply--I_0
maximum, will be simply
V_0/R.
 
It's as if L and
C are not there.
You've got them there but
they're not there.
They're not doing anything.
 
They neutralize each other.
 
So you can ask,
why put something in that
doesn't do anything?
 
Well, first of all,
it does something on other
frequencies.
 
Only at one magical frequency,
they go away.
But what is interesting is the
very sharp response you have.
So do you know where that comes
into play?
Student:  Resonance?
 
Prof: Resonance,
yeah, but which part of your
life?
 
Pardon me?
 
Student:  Radio.
 
Prof: Radio.
 
You guys know what that is,
because I don't.
You're always carrying some
recorded medium.
But if you listen to radio,
like in the old days,
you have this room full of
radio signals.
Everyone wants your attention.
 
All the radio stations are all
sending signals right now in
this room, and you want to pick
just one station that you like.
So what happens if that station
sends that information at a
certain
ω_0?
And if that's all you want,
you go to the store,
buy an LCR circuit,
with L and C
chosen so that LC is
1/ω_0
^(2),
then you will get a huge
response when you get the signal
from that station.
Now there's another station
with a different frequency.
You don't want to listen to
them, but you have to listen to
some of them,
because if their frequency is
here, your response to that
station is not 0.
It's a lot smaller than this,
but it's not 0,
and there may be yet another
station you can hear in the
background.
 
If you don't want to hear them,
you would like to get 0 signal
from them,
but you cannot kill this
function except at one
frequency,
but you can make it very sharp.
 
If R is very,
very small, this function will
be very large at resonance,
because V_0/R
will be a huge number.
 
It will also be very narrow,
so you can make this thing very
sharp.
 
And that's really controlled by
the ratio of R to
L.
 
So other stations do give you
weak signal, but you can make
them weaker and weaker by
picking a very finely tuned
oscillator.
 
Now the question is,
what if you changed your mind
and you want to listen to these
guys?
What should you do?
 
Can you go and buy one radio
for this station,
one radio for that station?
 
You know you do something,
right?
You fiddle with the dial,
but what do you think it does?
Yes?
 
Student:  Changes the
capacitance.
Prof: Changes the
capacitance.
It's not easy to change the
inductance, but it changes the
capacitance.
 
Nowadays if you open a radio,
I don't know what you will see.
It's all glued on to something.
 
But in the old days,
when all the parts were big,
you can open the radio and look
inside and you have a capacitor
which is called a variable
capacitor.
You show that with an arrow.
 
You can also have variable
inductors, I suppose,
but this is very common.
 
Now how do you vary the
capacitance?
Capacitance you know is
ε_0
A/d.
 
Yes?
 
Student:  Can you change
the surface area?
Prof: You can change the
surface area,
but how do you think you can go
change the surface area?
Student:  Having two
half circles and then the
current rotates around.
 
Prof: That's right.
 
You've got these interlocking
plates and you can have them
fully overlapped or not
overlapped.
If you draw the plates like
this, it's got many things.
And these plates can be either
really overlapping with the
other plates or pulled out.
 
When you turn the dial,
the two things move in and out,
and that varies the
capacitance.
That will give you a range of
frequencies and that's the range
you can hear.
 
That's one thing.
 
Another thing I want to mention
is,
this solution I wrote down,
I(t) is
V_0 over mod
Z cos(ωt −
Φ) has no
free parameters in it.
You tell me the time,
I tell you the current.
Whatever the voltage is,
you take that,
shift the phase by Φ
and divide by mod Z.
But you know that a second
order equation in time must have
two free parameters,
so where are those free
parameters going to come from?
 
Have you seen this before in
your math whatever?
No?
 
Okay, I'm going to give you a
clue and you have to think about
this clue and see what you get
out of that.
V_0
cosωt =
V_0
cosωt 0.
That's your clue.
 
What can you do with that clue?
 
Yes?
 
Student:  Set the
original equation equal to 0 to
find the complementary solution.
 
Prof: Yes,
so I will translate.
If you have a problem where you
apply one voltage,
V_1 and you
get a current
I_1,
and a second voltage
V_2 and you get
a current I_2,
just by adding the two
equations, left hand side to
left hand side and right hand
side to right hand side,
you can check that
V_1
V_2 drives a
current I_1
I_2.
 
If you don't see that,
you should do that.
Take that equation,
but any V of t,
get an answer due to
V_1,
call it I_1,
another due to
V_2,
call it I_2
and add them up.
 
Left hand side is clearly
V_1
V_2 and the right
hand side,
R times I_1
R times I_2
is R times
I_1
I_2 and so on.
 
So I_1
I_2 will satisfy
the equation when the driving
voltage is the sum of the two
voltages.
 
It all comes because there's
nothing nonlinear.
For example,
for the right hand side at
R times I^(2),
then the first equation will
have R times
I_1^(2),
second will have R times
I_2^(2).
When you add them,
you'll get R times
I_1^(2)
I_2^(2),
but what you want is really
R times I_1
I_2,
the whole thing squared.
So if you have nonlinear terms,
you cannot add solutions,
but if you have linear terms,
you can add solutions.
Therefore 0 is a hint.
 
It says, add the solution for 0
voltage.
You might say,
"There is no solution for
0 voltage,"
but I will have to remind you
that we did solve the problem
the other day of an LCR circuit
connected to nothing that had a
current flowing for a while.
But it won't do it if it's
completely inert.
But if you have put some charge
on this guy to begin with,
then you saw that the current
and the charge,
everything starts and
oscillates exponentially like
e to the - some number
t times some
cosω't - some other
phase,
call it χ.
 
And this thing had two free
parameters.
There's an area here and a
χ here.
Or you can write it as
e^(-)^(γ)
^(t) times
Acosω't
Bsinω't.
 
No matter how we write it,
this function,
this solution to 0 external
voltage is called a
complementary solution,
and the final solution for
I is really what I wrote,
V over mod Z
cos(ωt −
Φ),
this complementary function.
 
Because if you take this
complementary function and add
it to the equation,
you might think it will mess up
the equation,
but it won't,
because the I that I wrote down
will give me
V_0cosωt.
 
The complementary function,
when you take RI
LdI/dt  1 over C I dt
for the complementary part,
you will get 0,
because the complementary guy
obeys the following equation:
it obeys the equation L
dI/dt R times IC 1
over C integral IC dt
is 0,
is the equation satisfied in
that loop.
 
If it has a non-zero solution,
you can add it to your
equation.
 
See, in elementary calculus,
you learn that if you solve an
equation,
say dF/dx =
x_2,
the answer is
x_2/2
[x^(3)/3]
a constant,
because when you add a
constant, you don't screw up the
solution,
because the d/dx kills the
constant.
Question is,
what can you add to this
differential equation?
 
You can add anything,
any function,
that's annihilated by that
differential equation.
If you put the function in,
you get 0 from all the
derivatives and integrals,
you can add that to anything.
Therefore the true answer is
this complementary function.
It will contain these two
parameters A and B and you pick
them depending on initial
conditions.
In other words,
if you took an LCR circuit and
took a generator and you hooked
it up,
from the instant you hooked it
up, it won't immediately assume
this form that I have here,
because for example,
at t = 0,
maybe the current is 0 in the
real problem.
 
This current doesn't vanish,
but you should add to this,
this function.
 
Choose A and B so that the
current vanishes at the = 0.
You can fit the actual
experiment to the solution by
choosing A and B to match the
initial current and say the
initial charge in the capacitor.
 
But we don't usually care about
this guy.
Do you know why we don't talk
about this too much?
Yes?
 
Student:  It dies off.
 
Prof: It dies off
exponentially,
so in real life,
it doesn't matter.
If you really want to know
what's happening,
in other words,
take an LCR circuit,
take an AC source.
 
Put a switch,
keep the switch open.
The minute you close it,
the current will be 0 by
continuity.
 
Current cannot jump.
 
So this function obviously is
not the whole answer.
It doesn't vanish at t =
0.
But this extra part you add in
with suitably chosen A and B
will fit those initial
conditions.
But after oscillating for a
while, e to the -
something t for a long
t will just go away.
Then it will settle down only
to this function.
This is called a steady state
current and this is called a
transient current.
 
So transients are important;
you cannot ignore them.
Maybe a transient will burn
your circuit,
but it doesn't matter after a
long time.
If you survive the early time,
it doesn't matter for the long
time.
 
It's a lot like this course.
 
Okay, all right,
so I want to take a minute to
tell you a little more about the
use of these complex numbers.
I do it for a lot of reasons.
 
Now I don't know what field you
guys are in.
You could be in art history.
 
You just don't know when
complex numbers are going to be
relevant to what you do,
you just don't know.
But I'm serious,
if you're doing engineering,
electrical engineering,
mechanical engineering,
it's all about complex numbers,
because everybody writes a
differential equation.
 
If it's a linear differential
equation, you solve it with
complex numbers.
 
And I like to also show you why
sometimes we should not resist
the mathematics so much,
because you might think in this
circuit I don't need complex
numbers.
After all, in the end,
the answer was a
cos(ωt −
Φ) and that is
cos times some number sine times
some number.
I can put the combination and
fiddle with it till it works,
that's true.
 
But suppose I gave you a more
complicated circuit.
Here's a more complicated
circuit.
I have maybe a resistor,
an inductor.
Then I come here,
I have a capacitor,
maybe another inductor.
 
Go like that,
let's call it R_1,
L_1,
C_2,
L_3 and this is
some V_0
cosωt.
 
How are you going to solve this
problem by guessing?
You have no idea how to guess.
 
You cannot solve this problem
by guesswork.
But I claim that if you replace
this fellow by V_0
e^(iωt),
you'll be able to guess the
answer,
you'll be able to solve the
problem.
 
So I want to take a minute to
tell you why that works.
So do you know what I'm trying
to do now?
I'm trying to tell you that an
arbitrarily complex AC circuit
built of resistors,
inductors, capacitors,
connected any way you like,
add whatever you want,
if you go back to the
differential equation,
it will drive you insane,
because there is a derivative
of this current,
derivative of that current,
integral of that current,
all coupled in some way.
How are you going to solve it?
 
You cannot guess anything
cosine and anything sine.
It's going to be a madness.
 
But I will show you that if you
use complex numbers,
you can reduce the problem to
something that looks like a DC
circuit.
 
That's all I want to do.
 
So I'm going to give you the
idea, but I don't want to do the
whole thing, because I don't
want to spend too much time on
algebra.
 
For those of you who want to
know where it comes from,
I want to give you a chance.
 
Those of you who don't want to
know exactly where everything
comes from,
in the end, I will give you the
so called bottom line,
stuff you must know,
then you'll forget all this.
 
So let's take that problem.
 
Let me take an easier one that
this one, because it's got too
many wrinkles in it.
 
But the point I'm making,
I can illustrate just as easily
with this guy.
 
So what did I call it?
 
R_1,
L_1,
C_2,
L_3.
Let the current here be
I_1.
Let the current here be
I_2,
current here is
I_3.
And this is V_0
cosωt.
Our job is to find those
currents if they're all
oscillatory functions of time,
in magnitude and in phase.
If you do that, you're done.
 
So how do we do it normally?
 
What are the fundamental
equations for a circuit?
The fundamental equations are
that at every branch,
the incoming current should be
equal to the outgoing current,
which for us means
I_1 = I_2
I_3.
 
So I_2 and
I_3 come down
here and they join and become
I_1 again,
so I'm going to assume that and
write I_1
there.
 
So I've got two unknown
currents, I_2
and I_3.
 
That's the first thing to
understand, how many free,
independent currents there are.
 
Generally, if it's a simple
problem, it's equal to the
number of loops you have.
 
That's a loop here,
that's a loop there.
You can draw many other loops,
like that one and that one,
but they will not give you any
new results.
So you've got to write voltage
equations that say if you go
around a loop,
and you come to where you
start, the change is 0.
 
The equation I would write
would be V_0
cosωt = (I'm
going to take this loop)
R_1
I_1
L_1
dI_1/dt 1 over
C integral
I_2 dt'. Back
to here is 0.
 
That's one equation.
 
For the second loop,
I can go around the other way
if you like, going round this
inductor, or I can take a loop
like this.
 
I just want to write that to
show you what it may look like.
That equation will say
L_3
dI_3/dt - 1 over
C_2 integral
I_2 dt = 0.
 
The - sign comes because if the
current is assumed to flow this
way,
and your loop goes counter to
the current,
you're going uphill,
so everything should be
subtracted,
but if you go down,
you can write it going down
this way.
 
So these are the three
equations to solve.
They're complicated because
there are differentials,
integrals and you know what.
 
But if I solve the following
problem: V_0
e^(iωt)--
so don't bother to write all of
this,
because you're not responsible
for this detail.
 
I just want you to know what's
going on.
So I scrap this problem and I
solve this problem,
and the current for that I take
to be the twiddle,
okay?
 
I say, let the
I˜'s be the answer
to this problem with
e^(iωt).
Can you see that if I took the
real part of the first equation,
I get the correct equation for
the physical currents;
that if I took the real part of
the second equation,
everywhere I replace
I˜ by the real part
which I call I,
and this becomes
V_0
cosωt,
I get a second voltage equation.
 
And here again you drop the
twiddles by taking the real
part,
that's the equation satisfied
by the physical currents,
I_2 and
I_3.
 
So it is still true,
even in this complicated
problem,
that once you've solved it,
you may take the real part of
the voltage and take the real
part of the current anywhere,
that's the answer to the
primordial problem.
 
So why does this make life easy
for you?
Because now you may take the
current.
I'll just take one of these as
an example.
Let the current be current
I˜_1,
let it be I˜_1,0
e^(iωt),
and I˜_2 =
I˜_2,0
e^(iωt),
and I˜_3
is I˜_3,0
e^(iωt).
Make that guess,
and what does this mean?
Think about it?
 
Go to the first equation.
 
Equate these twiddle guys and
all the e^(iωt)'s
cancel on all the three terms,
so you get an equation that
says I˜_1,0 =
I˜_2,0
I˜_3,0.
 
No time dependence anywhere.
 
It looks like a DC equation.
 
And let's go to this equation
here.
You will find V_0
= R times
I˜_1_
 iωL_1
I˜_1
1/iωC (I should
have called this
C_2)
ωC_2
times
I˜_2.
 
e^(iωt)'s have
been canceled everywhere.
So what do you find?
 
This again looks like an Ohm's
law problem where R_1
is the impedance of this guy,
iωL_1 is
the impedance of that guy,
so the whole thing looks like
some Z_1 times
I˜_1
Z_2 times
I˜_2.
In other words,
let me just write it below this
here.
 
The equation we get is,
V_0 =
Z_1 times
I˜_1
Z_2 times
I˜_2.
So the problem will look like
this: just put some black box
here,
Z_1,
another black box,
Z_2,
another black box,
Z_3,
join them, bring them out.
 
Call the current here as
I˜_1,0,
the current here is
I˜_2,0,
the current here is
I˜_3,0.
And these constant numbers,
they have no time dependence,
obey the same equations as the
real currents did in the real
problem.
 
So it's like a DC problem.
 
The only subtlety is,
the impedances are complex,
and the rule for impedance is
very simple.
If you open the box and you
find a resistor an inductor,
the impedance for that is
R_1
iωL_1.
 
That's it.
 
Apart from that variation,
it is just like DC circuits.
Yes?
 
Student:  Is there any
physical significance to the
imaginary part of the solution?
 
Prof: No.
 
If you want,
the imaginary part of the
solution is the answer to a
person whose voltage was
V_0
sinωt.
So you are solving two
problems, but we believe that
it's nothing new.
 
If you can do
cosωt,
the answer to
sinωt is the same,
except you shift by
Π/2.
Everything looks the same.
 
But it answers that question.
 
Okay, so this is what I want
you to know.
Given an AC circuit,
here's what you do.
You make this rule:
whenever you see this guy,
it is R.
 
Whenever you see this guy,
it is iωL and
whenever you see this guy,
it's 1/iωC.
Then you can combine impedances
in series by adding them,
or in parallel by doing the
reciprocals and adding them,
then taking the inverse of
that.
It's all like DC circuits.
 
What will you have at the end?
 
At the end, you will have found
out every current
I˜_
0,α.
α is 1,2 or 3 in
this problem.
You solve these DC like
equations and you will find
these DC like currents,
but they'll be complex.
What is the relation of this
guy to the actual current
flowing through the actual
circuit element?
The answer is
I_α(t)
will be I˜_
α times
e^(iωt),
then take the real part.
That's the algorithm.
 
That's the relation between
these time independent
functions, because what did you
do?
You stripped their time
dependence.
You wrote every current as
e^(iωt) times a
number that doesn't depend on
time.
It's those numbers that we have
solved for.
But to go back to the original
current,
first you've got to reinstate
the exponential you canceled
everywhere and remember to take
the real part because that was
the deal.
 
It's called the real deal,
real deal.
You take the real part of your
answer.
How about following?
 
You've solved a very
complicated circuit.
I don't even know what's going
on.
Here's one guy.
 
It's part of a big mess.
 
Its impedance is Z.
 
There is a current,
I˜_0
going through it,
meaning it's the complex
amplitude of that.
 
If I ask you,
what is the actual voltage
across this?
 
If I put a volt meter here,
what will I measure?
Answer will be,
again,
I˜_0
times Z.
That's in the fake Ohm's law
calculation, but you've got to
remember that these all had time
dependence, which I removed.
Then you've got to remember
that I should take the real part
of the answer.
 
That will be the actual
instantaneous voltage across
that circuit element at that
time t.
But remember,
I˜_0
could be a complex number.
 
The Z inside the box
could be a complex number.
When you multiply them,
you've got lots of complex
numbers.
 
You add on to get this guy and
take the real part.
I would recommend,
when you come to such problems,
that every complex number like
Z,
you write as an absolute value
times e to the i times some
phase.
 
That makes your life easier,
because then the whole thing
will be the absolute value of
this times the absolute value of
that,
times e^(iωt) the
phase of this the phase of that.
 
And the real part of it is the
cosine of ωt these
two phases.
 
You understand?
 
So if you wrote it as mod times
e^(i)^(Φ)
_1,
absolutely value of Z
e^(i)^(Φ)
_2,
times e^(iωt),
what is the real part of this
crazy number?
 
These are all real numbers?
 
You pull them out,
combine the exponentials,
you'll get e^(iωt
i)^(Φ)
_1 from
this one,
iΦ
_2 from
that one.
 
So the full current will be
cos(ωt Φ
_1
Φ
_2).
 
So the answer will be mod
I_0 mod
Z cos(ωt 
Φ
_1
Φ
_2).
 
So I don't mind waiting to give
you some time to digest this.
You should be able to at least
solve simple DC circuits with
more than one loop.
 
If you've got only one loop,
just do what we did earlier,
V over Z and all that.
 
But if you've got two loops,
you should learn how to handle
that problem,
because it's very useful.
You're going to deal with
circuits no matter what you do.
Yes?
 
Student:  In the initial
problem, if you were just trying
to find the total current
through the system,
impedances add,
like--
Prof: That's what I said.
 
That's correct.
 
So let us ask the following
question.
His question was,
what if I'd like to know the
current coming out of this guy?
 
Right?
 
Everybody with me?
 
That was his question?
 
So we do it as if everything
were real.
First we combine these two guys
into a single impedance,
which will be Z_2
Z_3 divided by
Z_2
Z_3,
just like you combine resistors
in parallel.
To that guy,
I will add
Z_1.
 
That whole thing is the
impedance seen by this thing.
Divide the voltage by this
total effective impedance,
that will give you the current
coming here.
Yes?
 
Take that current,
multiply by
Z_1 times
e^(iωt),
take the real part.
 
That's the instantaneous
voltage on this one.
Here's one more thing:
when that current comes here,
how is it going to branch
between these two?
In DC circuits,
if this was
R_2,
this is R_1,
and the current comes here,
its propensity to go on this
side is proportional to that
resistor,
because if that is more
resistive, it's more likely to
go here.
 
So the fraction going to this
side will be
R_2 divided by
R_1
R_2.
 
The fraction going to this
branch will be
R_1 divided by
R_1 
R_2.
 
You replace all of the
R's with Z's and
you get the same result.
 
So you can do everything like
before, except you've got to get
used to complex numbers.
 
Here's the problem:
take the circuit assigned to
each element,
these impedances.
Add them like in the old days,
combine them like in the old
days, solve for the current.
 
To get the physical current,
multiply by e^(iωt)
and take the real part.
 
To find the drop against any
entry,
any element,
take the current there times
the impedance,
times e^(iωt),
then take the real part.
 
That's it.
 
I cannot say it anymore because
I don't know another way to say
it.
 
You've just got to do problems
so you get a feeling for how
it's done.
 
I wanted you to understand,
sometimes you take a problem,
you embed it in a bigger family
of problems that looks more
difficult,
but is actually easier.
It's even true,
there are many integrals you
try to do which are very
difficult.
But if you think of it as an
integral in the complex plane,
sometimes that problem's easier
to solve than the real integral,
so this happens a lot.
 
You take a problem,
you generalize it.
Sometimes the generalized
problem is easier than the
original one.
 
Now I'm going to give you one
counter-example,
let me see where to pick it,
where this rule about taking
the real part at the end of the
day fails.
So far the rule was,
do everything with a complex
thing.
 
At the end of the day when
you've found something,
you take the real part.
 
Here is a place where it does
not work, so let me tell you
where it is.
 
Take the LCR circuit,
where we found that the voltage
was V_0
cosωt,
and the current was
V_0 over
absolute value of
Zcos(ωt -
Φ),
where Φ is all that
stuff.
 
I don't want to repeat it.
 
What is the instantaneous power
generated by the power supply?
The instantaneous power for any
voltage source equals the
voltage at that instant times
the current at that instant.
And all of these are real
voltages and real currents.
I'm not doing any games right
now.
This is the real thing.
 
So what do you get here?
 
You get
V_0^(2) over
mod Z cosωt
times cos(ωt -
Φ).
 
That's what it is, right?
 
One voltage going like cos and
the current is going like
cos(ωt -
Φ).
So I want to write this out for
you as follows:
this is cos^(2)ωt
cosΦ
sinωt
cosωt
cosΦ.
 
This is just some high school
trig identities.
Cos of A B is cos
A cos B sine
A sine B.
 
So what do you notice?
 
CosΦ and
sinΦ are some
constants, but these are
functions of time and they're
all oscillating.
 
The reason they are oscillating
is that when you drive a current
in these circuits,
sometimes the L and
C are drawing energy from
the source,
sometimes they're giving it
back.
So sometimes it's not
monotonic, it's not always
drawing energy.
 
Sometimes it's taking it,
sometimes it's giving it back.
That's all the oscillatory
terms.
So what one likes to study is
called the average power over a
full cycle.
 
That means you take this time
dependent function,
you integrate it over time,
over a full cycle at frequency
ω and you divide by the
time.
And I ask what you get.
 
So let me write the obvious
parts, V_0^(2)
over mod Z.
 
Cosine square is a positive
definite number.
Its average over a cycle is 1
half.
I don't know how you want me to
show this.
One is to say
cos^(2)θ is 1
cos2θ/2.
 
And if you integrate that guy
over a full cycle,
the cosine 2θ is
periodic.
It gives you 0, you get 1 half.
 
Sinωt
cosωt,
if you put a 2 and divide by 2,
is proportional to
sin2ωt.
 
That guy completes 2 cycles in
one period.
Its average is definitely 0.
 
So the only thing that survives
then from all of this is the
cosine Φ term and the
1 over 2,
because average of cosine
squared is half.
So the average power in the
circuit looks like
V^(2)/2 mod Z
cosΦ.
That's the correct result.
 
In fact, we can see it in
another way.
Let's write it as
V_0 over mod
Z squared times mod
Z cosine Φ
over 2.
 
What is mod Z
cosΦ?
Can you tell from some picture
I drew of Z?
I don't know if I have any
picture.
I've hidden all of them.
 
But here is what Z looks
like.
That is Z,
that's Φ,
this is R,
this is ωL −
1/ωC.
 
What is mod Z
cosΦ?
Yes?
 
Student:  Shouldn't it
just be R?
Prof: It is just the
resistance.
So this is a fancy way for
R.
So this is just 1 over 2.
 
If you want,
it's just the amplitude of the
current squared times R.
 
That means the real energy loss
is taking place only in the
resistor.
 
You can write that as mod
Z cosine Φ if
you like.
 
The 1 half is new to AC
circuits.
In a DC circuit,
it's just I^(2)R,
because it's constant.
 
Here it's oscillating like a
cosine squared and the average
of that is 1 half.
 
So what people sometimes like
to do is to find something
called the RMS current,
which is
I_0/√2,
and to define the RMS voltage,
which is
V_0/√2.
Then the whole thing looks like
I_0I
_rms^(2)R,
or if you like,
V_rms times
I_rms,
period.
 
Now RMS means root mean squared.
 
There's a reason why for a
trigonometric function,
it's got a number which is 1
over root 2 times the amplitude,
but I don't want to go into
that, because we don't use it
again.
 
The main thing to know is that,
if you want,
you can redefine a new
quantity,
V_0/√2
and
I_0/√2,
so that these factors of 2
disappear,
and things look like the good
old days of resistive circuits.
 
So very often when they say
this is a 110 volt power supply,
they really mean the RMS value
110 volts.
That means
V_0/√2 is
110.
 
V_0 itself is
110 √2.
So the actual voltage goes up
and down from 110 √2 to
-110 √2.
 
Because for computation of
power, the peak amplitude
doesn't control it.
 
It's this RMS value that
controls the average power
consumption.
 
Now there is a way to get the
power using the complex numbers,
if you like.
 
I'm going to assign that as a
homework problem,
rather than do it here.
 
But I want you to think about
why does it fail here?
Why is it that when you come to
the power,
this is not simply--in other
words,
what I'm telling you is the
power is not simply the real
part of I˜t and
V˜t.
V˜ is a 1 with
V_0
e^(iωt).
 
Why is it that taking the real
part of this guy doesn't work?
Everywhere else,
we took the real part of the
current,
we got the right answer,
real part of the voltage,
we got the right answer,
but not when you take the real
part of this.
Yes?
 
Student:
Yes, he's saying you're
multiplying but not adding.
 
So let me say the following:
suppose you had a complex
number Z_1,
it's x_1
Iy_1,
and another complex number
Z_2,
which is x_2
Iy_2.
 
If you take the real part of
Z_1 you get
x_1.
 
You take the real part of
Z_2 you get
x_2.
 
But suppose you wanted
x_1x
_2 for some
reason.
That is the real part of
Z_1 times the
real part of
Z_2.
That does not = the real part
of Z_1Z
_2.
 
Because the real part of
Z_1Z
_2 has got
x_1x
_2,
that's pretty obvious,
but it's always got
-y_1
y_2,
because when these two
imaginaries multiply,
they can contribute a real part.
But what you wanted is the
analog of this one.
So you can fix that.
 
I mean, it's not that you can
never extract x_1x
_2.
 
The correct solution is
x_1 is Z
Z* over 2 x2
Z_2
Z_2* over 2,
and you can multiply it all out
and see what happens.
 
Then you can write the answer
in terms of these guys,
if you like,
but I'd rather just leave it
this way,
because it's not going to be
used extensively.
 
I just want you to be careful.
 
This is an example of something
that is quadratic in the
interesting quantities.
 
You cannot take the real part
at the end of the calculation,
only for things which are
linear.
Every time, the real and
imaginary parts do not talk to
each other throughout the whole
calculation.
You can follow them through and
at the end take the real part as
what's interesting.
 
But if you're going to multiply
two things,
where you have added on a
complex part to this guy and a
complex part to this guy,
in the product,
you've added on something real,
and that was not your
intention, so you've got to take
that out.
That's what makes it
complicated.
Therefore I tell you that when
you do these AC circuits,
when it comes to power,
forget any of the gimmicks you
learned.
 
You will be usually asked to
find the power only for LCR
circuit, the single loop like
that, then just go back to this
one.
 
It's just I^(2)R written
in this language.
And cosine Φ is
called the power factor.
Okay, so this completes one
chunk of the course.
I'm really going to the
finishing line for
electromagnetic theory.
 
There's only one big stuff
left, then we'll do optics and
then we'll do quantum mechanics.
 
Now the nice thing about
quantum mechanics is that you
don't have to worry about
whether you will get it,
because nobody gets it.
 
My idea is to take a class
where only I don't get it,
and turn it into a class where
everybody don't get it.
That's the plan.
 
When I say "get it,"
I think I talked to one of you
guys who called me and said
something which is quite true.
They said, "In the first
part of the course,
we could visualize what you
were doing.
There were masses rolling down
and things colliding.
With electricity and magnetism,
we're not afraid of the math,
but we don't have an intuitive
feeling for these things."
So I agree.
 
But if you work long enough
with electricity and magnetism,
you get used to them.
 
The only things born with the
knowledge of electricity and
magnetism are some creatures,
like ducks.
Apparently, ducks can feel the
earth's magnetic field and they
are just going through and they
know which way to go.
They don't solve Maxwell's
equation, but they know how to
travel.
 
And some bees are sensitive to
the polarization of light,
so they do respond to certain
things in an intrinsic way
people don't.
 
So you have to get used to it
only by doing more problems.
But in quantum theory,
any intuitive feeling you have
is actually a detriment.
 
It will get you in trouble,
because nothing is the way you
imagine it.
 
So no one has an advantage over
anybody else.
So it's better not to be well
informed.
So you may be saying,
"Hey, here is the course
for me,"
right?
So the less you know,
the better off you are in
quantum mechanics.
 
So don't worry about that.
 
It's a very strange world and I
want to give you an introduction
to that.
 
Anyway, they are the two things.
 
So what's left now is just
electromagnetic waves,
and we're sort of building up
to electromagnetic waves.
So I'm going to start by
writing here the equations that
we know about electromagnetism
as of now.
So the surface integral of the
electric field on any surface is
the charge enclosed divided by
ε_0.
That's called Gauss's law.
 
The line integral of the
electric field on a closed loop
used to be 0 when things are
static.
But when things are changing
with time, it's the rate of
change of the magnetic flux.
 
The surface integral of the
magnetic field,
B⋅dA
is just 0 all the time,
because there are no magnetic
monopoles.
There's nothing from which
lines come out.
So if you integrate any
configuration,
the lines don't start and end
anywhere, so whatever enters the
surface leaves the surface.
 
The surface integral of
B is always 0.
B⋅dA
is 0.
The line integral of
B⋅
dA--b
⋅dl is
μ_0I.
 
That's the old Ampere's law.
 
These are the four equations we
have.
Now it turns out this is not
still the end.
It's not the end.
 
There's one more fiddling you
have to do.
That's the last part I want to
talk about.
Remember, this itself came from
new experiments where you
started moving magnets and so
on,
found a current whenever
there's a changing magnetic
flux.
 
That's the induced electric
field due to changing flux.
So here is the thought process
that led Mr. Maxwell to a
little paradox.
 
So here is some circuit.
 
We don't know where it begins
or ends;
we don't care.
 
Here's the circuit.
 
There's a current flowing
through a capacitor.
When I say flowing through a
capacitor, I hope you know what
it means.
 
I want you to be very clear
about this.
Suppose you connect it to some
AC source.
I want you all to know what's
going on.
Nothing really flows through
the gap in the capacitor.
It's not possible.
 
What happens is,
for a while,
charges rush here and - charges
rush the other plate,
because that's the polarity of
my AC source.
Later on it's reversed,
then - charges go here and
charges go there.
 
So there is an alternating
current in the circuit,
but there is no current flowing
right through in one sense.
But there's nothing to keep it
from going this way and that way
and this way and that way,
because they don't have to jump
the gap to do that.
 
Yes?
 
Student:  It's sort of a
random question,
but you know how we were
talking about jumping the gap.
When would it actually jump the
gap?
Prof: It will jump the
gap if the--here is what
happens.
 
It depends on what's in the
medium.
If you put air,
for example,
the air molecules are all
neutral,
so there is no way for the
charges to ride them and go to
the - charge,
to the - terminal.
But if the electric fields,
this certainly will produce
electric field,
they're so strong that they rip
out the and - charges,
that suddenly you've got a
bunch of free carriers,
then the - charges can
immediately go and neutralize
this one,
can go and neutralize that one.
 
That's like a lightning strike
inside the little world.
That's what happens to us
during lightning.
We are in a capacitor.
 
The lower plate is the earth,
upper plate is the clouds.
And after a while,
there's a heavy charging.
At some point,
they cannot take it anymore,
and they ionize the air and
they create a little path and
the current flows through that,
and the clouds are discharged.
Okay, so here is a problem that
Maxwell had.
He says, let's look at this
equation,
B⋅dl
in a closed loop =
μ_0I,
where I is the current passing
through any surface with that
loop as the boundary.
Remember that?
 
So if you draw a loop here like
that, then some current is
crossing that shaded surface.
 
And the line integral of
B,
you know B fields go
around the current and the line
integral of B around a
loop will be the current
crossing that shaded surface.
 
That's how we get Ampere's law.
 
But now you say,
that's not the only surface
with that loop as the boundary.
 
I should be able to draw any
surface with the loop as the
boundary.
 
So you say let me take that
surface.
You are still okay,
because whatever current passes
this phase also passes through
that phase.
The law is still good.
 
So giddy with success,
you say, why not this?
That's my new surface.
 
It goes all the way around one
of the plates of the capacitor
and comes back.
 
Now I have a problem.
 
I have a problem because there
is no current passing that
surface.
 
Nothing going on between the
plates in terms of current.
If I draw an even bigger one
like this one,
I am again okay,
because now that same current
is passing through this one.
 
It's this surface in between
that things don't work.
You follow that?
 
We have a crisis where if you
say it's
μ_0I and you
say it's on any surface with
this loop as the boundary,
it's the boundary of the
surface, then this kind of
surface,
half in and half out,
has a problem.
So what will you do?
 
You realize you have to modify
your equations.
It's quite often how people
modify equations.
Sometimes they do experiments,
sometimes they do thought
experiments.
 
Einstein loved doing these
experiments, which are called
Gedanken experiments.
 
You don't really do the
experiment, but you say,
"If I did this,
what will happen?"
There's a little paradox,
and you have to modify your
theory.
 
So there are two reasons to
modify the theory.
One is experiments tell you is
wrong;
other is, theory tells you
there's a problem.
So you have to add something to
this.
That something I add should not
have any contribution from this
surface,
but on this surface,
it should make exactly the same
contribution as the physical
current made on this surface,
so that no matter which surface
I take,
I get the same answer.
And I'm going to do that by
the--there are many ways to do
this.
 
If you know more math,
there are more ways to get
this, but this is one that's
good enough.
I'm going to do the following?
 
I'm going to rewrite this as
follows.
You see in the region between
the plates, I have no current,
I agree.
 
I gave up.
 
I do have something between the
plates I don't have in the wire.
You know what that is?
 
Pardon me?
 
Student:  Electric field.
 
Prof: You have an
electric field.
There's something non 0 in
between the plates.
So I'm going to write this
μ_0I somehow
in terms of electric field
between the plates and let's see
how it goes.
 
So I'm rewriting the very same
term, I is dQ/dt.
Q is the charge of the
capacitor.
And let me write it as
Q/ε_0 d
by dt.
 
Now let me write it as
μ_0ε
_0Ad by dt of
Q/A divided by
ε_0.
 
That =
μ_0ε
_0A times
d by dt of the
electric field between the
plates.
Why is that?
 
Because the electric field is
σ/
ε_0 and
σ is just
Q/A.
 
Now bring that A in here
and you get μ_0
ε_0 d by
dt of electric field
times A.
 
But that we can write as
μ_0ε
_0,
the surface integral of the
electric field over the surface,
and that's what we call
μ_0ε
_0 electric flux,
d by dt.
 
Therefore my modified law is
going to be
B⋅dl =
μ_0I
μ_0
ε_0 d by
dt of Φ
_electric.
That's the bottom line.
 
You can fill in these blanks if
you have trouble writing and
listening.
 
I'm just saying,
go to the region between the
plates.
 
Once you start with
dQ/dt,
turn Q into the
σ,
the charge density on the
plates.
That's simply the electric
field, then you'll find,
this is simply the electric
flux, rate of change times
μ_0ε
_0.
This extra term is called the
displacement current.
I don't know why it's called
that name, but you know that you
have to put that extra term.
 
But now look at this equation
with that extra term.
Notice it works everywhere.
 
If you took a surface like the
first one I took,
that slices through the wire
here, there's no electric field
anywhere,
inside a perfect conductor or
anything,
there's going to be no electric
field.
 
Then μ_0I
is going to contribute.
If I go to the region between
the plates, there is no I
there, but there's the rate of
change of electric flux.
Therefore the last Maxwell
equation,
the Ampere's law,
is going to be modified with an
extra term,
μ_0
ε_0
dΦ
_electric
/dt.
And it has a nice symmetry now,
that the line integral of the
electric field is the rate of
change of magnetic flux.
The line integral of the
magnetic field contains a rate
of change of electric flux.
 
It's got other stuff,
but it's got that extra term.
So I'm going to come next time
and start looking at these
equations.
 
And remarkably,
these equations imply there are
electromagnetic waves.
 
Without the benefit of any
charge, without the benefit of
any current, there'll be
electric and magnetic fields.
That's the interesting part.
 
We know electric fields can be
produced by charges,
and magnetic fields can be
produced by currents,
but I'm saying go light years
from everything.
No ρ, no charge density,
no current density.
There'll be non-zero E
and B just moving on
their own, and you want to
understand how that happens.
First of all,
it's amazing it's predicted by
this stuff.
 
So basically,
this is a very important day in
your life, because now you know,
all of electromagnetism is that
equation that equation.
 
This is it.
 
There is no more stuff any of
us knows, at least in classical
theory.
 
In quantum theory,
there's new stuff.
Classical electromagnetism is
only that.
So you don't have to pack your
head with all kinds of results.
You can derive everything from
these.
Okay, so let's do that on
Wednesday.
 
 
