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YEN-JIE LEE: OK, happy
to see you again.
Welcome back to 8.03.
Today, as you see
on the slide, we're
going to continue the discussion
of dispersive medium--
how the waves and
vibration should
be sent through this medium.
And also, we will learn about
uncertainty principle today.
Kind of interesting.
That is connected back here
to what we discuss here.
And finally, if we
have time, we'll
move to two-dimensional system
and three-dimensional system
to look at two-dimensional waves
and three-dimensional waves.
OK, that's the plan for today.
Just a quick review about
what we have learned so far.
Last time, we
discussed about shaking
one end of this dispersive
medium which is actually
a string with stiffness.
And basically you would see that
the strategy that we have been
following is to do a Fourier
transform to actually decompose
the motion of the hand,
which is actually holding
one end of the string, and
then decompose that into wave
population in frequency space.
OK, so that's what
we have been doing.
And then, we know based on
the property of this medium,
the dispersion relation, which
is omega as a function of k,
we can propagate waves
with different frequency
at different speeds.
Then we can see how
this system will
evolve as a function of time.
That's the whole
idea and the strategy
we approach this
interesting problem.
Last time, we also
introduced AM radio.
As we discussed before, if
we have a very simple-minded
strategy to just
send the pulse--
which is containing
information-- directly
through this medium,
due to the dispersion
relation which we
have this medium,
different component would be
traveling at different speed.
Therefore, the
information is smeared out
after it travels
through a long distance.
OK?
That's the problem.
And then the solution was
to use this approach, which
is amplitude modulation mixer.
That's actually
how AM radio works.
So basically, we have a slowly
oscillating message or signal
like music or voice
which we want to send,
and then as we multiply that
by a really fast oscillating
cosine tan.
If we do this, assuming
that omega of 0
is actually much, much higher
or larger than the typical scale
of your signal--
which is omega s--
then, what is going to
happen is the following.
Up to all the calculations
we have done last time,
we found that the
resulting wave which
is the amplitude as a
function of time and space,
you can see that this can
factorize into two components.
The first component is
virtually the original signal
you are trying to send.
Since you're traveling at
the speed of group velocity,
and finally, the right hand
side-- the second component--
is actually the contribution,
the really small structure
of these high
frequency oscillation.
We call it carrier,
and the carrier
is still traveling at
the of face velocity.
That's how we actually
finally understand
what is the meaning
of group velocity
and the face velocity
through this example.
What I am going to do
today is to guide you
through another
example which will
ensure we can learn some more
insight from this calculation.
Today, we are going to have
another test of function,
which actually I can do Fourier
transforms really easily.
And this function I'm
trying to introduce here,
I have this functional
form exponential
minus gamma times
absolute value of t, OK?
The reason why I choose
absolute value of t
is because I would like to
make it symmetric around 0.
I can now do the usual
Fourier transform
and then to extract
the wave population.
The function of angular
frequency, c omega.
c as a function of omega.
And according to
the formula here,
which we introduced
last time, we
can quickly write
it down like this.
Basically you get 1 over 2
pi integration from minus
to infinity to infinity
integrating over time.
This is the original
function, f of t.
And multiply that by
exponential I omega t.
And that's the way
we extract c omega.
OK?
Since we have this
absolute value
here, basically the trick
is to change the interval,
split the interval
into two pieces.
So, y is actually
the negative t part,
therefore, you get the
exponential plus t here,
and the other part is
from 0 to infinity.
Then the absolute value
doesn't change side.
You have the
original exponential
minus gamma times t.
Then you can go ahead
and do with integration,
and you get the two turns
and you get the functional
form, which is c
omega equal to gamma
over pi times gamma
square plus omega square.
OK?
From this simple exercise,
they are interesting things
which we can learn from here.
If I go ahead and draw f
of t as a function of time,
this is what you will get.
Suppose I set gamma
to be equal to 0.1.
And I would like to
visualize this function
and that's what we did here.
You can see that from
the left hand side here,
is f of t as a function of time.
And you can see that
this like exponential
of t k but symmetric that
mirror at the t equal to 0.
And with a small
gamma value I choose,
that means this
exponential decay will
be really slow, therefore, you
have a pretty wide distribution
as a function of time.
However, if you look at the
right hand side, what did
I show you in the
right hand side?
Right hand side is c omega,
c as a function of omega,
it's the population
in frequency space.
And you can see that, if I
plug in gamma equal to 0.1
into that equation,
then you would
get a distribution
which is actually
pretty narrow, around 0.
That's actually
quite interesting.
And now, if I change gamma,
I increase the gamma slowly
so it changes to 0.2, you see
aha, that's what I expect--
the f function graphed
in the coordinate space
becomes narrower.
But, on the other
hand, you pay the price
that the wave population in the
frequency space becomes wider.
OK, the distribution
become wider.
I can increase and increase.
Now it's gamma equal to 0.5.
Gamma equal to 1.
And now I have a
rather large gamma.
Now it says 2.0, and you can
see that as a function of gamma,
if I set the gamma to be 5,
and you can see that the wave,
or say the waves of the wave
in the coordinate space,
becomes really small.
But if you look at the
corresponding c function,
you can see that waves
becomes really large.
This seems to be telling
us something interesting.
It seems to me that I could
not choose a gamma value which
simultaneously make waves
in a coordinate space
narrow and those wave
populations in the frequency
space narrow at the same time.
I cannot actually do that based
on this simple-minded exercise.
And what you are going
to do in your p-set
is to go through another
parameterization, which
is a Gaussian distribution.
And you will see
very similar, hope
for these very similar
conclusion from your exercise.
So what is going on?
And how do we
interpret this result?
And why is this result
actually related
to uncertainty principle?
That's the first part
of the lecture, which
we are going to discuss today.
We can demonstrate this in
fact by one example of f
of t, which is showing here.
And we go through and change
the waves of this distribution.
Of course, we can
also try to show
this in a much more precise
mathematical definition.
That's what we are
going to do now.
The first thing which
we would need to do
is to define how to
quantify the waves
of the distribution in frequency
space and in coordinate space.
First, we define that the
intensity of the signal
is proportional
to f of t squared.
OK, that is the to estimate
the size of the intensity.
It kind of makes sense
because, for example,
the energy of the
electromagnetic wave
is actually proportional to
the wave function squared.
That's kind of reasonable
to choose this definition.
And then, once we have that
the definition of intensity,
then I can now calculate
the average of some operator
function.
For example, I can
calculate g of t
is a average of the g function.
And in this definition
of intensity,
how to calculate the
average is to do integration
over minus infinity
to infinity over t.
And this g function
is put right there
and all the components
are weighted
by this intensity estimator,
which f of t squared.
Of course, since we are actually
calculating the average,
we need to take out the
sum of the intensity.
So, the sum of all
the intensity is
an integration from minus
infinity to infinity--
dt f of t squared.
With this definition, we
can calculate the average.
OK and don't forget our goal is
to have an estimator estimate
the waves of sum distribution.
Therefore, you are probably
very familiar with that.
We use standard deviation.
So basically that's also usually
associated with the exam,
but this time it's associated
with some physical quantity.
what is the estimator
of spread of time?
Right.
We can actually make
use of this definition
and I can write
this notation that
t-squared to be a quantity
which is associated
with the size of spread in time.
And now as you define to be the
average of t minus average of t
squared.
Basically you calculate
that difference with respect
to the mean value, square it,
and then do the two averages
again.
All right, everybody
is following?
Any questions?
OK.
If I have this so-called
standard deviation or spread
of time definition here, then
I can write it down explicitly,
and this will become minus
infinity to infinity,
to disintegration over t, and
I have t minus n value of t,
half of t squared.
And of course I would take
out that normalization, which
is a minus infinity to
infinity dt f of t squared
And I can also do a similar
exercise for the frequency
space.
Basically, I can
define the spread
of the frequency spectrum.
And that I define
it to be delta omega
squared, and this
will be defined
as the average of omega
minus the mean value of omega
squared.
And with this definition,
we have an estimator
of this spread of time,
and we have an estimator
of spread the frequency.
The phenomenon which we see
from here, from this exercise,
going from low gamma value to
a large gamma value is that it
seems to us that the spread of
the time or in coordinate space
and the spread of the
distribution in the frequency
space cannot
simultaneously be small.
OK.
Therefore, based on this
mathematical definition,
our goal is now to
show that we can
prove that delta
omega times delta t
will be larger or equal to 1/2.
That is an interesting
consequence based
on this definition of spread.
We can actually achieve
the lecture today.
That's our goal.
And we are going to try
to achieve this goal.
Before we go ahead and
prove this relation
delta omega times delta t
greater or equal to 1/2,
we also realize
that when we discuss
this spread of the
frequencies spectrum,
if I write it down
here, if I try
to calculate the
average of omega,
then what I'm going to do is
to do the integration from
minus infinity to
infinity, d omega.
because now I'm
trying to calculate
the mean value of omega.
I have the omega times c omega.
And the exponential
is i omega t.
If I go ahead and
evaluate this integral,
I integrate over omega, and
I have omega times c omega
times exponential i omega t.
And you can see that this
omega can actually be extracted
from this exponential function.
If I do differentiation,
which is spread through time,
then I can actually
extract one omega out
of the initial function.
Therefore, what I'm going
to get is this will be equal
to i partial t minus infinity
to infinity d omega, c omega,
exponential minus i omega t.
So you can see that
this is the design
if I do a partiality relative
to with respect to t,
then I take minus i omega out
of this exponential function
and this i will make
minus i become 1.
Therefore, you can see that this
integral, which I construct,
is equal to i
partial, partial t.
This function.
OK, and you can quickly
realize that we know what
this integral is doing right.
According to the form which
I just did here, f of t
is equal to this integral
which I actually just highlight
there.
Therefore, this is just f of t.
that's kind of interesting
because that would give me i
partial, partial t, f of t.
Basically you can see that I
don't need to deal with omega,
I can actually do a partial
relative with respect to time,
then I can take one omega
out of the function which
I have constructed.
Any questions?
All right, now I can calculate
what will be the mean omega.
What would be the mean omega?
The mean omega, according
to this definition here,
this is how we calculate
the mean of some quantity,
mean omega will be equal to
minus infinity to infinity,
tt f star, t i partial,
partial t, f of t.
OK sorry that this is
kind of close to here.
The original definition I should
put omega got here, right?
But instead of
putting omega there,
I used the trick that this
i times partial partial t
can generate an omega for me.
Therefore, instead of
putting omega explicitly
into the integral, I
put i partial partial t
into the integral, then
I get 1 omega out of it,
and that's equivalent
to the calculation
with g of t equal to omega.
OK, everybody's following?
Therefore, I of
course still need
to normalize the calculation.
This is the denominator, which
is minus infinity to infinity,
integral over tt f of t squared.
OK, you can see that instead
of using omega directly,
the I used this trick to use
i partial partial t to extract
1 omega and I can calculate
the mean value of omega.
Therefore I can also
calculate explicitly
what would be the delta omega
square based on the definition
which I outlined before.
This would be the average value
of omega minus mean omega.
Mean omega is a number,
and if I'd write it down
explicitly I get minus
infinity to infinity, tt, i
partial partial t, minus
average value omega, f
of t squared divided
by minus infinity
to infinity disintegration
over dt f of t squared.
The take home
message is that I'm
using this trick to
replace all the omega
by i partial partial t.
Therefore, in my
formula, you will
see that originally, this
is supposed to be omega
and now we were
using that trick.
Therefore, it can be written
as pi partial partial t.
And you'll realize what
this is used for afterwards.
All right, so those
are just preparation.
What we have done
is that my goal
is to show that delta
omega times delta t
is greater than or equal to 1/2.
OK, that's my goal and
I'm preparing for that.
And I have that definition
of delta t and delta omega.
Yes?
AUDIENCE: What do
you think [INAUDIBLE]
YEN-JIE LEE: Oh sorry,
there should be--
it should be like this.
So I am taking partial
partial t out of f.
OK, sorry.
Good question.
Any other mistakes?
Very good.
Not yet?
All right.
So now you can see that I have
the definition in my hand,
and I am almost
there to show you
that delta omega
times delta t is going
to be greater or equal to 1/2.
And what I'm going
to do after this--
maybe you will be
even more mad at me--
is to use exactly
the same trick which
would be used to show
Heisenberg's Uncertainty
Principle in quantum mechanics.
basically what I'm going to do
is to consider a function which
is r of t.
r as a function kappa and t.
and the definition of this
r function is like this.
I define this r function to be t
minus average t minus i kappa i
partial partial t minus omega.
f of t.
If you don't know where is
this relationship coming from,
don't be worried because
you don't really need to.
This is just to guide us through
this mathematical calculation.
But if you can see directly
how this will help,
the maybe you are Heisenberg.
Maybe.
So that's very nice.
It's a test.
What I am going to do
is to employ this r
as a function of kappa of t.
And the 2 for our purpose
to show that the delta omega
and delta t greater than 1/2.
And first, to make
my life easier,
I would define this
to be capital T,
and I would define this
thing to be capital omega.
So that my mathematical
expression doesn't explode.
Now I can consider this
ratio function r of kappa.
This is defined
as minus infinity
to infinity integrating
over t r kappa t divided
by minus infinity to
infinity dt f of t squared.
This is r function which is the
ratio of the area of r function
and the area of the f function.
You may say that, professor,
this is really crazy.
Today is telling about
all the crazy things,
but that is because
I would like to let
you know that we are going to
see a very interesting result.
So that's why I'm doing this.
And if I construct
this r function,
this r function will have
an interesting property.
What is the
interesting property?
I entered an integral
over something
squared in the numerator
and the denominator.
Now it means, what would be
the value of this r function?
The r function would
be always positive.
Right?
Because this is a square,
this is a square, therefore,
r is going to be positive.
That means r is going to be
greater than or equal to 0.
That's why we have
this r function.
And the miracle will happen
because if I go ahead
and calculate this r--
before I calculate this
capital R function-- what's
the function of kappa, I need to
actually deal with this small r
as a function of
kappa and t squared.
If I extract this component
and then calculate that,
r kappa t squared.
What I am going to do is
to use this expression r
is equal to t, capital T minus
i kappa omega times f of t.
So that my life would be easier.
Then basically you get
t minus i kappa omega f.
And then you need tje
complex conjugate.
Basically, you get T cross
i kappa omega star f star.
You can have T star,
but T is a real number.
Therefore, it
doesn't do anything.
Then, I can now go ahead
and collect all the terms.
Then the first terms
which I can collect
is everything
related to T times f.
Then basically you
get the T f squared.
That is coming from this
T times f times T times f.
This term times this
term times this term.
to give you the first term.
And you also you can
connect another term
which omega f squared.
Right.
Basically, you can
find that contribution.
Use should have a kappa
square in front of it.
Any questions so far?
Basically, I collect the
terms related to omega times f
and put it here.
Finally, you have
the third term,
which is i kappa T f omega star
f star minus omega f T f star.
Basically, this small
r function squared
can be written in
this functional form.
We are almost there.
What I'm going to discuss
first is that now I
have these three terms.
Number one, number
two, and number three.
I can now attack
number three first.
Number three, I'm going to
get i kappa Tf minus i partial
partial t minus omega f star.
Basically what I'm doing
is to take this omega here.
This is omega star.
And then use that
definition, write down
the expression for omega--
typical omega-- explicitly.
Since I am writing omega star,
therefore, you get a minus i
partial partial t minus
average omega out of it.
That's why here you
have this expression
and then multiple it by f, which
is the original expression.
I also write this omega
capital Omega explicitly.
I partial partial t
minus average Omega.
f t f star.
And you can immediately
realize that--
OK, this whole thing is
multiplied by i times kappa.
You can immediately recognize
that this term actually
canceled because they are--
actually they are
literally the same.
And then what is
actually left over
is the two terms,
which is in the middle.
So basically, you are
going to get now I
can multiply i and
cancel this minus i.
Basically what you
get is kappa time
T equals-- both terms have a
T, so I can extract this T out
of it.
f partial f star partial T cross
partial f partial T f star.
After all those works,
you can see that this one
looks pretty nice.
This says what?
This is not bad at all after
all those calculations basically
these will be equal to kappa
T partial partial t f f star.
Everybody's following or
everybody already lost?
We are almost there.
All right.
Now, we have these three.
Three originally is a beast.
Looks really horrible and after
I write it down explicitly,
it looks OK, not perfect.
Yes?
AUDIENCE: [INAUDIBLE]
YEN-JIE LEE: The complex
conjugate of the f function.
All right.
Now I can put one, two, and
three into this integral.
Then we are done.
Now let's put numbers 3
into the integral first.
I do a minus infinity to
infinity, number three, dt.
What is going to happen?
This will give you minus
infinity to infinity kappa T
partial partial t f f star.
And I can use integration by
parts so what I'm going to get
is kappa T f f star
evaluating minus infinity
and then plus infinity
minus kappa minus
infinity to infinity
f square partial t
partial capital T partial t d t.
Let's look at this.
Basically, what I'm doing is to
put in the numbers written back
into this integral and then
use integration by parts.
Basically you can see that
this is what you would expect.
The interesting thing
is that this function
is evaluated at crossing at
infinity and minus infinity.
If you assume that your
f function is localized--
it's confined in some
specific range of time,
instead of spreading out
over the whole universe.
That means this term
will be equal to 0
because it's evaluated at
plus infinity time and minus
infinity time.
If the f function is localized,
then at the boundary of time,
you are going to get 0.
This term disappears.
Very good.
We've solved one problem.
And this looks horrible, but
partial capital T, partial t,
what is capital T?
T is small t minus average of t.
Average of t is a
number and t is just t.
Therefore, partial t
partial small d is just 1.
You can see that
there are hopes,
things are becoming
simpler and simpler.
Therefore, what I'm
going to get is this--
minus kappa minus infinity
to infinity t t f squared.
And then if you divide this by
this term, you can see that 3--
number 3 term-- will give you
a contribution of minus kappa.
That's all.
Because once you plug
this integral back
into this function, the
third term contribution
gives you minus kappa.
That's a very good news because
it's actually pretty simple.
Any questions?
AUDIENCE: [INAUDIBLE]
YEN-JIE LEE: Oh,
you mean this one?
AUDIENCE: No.
YEN-JIE LEE: This one?
AUDIENCE: To the left.
YEN-JIE LEE: Oh, yeah.
You are right.
I missed a dt.
Thank you very much.
Very good.
Yeah.
Basically what I'm trying to
do is plug in the expression
here into the integral.
You can see that
the contribution
from the third term that
number 2 is rather simple.
It's just minus kappa.
Let's also take a look
at the computation
from the first and the second.
Wife Number one, will give you
minus infinity to infinity t
minus average of t
squared f of t squared dt.
And this is divided
by minus infinity
to infinity dt, f of t.
This is not crazy at all
because this just the definition
of delta t squared.
Just a reminder that the
definition of delta t squared
is written here.
Therefore, this is
just delta t squared--
the first term, which
looks really strange there,
but in reality, it's
actually very simple.
Let's look at the second term.
This is kappa squared minus
infinity to infinity i
partial partial t minus
average of omega f of t.
And then square that.
Divide it by minus infinity to
infinity dt, f of t squared.
And that will give you kappa
squared delta omega squared.
Basically, our conclusion
that this r function
is a function of kappa.
Essentially equal to the first
terms here delta t squared,
the second term is plus kappa
squared of delta omega squared
.
And finally, the
third term is there.
Minus kappa.
And this would be
greater or equal to 0.
Because what I am
doing is just summing
all those positive functions.
Then, take the rest.
.
Any questions?
AUDIENCE: Why does the
integral from negative infinity
to infinity dt f squared equal?
YEN-JIE LEE: This one?
This one?
This is equal to zero, right?
Oh, here?
AUDIENCE: Yeah.
Why does that--
YEN-JIE LEE: Oh, I see.
I see your point.
This is an integrated minus
infinity to infinity number
3 dt.
It's the contribution here.
Then, if I take a ratio between
this term and that term,
then this is canceled
by the denominator.
Therefore, what is actually
left over is minus kappa.
AUDIENCE: OK.
YEN-JIE LEE: This 3, the
contribution of 3 in green
is already taking the ratio
when I evaluate the capital R
function.
Good question.
Now you can see
that you can safely
ignore what I have said so far.
Everything you can ignore.
Those are just
mathematics tricks.
But what is very important is
that now I have this relation--
delta t squared plus kappa
square plus delta omega
squared minus k.
This is a function of k.
And I can actually minimize it.
I can minimize R if I
carefully choose a kappa value.
This kappa equal
to kappa mean value
which makes the minimize
the R function is
equal to 1/2 delta omega
squared, which I would not
go over this calculation because
this is just a minimization
problem.
That means if plug that
in, what I'm getting
is R kappa min will be equal
to delta T squared minus 1
over 4 delta omega squared.
That is greater or equal to.
0.
We arrive there.
If I multiple both
sides by 4 delta omega
squared you get delta t
squared delta omega squared
greater or equal to 1 over 4.
If you take the
square root of that,
the you get delta t delta
omega greater or equal to 1/2.
That's actually what we
started to try to prove right?
You can see that
after all those works
a lot of complicated
mathematic calculations,
you can see that we
make no assumption,
we are just using the
definition of the spread of time
and the spread of frequency.
We follow that definition and
the use of mathematical trick
which we used to prove
Heisenberg's Uncertainty
Principle and we arrive there.
This means that this is
an intrinsic property
of wave function.
Intrinsic property means that
it's a mathematic property
of wave function.
What do I mean by this equation,
which we finally did right?
After all those
hard work, we have
to enjoy what we
have learned right
from all of those crazy things.
What do we learn?
Look at this function.
Delta t times delta omega,
greater or equal to 1/2.
That means if I construct
a function, which
is how I oscillate the
stream as a function of time,
if I construct a really
narrow one to this very fast
and then I stop--
very narrow-- then you will
have a very small delta t.
Now it sounds really nice.
I produce a delta
function, delta t,
but the delta omega space
is going to be a mess.
It's going to be a super wide
distribution because delta t is
really very, very small.
That means you have
to compensate that
by a rather large delta omega
because if you multiple delta t
times delta omega, that is going
to be great or equal to 1/2.
And is the consequence
of this, for example,
for the discussion of AM radio.
If I have an AM radio with
bandwidth delta omega.
This is 2 pi delta nu and that
is something like 3 times 10
to the 4 Hz.
If I have some kind
of bandwidth which
is actually roughly this value.
I can now immediately
calculate what
will be the resulting delta t.
The resulting delta t
will be a few times 10
to the minus 5 seconds
based on this equation.
This means that if I'm trying
to send two signals in sequence
through this AM radio.
that mean if the delta t--
the time difference
between the first
and the second information--
if the time difference
is large, if that delta-t
between these two much, much
larger than 10 to the minus
to the minus 5 seconds.
Then I can actually easily
separate these two signals.
On the other hand, if
I send then really,
the two signal really
close to each other,
if it looks like this,
then the receiver,
the ones who will
receive the signal,
will not be able to
separate, if this is just
one signal or two signals,
or one pulse or two pulse
which you are trying to send.
Any questions so far?
So you can see that
we can actually
quantify what will be the
limitation in the resolution,
tiny resolution, due to the
limitation of bandwidth delta
omega.
Before we take a
break, I would like
to make a connection
to quantum physics.
So if I look at this delta t
times delta omega greater than
or equal to 1 over 2, this
expression, I can rewrite it.
I can multiply t by velocity
v. And I get v times velocity
and I can have
omega divided by v.
And this would be better
or equal to 1 over 2.
So I just multiply
v and divide by v,
then actually you can solve.
And that means this
will become delta x.
And that, the second
term, will become delta k.
And that would be greater
or equal to 1 over 2.
In the quantum physics, momentum
is equal to h bar times k.
Momentum will be equal
to h bar times k.
And h bar is actually
the Planck constant.
So that, actually you will
see that a few times in L4.
OK.
So if I have p equal
to h bar times k,
that means I have delta x times
delta p greater or equal to h
bar over 2.
That is exactly the uncertainty
principle, which was actually
introduced by Heisenberg.
And what is actually
the meaning of this?
So if we describe
all those particles
we see by quantum
mechanical waves,
if I have momentum p, now it
corresponds to a wave function,
with wave number k.
And the constant, which is
associated with p and the k
is the Planck constant.
So this means that if I
measure one particle really,
really precisely
in a position, due
to the nature of
wave function that
means I will not have
a lot of information
about the momentum
of that particle.
And where this uncertainty
principle is coming from,
it's coming from purely the
mathematics related to waves.
As you can see there,
there's really nothing
to do with quantum so far.
Quantum I'm saying
actually only goes in
after we prove the uncertainty
principle, delta omega
times delta t.
You can cannot have a very
precise frequency and a very
precise position in a
coordinated space over time
at the same time.
And that actually has
direct consequence.
That means if you are considered
in quantum mechanics, that
is essentially the limitation
which will be posted,
the uncertainty principle.
So we will take a
five minute break.
And we come back and we take a
look at 2-3 dimensional waves.
And let me know if you
have any questions.
So welcome, back everybody.
So before we actually moved
to 2-3 dimensional waves,
we will discuss a very
interesting topic,
which is related
to the dispersion
relation of the light actually.
So if you use
spatial relativity,
basically you can relate energy
to momentum and the mass.
So E square will be equal to a p
square c square plus m square c
to the 4.
And you actually interpret
light as a photon,
then basically E is actually
equal-- to the energy
of the photon will be
equal to h bar times omega.
So we are actually really
going really forward a bit.
Because maybe some
of you actually
haven't seen this before.
But if you just believe
what I have said,
basically you can
actually divide everything
from the first formula, which
is the spatial relativity
formula, by h bar square.
Then you will be able
to derive and arrive
the second formula, which
is omega square equal to c
square k square
plus omega 0 square.
And the omega 0 is
actually defined
as mc square over h bar,
just for simplicity.
So if we look at this
equation, this is essentially
a dispersion relation.
Now you have seen
this so many times.
And this omega square equal to
c square k square plus omega 0
square, this formula is
actually reminding you
that this is actually
a dispersion relation.
So what I mean by a
photon having mass here?
That means the m term in this
special relativity formula
is not 0.
Therefore omega 0
will be non-zero.
What is going to happen?
That means the space
of velocity of light
is going to be different.
It depends on what
value of k you choose.
That's kind of interesting.
Because that means light
with different frequency
or different
wavelengths is going
to be traveling through the
vacuum at different speeds,
if that's true.
Everybody get it?
Very good.
So how do we actually test this?
So that means I need a light
source, which are very,
very far away from earth.
Then I would like to
measure the delta t
as a function of frequency,
for example, and analyzing.
So how do we do that?
So this is actually
possible if you actually
use a natural light source,
which is the pulsar.
So what is actually a pulsar?
So what we actually
use, essentially a
millisecond pulsar.
So those are actually coming
from rapidly rotating neutron
stars, and that those
rotating neutron stars will
emit pulses of radiation
like x-ray and radio waves,
at regular intervals.
Because it's
essentially rotating,
rotating, rotating
again and again.
Based on this movie, basically
what it's showing here
is a very old neutron star.
It's actually in
a binary system.
And this neutron star
can absorb the material
from the other partner.
So that actually is--
the rotation speed
actually increased.
And finally at the speed
of a millisecond per turn.
So this actually
really happened.
And we can actually
observe this.
And if we are lucky,
the earth is essentially
somehow in a spatial direction
such that the emitting radio
wave actually pointing
from the pulsar to earth,
then I can see the pulsar,
the amplitude of the light
from pulsar essentially changing
rapidly as a function of time.
And another very good news is
that typically those pulsars
are really far away.
For example, in this
example, pulsar B1937+21,
this is essentially a pulsar
with rotation period of just
1.6 milliseconds.
And this is actually
something which is really
happening really far away from
the Earth, which essentially
is 16,000 light years away.
And that we can
actually observe this.
This is actually pretty
close to Sagitta,
and you can actually
see this pulsar.
And how does that
actually associate
with the original
question we were posting?
The original question
is, does the light
with different frequency
travel at different speed.
And this is essentially
a very nice tool.
Right?
Because it is emitting
the radio wave.
And now I can just measure the
spectra as a function of time.
And I will be able
to see if we actually
can observe different speed.
Because we know the rotation
in the world, and et cetera.
And it also emits a wide
spectra of the frequency,
the light frequency.
Therefore, I can use
this as a light source
far, far away from the Earth,
to see what will happen.
So somebody actually
did this measurement,
and this is that
what they found.
They found a non-zero omega 0.
A non-zero omega 0 was found.
So that means the mass
will be 1.3 times 10
to the minus 49 gram.
That sounds really small.
But it's not small at all.
That's actually destroying the
whole understanding of light.
What is going on?
So we are in trouble.
So after all this
discussion, et cetera,
and also other
measurements which
are sensitive to photon
mass, they actually
threw out this
possible contribution.
This is essentially is
just simply too large
based on, for example,
measurement of magnetic field
in the galaxy, et cetera.
It doesn't really work.
So what essentially
is really happening?
The explanation is that the path
from the pulsar to the earth
it's really not vacuum.
There are a lot of--
not a lot, but we have very
few or very dilute electrons,
very diluted free electrons
all over the place.
And that will change the
frequency and the speed
of light slightly.
Therefore you observe
the interesting--
observe the effect.
And we are going to
actually also talk
about how the material
actually changes
the behavior of the
electromagnetic wave
in the coming lectures.
I hope you find
this interesting.
Any questions?
All right.
So we are going to move on.
So far what we have
been discussing
is always 1-dimensional waves.
So for example, a string,
and also the sound
save in a tube, et cetera.
We always discuss things
which are in one dimension.
But we are actually not
one dimensional animal.
We are 3-dimensional And
of course, for example,
these objects the
surface is 2-dimensional
So there are many,
many things which
are more than one dimension.
So can-- the question
that I'm trying
to ask is, can we actually
understand this kind of object,
and how actually to
understand those objects
and how do we actually
derive the normal amounts,
and how do we
actually write down
the general solution, which
describes a 2-dimensional
or a 3-dimensional wave.
That's actually the next topic
which I would like to discuss.
So that's actually gets
started with a plate like this.
So basically that plate is
actually a 2-dimensional.
And assuming that this plate is
infinitely long, for a moment,
very, very long.
So what does that mean?
This means that if I define
my x and y-coordinate, which
is actually used to describe
the position of a specific point
on this plate,
then basically you
will see that they are
beautiful symmetries, which
you can actually identify
from this simple example.
What is actually the symmetry
which we can identify?
Can anybody help me with that?
AUDIENCE: x and y.
YEN-JIE LEE: Yeah.
So yeah, x and y
are symmetric, yes.
And the other function of x,
what kind of symmetry to you
have?
AUDIENCE: Reflection.
YEN-JIE LEE: Yeah.
Also reflection,
and what I'm looking
for is if I change x and
change y, what kind of symmetry
do you have?
AUDIENCE: Translation.
YEN-JIE LEE:
Translation symmetry.
Well, all of you are correct.
But what I am trying
to focus on now
is the translation symmetry.
So if I use translation
symmetry, what I'm going to get
is that I can already
know the functional
form of the normal mode.
Because essentially if
it's translation symmetric,
as a function of x, it's
translation symmetric
as a function of y.
Then I can say is in the x
direction will be proportional
to exponential iKxX.
K underscore x is essentially
the wave number associated
with the wave in
the x direction.
So that's essentially
one consequence
which we actually learned from
the discussion of symmetry.
And in the y direction,
I can conclude also
that the normal mode will be
proportional to exponential iKy
times Y. Therefore,
I already know
what will be the function
form of the normal mode
of this highly symmetric system.
What is that?
The psi xy will be equal
to A times exponential iKx
times X, exponential iKyY.
So you can see that.
And also I need to
take the real part.
Something like this will
be possible in normal mode.
Therefore without going
into detail basically,
we will see that the
expected behavior of psi
as a function of x and
y will be something
like a sine Kx times
x, sin Ky times y.
So that's actually the
kind of normal mode, which
we will expect based on
the argument of translation
symmetry.
And of course if I now go back
from infinitely long system
to a finite system, then you
can use the boundary condition
to determine what
would be the K value,
Kx value, and allow the Kx value
and allow the Ky value using
boundary conditions.
So actually without
doing any calculation,
we can already find
that, so now if I
have a plate with finite
size, basically you
expect that I can have some
kind of normal mode, which this
is the amplitude, a
projection in the x direction,
it can be a sine function.
And that it can become 0
at the left-hand side edge
and the right-hand side edge.
And in the y
direction it has to be
also some kind of sine
wave as a function of y.
And of course it goes
to 0 at the edge.
Because if those are
actually the fixed boundary,
for example.
And if those are actually
not fixed boundary,
then you expect that--
like open-end solution.
So you expect that
the distribution
will be more like
a cosine function
for the first normal mode.
And if you look at
this, the structure
of this kind of solution,
it looks really complicated.
Because you have x direction
and you also have y direction.
Both of them are
actually sine functions.
And how do we actually visualize
this kind of sine function?
And here is a demonstration,
which I have prepared.
It's really a
2-dimensional plate.
And as you can see
that under this plate,
I have a loudspeaker which
actually produces a sound wave
to try to excite one
of the normal mode.
And the one I am going to do is
to turn on this loud speaker.
You can hear the sound.
And I would like to
see the normal mode.
But it's very hard to see
that, without doing anything.
Because it's vibrating, but
its so fast that it is really
very difficult to see it.
So what I am going to do is to
pour some sand on the surface,
and see what is going to happen.
And if we look at this,
I am putting sand on it.
And you can see that, there
is something happening.
If I change the frequency to one
of the normal mode frequencies,
you can see that now we are
reaching some kind of resonance
and exciting one
of the normal mode.
And you can see that
the sand actually it
doesn't like to stay
on some of the plate.
Because it's
vibrating like crazy
and it's not very
comfortable to sit there.
So the sand, where will
the sand actually sit?
They will set at the place where
you don't have any vibration.
Because what we
are talking here,
is essentially some
kind of sine wave times
sine wave or cosine
wave times cosine wave.
That means there will
be nodes on the plate.
And those are
2-dimensional nodes.
In the 1-dimensional case,
we are talking about nodes,
it's actually the place where
you have zero amplitude.
And now I have
cosine times cosine.
Therefore, there will be a
complicated pattern appearing
which is essentially the place
the plate is not actually
moving at all as a
function of time.
And you can see that now
I can actually excite one
with the normal node.
And you can see a really
beautiful pattern.
And allow me to do this
and increase the frequency.
So that if we see if I can
excite another normal mode.
Look at what is happening.
So now you see that the number
of lines actually increased.
So this is actually
so-called Chladni figures.
Basically those
figures are actually
produced by this trying to
excite one of the normal mode.
And basically the sand will be
collected in the nodal lines.
And you can see that this higher
frequency input sound wave.
You can excite the higher
frequency in normal mode.
And of course I can continue to
increase and see what happens.
Now I'm increasing the frequency
even higher and higher.
You can see that now the
sound is actually rather loud.
And I am actually
putting more sand.
You can see that there are
more and more patterns.
Because now I am
increasing the frequency,
so that actually the higher
frequency normal modes
are excited.
And you will expect more nodes
for higher frequency ones.
And now I can even
go even higher
to see if I find success.
It's not easy now.
Look.
Probably this is a
very good way to design
the pattern of your t-shirt.
OK.
So how do we actually
understand all those patterns?
And we have already started.
This is actually something
related to cosine and sine
multiplied to each other.
And the next time
we are going to do
a more detailed calculation
and show you a few more demos
and see what we
can actually learn
from the 2-dimensional case.
Thank you very much.
I hope you enjoyed
the lecture today.
And if you have any
questions, let me know.
And you can actually
come forward and play
with those demos if you want.
