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PROFESSOR: So last
time we started

00:00:22.590 --> 00:00:25.620
talking about superfluid helium.

00:00:25.620 --> 00:00:32.070
And we said that the
phase diagram of helium-4,

00:00:32.070 --> 00:00:36.710
the isotope that is a boson
has the following interesting

00:00:36.710 --> 00:00:37.700
properties.

00:00:37.700 --> 00:00:40.530
First of all, helium
stays a liquid

00:00:40.530 --> 00:00:43.000
all the way down
to 0 temperature

00:00:43.000 --> 00:00:45.990
because of the combinations
of its light mass and heat

00:00:45.990 --> 00:00:47.960
interactions.

00:00:47.960 --> 00:00:53.970
And secondly, that [INAUDIBLE]
to cool down helium

00:00:53.970 --> 00:00:57.580
through this process
of evaporated cooling,

00:00:57.580 --> 00:00:59.550
one immediately observes
something interesting

00:00:59.550 --> 00:01:03.200
happening at temperatures
below 2 degrees

00:01:03.200 --> 00:01:07.020
Kelvin, where it becomes
this superfluid that

00:01:07.020 --> 00:01:10.000
has a number of
interesting properties.

00:01:10.000 --> 00:01:13.430
And in particular as
pertaining to viscosity,

00:01:13.430 --> 00:01:15.080
we made two observations.

00:01:15.080 --> 00:01:18.400
First of all, you can
make these capillaries--

00:01:18.400 --> 00:01:20.880
and I'll show you been
movie in more detail later

00:01:20.880 --> 00:01:24.710
on where it flows through
capillaries as if there

00:01:24.710 --> 00:01:28.440
is no resistance
and there is nothing

00:01:28.440 --> 00:01:31.010
that sticks to the walls
of the capillaries.

00:01:31.010 --> 00:01:33.770
It flows without
viscosity, whereas there

00:01:33.770 --> 00:01:37.190
was this experiment of
Andronikashvili, in which you

00:01:37.190 --> 00:01:40.080
had something that
was oscillating

00:01:40.080 --> 00:01:44.760
and you were calculating
how much of the helium

00:01:44.760 --> 00:01:47.710
was stuck to the plates
of the container.

00:01:47.710 --> 00:01:50.920
And the result was
something like this,

00:01:50.920 --> 00:01:58.080
that is there was a decrease
in the amount of fluid

00:01:58.080 --> 00:01:59.460
that is stuck to the plates.

00:01:59.460 --> 00:02:02.705
But it doesn't go
immediately down to 0.

00:02:02.705 --> 00:02:04.780
It has a kind of
form such as this

00:02:04.780 --> 00:02:10.560
that I will draw more
clearly now and explain.

00:02:10.560 --> 00:02:16.250
So what we did last time was
to note that people observed

00:02:16.250 --> 00:02:17.860
that there were
some similarities

00:02:17.860 --> 00:02:20.010
between this
superfluid transition

00:02:20.010 --> 00:02:22.640
and Bose-Einstein condensation.

00:02:22.640 --> 00:02:26.360
But what I would
like to highlight

00:02:26.360 --> 00:02:28.200
in the beginning
of this lecture is

00:02:28.200 --> 00:02:32.450
that there are also very
important differences.

00:02:32.450 --> 00:02:40.310
So let's think about
these distinctions

00:02:40.310 --> 00:02:48.126
between Bose-Einstein condensate
add the superfluid helium.

00:02:58.690 --> 00:03:05.120
One set of things we would
like to take from the picture

00:03:05.120 --> 00:03:12.660
that I have over there, which
is diffraction of the fluid that

00:03:12.660 --> 00:03:15.370
is stuck to the plates
and in some sense

00:03:15.370 --> 00:03:19.030
behaves like a normal fluid.

00:03:19.030 --> 00:03:24.050
Now let me make the analogy
to Bose-Einstein condensation.

00:03:24.050 --> 00:03:27.400
You know that in the
Bose-Einstein condensation

00:03:27.400 --> 00:03:32.260
there was also this phenomenon
that there was a separation

00:03:32.260 --> 00:03:36.230
into two parts of
the total density.

00:03:36.230 --> 00:03:39.650
And be regarded as a
function of temperature

00:03:39.650 --> 00:03:45.750
some part of the density as
belonging to the normal state.

00:03:45.750 --> 00:03:50.840
So when you are above
Tc of n, everything

00:03:50.840 --> 00:03:53.860
is essentially normal.

00:03:53.860 --> 00:04:02.600
And then what happens is
that when you hit Tc of n

00:04:02.600 --> 00:04:07.530
you can no longer put
all of the particles

00:04:07.530 --> 00:04:11.320
that you have in
the excited states.

00:04:11.320 --> 00:04:13.870
So the fraction that goes
in the excited states

00:04:13.870 --> 00:04:20.590
goes down and eventually
goes to 0 at 0 temperature.

00:04:23.460 --> 00:04:31.770
And essentially, this would
be the reverse of the curve

00:04:31.770 --> 00:04:34.180
that we have in that
figure over there.

00:04:37.320 --> 00:04:39.610
Basically, there's
a portion that

00:04:39.610 --> 00:04:42.940
would be the normal part that
would be looking like this.

00:04:46.040 --> 00:04:52.430
Now, the way that we
obtained this result

00:04:52.430 --> 00:04:54.860
was that basically there
was a fraction that

00:04:54.860 --> 00:04:57.100
was in the normal state.

00:04:57.100 --> 00:05:00.040
The part that was
excited was described

00:05:00.040 --> 00:05:05.500
by this simple formula that was
g over lambda cubed [INAUDIBLE]

00:05:05.500 --> 00:05:07.410
of 3/2.

00:05:07.410 --> 00:05:10.960
So it went to 0 as T to the 3/2.

00:05:10.960 --> 00:05:12.990
So basically, the
proportionality

00:05:12.990 --> 00:05:16.040
here is T to the 3/2.

00:05:16.040 --> 00:05:17.950
And then basically,
the curve would

00:05:17.950 --> 00:05:22.270
come down here and
go to 0 linearly.

00:05:22.270 --> 00:05:26.260
Now what is shown
in the experiment

00:05:26.260 --> 00:05:30.000
is that the curve
actually goes to 0

00:05:30.000 --> 00:05:33.600
in a much more sharp fashion.

00:05:33.600 --> 00:05:40.250
And actually, when people try
to fit a curve through this,

00:05:40.250 --> 00:05:44.010
the curve looks something like
Tc minus T to the 2/3 over.

00:05:48.410 --> 00:05:54.140
But also it goes to 0 much
more rapidly than the curve

00:05:54.140 --> 00:05:57.100
that we have for
Bose-Einstein condensation.

00:05:57.100 --> 00:05:59.900
Indeed, it goes through
0 proportionately to T

00:05:59.900 --> 00:06:02.650
to the fourth.

00:06:02.650 --> 00:06:06.100
And so that's something that we
need to understand and explain.

00:06:12.200 --> 00:06:18.070
Now all of the properties of
the Bose-Einstein condensate

00:06:18.070 --> 00:06:21.840
was very easy to
describe once we realized

00:06:21.840 --> 00:06:26.010
that all of the things that
correspond to excitation,

00:06:26.010 --> 00:06:28.500
such as the energy,
heat capacity,

00:06:28.500 --> 00:06:32.350
pressure, come from
this fraction that

00:06:32.350 --> 00:06:35.210
is in the excited state.

00:06:35.210 --> 00:06:39.010
And we can calculate,
say, the contribution

00:06:39.010 --> 00:06:42.060
to energy heat
capacity, et cetera.

00:06:42.060 --> 00:06:45.730
And in particular, if we look
at the behavior of the heat

00:06:45.730 --> 00:06:50.400
capacity as a function
of temperature,

00:06:50.400 --> 00:06:57.240
for this Bose-Einstein
condensate, the behavior

00:06:57.240 --> 00:07:01.090
that we had was that again
simply at low temperatures

00:07:01.090 --> 00:07:08.000
it was going proportionately
to T to the 3/2,

00:07:08.000 --> 00:07:12.160
because this was the number
of excitations that you had.

00:07:12.160 --> 00:07:15.090
So these two T to the
3/2 are very much related

00:07:15.090 --> 00:07:16.960
to each other.

00:07:16.960 --> 00:07:22.300
And then this curve would
basically go along its way

00:07:22.300 --> 00:07:26.450
until it hit Tc of
n at some point.

00:07:26.450 --> 00:07:28.930
And we separately
calculated the behavior

00:07:28.930 --> 00:07:30.640
coming from high temperatures.

00:07:30.640 --> 00:07:33.140
And the behavior from
high temperatures

00:07:33.140 --> 00:07:35.480
would start with
the classical result

00:07:35.480 --> 00:07:39.500
that the heat capacity
is 3/2 per particle

00:07:39.500 --> 00:07:42.800
due to the kinetic energy that
you can put in these things.

00:07:42.800 --> 00:07:46.327
And then it would rise, and it
would then join this curve over

00:07:46.327 --> 00:07:49.189
here.

00:07:49.189 --> 00:07:54.760
Now when you look at the
actual heat capacity,

00:07:54.760 --> 00:07:57.380
indeed the shape up
the heat capacity

00:07:57.380 --> 00:08:01.370
is the thing that gives this
transition the name of a lambda

00:08:01.370 --> 00:08:02.180
transition.

00:08:02.180 --> 00:08:03.805
It kind of looks like a lambda.

00:08:09.090 --> 00:08:13.390
And at Tc there are
divergences approaching

00:08:13.390 --> 00:08:17.600
from the two sides that
behave like the log.

00:08:24.450 --> 00:08:27.680
And again more
importantly, what we

00:08:27.680 --> 00:08:30.770
find is that at
0 temperature, it

00:08:30.770 --> 00:08:34.730
doesn't go through 0, the
heat capacity as T to the 3/2,

00:08:34.730 --> 00:08:38.909
but rather as T to
the third power.

00:08:38.909 --> 00:08:46.060
So the red curve
corresponds to superfluid,

00:08:46.060 --> 00:08:51.710
the green curve corresponds
to Bose-Einstein condensate.

00:08:51.710 --> 00:08:56.550
And so they're clearly
different from each other.

00:08:56.550 --> 00:09:00.980
So that's what we would
like you to understand.

00:09:00.980 --> 00:09:07.010
Well the thing that is easiest
to understand and figure out

00:09:07.010 --> 00:09:10.780
is the difference between
these heat capacities.

00:09:10.780 --> 00:09:13.510
And the reason for that
is that we had already

00:09:13.510 --> 00:09:17.400
seen a form that was of
the heat capacity that

00:09:17.400 --> 00:09:18.600
behaved like T cubed.

00:09:18.600 --> 00:09:23.050
That was when we were looking
at phonons in a solid.

00:09:23.050 --> 00:09:25.880
So let's remind
you why was it that

00:09:25.880 --> 00:09:28.740
for the Bose-Einstein
condensate we

00:09:28.740 --> 00:09:34.010
were getting this T
to the 3/2 behavior?

00:09:34.010 --> 00:09:38.520
The reason for that was that
the various excitations,

00:09:38.520 --> 00:09:46.612
I could plot as a function
of k, or p, which is h bar k.

00:09:46.612 --> 00:09:49.450
They're very much
related to each other.

00:09:49.450 --> 00:09:53.890
And for the
Bose-Einstein condensate,

00:09:53.890 --> 00:09:58.310
the form was simply
a parabola, which

00:09:58.310 --> 00:10:03.144
is this p squared over
2 mass of the helium.

00:10:03.144 --> 00:10:06.330
Let's say, assuming that
what we are dealing with

00:10:06.330 --> 00:10:12.620
is non-interacting particles
with mass of helium.

00:10:12.620 --> 00:10:17.390
And this parabolic
curve essentially

00:10:17.390 --> 00:10:28.050
told us that various quantities
behave as T to the 3/2.

00:10:28.050 --> 00:10:32.920
Roughly, the idea is that
at some temperature that

00:10:32.920 --> 00:10:37.520
has energy of the order
of kT, you figure out

00:10:37.520 --> 00:10:41.800
how far you have excited things.

00:10:41.800 --> 00:10:45.000
And since this
form is a parabola,

00:10:45.000 --> 00:10:48.700
the typical p is going to
scale like T to the 1/2.

00:10:51.220 --> 00:10:55.810
You have a volume in three
dimensions in p space.

00:10:55.810 --> 00:10:59.110
If the radius goes
like T to the 1/2,

00:10:59.110 --> 00:11:02.030
the volume goes
like T to the 3/2.

00:11:02.030 --> 00:11:04.880
That's why you have all kinds
of excitations such as this.

00:11:08.150 --> 00:11:13.090
And the reason for the
Bose-Einstein condensation

00:11:13.090 --> 00:11:19.790
was that you would start to fill
out all of these excitations.

00:11:19.790 --> 00:11:25.430
And when you were adding all
of the mean occupation numbers,

00:11:25.430 --> 00:11:27.783
the answer was not
coming up all the way

00:11:27.783 --> 00:11:29.800
to the total number
of particles.

00:11:29.800 --> 00:11:32.400
So then you have to put
an excess at p close

00:11:32.400 --> 00:11:35.880
to 0, which corresponds to the
ground state of this system.

00:11:38.680 --> 00:11:41.090
Now of course when
you look at helium,

00:11:41.090 --> 00:11:44.310
helium molecules, helium
atoms have the interactions

00:11:44.310 --> 00:11:46.440
between them that we discussed.

00:11:46.440 --> 00:11:50.130
In particular, you can't really
put them on top of each other.

00:11:50.130 --> 00:11:53.150
There is a hard exclusion
when you bring things

00:11:53.150 --> 00:11:55.460
close to each other.

00:11:55.460 --> 00:11:58.470
So the ground
state of the system

00:11:58.470 --> 00:12:00.760
must look very different
from the ground state

00:12:00.760 --> 00:12:04.180
of the Bose-Einstein condensate
in which the particles freely

00:12:04.180 --> 00:12:08.000
occupy the entire box.

00:12:08.000 --> 00:12:12.030
So there is a very
difficult story

00:12:12.030 --> 00:12:15.020
associated with
figuring out what

00:12:15.020 --> 00:12:18.930
the ground state of this
combination of interacting

00:12:18.930 --> 00:12:22.880
particles that make
up liquid helium is.

00:12:22.880 --> 00:12:23.930
What is the behavior?

00:12:23.930 --> 00:12:26.260
What's the many-body wave
function at 0 temperature?

00:12:29.470 --> 00:12:34.630
Now as we see here, in order
to understand the heat capacity

00:12:34.630 --> 00:12:37.130
we really don't
need to know what

00:12:37.130 --> 00:12:40.060
is happening at
the ground state.

00:12:40.060 --> 00:12:43.350
What we need to know in
order to find heat capacity

00:12:43.350 --> 00:12:47.170
is how to put more energy in the
system above the ground state.

00:12:47.170 --> 00:12:51.090
So we need to know
something about excitations.

00:12:51.090 --> 00:12:54.930
And so that's the
perspective that Landau took.

00:12:54.930 --> 00:12:58.900
Landau said, well, this is
the spectrum of excitations

00:12:58.900 --> 00:13:03.470
if you had [? plain ?] particles
without any interactions.

00:13:03.470 --> 00:13:06.120
Let's imagine what
happens if we gradually

00:13:06.120 --> 00:13:08.220
tune in the interactions,
the particles

00:13:08.220 --> 00:13:10.960
start to repel each
other, et cetera.

00:13:10.960 --> 00:13:13.120
This non-interacting
ground state

00:13:13.120 --> 00:13:16.480
that we had in which the
particles were uniformly

00:13:16.480 --> 00:13:18.590
distributed across
the system will

00:13:18.590 --> 00:13:22.910
evolve into some complicated
ground state I don't know.

00:13:22.910 --> 00:13:25.210
And then, presumably
there would be a spectrum

00:13:25.210 --> 00:13:30.260
of excitations around
that ground state.

00:13:30.260 --> 00:13:34.760
Now the excitations around
the non-interacting ground

00:13:34.760 --> 00:13:39.250
state we can label by
this momentum peak.

00:13:39.250 --> 00:13:40.990
And it kind of
makes sense that we

00:13:40.990 --> 00:13:45.080
should be able to have a
singular label for excitations

00:13:45.080 --> 00:13:48.850
around the ground state of
these interacting particles.

00:13:48.850 --> 00:13:50.550
And this is where
you sort of needed

00:13:50.550 --> 00:13:53.330
a little bit of Landau's
type of insight.

00:13:53.330 --> 00:13:55.590
He said, well,
presumably what you

00:13:55.590 --> 00:13:59.770
do when you have
excitations of momentum p

00:13:59.770 --> 00:14:03.940
is to distort the wave
function in a manner that

00:14:03.940 --> 00:14:10.330
is consistent with having these
kind of excitations of momentum

00:14:10.330 --> 00:14:11.580
p.

00:14:11.580 --> 00:14:14.680
And he said, well
we typically know

00:14:14.680 --> 00:14:17.990
that if you have
a fluid or a solid

00:14:17.990 --> 00:14:21.850
and we want to impart some
momentum above the ground

00:14:21.850 --> 00:14:24.870
state, if will go in
the form of phonons.

00:14:24.870 --> 00:14:28.870
These are distortions in
which the density will

00:14:28.870 --> 00:14:35.060
vary in some sinusoidal or
cosine way across the system.

00:14:35.060 --> 00:14:37.920
So he said that maybe
what is happening

00:14:37.920 --> 00:14:40.370
is that for these
excitations you

00:14:40.370 --> 00:14:43.730
have to take what ever this
interacting ground state is--

00:14:43.730 --> 00:14:47.590
which we don't know and
can't write that down--

00:14:47.590 --> 00:14:49.900
but hope that
excitations around it

00:14:49.900 --> 00:14:52.950
correspond to these
distortions in density.

00:14:52.950 --> 00:14:56.410
And that by analogy with
what happens for fluids,

00:14:56.410 --> 00:15:01.350
that the spectrum of excitations
will then become a linear.

00:15:01.350 --> 00:15:04.240
You would have something
like a sound wave

00:15:04.240 --> 00:15:07.060
that you would have in
a liquid or a solid.

00:15:07.060 --> 00:15:14.290
So if you do this, if you
have a linear spectrum

00:15:14.290 --> 00:15:16.895
then we can see what happen.

00:15:16.895 --> 00:15:21.540
For a particular energy
of the order of kT,

00:15:21.540 --> 00:15:24.670
we will go here,
occupied momenta that

00:15:24.670 --> 00:15:27.160
would be of the order
of kT over H bar.

00:15:30.920 --> 00:15:35.560
The number of excitations would
be something like this cubed,

00:15:35.560 --> 00:15:37.300
and so you would
imagine that you

00:15:37.300 --> 00:15:41.000
would get a heat capacity that
is proportional to this times

00:15:41.000 --> 00:15:42.870
kB.

00:15:42.870 --> 00:15:45.570
And if you do things
correctly, like really

00:15:45.570 --> 00:15:47.840
for the case of
phonons or photons,

00:15:47.840 --> 00:15:52.010
you can even figure out what
the numerical prefactor is here.

00:15:52.010 --> 00:15:56.240
And there's a velocity
here because this curve

00:15:56.240 --> 00:15:59.280
goes like H bar vP.

00:16:05.980 --> 00:16:09.010
So then you can
compare what you have

00:16:09.010 --> 00:16:15.330
over here with the coefficient
of the T cubed over here,

00:16:15.330 --> 00:16:19.140
and you could even figure
out what this velocity is.

00:16:19.140 --> 00:16:23.030
And it turns out to be of the
order of 240 meters per second.

00:16:27.640 --> 00:16:31.450
A typical sound wave that
you would have in a fluid.

00:16:31.450 --> 00:16:35.390
So that's kind of
a reasonable thing.

00:16:35.390 --> 00:16:38.700
Now of course, when you go
to higher and higher momenta,

00:16:38.700 --> 00:16:41.420
it corresponds to essentially
shorter and shorter

00:16:41.420 --> 00:16:42.520
wavelengths.

00:16:42.520 --> 00:16:44.880
You expect that when
you get wavelengths

00:16:44.880 --> 00:16:49.860
that is of the order of
the interatomic spacing,

00:16:49.860 --> 00:16:52.670
then the interactions become
less and less important.

00:16:52.670 --> 00:16:55.080
You have particles
rattling in a cage

00:16:55.080 --> 00:16:57.230
that is set up by
everybody else.

00:16:57.230 --> 00:17:00.320
And then you should regain
this kind of spectrum

00:17:00.320 --> 00:17:03.120
at high values of momentum.

00:17:03.120 --> 00:17:06.920
And so what Landau
did was he basically

00:17:06.920 --> 00:17:10.270
joined these things
together and posed

00:17:10.270 --> 00:17:14.760
that there is a spectrum
such as this that

00:17:14.760 --> 00:17:17.310
has what is called
a phonon part, which

00:17:17.310 --> 00:17:24.359
is this linear part where
energy goes like H bar,

00:17:24.359 --> 00:17:30.290
like the velocity
times the momentum.

00:17:30.290 --> 00:17:35.010
And it has a part that in
the vicinity of this point,

00:17:35.010 --> 00:17:37.930
you can expand parabolically.

00:17:37.930 --> 00:17:38.960
And it's called rotons.

00:17:41.900 --> 00:17:46.690
There is a gap
delta, and then H bar

00:17:46.690 --> 00:17:52.390
squared over 2, some effective
mass, k minus k0 squared.

00:17:54.970 --> 00:17:58.110
This k0 turns out to
be roughly of the order

00:17:58.110 --> 00:18:00.930
of the inverse
[INAUDIBLE], two Angstrom

00:18:00.930 --> 00:18:04.470
inverse, between particles.

00:18:04.470 --> 00:18:09.430
This mu is of the order
of mass of [INAUDIBLE].

00:18:14.580 --> 00:18:21.330
So about 10 years
or so after Landau,

00:18:21.330 --> 00:18:25.540
people are able to get to this
whole spectrum of excitation

00:18:25.540 --> 00:18:29.330
through neutron scattering
and other scattering

00:18:29.330 --> 00:18:30.700
types of experiments.

00:18:30.700 --> 00:18:33.160
And so this picture
was confirmed.

00:18:39.710 --> 00:18:41.344
So, yes?

00:18:41.344 --> 00:18:43.320
AUDIENCE: What is a
roton? [INAUDIBLE]

00:18:49.874 --> 00:18:51.540
PROFESSOR: Over here
what you are seeing

00:18:51.540 --> 00:18:56.500
is essentially particles
rattling in the cage.

00:18:56.500 --> 00:18:59.840
It is believed that
what is happening here

00:18:59.840 --> 00:19:02.700
are collections of
three or four atoms that

00:19:02.700 --> 00:19:06.300
are kind of rotating
in a bigger cage.

00:19:06.300 --> 00:19:08.750
So something, the
picture that people draw

00:19:08.750 --> 00:19:12.192
is three or four
particles rotating around.

00:19:12.192 --> 00:19:13.178
Yes?

00:19:13.178 --> 00:19:15.643
AUDIENCE: Is there
some [INAUDIBLE] curve

00:19:15.643 --> 00:19:17.615
where energy's
decreasing [INAUDIBLE]?

00:19:21.559 --> 00:19:26.060
Does the transition between
photon and roton [INAUDIBLE]?

00:19:26.060 --> 00:19:29.630
PROFESSOR: There is no
thermodynamic or other rule

00:19:29.630 --> 00:19:32.746
that says that the energy should
be one of [INAUDIBLE] momentum

00:19:32.746 --> 00:19:36.040
that I know.

00:19:36.040 --> 00:19:37.290
Yes?

00:19:37.290 --> 00:19:41.877
AUDIENCE: Is there an
expression for that k0

00:19:41.877 --> 00:19:47.416
in terms of temperature and
other properties of the system?

00:19:47.416 --> 00:19:49.880
PROFESSOR: This
curve of excitations

00:19:49.880 --> 00:19:53.020
is supposed to be property
of the ground state.

00:19:53.020 --> 00:19:56.780
That is, you take this
system in its ground state,

00:19:56.780 --> 00:19:58.970
and then you create
an excitation that

00:19:58.970 --> 00:20:01.660
has some particular
momentum and calculate

00:20:01.660 --> 00:20:04.930
what the energy of that is.

00:20:04.930 --> 00:20:08.300
Actually, this whole
curve is phenomenological,

00:20:08.300 --> 00:20:12.650
because in order to get
the excitations you better

00:20:12.650 --> 00:20:16.430
have an expression for
what the ground state is.

00:20:16.430 --> 00:20:19.910
And so writing a
kind of wave function

00:20:19.910 --> 00:20:23.590
that describes the ground state
of this interacting system

00:20:23.590 --> 00:20:26.270
is a very difficult task.

00:20:26.270 --> 00:20:30.800
I think Feynman has some
variational type of wave

00:20:30.800 --> 00:20:33.490
function that we can start
and work with, and then

00:20:33.490 --> 00:20:37.760
calculate things approximately
in terms of that.

00:20:37.760 --> 00:20:38.350
Yes?

00:20:38.350 --> 00:20:40.058
AUDIENCE: What inspired
Landau to propose

00:20:40.058 --> 00:20:42.510
that there was a [INAUDIBLE]?

00:20:42.510 --> 00:20:46.460
PROFESSOR: Actually,
it was not so much

00:20:46.460 --> 00:20:49.300
I think looking at
this curve, but which

00:20:49.300 --> 00:20:51.090
I think if you
want to match that

00:20:51.090 --> 00:20:53.890
and that, you have to
have something like this.

00:20:53.890 --> 00:20:58.610
But this was really that the
whole experimental version

00:20:58.610 --> 00:21:01.440
of the heat capacity,
it didn't seem

00:21:01.440 --> 00:21:04.620
like this expression
was sufficient.

00:21:04.620 --> 00:21:08.800
And then there was some amount
of excitation and energy

00:21:08.800 --> 00:21:12.710
at the temperatures that were
experimentally accessible

00:21:12.710 --> 00:21:17.100
that heated at the presence
of the rotons in the spectrum.

00:21:17.100 --> 00:21:17.860
Yes?

00:21:17.860 --> 00:21:21.270
AUDIENCE: So
continuing [INAUDIBLE],

00:21:21.270 --> 00:21:26.230
if you raise the thermal energy
kBT, [INAUDIBLE] level where

00:21:26.230 --> 00:21:28.256
you have multiple
roots of this curve.

00:21:28.256 --> 00:21:29.030
PROFESSOR: Yes.

00:21:29.030 --> 00:21:33.010
AUDIENCE: So you will be
able to excite some states

00:21:33.010 --> 00:21:38.780
and have some kind of gap, like
a gap of momenta which are not.

00:21:38.780 --> 00:21:42.240
PROFESSOR: OK, so at any
finite temperatures-- and I'll

00:21:42.240 --> 00:21:45.380
do the calculation for
you shortly-- there

00:21:45.380 --> 00:21:50.430
is a finite probability for
exciting all of these states.

00:21:50.430 --> 00:21:52.280
What you are saying
is that when there

00:21:52.280 --> 00:21:56.070
is more occupation at this
momentum compared to that,

00:21:56.070 --> 00:21:59.420
but much less compared to this.

00:21:59.420 --> 00:22:06.600
So that again does not
violate any condition.

00:22:06.600 --> 00:22:13.570
So it is like, again, trying to
shake this system of particles.

00:22:13.570 --> 00:22:16.170
Let's imagine that
you have grains,

00:22:16.170 --> 00:22:17.780
and you are trying
to shake them.

00:22:17.780 --> 00:22:20.900
And it may be that at
some shaking frequencies,

00:22:20.900 --> 00:22:22.820
then there are things
that are taking place

00:22:22.820 --> 00:22:25.702
at short distances in
addition to some waves

00:22:25.702 --> 00:22:26.660
that you're generating.

00:22:32.361 --> 00:22:32.860
Yeah?

00:22:32.860 --> 00:22:34.901
AUDIENCE: I have a question
about the methodology

00:22:34.901 --> 00:22:36.880
of getting this
spectrum, because if we

00:22:36.880 --> 00:22:40.290
have a experimental result
of the [INAUDIBLE] capacity,

00:22:40.290 --> 00:22:42.700
then if we assume
there's a spectrum,

00:22:42.700 --> 00:22:44.840
there has to be
this one, because it

00:22:44.840 --> 00:22:47.250
gives a one to one
correspondence.

00:22:47.250 --> 00:22:54.840
So we can get the spectrum
directly from the c.

00:22:54.840 --> 00:22:55.808
So--

00:22:55.808 --> 00:23:00.460
PROFESSOR: I'm not sure,
because in reality this

00:23:00.460 --> 00:23:04.680
is going to be spectrum in
three-dimensional space.

00:23:04.680 --> 00:23:07.870
And there is certainly
an expression

00:23:07.870 --> 00:23:12.950
that relates the heat capacity
to the excitation spectrum.

00:23:12.950 --> 00:23:15.340
What I'm not sure is
whether mathematically

00:23:15.340 --> 00:23:18.840
that expression
uniquely invertible.

00:23:18.840 --> 00:23:22.840
It is given an epsilon-- c,
you have a unique epsilon of p.

00:23:22.840 --> 00:23:27.090
Certainly, given an epsilon
of p, you have a unique c.

00:23:27.090 --> 00:23:27.932
Yes?

00:23:27.932 --> 00:23:30.307
AUDIENCE: But if the excitation
spectrum only depends on,

00:23:30.307 --> 00:23:33.344
let's say, k squared, not on
the three-dimensional components

00:23:33.344 --> 00:23:36.296
of k, then maybe
it's much easier

00:23:36.296 --> 00:23:39.531
to draw a one to
one correspondence?

00:23:39.531 --> 00:23:41.030
PROFESSOR: I don't
know, because you

00:23:41.030 --> 00:23:43.230
have a function of
temperature and you

00:23:43.230 --> 00:23:47.130
want to convert it to
a function of momentum

00:23:47.130 --> 00:23:49.490
that after some
integrations will give you

00:23:49.490 --> 00:23:51.950
that function of temperature.

00:23:51.950 --> 00:23:56.530
I don't know the difficulty
of mathematically doing that.

00:23:56.530 --> 00:24:01.300
I know that I can't off my head
think of an inversion formula.

00:24:01.300 --> 00:24:03.426
It's not like the function
that you're inversing.

00:24:13.640 --> 00:24:20.940
So the Landau spectrum
can explain this part.

00:24:20.940 --> 00:24:24.800
It turns out that the Landau
spectrum cannot explain this

00:24:24.800 --> 00:24:26.470
logarithmic divergence.

00:24:26.470 --> 00:24:27.334
Yes?

00:24:27.334 --> 00:24:29.250
AUDIENCE: Sorry, one
more question about this.

00:24:32.040 --> 00:24:34.538
The allowed values of
k, do they get modified,

00:24:34.538 --> 00:24:38.410
or are they thought
to be the same?

00:24:38.410 --> 00:24:39.010
PROFESSOR: No.

00:24:39.010 --> 00:24:43.990
So basically at some point,
I have to change perspective

00:24:43.990 --> 00:24:49.520
from a sum over k to an integral
over k or an integral over p.

00:24:49.520 --> 00:24:52.120
The density of
states in momentum

00:24:52.120 --> 00:24:57.390
is something that is
kind of invariant.

00:24:57.390 --> 00:25:00.920
It is a very slight
function of shape.

00:25:00.920 --> 00:25:03.540
So the periodic
boundary conditions

00:25:03.540 --> 00:25:06.930
and the open boundary
conditions, et cetera,

00:25:06.930 --> 00:25:10.882
give you something slightly
different over here.

00:25:10.882 --> 00:25:14.190
But by the time you
go to the continuum,

00:25:14.190 --> 00:25:17.520
it's a property
of dimension only.

00:25:17.520 --> 00:25:20.673
It doesn't really depend
on the underlying shape.

00:25:20.673 --> 00:25:22.048
AUDIENCE: So we
still change sums

00:25:22.048 --> 00:25:23.650
for integrals with
the same rules?

00:25:23.650 --> 00:25:24.275
PROFESSOR: Yes.

00:25:26.550 --> 00:25:30.780
It's a sort of general
density of state property.

00:25:30.780 --> 00:25:34.060
So there's some nice
formula that tells you

00:25:34.060 --> 00:25:38.110
what the density of state
is for an arbitrary shape,

00:25:38.110 --> 00:25:41.740
and the leading term is
always proportional to volume

00:25:41.740 --> 00:25:44.260
or area [INAUDIBLE]
the density of state

00:25:44.260 --> 00:25:46.180
that you have been calculating.

00:25:46.180 --> 00:25:47.910
And then there are
some leading terms

00:25:47.910 --> 00:25:52.140
that are proportional to
if it is volume to area,

00:25:52.140 --> 00:25:55.260
or number of edges, et cetera.

00:25:55.260 --> 00:25:58.626
But those are kind of subleading
the thermodynamic sense.

00:26:04.500 --> 00:26:09.050
So I guess Feynman
did a lot of work

00:26:09.050 --> 00:26:15.100
on formalizing these
ideas of Landau getting

00:26:15.100 --> 00:26:19.600
some idea of what the ground
state does, is, and excitations

00:26:19.600 --> 00:26:22.090
that you can have
about the spectrum.

00:26:22.090 --> 00:26:24.890
And so he was very
happy at being

00:26:24.890 --> 00:26:28.990
able to explain this,
including the nature of rotons,

00:26:28.990 --> 00:26:29.920
et cetera.

00:26:29.920 --> 00:26:31.890
And he was worried
that somehow he

00:26:31.890 --> 00:26:35.060
couldn't get this
logarithmic divergence.

00:26:35.060 --> 00:26:37.080
And that bothered
him a little bit,

00:26:37.080 --> 00:26:39.990
but Onsager told him
that that's really

00:26:39.990 --> 00:26:42.280
a much more fundamental
property that

00:26:42.280 --> 00:26:45.960
depends on critical phenomena,
and for resolving that issue,

00:26:45.960 --> 00:26:49.700
you have to come to 8.334.

00:26:49.700 --> 00:26:52.460
So we will not discuss
that, nor will we

00:26:52.460 --> 00:26:57.165
discuss why this is Tc minus
T to the 2/3 power and not

00:26:57.165 --> 00:26:58.820
a linear dependence.

00:26:58.820 --> 00:27:01.910
It again is one of these
critical properties.

00:27:01.910 --> 00:27:06.180
But we should be able to
explain this T to the fourth.

00:27:06.180 --> 00:27:07.900
And clearly, this
T to the fourth

00:27:07.900 --> 00:27:13.880
is not as simple as saying this
exponent changed from 3/2 to 3,

00:27:13.880 --> 00:27:16.060
this 3/2 should
also change the 3.

00:27:16.060 --> 00:27:21.420
No, it went to T to the fourth,
so what's going on over here?

00:27:21.420 --> 00:27:24.580
So last time at the
end of the lecture

00:27:24.580 --> 00:27:35.100
I wrote a statement that
the BEC is not superfluid.

00:27:40.310 --> 00:27:44.110
And what that really
means is that it

00:27:44.110 --> 00:27:48.646
has too many excitations,
low energy excitations.

00:27:56.410 --> 00:27:59.630
So imagine the
following, that maybe we

00:27:59.630 --> 00:28:03.650
have a container-- I don't
know, maybe we have a tube--

00:28:03.650 --> 00:28:13.480
and we have our superfluid going
through this with velocity v

00:28:13.480 --> 00:28:15.390
sub s.

00:28:15.390 --> 00:28:21.660
We want it to maintain that
velocity without experiencing

00:28:21.660 --> 00:28:23.870
friction, which it
seems to do in going

00:28:23.870 --> 00:28:25.050
through these capillaries.

00:28:25.050 --> 00:28:29.580
You don't have to push it, it
seems to be going by itself.

00:28:29.580 --> 00:28:36.480
And so the question is,
can any of these pictures

00:28:36.480 --> 00:28:42.230
that we drew for excitations
be consistent with this?

00:28:42.230 --> 00:28:45.690
Now, why am I talking about
excitations and consistency

00:28:45.690 --> 00:28:47.810
with superfluid?

00:28:47.810 --> 00:28:51.440
Because what can
happen in principle

00:28:51.440 --> 00:28:57.190
is that within your system,
you can spontaneously

00:28:57.190 --> 00:29:00.120
generate some excitation.

00:29:00.120 --> 00:29:04.120
This excitation will
have some momentum p

00:29:04.120 --> 00:29:08.410
and some energy epsilon of p.

00:29:08.410 --> 00:29:13.500
And if you spontaneously can
create these excitations that

00:29:13.500 --> 00:29:19.310
would take away energy from this
kinetic energy of the flowing

00:29:19.310 --> 00:29:23.950
superfluid, gradually the
superfluid will slow down.

00:29:23.950 --> 00:29:27.420
Its energy will be
dissipated and the superfluid

00:29:27.420 --> 00:29:29.997
itself will heat up
because you generated

00:29:29.997 --> 00:29:31.080
these excitations with it.

00:29:34.000 --> 00:29:38.690
So let's see what happens.

00:29:38.690 --> 00:29:42.750
If I were to create
such an excitation,

00:29:42.750 --> 00:29:51.320
actually I have to worry
about momentum conservation

00:29:51.320 --> 00:29:56.120
because I created something
that carried momentum p.

00:29:56.120 --> 00:30:04.690
Now initially, let's say
that this whole entity, all

00:30:04.690 --> 00:30:08.330
of the fluid that are
superflowing with velocity Vs

00:30:08.330 --> 00:30:13.055
have mass M. So the initial
momentum would be MVs.

00:30:16.280 --> 00:30:21.280
Now, I created some
excitation that

00:30:21.280 --> 00:30:24.970
is carrying away
some momentum p.

00:30:24.970 --> 00:30:28.740
So the only thing that
can ensure this happens

00:30:28.740 --> 00:30:32.510
is that I have to
slightly change

00:30:32.510 --> 00:30:34.285
the velocity of the fluid.

00:30:37.240 --> 00:30:40.425
Now this change in
velocity is infinitesimal.

00:30:40.425 --> 00:30:48.470
It is Vs minus p
divided by M. M is huge,

00:30:48.470 --> 00:30:53.292
so why bother
thinking about this?

00:30:53.292 --> 00:30:57.040
Well, let's see what
the change in energy is.

00:31:02.250 --> 00:31:07.160
Delta E. Let's say, well,
you created this excitation

00:31:07.160 --> 00:31:09.280
so you have energy epsilon of p.

00:31:12.310 --> 00:31:15.295
But I say, in
addition to that there

00:31:15.295 --> 00:31:20.500
is a change in the kinetic
energy of the superfluid.

00:31:20.500 --> 00:31:23.830
I'm now moving at
Vs prime squared,

00:31:23.830 --> 00:31:28.130
whereas initially when this
excitation was not present,

00:31:28.130 --> 00:31:32.650
I was moving at Vs.

00:31:32.650 --> 00:31:36.300
And so what do you have here?

00:31:36.300 --> 00:31:44.566
We have epsilon of p, M
over 2 Vs minus p over M,

00:31:44.566 --> 00:31:49.920
this infinitesimal change
in velocity squared minus M

00:31:49.920 --> 00:31:53.760
over 2 Vs squared.

00:31:53.760 --> 00:31:56.830
We can see that the leading
order of the kinetic energy

00:31:56.830 --> 00:32:01.780
goes away, but that there
is a cross term here

00:32:01.780 --> 00:32:05.660
in which the M contribution--
the contribution of the mass--

00:32:05.660 --> 00:32:07.460
goes away.

00:32:07.460 --> 00:32:15.140
And so the change in energy is
actually something like this.

00:32:18.020 --> 00:32:25.330
So if I had a system that
when stationary, the energy

00:32:25.330 --> 00:32:30.830
to create an excitation of
momentum p was epsilon of p,

00:32:30.830 --> 00:32:35.960
when I put that in a frame that
is moving with some velocity

00:32:35.960 --> 00:32:40.250
Vs, you have the
ability to borrow

00:32:40.250 --> 00:32:45.040
some of that kinetic energy
and the excitation energy

00:32:45.040 --> 00:32:47.045
goes down by this amount.

00:32:58.610 --> 00:33:07.330
So what happens if I take
the Bose-Einstein type

00:33:07.330 --> 00:33:10.810
of excitation spectrum
that is p squared over 2M

00:33:10.810 --> 00:33:15.860
and then subtract
a v dot p from it?

00:33:15.860 --> 00:33:20.520
Essentially there is a
linear subtraction going on,

00:33:20.520 --> 00:33:24.870
and I would get a
curve such as this.

00:33:24.870 --> 00:33:28.630
So I probably exaggerated
this by a lot.

00:33:28.630 --> 00:33:31.150
I shouldn't have
subtracted so much.

00:33:31.150 --> 00:33:37.590
Let me actually not
subject so much,

00:33:37.590 --> 00:33:40.260
because we don't want to go
all the way in that range.

00:33:45.860 --> 00:33:50.590
But you can see that there
is a range of momenta where

00:33:50.590 --> 00:33:56.710
you would spontaneously gain
energy by creating excitations.

00:33:56.710 --> 00:34:00.630
If the spectrum was
initially p squared over 2M,

00:34:00.630 --> 00:34:05.430
basically you just have too
many low energy excitations.

00:34:05.430 --> 00:34:07.960
As soon as you
start moving it, you

00:34:07.960 --> 00:34:11.280
will spontaneously
excite these things.

00:34:11.280 --> 00:34:14.080
Even if you were initially
at 0 temperature,

00:34:14.080 --> 00:34:18.340
these phonon excitations
would be created spontaneously

00:34:18.340 --> 00:34:19.159
in your system.

00:34:19.159 --> 00:34:20.840
They would move
all over the place.

00:34:20.840 --> 00:34:22.610
They would heat up your system.

00:34:22.610 --> 00:34:25.800
There is no way
that you can pass

00:34:25.800 --> 00:34:28.889
the Bose-Einstein
condensate-- actually,

00:34:28.889 --> 00:34:32.350
there's no way that you can even
move it without losing energy.

00:34:35.300 --> 00:34:38.900
But you can see that
this red curve does not

00:34:38.900 --> 00:34:40.830
have that difficulty.

00:34:40.830 --> 00:34:44.580
If I were to shift this curve
by an amount that is linear,

00:34:44.580 --> 00:34:45.520
what do I get?

00:34:45.520 --> 00:34:47.202
I will get something like this.

00:34:53.880 --> 00:35:00.030
So the Landau spectrum
is perfectly fine

00:35:00.030 --> 00:35:04.980
as far as excitations
is concerned.

00:35:04.980 --> 00:35:11.010
At zero temperature, even if the
whole fluid is moving then it

00:35:11.010 --> 00:35:14.440
cannot spontaneously
create excitations,

00:35:14.440 --> 00:35:17.490
because you would increase
energy of the system.

00:35:20.230 --> 00:35:22.530
So there's this difference.

00:35:22.530 --> 00:35:25.430
Ultimately, you would
say that the first time

00:35:25.430 --> 00:35:32.830
you would get excitation is
if you move it fast enough

00:35:32.830 --> 00:35:36.690
so that some portion of
this curve goes to 0.

00:35:36.690 --> 00:35:39.890
And indeed, if you were
to try to stir or move

00:35:39.890 --> 00:35:43.140
a superfluid fast
enough, there's

00:35:43.140 --> 00:35:45.240
a velocity at which
it breaks down,

00:35:45.240 --> 00:35:47.860
it stops being a superfluid.

00:35:47.860 --> 00:35:52.040
But it turns out that that
velocity is much, much smaller

00:35:52.040 --> 00:35:56.430
than you would predict based on
this roton spectrum going down.

00:35:56.430 --> 00:35:59.530
There are some other
many body excitations

00:35:59.530 --> 00:36:04.830
that come before and cause
the superfluid to lose energy

00:36:04.830 --> 00:36:06.650
and break down.

00:36:06.650 --> 00:36:12.050
But the genetic idea as
to why a linear spectrum

00:36:12.050 --> 00:36:16.140
for k close to 0 is
consist with superfluidity

00:36:16.140 --> 00:36:18.942
but the quadratic one
is not remains correct.

00:36:25.310 --> 00:36:31.910
Now suppose I am
in this situation.

00:36:31.910 --> 00:36:34.410
I have a moving super
fluid, such as the one

00:36:34.410 --> 00:36:36.950
that I have described over here.

00:36:36.950 --> 00:36:40.730
The spectrum is going to
be somewhat like this,

00:36:40.730 --> 00:36:43.450
but I'm not at 0 temperature.

00:36:43.450 --> 00:36:46.365
I want to try to describe
this T to the fourth behavior,

00:36:46.365 --> 00:36:49.880
so I want to be at some
finite temperature.

00:36:49.880 --> 00:36:52.480
So if I'm at some
finite temperature,

00:36:52.480 --> 00:36:57.070
there is some probability to
excite these different states,

00:36:57.070 --> 00:37:03.380
and the number that would
correspond to some momentum p

00:37:03.380 --> 00:37:07.170
would be given by this
general formula you have,

00:37:07.170 --> 00:37:14.490
1 over Z inverse e to the
beta epsilon of p minus 1.

00:37:14.490 --> 00:37:22.230
Furthermore, if I think that
I am in the regime where

00:37:22.230 --> 00:37:27.430
the number of excitations
is not important because

00:37:27.430 --> 00:37:32.840
of the same reason that I had
for Bose-Einstein condensate,

00:37:32.840 --> 00:37:36.740
I would have this formula
except that I would use epsilon

00:37:36.740 --> 00:37:41.540
of p that is appropriate
to this system.

00:37:41.540 --> 00:37:44.800
Actually, what is it
appropriate to this system

00:37:44.800 --> 00:37:53.160
is that my epsilon of
p was velocity times p.

00:37:53.160 --> 00:37:58.056
But then I started to move
with this superfluid velocity.

00:37:58.056 --> 00:37:59.850
Actually, maybe I'll
call this c so that I

00:37:59.850 --> 00:38:04.580
a distinction between c, which
is the linear spectrum here,

00:38:04.580 --> 00:38:06.430
and the superfluid
velocity dot p.

00:38:17.580 --> 00:38:21.565
No, this is actually
a vectorial product.

00:38:25.390 --> 00:38:28.570
And because of that,
I only drew one part

00:38:28.570 --> 00:38:31.920
of this curve that corresponds
to positive momentum.

00:38:31.920 --> 00:38:33.590
If I had gone to
negative momentum,

00:38:33.590 --> 00:38:37.460
actually, this curve
would have continued

00:38:37.460 --> 00:38:41.120
and whereas one branch
the energy is reduced,

00:38:41.120 --> 00:38:45.600
if I go to minus p,
the energy goes up.

00:38:45.600 --> 00:38:50.740
So whereas if the
superfluid was not moving,

00:38:50.740 --> 00:38:54.190
I can generate as many
excitations with momentum p

00:38:54.190 --> 00:38:56.510
as momentum minus p.

00:38:56.510 --> 00:39:02.450
Once the superfluid is
moving, there is a difference.

00:39:02.450 --> 00:39:04.100
One of them has a v dot p.

00:39:04.100 --> 00:39:06.660
The other has minus v dot p.

00:39:06.660 --> 00:39:11.020
So because of that,
there is a net momentum

00:39:11.020 --> 00:39:14.180
that is carried by
these excitations.

00:39:14.180 --> 00:39:16.660
This net momentum is
obtained by summing

00:39:16.660 --> 00:39:19.600
over all of these
things, multiplying

00:39:19.600 --> 00:39:20.850
with appropriate momentum.

00:39:20.850 --> 00:39:31.230
So I have beta is CP
minus v dot p minus 1.

00:39:31.230 --> 00:39:33.920
This is the momentum
of the excitation.

00:39:33.920 --> 00:39:39.080
This is the net momentum of
the system for one excitation.

00:39:39.080 --> 00:39:43.060
But then I have to sum
over all possible P's.

00:39:43.060 --> 00:39:45.100
Sum over P's, as
we've discussed,

00:39:45.100 --> 00:39:48.550
I can replace with an integral.

00:39:48.550 --> 00:39:54.960
And sum over k be replaced with
V. Integral over K, K and P

00:39:54.960 --> 00:39:58.540
are simply related
by a factor of h bar.

00:39:58.540 --> 00:40:02.930
So whereas before for k I
had 2 pi cubed for v cubed,

00:40:02.930 --> 00:40:04.135
I have 2 pi h bar cubed.

00:40:07.510 --> 00:40:11.360
So this is what I
have to calculate.

00:40:11.360 --> 00:40:13.160
Now what happens for small v?

00:40:18.760 --> 00:40:23.360
I can make an expansion
in vs. The 0 order term

00:40:23.360 --> 00:40:25.090
in the expansion
is what we would

00:40:25.090 --> 00:40:27.490
have for non-moving fluid.

00:40:27.490 --> 00:40:29.550
Momenta in the two
directions are the same,

00:40:29.550 --> 00:40:32.190
so that contribution goes away.

00:40:32.190 --> 00:40:35.740
The first contribution
that I'm going to get

00:40:35.740 --> 00:40:47.760
is going to come from expanding
this to lowest order in P.

00:40:47.760 --> 00:40:52.390
So there is a P that
is sitting out front.

00:40:52.390 --> 00:40:54.240
When I make the
expansion, I will

00:40:54.240 --> 00:40:59.000
get a vs dot P
times the derivative

00:40:59.000 --> 00:41:02.410
of the exponential
function gives me

00:41:02.410 --> 00:41:08.760
a factor of beta
e to the beta CP.

00:41:08.760 --> 00:41:15.410
And down here I will have e to
the beta CP minus 1 squared.

00:41:22.970 --> 00:41:25.890
Now, in the problem set you
have to actually evaluate

00:41:25.890 --> 00:41:27.260
this integral.

00:41:27.260 --> 00:41:28.660
It's not that difficult.

00:41:28.660 --> 00:41:31.020
It's related to Zeta functions.

00:41:31.020 --> 00:41:33.990
But what I'm really
only interested in what

00:41:33.990 --> 00:41:37.170
is the temperature dependence?

00:41:37.170 --> 00:41:43.660
You can see that I can rescale
this combination, call it x.

00:41:46.250 --> 00:41:50.540
Essentially what it says is that
whenever I see a factor of p,

00:41:50.540 --> 00:41:55.590
replace it by Kt over Cx.

00:41:55.590 --> 00:41:57.562
And how many P's do I have?

00:41:57.562 --> 00:42:01.715
I have three, four, five.

00:42:04.610 --> 00:42:10.740
So I have five factors of P. So
I will have five factors of Kt.

00:42:10.740 --> 00:42:13.910
One of them gets
killed by the beta,

00:42:13.910 --> 00:42:16.450
so this whole thing
is proportional to T

00:42:16.450 --> 00:42:17.380
to the fourth power.

00:42:23.170 --> 00:42:25.480
So what have we found?

00:42:25.480 --> 00:42:31.820
We have found that as
this fluid is moving

00:42:31.820 --> 00:42:35.240
at some finite
temperature T, it will

00:42:35.240 --> 00:42:38.220
generate these excitations.

00:42:38.220 --> 00:42:41.200
And these excitations
are preferably

00:42:41.200 --> 00:42:44.520
along the direction
of the momentum.

00:42:44.520 --> 00:42:49.610
And they correspond to
an additional momentum

00:42:49.610 --> 00:42:53.460
of the fluid that is
proportional to the volume.

00:42:57.210 --> 00:43:02.400
It's proportional to temperature
to the fourth and something.

00:43:02.400 --> 00:43:05.990
And of course, proportional
to the velocity.

00:43:09.200 --> 00:43:14.830
Now, we are used to thinking of
the proportionality of momentum

00:43:14.830 --> 00:43:18.050
and velocity to be
some kind of a mass.

00:43:21.280 --> 00:43:24.040
If I divide that
mass by the volume,

00:43:24.040 --> 00:43:28.710
I have a density of
these excitations.

00:43:28.710 --> 00:43:31.480
And what we have established
is that the density

00:43:31.480 --> 00:43:35.000
of those excitations
is proportional to t

00:43:35.000 --> 00:43:36.430
to the fourth.

00:43:36.430 --> 00:43:40.190
And what is happening in this
Andronikashvili experiment

00:43:40.190 --> 00:43:45.170
is that as these plates are
moving, by this mechanism

00:43:45.170 --> 00:43:47.230
the superfluid
that is in contact

00:43:47.230 --> 00:43:50.740
with them will
create excitations.

00:43:50.740 --> 00:43:53.720
And the momentum of
those excitations

00:43:53.720 --> 00:43:57.130
would correspond to some
kind of a density that

00:43:57.130 --> 00:44:00.540
vanishes as T to
the fourth, again

00:44:00.540 --> 00:44:06.370
in agreement with
what we've seen here.

00:44:06.370 --> 00:44:07.790
OK?

00:44:07.790 --> 00:44:08.420
Yes?

00:44:08.420 --> 00:44:10.825
AUDIENCE: So in an
integral expression

00:44:10.825 --> 00:44:13.086
you have vs as part
of a dot product.

00:44:13.086 --> 00:44:13.711
PROFESSOR: Yes.

00:44:13.711 --> 00:44:16.597
AUDIENCE: And then in the
next line [INAUDIBLE].

00:44:16.597 --> 00:44:18.490
So it's in that direction.

00:44:18.490 --> 00:44:23.530
PROFESSOR: So let's give these
indices p in direction alpha.

00:44:23.530 --> 00:44:26.520
This is p in direction alpha.

00:44:26.520 --> 00:44:29.250
This is p in direction alpha.

00:44:29.250 --> 00:44:32.330
This is v in direction
beta, p in direction

00:44:32.330 --> 00:44:33.590
beta, sum over beta.

00:44:36.220 --> 00:44:40.880
Now, I have to do an
angular integration that

00:44:40.880 --> 00:44:43.440
is spherically symmetric.

00:44:43.440 --> 00:44:48.170
And then somewhere inside
there it has a p alpha p beta.

00:44:48.170 --> 00:44:54.120
That angular
integration will give me

00:44:54.120 --> 00:44:59.080
a p squared over three delta
alpha beta, which then converts

00:44:59.080 --> 00:45:02.770
this v beta to a
b alpha, which is

00:45:02.770 --> 00:45:04.284
in the direction
of the momentum.

00:45:11.280 --> 00:45:11.780
Yes?

00:45:11.780 --> 00:45:13.155
AUDIENCE: Will
you say once again

00:45:13.155 --> 00:45:15.130
what happened to the
integral dimension?

00:45:15.130 --> 00:45:15.760
PROFESSOR: OK.

00:45:15.760 --> 00:45:24.020
So when we are in the
Bose-Einstein condensate,

00:45:24.020 --> 00:45:28.050
as far as the
excitations of concerned

00:45:28.050 --> 00:45:30.980
we have zero chemical potential.

00:45:30.980 --> 00:45:33.090
Whatever number of
particle that we

00:45:33.090 --> 00:45:36.560
have in excess of what
can be accommodated

00:45:36.560 --> 00:45:40.920
through the excitations we put
together in the ground state.

00:45:40.920 --> 00:45:43.980
So if you like the ground
state, the kp equals to 0

00:45:43.980 --> 00:45:47.430
or k equals to 0,
is a reservoir.

00:45:47.430 --> 00:45:49.410
You can add as many
particles there

00:45:49.410 --> 00:45:52.230
or bring as many particles
out of it as you like.

00:45:52.230 --> 00:45:55.000
So effectively, you have
no conservation number

00:45:55.000 --> 00:45:56.540
and no need for a [? z. ?]

00:46:03.350 --> 00:46:05.880
Of course, that we
only know for the case

00:46:05.880 --> 00:46:08.420
of the true
Bose-Einstein condensate.

00:46:08.420 --> 00:46:10.470
We are kind of
jumping and giving

00:46:10.470 --> 00:46:13.880
that concept relevance for
the interacting superfluid.

00:46:18.960 --> 00:46:21.407
OK?

00:46:21.407 --> 00:46:22.240
Any other questions?

00:46:26.230 --> 00:46:29.660
So this is actually
the last item

00:46:29.660 --> 00:46:34.280
I wanted to cover for
going on the board.

00:46:34.280 --> 00:46:37.170
The rest of the hour,
we have this movie

00:46:37.170 --> 00:46:39.846
that I had promised you.

00:46:39.846 --> 00:46:42.660
I will let that movie run.

00:46:42.660 --> 00:46:47.240
I also have all the connection
of problem sets, and exams,

00:46:47.240 --> 00:46:49.760
and test that you
have not picked up.

00:46:49.760 --> 00:46:53.450
So while the movie runs, you
are welcome to sit and enjoy it.

00:46:53.450 --> 00:46:54.940
It's very nice.

00:46:54.940 --> 00:46:58.510
Or you can go and take your
stuff and go your own way

00:46:58.510 --> 00:47:02.020
or do whatever you like.

00:47:02.020 --> 00:47:05.025
So let's go back.

00:47:29.244 --> 00:47:32.730
[VIDEO PLAYBACK]

00:48:29.314 --> 00:48:30.855
PROFESSOR: There
will be more action.

00:49:58.968 --> 00:50:00.960
-We just made a transfer
from liquid helium

00:50:00.960 --> 00:50:06.438
out of the storage tank into
our own experimental equipment.

00:50:06.438 --> 00:50:08.928
It is a remarkable [INAUDIBLE].

00:50:08.928 --> 00:50:11.418
It has two different and
easily distinguishable

00:50:11.418 --> 00:50:15.402
liquid phases-- a
warmer and a colder one.

00:50:15.402 --> 00:50:19.137
The warmer phase is called
liquid helium I and the colder

00:50:19.137 --> 00:50:21.876
phase liquid helium II.

00:50:21.876 --> 00:50:25.860
The two stages are separated
by a transition temperature,

00:50:25.860 --> 00:50:28.570
known as the lambda point.

00:50:28.570 --> 00:50:31.725
When liquid helium is pulled
down through the lambda point,

00:50:31.725 --> 00:50:36.610
a transition from helium I to
helium II is clearly visible.

00:50:36.610 --> 00:50:39.830
We will show it to you
later in this film.

00:50:39.830 --> 00:50:44.720
The two liquids behave nothing
like any other known liquid,

00:50:44.720 --> 00:50:48.620
although it could be said that
helium I, the warmer phase,

00:50:48.620 --> 00:50:52.140
approximates the behavior
of common liquids.

00:50:52.140 --> 00:50:55.620
But it is helium II,
the colder phase,

00:50:55.620 --> 00:50:57.690
which is truly different.

00:50:57.690 --> 00:51:01.676
Because of this, it is
called a superfluid.

00:51:01.676 --> 00:51:04.730
The temperatures involved when
working with liquid helium

00:51:04.730 --> 00:51:06.730
are quite low.

00:51:06.730 --> 00:51:10.154
Helium boils at
4.2 degrees Kelvin

00:51:10.154 --> 00:51:12.850
under conditions of
atmospheric pressure.

00:51:12.850 --> 00:51:17.400
And the lambda point lies
at roughly 2.2 degrees.

00:51:17.400 --> 00:51:21.810
Note that this
corresponds to minus 269

00:51:21.810 --> 00:51:26.514
and minus 271
degrees centigrade.

00:51:26.514 --> 00:51:29.390
The properties of liquid
helium that I have just

00:51:29.390 --> 00:51:31.738
been telling you about
are characteristic

00:51:31.738 --> 00:51:36.202
of the heavy isotope
if helium, helium-4.

00:51:36.202 --> 00:51:40.178
The element occurs in the
form of two stable isotopes.

00:51:40.178 --> 00:51:43.330
[INAUDIBLE] The second
and lighter one,

00:51:43.330 --> 00:51:46.620
helium-3, is very rare.

00:51:46.620 --> 00:51:51.738
Its abundance is only
about 1 part of 10 million.

00:51:51.738 --> 00:51:55.126
Pure liquid helium-3
is the subject

00:51:55.126 --> 00:51:57.950
of intensive study
at the present time,

00:51:57.950 --> 00:52:02.690
but so far no second
superfluid liquid phase

00:52:02.690 --> 00:52:04.868
has been found to
exist for helium-3.

00:52:08.330 --> 00:52:10.670
The low temperature at
which we'll be working

00:52:10.670 --> 00:52:12.980
calls for well-insulated
containers.

00:52:12.980 --> 00:52:15.045
The dewar meets
our requirements.

00:52:15.045 --> 00:52:17.620
The word "dewar" is
a scientific name

00:52:17.620 --> 00:52:19.758
given to a double-walled
vessel with the space

00:52:19.758 --> 00:52:22.500
between the walls evacuated.

00:52:22.500 --> 00:52:24.000
When these dewars
are made of glass,

00:52:24.000 --> 00:52:26.800
the surface of this
inner space is usually

00:52:26.800 --> 00:52:30.045
filtered to cut down heat
transfer by radiation.

00:52:30.045 --> 00:52:33.920
However, our dewars will
have to be transparent

00:52:33.920 --> 00:52:37.280
so that we can look at
what's going on inside.

00:52:37.280 --> 00:52:40.420
Now, liquid helium is commonly
stored in double dewars.

00:52:40.420 --> 00:52:44.035
The design is quite
simple, just put one

00:52:44.035 --> 00:52:49.676
inside the other like this.

00:52:49.676 --> 00:52:52.360
In the inner dewar, we
put the liquid helium,

00:52:52.360 --> 00:52:54.820
and in the space between
the inner and outer dewar,

00:52:54.820 --> 00:52:57.240
we maintain a supply
of liquid air.

00:53:00.300 --> 00:53:02.955
Here is a double dewar
exactly like the one

00:53:02.955 --> 00:53:07.530
we will be using in our
demonstration experiment.

00:53:07.530 --> 00:53:10.700
The inner dewar is filled
with liquid helium.

00:53:10.700 --> 00:53:12.910
The outer dewar
contains liquid air.

00:53:16.530 --> 00:53:19.450
The normal boiling
temperature of liquid air

00:53:19.450 --> 00:53:23.740
is about 80 degrees
Kelvin, 75 or more degrees

00:53:23.740 --> 00:53:26.190
hotter than liquid helium.

00:53:26.190 --> 00:53:28.620
The purpose of the
liquid air is twofold.

00:53:31.710 --> 00:53:34.450
First, we put the liquid air
in the outer dewar well ahead

00:53:34.450 --> 00:53:37.810
of putting liquid helium
in the inner dewar.

00:53:37.810 --> 00:53:42.003
In this way, the inner
dewar is pre-cooled.

00:53:42.003 --> 00:53:44.861
Secondly, we maintain
a supply of liquid air

00:53:44.861 --> 00:53:47.240
in the outer dewar
because it provides

00:53:47.240 --> 00:53:49.910
an additional
[INAUDIBLE] of insulation

00:53:49.910 --> 00:53:53.010
now that the liquid helium
is in the inner dewar.

00:53:53.010 --> 00:53:55.480
The [INAUDIBLE] liquid
air attests to the fact

00:53:55.480 --> 00:53:57.510
that it is absorbing
some of the heat which

00:53:57.510 --> 00:53:58.510
enters the double dewar.

00:54:01.410 --> 00:54:03.850
Even with the boiling
of the liquid air,

00:54:03.850 --> 00:54:06.690
the liquid helium
is clearly visible.

00:54:06.690 --> 00:54:09.140
Later, we will use
liquid air cooled

00:54:09.140 --> 00:54:12.228
below its boiling temperature
to reduce or eliminate

00:54:12.228 --> 00:54:13.811
the air bubbles for
better visibility.

00:54:17.870 --> 00:54:20.050
Now the liquid
air is cooled down

00:54:20.050 --> 00:54:22.540
and we have eliminated boiling.

00:54:22.540 --> 00:54:25.340
The smaller bubbles of
the boiling liquid helium

00:54:25.340 --> 00:54:27.424
are clearly visible.

00:54:27.424 --> 00:54:32.210
The cover over the inner dewar
has a port, at present open.

00:54:32.210 --> 00:54:35.750
The liquid helium is at
atmospheric pressure,

00:54:35.750 --> 00:54:39.200
so its temperature is
4.2 degrees Kelvin.

00:54:43.200 --> 00:54:45.546
In other words, what
we have in here now

00:54:45.546 --> 00:54:49.800
is liquid helium one, the
warmer of the two phases.

00:54:49.800 --> 00:54:53.740
Before we cool it down to take
a look at the superfluid phase,

00:54:53.740 --> 00:54:57.710
I want to dwell greatly on the
properties of helium I. I've

00:54:57.710 --> 00:55:00.665
told you before
that even helium I

00:55:00.665 --> 00:55:04.361
is different from
the normal liquids.

00:55:04.361 --> 00:55:06.610
The distance between neighboring
atoms and this liquid

00:55:06.610 --> 00:55:08.186
is quite large.

00:55:08.186 --> 00:55:10.150
The atoms are not
as closely packed

00:55:10.150 --> 00:55:11.700
as in the classical liquids.

00:55:11.700 --> 00:55:14.890
The reason for this
is quantum mechanics.

00:55:14.890 --> 00:55:17.700
The zero point energy is
relatively more important

00:55:17.700 --> 00:55:19.270
here than in any other liquid.

00:55:21.817 --> 00:55:23.275
As a consequence,
liquid helium has

00:55:23.275 --> 00:55:29.460
a very low mass density, only
about 13% the density of water,

00:55:29.460 --> 00:55:31.430
and a very low optical density.

00:55:31.430 --> 00:55:35.455
The index of refraction
is quite close to 1.

00:55:35.455 --> 00:55:38.510
This makes its surface hard
to see with the naked eye

00:55:38.510 --> 00:55:42.160
under ordinary
lighting conditions.

00:55:42.160 --> 00:55:44.434
You are no doubt
familiar with the fact

00:55:44.434 --> 00:55:49.150
that the helium atom has
closed shell atomic structure.

00:55:49.150 --> 00:55:52.500
This explains why helium is
a chemically inert element.

00:55:52.500 --> 00:55:55.820
It also accounts for the fact
that the force of attraction

00:55:55.820 --> 00:55:58.905
between neighboring helium
atoms, the so-called van der

00:55:58.905 --> 00:56:02.760
Waals force, is small.

00:56:02.760 --> 00:56:05.580
It takes little energy to
pull two helium atoms apart,

00:56:05.580 --> 00:56:10.170
as for example in evaporation.

00:56:10.170 --> 00:56:12.550
This gives liquid helium
a better small latent

00:56:12.550 --> 00:56:14.660
heat of vaporization.

00:56:14.660 --> 00:56:18.242
Only five calories are
needed to evaporate one gram.

00:56:18.242 --> 00:56:21.900
Compare this with water,
where evaporation requires

00:56:21.900 --> 00:56:24.105
between 500 and 600
calories per gram.

00:56:26.780 --> 00:56:30.745
The low van der Waals force
combined with a large zero

00:56:30.745 --> 00:56:34.200
point energy also accounts for
the fact that liquid helium

00:56:34.200 --> 00:56:38.150
does not freeze, cannot
be solidified at ordinary

00:56:38.150 --> 00:56:41.490
pressure, no matter
how far we cool it.

00:56:41.490 --> 00:56:44.700
However, liquid helium has been
solidified at high pressure.

00:56:47.450 --> 00:56:51.465
The liquid helium in the
dewar is at 4,2 degrees.

00:56:51.465 --> 00:56:54.720
We now want to cool it
down to the lambda point

00:56:54.720 --> 00:56:58.396
and show you the transition
to the [INAUDIBLE].

00:56:58.396 --> 00:57:02.132
Our method will be
cooling by evaporation

00:57:02.132 --> 00:57:04.562
using a vacuum pump.

00:57:04.562 --> 00:57:09.960
Now, the lambda point lies at
2.2 degrees, only 2 degrees

00:57:09.960 --> 00:57:14.500
colder than the [INAUDIBLE]
temperature of the liquid.

00:57:14.500 --> 00:57:17.020
What's more, not
very much heat has

00:57:17.020 --> 00:57:19.105
been removed from
the liquid helium

00:57:19.105 --> 00:57:22.455
now in the dewar to bring
it to the lambda point.

00:57:22.455 --> 00:57:27.305
It amounts to only
about 250 calories.

00:57:27.305 --> 00:57:31.013
Nevertheless, don't get the
idea that this cooling process

00:57:31.013 --> 00:57:32.456
is easy.

00:57:32.456 --> 00:57:35.830
On the contrary,
it's quite difficult.

00:57:35.830 --> 00:57:39.240
More than 1/3 of the liquid
helium now in the dewar

00:57:39.240 --> 00:57:41.790
has to be knocked
away in vapor form

00:57:41.790 --> 00:57:46.830
before we can get what remains
behind to the lambda point.

00:57:46.830 --> 00:57:49.763
That requires an
awful lot of pumping

00:57:49.763 --> 00:57:54.020
and explains why we use this
large and powerful vacuum

00:57:54.020 --> 00:57:55.510
pump over here.

00:58:00.812 --> 00:58:03.080
Even with this pump,
the cooling process

00:58:03.080 --> 00:58:04.960
takes a considerable
amount of time.

00:59:09.272 --> 00:59:11.268
Let me explain why
it is so difficult

00:59:11.268 --> 00:59:14.262
to cool liquid helium
to the lambda point.

00:59:14.262 --> 00:59:17.006
I have already mentioned
that liquid helium has

00:59:17.006 --> 00:59:20.250
a remarkably small
[INAUDIBLE] vaporization,

00:59:20.250 --> 00:59:23.244
only five calories per gram.

00:59:23.244 --> 00:59:26.737
At the same time, liquid
helium at 4.2 degrees

00:59:26.737 --> 00:59:31.750
has a high specific heat,
almost calorie per gram.

00:59:31.750 --> 00:59:34.355
Therefore, 1 gram of
the vapor pumped away

00:59:34.355 --> 00:59:36.830
carries with it
an amount of heat

00:59:36.830 --> 00:59:42.230
which can cool only 5 or 6 grams
of liquid helium by 1 degree.

00:59:42.230 --> 00:59:44.460
That's not very much cooling.

00:59:44.460 --> 00:59:46.930
It is less by a
factor of almost 100

00:59:46.930 --> 00:59:51.110
than when we cool
water by evaporation.

00:59:51.110 --> 00:59:53.710
The situation gets even
worse as cooling progresses

00:59:53.710 --> 00:59:57.588
below 4.2 degrees because the
specific heat of liquid helium

00:59:57.588 --> 01:00:00.570
rises astonishingly.

01:00:00.570 --> 01:00:05.280
As we approach 2,17
degrees, the lambda point.

01:00:05.280 --> 01:00:07.440
The heat of vaporization,
on the other hand,

01:00:07.440 --> 01:00:10.190
remains roughly the same.

01:00:10.190 --> 01:00:12.460
So a given amount
of vapor carried

01:00:12.460 --> 01:00:14.960
off produces less
and less cooling

01:00:14.960 --> 01:00:20.240
as we approach 2.17 degrees,

01:00:20.240 --> 01:00:23.130
Our thermometer here
is a low pressure gauge

01:00:23.130 --> 01:00:26.520
connected to the space
above the liquid helium.

01:00:26.520 --> 01:00:29.272
The needle registers
the pressure there.

01:00:29.272 --> 01:00:33.240
It is the saturated vapor
pressure of liquid helium.

01:00:33.240 --> 01:00:37.430
The gauge is calibrated for
the corresponding temperature.

01:00:37.430 --> 01:00:41.412
We call it a vapor
pressure thermometer.

01:00:41.412 --> 01:00:43.620
As we approach 2.17
degrees, boiling

01:00:43.620 --> 01:00:44.885
becomes increasingly violent.

01:00:48.662 --> 01:00:51.142
Suddenly it stops.

01:00:51.142 --> 01:00:55.120
This was the transition.

01:00:55.120 --> 01:00:57.732
The liquid you now
see is helium II.

01:00:57.732 --> 01:01:02.500
Even though evaporation does
continue, there is no boiling.

01:01:02.500 --> 01:01:06.309
The normal liquids, such
as the water in this basin,

01:01:06.309 --> 01:01:10.660
boil because of their relatively
low heat conductivity.

01:01:10.660 --> 01:01:13.199
Before heat,
[INAUDIBLE] at one point

01:01:13.199 --> 01:01:15.075
can be carried away
to a cooler place

01:01:15.075 --> 01:01:19.300
in the liquid bubbles
of the vapor form.

01:01:19.300 --> 01:01:23.390
Helium I behaves like a
normal liquid in this respect.

01:01:23.390 --> 01:01:25.470
The absence of
boiling in helium II

01:01:25.470 --> 01:01:27.790
reveals that this
phase acts as if it

01:01:27.790 --> 01:01:31.546
had a large heat conductivity.

01:01:31.546 --> 01:01:34.844
As a matter of fact, as
the liquid helium passed

01:01:34.844 --> 01:01:36.302
through the lambda
point transition

01:01:36.302 --> 01:01:38.926
you just saw, its
heat conductivity

01:01:38.926 --> 01:01:41.290
increased by the fantastic
factor of one million.

01:01:43.880 --> 01:01:47.340
The heat conductivity of
helium II is many times

01:01:47.340 --> 01:01:50.270
greater than in the
metals silver and copper,

01:01:50.270 --> 01:01:52.870
which are among the best
solid heat conductors.

01:01:52.870 --> 01:01:55.265
And yet here we
deal with a liquid.

01:01:55.265 --> 01:01:57.220
For this alone,
helium II deserves

01:01:57.220 --> 01:01:59.500
the name of superfluid.

01:01:59.500 --> 01:02:02.610
Actually, the way in
which helium II transports

01:02:02.610 --> 01:02:05.304
such large quantities
of heat so rapidly

01:02:05.304 --> 01:02:08.750
is totally different from
the classical concept

01:02:08.750 --> 01:02:11.040
for heat conduction.

01:02:11.040 --> 01:02:12.550
I'll come back to
the subject later

01:02:12.550 --> 01:02:15.440
in connection with an
experiment demonstrating

01:02:15.440 --> 01:02:20.260
the phenomenon of second
sound in helium II.

01:02:20.260 --> 01:02:23.710
Remember that this great
change in heat conductivity

01:02:23.710 --> 01:02:26.880
occurred at a single
and fixed transition

01:02:26.880 --> 01:02:29.246
temperature, the lambda point.

01:02:29.246 --> 01:02:32.630
We do indeed deal with
a change in phase,

01:02:32.630 --> 01:02:38.440
only here it is a change from
one liquid to another liquid.

01:02:38.440 --> 01:02:42.050
As we told you before, the
specific heat of liquid helium

01:02:42.050 --> 01:02:44.590
is very large as a lambda point.

01:02:44.590 --> 01:02:48.900
In fact, it behaves abnormally
even below the lambda point

01:02:48.900 --> 01:02:53.280
and falls again very rapidly
with the temperature.

01:02:53.280 --> 01:02:55.880
This discontinuity
in specific heat

01:02:55.880 --> 01:02:57.570
is another reflection
of the fact

01:02:57.570 --> 01:02:59.440
that we are dealing
with a change

01:02:59.440 --> 01:03:01.720
in the phase of the substance.

01:03:01.720 --> 01:03:05.415
By the way, the curve resembles
the Greek letter lambda.

01:03:05.415 --> 01:03:07.760
The transition
temperature got its name

01:03:07.760 --> 01:03:08.970
from the shape of this curve.

01:03:12.881 --> 01:03:14.370
[INAUDIBLE]

01:03:14.370 --> 01:03:19.200
The next one has to do with
the viscosity of liquid helium.

01:03:19.200 --> 01:03:21.285
When a normal liquid
flows through a tube,

01:03:21.285 --> 01:03:23.270
it will resist the flow.

01:03:23.270 --> 01:03:26.460
In this experiment, we
shall cause some glycerin

01:03:26.460 --> 01:03:29.800
to flow to a tube
under its own weight.

01:03:29.800 --> 01:03:32.043
The top layer is
colored glycerin.

01:03:36.960 --> 01:03:41.140
The liquid layer closest to
the tube wall adheres to it.

01:03:41.140 --> 01:03:45.040
The layer next in from the one
touching the wall flows by it

01:03:45.040 --> 01:03:49.530
and is retarded as it flows due
to the interatomic, the van der

01:03:49.530 --> 01:03:51.850
Waals force of attraction.

01:03:51.850 --> 01:03:54.930
The second layer in
turn drags on the third,

01:03:54.930 --> 01:03:59.260
and so on inward from the
wall, producing fluid friction,

01:03:59.260 --> 01:03:59.820
or viscosity.

01:04:07.770 --> 01:04:09.910
The narrower the
tube, the slower

01:04:09.910 --> 01:04:11.972
the liquid rate
of flow through it

01:04:11.972 --> 01:04:14.750
under a given head of pressure.

01:04:14.750 --> 01:04:17.680
Here I have a beaker with
an unglazed ceramic bottom

01:04:17.680 --> 01:04:21.600
of ultra-fine [INAUDIBLE].

01:04:21.600 --> 01:04:24.420
Many capillary channels run
through this ceramic disk.

01:04:24.420 --> 01:04:28.360
The diameter is quite small,
about one micron which is

01:04:28.360 --> 01:04:32.390
1/10,000 of a centimeter.

01:04:32.390 --> 01:04:34.410
There is liquid
helium in the beaker.

01:04:34.410 --> 01:04:39.611
It is 4.2 degrees Kelvin,
helium I, the normal phase.

01:04:39.611 --> 01:04:41.916
The capillaries in the
disk are fine enough

01:04:41.916 --> 01:04:44.845
to prevent the liquid now in
the beaker from flowing through

01:04:44.845 --> 01:04:46.930
under its on weight.

01:04:46.930 --> 01:04:49.750
Clearly, helium I is viscous.

01:04:49.750 --> 01:04:52.730
To be sure, its
viscosity is very small.

01:04:52.730 --> 01:04:55.210
That's why we had to choose
extremely fine capillaries

01:04:55.210 --> 01:04:56.584
to demonstrate it.

01:05:00.560 --> 01:05:04.110
Here you see the lambda
point transition.

01:05:08.770 --> 01:05:11.700
The helium II all poured out.

01:05:11.700 --> 01:05:15.070
The rate of pouring would
not be noticeably slower

01:05:15.070 --> 01:05:18.770
if the [INAUDIBLE]
were made yet finer.

01:05:18.770 --> 01:05:20.938
We call this kind
of flow a superflow.

01:05:24.842 --> 01:05:30.220
The temperature is
now at 1.6 degrees.

01:05:30.220 --> 01:05:33.304
The superflow is even faster.

01:05:33.304 --> 01:05:36.200
The viscosity of helium
II in this experiment

01:05:36.200 --> 01:05:38.532
is so small that it has
not been possible to find

01:05:38.532 --> 01:05:40.620
a value for it.

01:05:40.620 --> 01:05:43.990
It is less than the experimental
uncertainty incurred

01:05:43.990 --> 01:05:46.760
in attempts to measure it.

01:05:46.760 --> 01:05:50.064
We now believe that
helium II, the superfluid,

01:05:50.064 --> 01:05:54.306
has zero viscosity, although
we should be more precise here.

01:05:54.306 --> 01:06:00.675
We believe its viscosity is zero
when observing capillary flow.

01:06:00.675 --> 01:06:03.280
Bear this statement
in mind, for we

01:06:03.280 --> 01:06:05.720
will come up with a
contradiction to it

01:06:05.720 --> 01:06:07.916
in the next experiment,
where we will

01:06:07.916 --> 01:06:11.576
look for viscosity by
a different method.

01:06:11.576 --> 01:06:14.016
There is a copper cylinder
in the liquid helium,

01:06:14.016 --> 01:06:18.896
so mounted as we can turn
it about a vertical axis.

01:06:18.896 --> 01:06:22.090
In order to turn it smoothly
and with as little vibration as

01:06:22.090 --> 01:06:25.110
possible, we laid the
cylinder into the [INAUDIBLE]

01:06:25.110 --> 01:06:27.542
of a simple induction
motor energized

01:06:27.542 --> 01:06:29.024
from outside the dewar.

01:06:29.024 --> 01:06:31.494
The four horizontal
coils you see

01:06:31.494 --> 01:06:34.458
provide the torque which
turns the cylinder.

01:06:34.458 --> 01:06:37.230
The liquid helium is
electrically non-conducting.

01:06:37.230 --> 01:06:40.410
The coil exerts no
torque on it directly.

01:06:40.410 --> 01:06:44.350
Yet as we turn on our
motor, the liquid layer

01:06:44.350 --> 01:06:47.048
bounding the cylinder is
dragged along behind it.

01:06:47.048 --> 01:06:50.280
The boundary layer in turn
drags on the next layer,

01:06:50.280 --> 01:06:51.310
and so on outward.

01:06:58.955 --> 01:07:01.640
Finally a circulation
showing up in the helium

01:07:01.640 --> 01:07:05.432
due to its own viscosity and
the wooden panels we [INAUDIBLE]

01:07:05.432 --> 01:07:06.358
is turned along.

01:07:08.987 --> 01:07:10.820
What we have just seen
occurred in helium I,

01:07:10.820 --> 01:07:15.290
the normal phase at
4.2 degrees Kelvin.

01:07:15.290 --> 01:07:17.952
That is to say,
this demonstration

01:07:17.952 --> 01:07:22.470
is consistent with our results
for helium I by capillary flow.

01:07:22.470 --> 01:07:26.254
Helium I is viscous.

01:07:26.254 --> 01:07:28.470
Here you see the liquid
cooled down and passing

01:07:28.470 --> 01:07:30.400
into the superfluid
phase, helium II.

01:07:42.960 --> 01:07:44.020
Let's turn on the motor.

01:07:55.428 --> 01:07:58.404
The paddle wheel starts again.

01:07:58.404 --> 01:08:01.248
What does this mean?

01:08:01.248 --> 01:08:04.980
First of all, let me
emphasize that, like helium I,

01:08:04.980 --> 01:08:08.560
helium II is also non-conducting
in the electrical sense.

01:08:08.560 --> 01:08:10.710
In other words, the
circulation in the experiment

01:08:10.710 --> 01:08:15.910
can only have been caused
through viscous drag.

01:08:15.910 --> 01:08:19.250
So we conclude from the
rotating cylinder observations

01:08:19.250 --> 01:08:22.499
that helium II is viscous
and from the method

01:08:22.499 --> 01:08:26.370
of capillary flow that
it has zero viscosity.

01:08:26.370 --> 01:08:29.649
Our experimentation has
come up with a paradox.

01:08:29.649 --> 01:08:32.415
No normal classical
liquid is known

01:08:32.415 --> 01:08:35.860
to behave so inconsistently, in
capillary flow on the one hand

01:08:35.860 --> 01:08:39.158
and in bulk flow on the other.

01:08:39.158 --> 01:08:42.180
This state of affairs
forces us to think

01:08:42.180 --> 01:08:46.560
of helium II, the
superfluid, not as a single,

01:08:46.560 --> 01:08:48.996
but as a dual liquid.

01:08:48.996 --> 01:08:52.390
It appeared as if helium
II had two separate and yet

01:08:52.390 --> 01:08:54.817
interpenetrating
component liquids.

01:08:54.817 --> 01:08:56.813
We shall call one
component normal.

01:08:56.813 --> 01:09:00.306
It is this component
which we call

01:09:00.306 --> 01:09:02.240
responsible for the
appearance of viscosity

01:09:02.240 --> 01:09:05.565
below the lambda point in the
rotating cylinder experiment.

01:09:05.565 --> 01:09:09.016
The normal component,
as the name suggests,

01:09:09.016 --> 01:09:14.439
behaves like a normal liquid,
and therefore as viscosity.

01:09:14.439 --> 01:09:17.910
It is the one which the cylinder
drags along as its turned.

01:09:17.910 --> 01:09:20.910
But the normal components
cannot flow through the narrow

01:09:20.910 --> 01:09:25.410
channels of the ceramic disc
because of its viscosity.

01:09:25.410 --> 01:09:28.410
The second component
has zero viscosity,

01:09:28.410 --> 01:09:30.910
and it's called the
superfluid component.

01:09:30.910 --> 01:09:33.910
We think that it does
not participate at all

01:09:33.910 --> 01:09:36.810
in the rotating cylinder
experiment below the lambda

01:09:36.810 --> 01:09:37.639
point.

01:09:37.639 --> 01:09:40.034
It stays at rest.

01:09:40.034 --> 01:09:43.546
On the other hand, it can flow
through channels of one micron

01:09:43.546 --> 01:09:45.864
diameter with the greatest
of ease and countering

01:09:45.864 --> 01:09:50.834
no resistance whatever
because it has no viscosity.

01:09:50.834 --> 01:09:53.819
As we'll see later, this
flow is not repeated even

01:09:53.819 --> 01:09:56.070
when the capillary
diameters are made

01:09:56.070 --> 01:09:58.720
far smaller than one micron.

01:09:58.720 --> 01:10:01.822
This [INAUDIBLE] construction
is called the two fluid

01:10:01.822 --> 01:10:04.237
model for liquid helium II.

01:10:04.237 --> 01:10:05.927
Whether it is correct
or not depends

01:10:05.927 --> 01:10:08.190
on further tests
comparing the theory

01:10:08.190 --> 01:10:13.750
based on this model with
experimental results.

01:10:13.750 --> 01:10:17.120
We now go on to another
phenomenon, the fountain

01:10:17.120 --> 01:10:17.620
effect.

01:10:17.620 --> 01:10:20.470
What you see here
is a tube which

01:10:20.470 --> 01:10:23.125
narrows down and then
opens into a bulb.

01:10:23.125 --> 01:10:26.240
A small piece of cotton is
stuffed into the [INAUDIBLE]

01:10:26.240 --> 01:10:27.930
section between the
tube and the bulb.

01:10:27.930 --> 01:10:31.311
And the bulb has been tightly
packed with one of the finest

01:10:31.311 --> 01:10:35.660
powders available, [INAUDIBLE].

01:10:35.660 --> 01:10:39.388
And second wad of cotton
keeps the powder in the bulb.

01:10:39.388 --> 01:10:44.278
This powder presents extremely
fine capillary channels.

01:10:44.278 --> 01:10:47.020
Their average diameter is a
small fraction of 1 micron.

01:10:51.000 --> 01:10:53.590
This device has been
placed in the dewar.

01:10:53.590 --> 01:10:57.590
The liquid helium is
below the lambda point.

01:10:57.590 --> 01:11:01.110
We submerge the bulb, and then
we'll send a beam of light

01:11:01.110 --> 01:11:05.610
from this lamp to a
point near the top.

01:11:05.610 --> 01:11:09.375
You will see the light beam
when the lamp is turned on.

01:11:09.375 --> 01:11:13.080
It focuses some heat in the
form of infrared radiation

01:11:13.080 --> 01:11:15.000
on the point in question.

01:11:15.000 --> 01:11:17.880
The temperature will rise
above the temperature

01:11:17.880 --> 01:11:20.760
of the rest of the apparatus.

01:11:20.760 --> 01:11:21.720
Let us turn it on.

01:11:29.600 --> 01:11:33.102
Liquid helium flows through the
hole in the bottom of the bulb,

01:11:33.102 --> 01:11:35.940
through the fine
powder, and rises

01:11:35.940 --> 01:11:38.310
above the level of
liquid helium outside.

01:11:38.310 --> 01:11:40.225
The height to which
it will go depends

01:11:40.225 --> 01:11:42.205
on the temperature
increase produced

01:11:42.205 --> 01:11:45.700
bu the lamp focused on the bulb.

01:11:45.700 --> 01:11:48.422
We can very well ask, where
does the mechanical energy

01:11:48.422 --> 01:11:51.866
come from that does the
work necessary to pump

01:11:51.866 --> 01:11:55.310
the liquid above
the ambient level?

01:11:55.310 --> 01:11:57.662
Before we attempt to
discuss this question,

01:11:57.662 --> 01:12:00.690
there are two other facts
that should be noted.

01:12:00.690 --> 01:12:03.780
The first is by now obvious.

01:12:03.780 --> 01:12:07.780
The upward flow through the bulb
must clearly be a superfluid.

01:12:07.780 --> 01:12:13.280
Only the superfluid component
of helium II could get through.

01:12:13.280 --> 01:12:15.780
The second fact is
more significant.

01:12:15.780 --> 01:12:19.700
Let me explain it this way, the
superfluid flows spontaneously

01:12:19.700 --> 01:12:23.440
from a to b,from a
cooler to a warmer place.

01:12:23.440 --> 01:12:25.970
Point a is in the
cold liquid, but b

01:12:25.970 --> 01:12:28.139
is being heated
with infrared rays.

01:12:28.139 --> 01:12:32.576
The second law of thermodynamics
positively says that heat

01:12:32.576 --> 01:12:36.273
cannot of itself flow from
a point of lower to a point

01:12:36.273 --> 01:12:38.492
of higher temperature.

01:12:38.492 --> 01:12:40.628
What does this mean
to us here, knowing

01:12:40.628 --> 01:12:42.436
as we do that the
superfluid is flowing

01:12:42.436 --> 01:12:45.790
from a colder to a warmer spot?

01:12:45.790 --> 01:12:51.272
Simply this, it carries no
heat, no thermal energy.

01:12:51.272 --> 01:12:55.060
Any internal energy
[INAUDIBLE] is no longer

01:12:55.060 --> 01:12:56.964
thermally available.

01:12:56.964 --> 01:13:01.724
To say it precisely,
it has zero entropy.

01:13:01.724 --> 01:13:03.730
We have discovered another
remarkable property

01:13:03.730 --> 01:13:05.248
of helium II.

01:13:05.248 --> 01:13:09.072
Its superfluid component
not only is friction free,

01:13:09.072 --> 01:13:11.940
it also contains no heat.

01:13:11.940 --> 01:13:13.852
The heat energy
contained in helium II

01:13:13.852 --> 01:13:19.595
as a whole resides, all of
it, in the normal component.

01:13:19.595 --> 01:13:23.060
We may, of course, add heat
to the superfluid component,

01:13:23.060 --> 01:13:27.030
as we are doing when it passes
the spot heated by the lamp.

01:13:27.030 --> 01:13:28.910
But in doing so, we
are converting it

01:13:28.910 --> 01:13:32.908
into the normal component.

01:13:32.908 --> 01:13:36.360
Let me return briefly to
a question posed earlier.

01:13:36.360 --> 01:13:38.808
Mechanical work is done
in pumping the liquid

01:13:38.808 --> 01:13:40.804
above equilibrium level.

01:13:40.804 --> 01:13:42.800
Where does it come from?

01:13:42.800 --> 01:13:45.295
I cannot answer this
question here in full,

01:13:45.295 --> 01:13:49.287
but it suffice to tell you that
we are dealing here with a heat

01:13:49.287 --> 01:13:49.800
engine.

01:13:49.800 --> 01:13:51.508
The mechanical energy
comes from the heat

01:13:51.508 --> 01:13:54.304
added at the light spot.

01:13:54.304 --> 01:13:57.400
An amusing demonstration
of the same phenomenon

01:13:57.400 --> 01:14:00.522
again uses a bulb
packed with rouge,

01:14:00.522 --> 01:14:03.994
but this one opens
into a capillary.

01:14:03.994 --> 01:14:07.962
Light is beamed on a spot
just below the capillary,

01:14:07.962 --> 01:14:09.610
and it produces a
helium fountain.

01:14:14.273 --> 01:14:17.195
The phenomenon in this and
the previous experiment

01:14:17.195 --> 01:14:19.143
has become known as
the thermomechanical,

01:14:19.143 --> 01:14:23.526
or the fountain, effect.

01:14:23.526 --> 01:14:27.160
Below the lambda point,
the superfluid component

01:14:27.160 --> 01:14:29.980
of liquid helium creeps
up along the walls

01:14:29.980 --> 01:14:32.758
of its container in an
extremely thin film.

01:14:32.758 --> 01:14:34.740
It is known as a Rollin film.

01:14:37.674 --> 01:14:41.586
This creeping film is
a variety of superflow.

01:14:41.586 --> 01:14:45.752
It is difficult to make the film
itself directly visible to you.

01:14:45.752 --> 01:14:49.215
To show it indirectly,
we've put some liquid helium

01:14:49.215 --> 01:14:51.370
into a glass vessel.

01:14:51.370 --> 01:14:54.360
It is below the lambda point.

01:14:54.360 --> 01:14:57.140
There is no part porous
bottom in this vessel.

01:14:57.140 --> 01:14:59.920
The film rises along
the inside wall

01:14:59.920 --> 01:15:01.310
and comes down
along the outside,

01:15:01.310 --> 01:15:03.880
collecting in drops
at the bottom.

01:15:03.880 --> 01:15:06.265
The thickness of
this creeping film

01:15:06.265 --> 01:15:09.604
is only a small fraction of
1 micron and of the order

01:15:09.604 --> 01:15:12.480
of 200 to 300 angstrom.

01:15:12.480 --> 01:15:16.512
Its speed, while small just
below the lambda point,

01:15:16.512 --> 01:15:18.902
may reach a value as
high as 35 centimeters

01:15:18.902 --> 01:15:22.815
per second at
lower temperatures.

01:15:22.815 --> 01:15:27.120
Our next experiment deals with
the phenomenon of second sound.

01:15:27.120 --> 01:15:29.915
We are all familiar with wave
motion in elastic materials,

01:15:29.915 --> 01:15:34.120
be they solids,
liquids, or gases.

01:15:34.120 --> 01:15:37.630
Elastic energy of deformation,
carried away from its source

01:15:37.630 --> 01:15:40.690
in the form of waves with
a characteristic speed,

01:15:40.690 --> 01:15:43.842
the speed of sound.

01:15:43.842 --> 01:15:45.600
Liquid helium is an
elastic substance

01:15:45.600 --> 01:15:48.320
both above and below
the lambda point.

01:15:48.320 --> 01:15:51.300
Both helium one and two
support sound waves.

01:15:51.300 --> 01:15:54.640
Now helium II, the
superfluid phase,

01:15:54.640 --> 01:15:57.650
also conducts heat
in the form of waves.

01:15:57.650 --> 01:16:01.580
This remarkable property is
shared by no other substance.

01:16:01.580 --> 01:16:06.593
For better or for worse, it
has been called second sound.

01:16:06.593 --> 01:16:09.540
Normal heat conduction
is a diffusion process.

01:16:09.540 --> 01:16:12.400
The rate of flow of heat is
proportional to the temperature

01:16:12.400 --> 01:16:13.310
differences.

01:16:13.310 --> 01:16:18.104
But in helium II it
is a wave process.

01:16:18.104 --> 01:16:23.170
Heat flows through helium II
with a characteristic speed,

01:16:23.170 --> 01:16:26.210
the speed of second sound.

01:16:26.210 --> 01:16:30.360
We shall send small heat pulses
into helium II from a heater.

01:16:30.360 --> 01:16:32.682
They will spread away
from the heater uniformly,

01:16:32.682 --> 01:16:34.140
carrying the heat
energy with them.

01:16:39.712 --> 01:16:41.337
The speed of second
sound is small just

01:16:41.337 --> 01:16:43.960
below the lambda point.

01:16:43.960 --> 01:16:46.420
In the neighborhood
of 1.6 degrees Kelvin,

01:16:46.420 --> 01:16:49.210
it reaches a value of
roughly 20 meters per second,

01:16:49.210 --> 01:16:54.010
and it is in this range that
we will run our demonstration.

01:16:54.010 --> 01:16:57.142
The experimental
procedure is as follows.

01:16:57.142 --> 01:17:01.081
There are two disks
in the liquid helium.

01:17:01.081 --> 01:17:04.490
They are carbon resistors
with the carbon applied

01:17:04.490 --> 01:17:06.925
in thin layers on one
side of each disk.

01:17:06.925 --> 01:17:09.360
In this way, good
thermal contact

01:17:09.360 --> 01:17:13.256
is established between the
resistor and the liquid helium.

01:17:13.256 --> 01:17:15.920
The following resistor
will be used as a heater.

01:17:15.920 --> 01:17:18.580
Electric currents will
be sent through it

01:17:18.580 --> 01:17:22.435
in pulses from this pulse
generator by means of the cable

01:17:22.435 --> 01:17:24.295
you see here.

01:17:24.295 --> 01:17:26.620
The [INAUDIBLE] of
the generator is also

01:17:26.620 --> 01:17:31.490
connected via a second cable
to a dual-trace oscilloscope,

01:17:31.490 --> 01:17:34.760
where it will be recorded
on the bottom [? trip. ?]

01:17:34.760 --> 01:17:37.951
In other words, it will record
the heat pulse as it enters

01:17:37.951 --> 01:17:41.220
the liquid helium.

01:17:41.220 --> 01:17:43.790
The pulses have been turned on.

01:17:43.790 --> 01:17:45.770
They themselves trigger
the horizontal sweep

01:17:45.770 --> 01:17:50.214
of [INAUDIBLE], which
records [INAUDIBLE].

01:17:50.214 --> 01:17:53.060
It is calibrated
at 1 millisecond

01:17:53.060 --> 01:17:54.785
per unit on the scale.

01:17:54.785 --> 01:17:58.330
The pulses are 1
millisecond long.

01:17:58.330 --> 01:18:00.185
The pulses leave the
heater at the bottom

01:18:00.185 --> 01:18:02.420
in the form of
second sound and move

01:18:02.420 --> 01:18:05.790
up to where they strike the
carbon resistor at the top.

01:18:05.790 --> 01:18:10.375
Being heat pulses, they
greatly raise is temperature.

01:18:10.375 --> 01:18:13.550
The carbon resistor is
quite sensitive to changes

01:18:13.550 --> 01:18:14.175
in temperature.

01:18:14.175 --> 01:18:17.590
It acts as a thermometer.

01:18:17.590 --> 01:18:19.480
So the heat pulse
of second sound

01:18:19.480 --> 01:18:21.582
creates a pulse-like
change in the resistance

01:18:21.582 --> 01:18:24.300
of the [INAUDIBLE] up here.

01:18:24.300 --> 01:18:26.410
It isn't hard to
convert this resistance

01:18:26.410 --> 01:18:28.200
pulse into a [INAUDIBLE].

01:18:28.200 --> 01:18:31.260
What we will do is to
maintain a small DC

01:18:31.260 --> 01:18:33.211
current in the top resistor.

01:18:33.211 --> 01:18:36.650
It is supplied from a
battery in this metal box.

01:18:36.650 --> 01:18:38.757
The box shields the
circuits in order

01:18:38.757 --> 01:18:41.619
to reduce electronic noise.

01:18:41.619 --> 01:18:44.004
The voltage pulse is small.

01:18:44.004 --> 01:18:50.080
In this second box
we have an amplifier.

01:18:50.080 --> 01:18:53.527
The amplified output is fed
into the oscilloscope, where

01:18:53.527 --> 01:18:55.660
it will appear on
the upper trace.

01:18:55.660 --> 01:18:57.556
The horizontal
timescale on this trace

01:18:57.556 --> 01:19:00.860
is exactly the same as
for the bottom trace.

01:19:00.860 --> 01:19:03.710
However, the upper trace
records both exchanges

01:19:03.710 --> 01:19:06.520
as they occur in
the top resistor,

01:19:06.520 --> 01:19:09.852
a detector of second sound.

01:19:09.852 --> 01:19:36.172
The temperature of the liquid
is about 1.65 degrees Kelvin.

01:19:36.172 --> 01:19:42.110
The battery has been turned
on, and now the amplifier.

01:19:42.110 --> 01:19:45.515
Among noise and other
distortions in the upper trace,

01:19:45.515 --> 01:19:49.760
a clear-cut voltage pulse
appears about 4 and 1/2 units

01:19:49.760 --> 01:19:53.100
to the right, 4 and
1/2 milliseconds later

01:19:53.100 --> 01:19:57.687
than the pulse
entering the heater.

01:19:57.687 --> 01:20:02.724
This pulse in the upper trace is
also about 1 millisecond long.

01:20:02.724 --> 01:20:07.188
It is the second sound as it
arrives at the upper resistor.

01:20:07.188 --> 01:20:10.164
The upper trace also shows
a strong voltage pulse

01:20:10.164 --> 01:20:13.636
at the left, simultaneous
to the heater pulse.

01:20:13.636 --> 01:20:16.612
That's due to pick-up
by electromagnetic waves

01:20:16.612 --> 01:20:20.084
with the heater acting as
transmitter and the detector

01:20:20.084 --> 01:20:21.076
as receiver.

01:20:28.020 --> 01:20:30.996
We're moving the detector
toward the heater.

01:20:30.996 --> 01:20:33.476
The pulse moves
with it to the left.

01:20:33.476 --> 01:20:37.865
Notice the echos of second sound
which appear on the upper trace

01:20:37.865 --> 01:20:41.360
while the detector
is near the heater.

01:20:41.360 --> 01:20:44.270
They're caused by multiple
reflections between the two

01:20:44.270 --> 01:20:45.725
resistors.

01:20:45.725 --> 01:20:50.090
A total of three echos
is clearly visible.

01:20:50.090 --> 01:20:51.060
[END VIDEO PLAYBACK]

01:20:51.060 --> 01:20:55.400
PROFESSOR: OK, you can watch
the rest of it at home.