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PROFESSOR: Let's start.
Any questions?
We've been gradually building up thermodynamics, culminating
in what I call the most important fundamental relation,
which was the E is T dS plus Ji dxi.
And essentially we said that this
was some statement of conservation of energy.
where there's a heat component that goes into the system,
and there's a work component.
And actually, we ended up breaking the work component
into a mechanical part and a chemical part,
which we wrote as mu alpha dN alpha.
I guess if I don't write the sums,
the summation convention is assumed.
Now, this gave us an idea of what a thermodynamic system is,
in the sense of how many coordinates do we need in order
to place the system in equilibrium
somewhere in some coordinate space.
And we said that we can just count the number of variables
that we have on this side.
And indeed, for the case of energy,
the natural set of variables would be S, x, and N.
And given some place in the coordinate system that
is S, x, and N, you could find a point that
corresponds to that thermodynamic equilibrium
state.
Now, having gotten this fundamental relation,
we can manipulate and write it in different fashion.
And in particular, for example, you
can write the S is dE by dT, minus Ji dxi over T,
minus mu alpha dN alpha over T. So you could in principle
express everything rather than in terms of entropy-- which
in some circumstances may not be the right variable-- in terms
of the energy content of the system and the variables
that you need in order to make work type of changes
on the system.
And we saw that other versions of it is possible.
So for example, we could rearrange.
Rather than looking at E, we could
look at E minus TS, the quantity that we call F.
And then dF would look very much like the original equation.
Rather than T dS, we would get minus S dT.
And clearly, for this quantity F,
the natural variables are T, x, and N.
And there were other types of these functions
that we construct.
So basically at this stage, we have a formula that tells us
how to navigate our way in this space of things
that are in equilibrium.
And by taking various derivatives of equilibrium
functions with respect to each other,
we can express other ones.
And via this methodology, we can relate to many different things
to each other and have mathematical relations
between physical observables.
So following up on that, we then ask
what are some other mathematical forms that we can look at.
And one set of forms followed from extensivity.
And we saw that if we were under circumstances such
that we were to multiply all of the extensive variables
by some factor-- 2, 3, whatever-- and the energy
content would increase by that same factor,
then we could in fact integrate the equation
and drop the d's and write it in this fashion--
TS Jx plus mu N, regarding these as kind of vectors.
And importantly, that told us that the intensive variables
were not independent of each other.
And you had relationship that you-- S dT plus Ji xi dJi
plus N alpha d mu alpha is 0.
And last time, I did one calculation
using this Gibbs-Duhem relation that I repeat.
And that was for a gas isotherm.
Isotherm corresponds to dT goes to 0.
And the analogue of hydrostatic work for a gas was minus P
for J. So this becomes minus V dP plus N d mu.
If you have one component, it goes to 0.
And we said that this gives us the relationship
along an isotherm.
So for a fixed T, d mu by dP is V over N.
I'll come back to that shortly.
Towards the end of last lecture, I
mentioned that we can get a number
of other so-called Maxwell relations
by noting that mixed second derivatives are
independent of the order of taking derivatives.
So that if you have some df of x and y written as df by dx
at constant y dx, plus df by dy at constant x dy,
then the order of the second derivatives, which
would be obtained by taking either
a derivative of the first term with respect
to y or the second term with respect to x--
doesn't matter which order.
So let's start with our initial equation.
And for simplicity, let's think of cases
where we have a fixed number of particles
so that we can write dE as T dS plus Ji dxi.
So this looks like this dF that I wrote before for you.
And I can identify temperature as dE by dS at constant x,
and each one of the forces Ji as dE by dxi at constant S.
And all of the J's that are not equal to i kept fixed.
And then, since the second derivative
is irrespective of order, I can form the second derivative
with respect to the combination S xi into a different fashion.
I can either take a derivative over here-- I already
have derivative with respect to S giving me T,
so I will get dT by dxi, essentially at constant S--
or I can take the derivative of this object.
I already have the x derivative, so I take the S derivative.
So I have dJi with respect to S at constant x.
So this is an example of the Maxwell relation.
And just to remind you, we can always
invert these derivatives.
So I can invert this and say that actually,
if I had wondered how the entropy changes
as a function of a force such as pressure at constant x,
it would be related to how the corresponding displacement
to the force changes as a function of temperature.
So you would have a relationship between some observable
such as how the length of a wire changes
if you change temperature rapidly,
and how the entropy would change as a function of pressure,
or corresponding force, whatever it may be.
Now, what I would like to remind you is that the best
way to sort of derive these relationships
is to reconstruct what kind of second
derivative or mixed derivative you
want to have to give you the right result.
So for example, say somebody told me
that I want to calculate something similar-- dS by dJ,
let's say at constant temperature.
How do I calculate that?
A nice way to calculate that is if I somehow
manipulate so that S appears as a first derivative.
Now, S does not appear as a first derivative here,
but certainly appears as a first derivative here.
So who says I have to take E as a function of state?
I can look at the second derivative of F.
So if I look at F, which is E minus TS--
and actually, I don't even need to know
what the name of this entity is, whether it's F, G, et cetera.
All I know is that it will convert this to minus S dT--
the TS that I don't want, so that S appears now
as a first derivative.
Now I want to take a derivative of S with respect
to J. If I just stop here, the next term that I would have
is Ji dxi.
And the natural way that I would construct a second derivative
would give me dS with respect to xi.
But I want the S with respect to Ji.
So I say, OK, I'll do this function instead.
As I said, I don't really care what
the name of these functions is.
This will get converted to xi dJ.
Now I can say there is this first derivative-- d o
minus S with respect to Ji at constant T
is the same thing as d of minus y with respect
to T at constant J. So I can get rid of the minus sign
if I want, and I have the answer that I want.
It is dxi by dT at constant J-- at constant-- yeah, T-- J.
And you can go and construct anything else that you like.
Let me, for example, try to construct
this entity we already calculated.
So let's see.
What can I say about d mu by dP at constant temperature,
and I want to calculate that via a Maxwell relation?
So what do I know?
I know that dE is T dS minus pdV.
Good.
Oh, I want to have mu in the play,
so now I have to add mu dN.
That's good, because I have mu as a first derivative.
I have it over here.
But I have P as a first derivative,
whereas I want P to appear as the work element, something
like dP.
So what I will do is I will make this d of E plus pV.
This converts this to plus V dP plus mu dN.
That's fine, except that when I do things now,
I would have calculated things at constant S.
And I want to calculate things at constant T.
That's not really important.
I do a minus ST. This becomes minus S dT.
And then I have a function of state--
I don't care what it's called-- that has
the right format for me to identify that d mu
by dP at constant N and T, along an isotherm of a fixed
number of particles, is the same thing as dV
by dN at constant PN T. So that's my Maxwell relation.
So why didn't I get this result?
Did I make a mistake?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Louder.
AUDIENCE: Is this valid for extensive systems?
PROFESSOR: Yes, this is valid for extensive system.
I used extensively in the relation of this.
And here, I never used extensivity.
But if I have a system that I tell you its pressure
and temperature, how is its volume related
to the number of particles?
I have a box in this room.
I tell you pressure and temperature are fixed.
If I make the box twice as big, number of particles, volume
will go twice as big.
So if I were to apply extensivity,
I would have to conclude that this is the same thing as V
over N, and I would be in agreement
with what I had before.
Any questions?
Yes.
AUDIENCE: What about the constraint of constant PN T
[INAUDIBLE]?
PROFESSOR: OK.
Actually, constraint of constant PN T
is important because we said that said there
it is a kind of Gibbs-Duhem relation.
And the Gibbs-Duhem relation tells me
for the case of the gas that the S dT minus V dP plus N d mu
is 0.
And that constraint is the one that
tells me that once you have set P and T, which are the two
variables that I have specified here, you also know mu.
That is, you know all the intensive quantities.
So once you set PN T-- the constraints over there--
really, the only thing that you don't know
is how big the system is.
And that's what's the condition that I used over here.
But it is only-- so let me re-emphasize
it this way-- I will write dV by dN
is the same thing as V over N, only if I have PN T here.
If I had here PN S, for example, I wouldn't be able to use this.
I was going to build this equation
to say, look at the elements of this.
This part we got from the first law,
the temperature we got from the zeroth law,
and this we got from the second law, which is correct,
but it leaves out something, which
is that what the second law had was things
going some particular direction.
And in particular, we had that for the certain changes
that I won't repeat, the change in entropy to be positive.
So when you are in equilibrium, you
have this relationship between equilibrium state functions.
But in fact, the second law of thermodynamics
tells you a little bit more than that.
So the next item that we are going to look at
is, what are the consequences of having
underlying this equality between functions
of state some kind of an inequality about certain things
being possible and things not being possible?
And that relates to stability conditions.
And I find it easier to again introduce to you these
stability conditions by thinking of a mechanical analogue.
So let's imagine that you have some kind of a spring,
and that this spring has some kind of a potential energy
that I will call phi as a function of the extension x.
And let's say that it's some non-linear function
such as this.
There is no reason why we should use a hook and spring,
but something like this.
Now, we also discussed last time what
happens if I were to pull on this with some force J.
So I pull on this.
As a result, presumably it will no longer
be sitting at x, because what I need to do
is to minimize something else-- the function that we call H--
which was the potential energy of the spring
plus the entity that was exerting the force.
So if I were to make it in the vertical direction,
you can imagine this as being the potential energy
of the mass that was pulling on the spring.
So what that means is that in the presence of J,
what I need to do is to find the x that
minimizes this expression.
Once essentially the kinetic energy has disappeared,
my spring stops oscillating and goes to a particular value.
And mathematically, that corresponds
to subtracting a linear function from this.
Actually, what will it look like?
Let's not make this go flat, but go up in some linear fashion
so that at least I can draw something like this.
So what I need to do is to minimize this.
So I find that for a given J, I have
to solve d phi dx equals J. But especially if you have
a spring that has the kind of potential that I drew for you,
you start increasing the-- so basically,
what it says is that you end up at the place
where the slope is equal to the derivative
of the potential function.
What happens if you start increasing this slope,
if it is a non-linear type of thing, at some point
you can see it's impossible.
Essentially, it's like a very flat noodle.
You pull on it too much and then it starts to expand forever.
So essentially, this thing that we are looking for a minimum,
ultimately also means that you need
to have a condition on the second derivative.
d2 phi by dx squared better be positive actually,
I have to take d2 H y dx squared.
I took one derivative to identify
where the location of the best solution is.
If I take a second derivative, this
became essentially the J. J is a constant.
It only depends on the potential energy.
So it says independent of the force that you apply.
You know that if the shape of the potential energy is given,
the only places that are accessible
are the kinds of places where you
take the second derivative and the second
has a particular sign.
So this portion of the curve is in principle
physically accessible.
The portion that corresponds to essentially changing
the curvature and going the other way, there's no way,
there's no force that you can put
that would be able to access those kinds of displacements.
So there is this convexity condition.
This was for one variable.
Now suppose you have multiple variables-- x2, x3, et cetera.
So you will have the potential energy
as a function of many variables.
And the condition that you have is that the second derivative
of the potential with respect to all variables, which forms--
and if there are n variables, this would be an n by n matrix.
This matrix is positive definite.
What that means is that if you think
of this as defining for you the bottom of a potential,
in whatever direction you go, this
would describe the change of potential for you.
The first order change you already
set to 0 because of the force J. The second order change better
be positive.
So this is the condition that you would have for stability,
quite generally mechanically.
And again, this came from the fact
that if you were to exert a force,
you would change the energy by something like J delta
x, which is certainly the same thing that we have here.
This is just one component that I have been focusing on.
And so ultimately, I'm going to try
to generalize that to this form.
But one thing that I don't quite like in this form
is that I'm focusing too much on the displacements.
This is written appropriately using this form.
And we have seen over here that whether or not
I express things in terms of the displacements
or in terms of entropy, temperature, conjugate forces,
et cetera, these are all equivalent ways
of describing the equilibrium system.
So I'm going to do a slight manipulation of this
to make that symmetry between forces and displacements
apparent.
So the generalization of this expression
that we have over here is that once I
know the potential energy, the force
in this multi-component space is df by dxi.
So in this multi-component space, I have some J,
and there's a corresponding x.
They're related by this.
Now, if I were to make a change in J,
there will be a change in the position of the equilibrium.
And I can get that by doing a corresponding cange
in the derivative.
The corresponding change in the derivative
is the second derivative of phi with respect
to xi xj delta xj, sum over J. So what I'm saying
is that this is a condition that relates
the position to the forces.
If I make a slight change in the force,
there will be a corresponding slight change in the position.
And the changes in the position and the force given
by this equilibrium condition are related by this formula.
Now, you can see that that sum is already present here.
So I can rewrite the expression that
is over here using that formula as delta xi delta
Ji, sum over i-- better be positive.
So this expression is slightly different and better way
of writing essentially the same convexity condition, stability
condition, but in a manner that treats
the displacements and the forces equivalently.
And I'm going to apply that to this
and say that in general, there will
be more ways of manipulating the energy
of a thermodynamic system, and that having the system reach
to equilibrium requires the generalization of this
as a delta T delta S, plus sum over i delta Ji delta xi,
plus sum over alpha, delta mu alpha,
delta N alpha-- to be positive.
And you may say I didn't really derive this for you
because I started from mechanical equilibrium
and stability condition and generalized
to the thermodynamic one.
But really, if you go back and I use this condition
and do manipulations that are compatible to this,
you will come up with exactly this stability condition.
And I have that in the notes.
It is somewhat more mathematically complicated
than this, because the derivatives involve
additional factors of 1 over T. And you
have to do slightly more work, but you will ultimately
come with the same expression.
OK?
You say, why is that of any use?
It is possible that you've already
seen a particular use of this expression.
Let's look at the case of a gas-- one component.
You would say that the condition that we have
is that delta T delta S minus delta P delta V plus delta mu
delta N has to be positive.
Let's even be simpler.
Look at the case where delta T is 0,
delta N equals to 0, which means that I have a fixed
amount of gas at a particular temperature.
So I could have, for example, a box
in this room which is in equilibrium
at the same temperature as the temperature of the room.
But its volume I'm allowed to adjust
so that the pressure changes.
So in that process, at a particular temperature,
I will be finding a particular curve in the pressure volume
space.
And the reason that I drew this particular way
is because I have the condition that minus dP dV better
be positive, so that if I have P as a function of V,
then I know that dP is dP by dV delta V.
And if the product dP dV is positive,
then this derivative dP by dV has to be negative.
This has to be a decreasing function.
And another way of saying that is
that the compressibility-- kappa T, which is minus 1
over V, the inverse of this dV by dP at constant T--
has to be positive.
So this is a curve where dP by dV is negative.
Fine, you say.
Nothing surprising about that.
Well, it turns out that if you cool the gas,
the shape of this isotherm can change.
It can get started.
And there is a critical isotherm at which
you see a condition such as this, where there's
a point where the curve comes with 0 slope.
So you say that for that particular critical isotherm
occurring at TC, if I were to make this expansion,
the first term in this expansion is 0.
You say, OK, let's continue and write down the second term--
1/2 d2 P by dV squared at constant P times
delta V squared.
Now the problem is that if I multiply this
by delta V-- which would have made this into delta V squared
and then this into delta V cubed--
I have a problem because delta V can
be both positive or negative.
I can go one direction or the other direction.
And a term that is a cubic can change sign--
can sometimes be positive, can sometimes be negative,
depending on the sign of that.
And that is disallowed by this condition.
So that would say that if an analytical expression
for the expansion exists, the second derivative has to be 0.
And then you would have to go to the next term-- d
cubed P dV cubed.
And then delta V-- you would normally write delta V cubed,
but then I've multiplied by delta V on the other side.
We'd have delta V to the fourth.
Delta V to the fourth you would say is definitely positive.
Then I can get things done by having to make sure
that the third derivative is negative.
And you say, OK, exactly what have you done here?
What you have done here is you've
drawn a curve that locally looks like a cubic
with the appropriate sign.
And so that's certainly true that if you
have an isotherm there the compressibility diverges,
you have a condition on second derivatives.
And you probably have used this for calculating critical points
of van der Waals and other gases.
And we will do that ourselves.
But it's actually not the right way to go about,
because when you go and look at things experimentally
with sufficient detail, you find that these curves at this point
are not analytic.
They don't admit a Taylor expansion.
So while this condition is correct,
that the shape should be something like this,
it is neither cubic nor a fifth power,
but some kind of a non-analytic curve.
AUDIENCE: Could you repeat last statement?
PROFESSOR: That statement, in order
to fully understand and appreciate,
you have to come to next term, where
we talk about phase transitions.
But the statement is that the shape
of the function in the vicinity of something like this
is delta P is proportional to delta V
to some other gamma that is neither 3 nor 5 nor an integer.
It is some number that experimentally
has been determined to be of the order of, I don't know, 4.7.
And why it is 4.7 you have to do a lot of interesting field
theory to understand.
And hopefully, you'll come next semester and I'll explain it.
AUDIENCE: So your point is that gamma is not [INAUDIBLE].
PROFESSOR: Gamma is not-- so the statement is, in order
for me to write this curve as something delta V
plus some other thing-- delta V Squared--
plus some other thing-- delta V cubed--
which we do all the time, there is an inherent assumption
that an analytical expansion is possible around this point.
Whereas if I gave you a function y that is x to the 5/3,
around x of 0, you cannot write a Taylor series for it.
OK?
Essentially, this stability condition
tells you a lot of things about signs of response functions.
In the case of mechanical cases, it
is really that if you have a spring,
you better have a force constant that is positive,
so that when you pull on it, the change in displacement
has the right sign.
It wants to be contracted.
And similarly here, you are essentially
compressing the gas.
There's a thermal analogue of that sine which is worth making
sure we know.
Let's look at the case where we have
a fixed number of particles.
And then what we would have is that our dE
is T dS plus Ji dxi.
I am free to choose any set of variables.
The interesting set of variables-- well, OK.
So then stability implies that delta T delta
S plus delta Ji delta xi with the sum over i is positive.
Now, I can take a look at this expression
and choose any set of variables to express these changes
in terms of.
So I'm going to choose for my variables temperature and xi.
You see, the way that I've written it here,
S, T, J, x appear completely equivalently.
I could have chosen to express things
in terms of S and J, T and J, S and x.
But I choose to use T and x.
All right, so what does that mean?
It means that I write my delta S in terms of dS
by dT at constant x dT, plus dS by dxi at constant T
and other xj not equal to i, delta xi.
I also need an expression for delta Ji here.
so my delta Ji would be dJi by dT
at constant x, delta T, plus dJi with respect
to xj at constant T, an appropriate set of J's,
delta xi-- delta xj.
Sum over J implicit here.
I substitute these-- delta S and delta J--
in the general form of the stability condition
to get constraints applicable to combinations
involving delta T and delta x.
So I have to multiply delta S with delta T.
I have to multiply delta Ji with delta xi here.
Yes?
AUDIENCE: So here, J is not equal to i.
And then which one is J?
[INAUDIBLE]
PROFESSOR: OK.
So let's say that I have S that is a function of-- I
have x1 and x2.
So here, I would have derivatives
dS by dx1 times delta x1.
I would have derivatives dS by dx2 times delta x2.
In both cases, T is kept fixed.
In this case, additionally, x2 is kept fixed.
In this case, additionally, x1 is kept fixed.
That's all.
OK?
So let's do that multiplication.
We multiply a delta T here and we get dS by dT at constant x,
delta T squared.
We get here a term which is delta T delta xi, multiplying
dS by dxi at constant T, and the others
that I don't bother to write.
But I note that I will get another term that
is delta xi delta t from the second term, whose coefficient
is dJi by dT at constant x.
And I group those two terms together.
And finally, we have the last term here,
which is delta xi delta xj, dJi by dxj, constant T.
And the constraint for all of these stability conditions
that I have written, I have to be
in the minimum of a potential.
Any deviation that I make better lead
to an increase in this combination, which
is kind of equivalent to being at the minimum
of some kind of potential energy.
If you do some manipulations, this
is also equivalent to being at the maximum
of some entropy function.
Now, let's look at that manipulation a little bit.
Now, I happen to know that this form will
simplify a little bit.
And that's why I chose this combination of variables.
And the simplification that will happen
is that this entity is 0.
So let me show you why that entity is 0.
And that relies on one of these Maxwell relationships
that I had.
So let's find the Maxwell relation that
is applicable to dS by dxi at constant T, which
is what I have over here.
OK, let's again start with dE being T dS plus ji dxi.
I'm at fixed number of particles.
I don't need to write the other term-- UDN.
Unfortunately here, S is not the first derivative.
So what I will do is I will look at E minus TS.
That's going to turn this into minus S dT.
Good.
Ji dxi is exactly what I want because this will tell me
something about the derivative in this fashion.
dS by dxi at constant T is equal to what?
It's taking the things in this fashion,
which is dJi by dT at constant x.
Good.
So I can see that the sum total of these two terms is 0.
And that's exactly what I have here.
So this is 0 by Maxwell relation.
OK?
OK
So having set that to 0, I have two terms.
One of them is this entity that goes with delta T squared.
And the other is this entity-- sum over i and J, delta xi,
delta xj, dJi, dxj, constant T. And this--
for any choice of delta x's and delta T's-- for any choice,
it better be positive.
So in particular, given this nice form
that I have over here, I can certainly
choose all of the delta x's to be 0, because I do things
at constant displacement.
And then what I have is that the whole thing
is proportional to delta T squared and some coefficient.
So that coefficient better be positive.
So entropy should always be an increasing function
of temperature at constant displacement.
And the consequence of that is that heat capacities measure
at constant displacement.
So this is the heat capacity-- Cx.
The general definition is you put some amount of heat
into the system at constant displacement.
So I have to specify the procedure
by which heat is applied to the system
and see how much this temperature changes.
I can in principle do this sufficiently stability
so I can measure precisely the changes in temperature
infinitesimally.
And then I can use the relationship
between dQ and reversible heat-- so dS at constant x dT.
And dS by dT at constant x is precisely the entity
that we have looked at-- to be positive.
So in the same way that a spring is stable-- if you pull on it,
its extension should increase a little bit.
And so the coefficient that relates
the change in displacement corresponding
to the change in force has to be positive for stability.
Heat capacities must have this positivity in the sense
that if you add heat to a system,
its temperature should go up.
It it went the other way, it would actually also violate
the second law of thermodynamics.
Yes?
AUDIENCE: I'm a bit confused about the J notation
that we used.
The initial equation that we start
has only-- it deals with i, but where does J come from?
PROFESSOR: This equation.
Which equation?
Up here?
AUDIENCE: So index J is just a dummy variable?
[INAUDIBLE] that dE equals T dS equals Ji dSi.
PROFESSOR: OK.
Fine.
So let's start from here.
And part of the confusion is that I'm not really consistent.
Whenever I write an index that is repeated,
there is a so-called summation convention
that you may have seen in quantum mechanics
and other courses which says that the index that
is repeated twice is summed over.
So when I write ai bi, using the summation convention, what
I mean is, sum over all possible values of i, ai, bi.
So what I meant here is that there
may be multiple ways of doing work on the system.
So there could be pressure, there could be spring,
there could be magnetic work.
And if I do mechanical work on the system,
the change in energy is the sum total of all different ways
to do mechanical works, plus the heat
that is supplied to the system.
So that's statement number one.
Now, I did something else here, which
is that when I take a derivative of a function that depends
on multiple variables-- so f depends on x1, x2, x3,
xn-- then if I look at the change in this,
any one of these variables can change
by an amount that is dxi.
And if the second variable changes,
there is a corresponding contribution,
which is df with respect to the second variable.
And I have to add all of these things together.
So if I'm again inconsistent, this
is really what I mean, that when you
do a change in a function of multiple variables,
you would write it as dxi, df by dxi.
If you use the summation convention,
I have to really sum over all possibilities.
Now, suppose I took one of these things, like df by dx3.
So it's one of the derivatives here.
And I took another derivative, or I looked at the change
in this function.
This function depends also on variables x1 through xn.
So I have to look at the sum over all changes with respect
to xj of df by dx3 delta xj.
So you can see that automatically, I
have kind of assumed some of these things
as I write things rapidly.
And if it is not clear, you have to spend a little bit of time
writing things more in detail, because it's
after all kind of simple algebra.
Yes, did I make another mistake?
Yes?
AUDIENCE: Oh, not a mistake, just--
PROFESSOR: Yes?
AUDIENCE: When you were talking about mechanicals,
you had a non-linear spring.
You start pulling on it.
At some point, it just increases.
PROFESSOR: If it is a non-linear spring, yes, that's right.
AUDIENCE: Can you provide example of thermodynamic system
which, upon some threshold action upon it,
becomes unstable?
PROFESSOR: OK.
I think I more or less had that example.
Let's try to develop it a little bit further.
So I said that typically, if I look at the pressure volume
isotherms of a gas, it looks something like this.
And I told you that there is actually
some kind of a critical isotherm that looks something like this.
And you may ask, well, what do things
look below that for lower temperatures?
And again, you probably know this,
is that this is the temperature below which
you have a transition to liquid gas coexistence.
And so the isotherms would look like that.
Now, it turns out that while these are true equilibrium
things-- you would have to wait sufficiently long
so that the system explores these isotherms--
it is also possible in certain systems
to do things a little bit more rapidly.
And then when you go below, you will
see a kind of remnant of that isotherm which kind of looks
like this.
So that isotherm would have portions of it
that don't satisfy these conditions of stability.
And the reason that you still get some remnant of it--
not quite the entire thing.
You get actually some portion here,
some portion here-- is because you do things very rapidly
and the system does not have time
to explore the entire equilibrium.
So that's one example that I can think
of where some remnants of something that
is like that non-linear noodle takes place.
But even the non-linear noodle, you cannot really see
in equilibrium.
You can only see it when you're pulling it.
And you can see that it is being extended.
There is no equilibrium that you can see that has that.
So if I want to find an analogue of that non-linear noodle
showing instability, I'd better find some near-equilibrium
condition.
And this is the near-equilibrium condition that I can think of.
There are a number of cases.
And essentially, you've probably heard hysteresis
in magnets and things like that.
And by definition, hysteresis means
that for the same set of conditions,
you see multiple different states.
So not both of them can be or multiple of them
can be in equilibrium.
And over some sufficient time scale,
you would go from one behavior to another behavior.
But if you explore the vicinity of the curves that correspond
to these hysteretic behavior, you
would see the signatures of these kinds
of thermodynamic instabilities.
There was one other part of this that maybe I
will write a little bit in more detail, given
the questions that were.
Asked It is here I looked at this general condition
for the case where the delta x's were 0.
Let's look at the same thing for the case
where delta T equals to 0, because then I
have the condition that the Ji by the xj, delta xi,
delta xj should be positive.
And this was done at constant T.
And to be more explicit, there is a sum over i and J in here.
So if I'm doing things at constant temperature,
I really have the analogue of the instability condition.
What I have here, it is the same thing as this one.
And if you think about this in a little bit more detail--
let's say I have something like a gas.
We have two ways of doing work on the system.
I have dJ1 by dx1.
I have dJ2 by dx2.
I have dJ1 by dx2.
I have dj2 by dx1.
Essentially, this object, there are four of them.
And those four I can put into something like this.
And the statement that I have written
is that this 2 by 2 matrix, if it
acts on the displacement delta x1 delta x2 on the right,
and then gets contracted on the left by the same vector delta
x1, delta x2, for any choice of the vector delta
x1 delta x2, this better be positive.
This is for two, but in general, this expression
is going to be valid for any number.
Now, what are the constraints that a matrix must satisfy
so that irrespective of the choice of this displacement,
you will get a positive result?
Mathematically, that's caused a positive definite matrix.
So one of the conditions that is certainly immediately obvious
is that if I choose the case where delta x2 is 0, then
the diagonal term that corresponds
to that entity better be positive.
So a matrix is positive definite if for every one
of the diagonal elements, I have this condition.
So this is actually something that we already saw.
In the case of the gas, this would
be something like minus dP by dV at constant temperature--
be positive.
This is the nature of these isotherms always
going one way that I was drawing.
AUDIENCE: Question.
PROFESSOR: Yes.
AUDIENCE: You're not summing over alpha now, are you?
Or are you--
PROFESSOR: Yes, I'm not.
And I don't know how to write. [LAUGHS]
Let's call it delta J1, x1.
So all diagonal elements must be positive,
but that's not enough.
So I can write a matrix that has only positive elements
along the diagonal, but off-diagonal elements such
that the combination will give you
negative values of this culmination
for appropriate choices of delta x1 and delta x2.
The thing that you really need is
that all eigenvalues must be positive.
It is not that easy to write down mathematically
in general what that means in terms of multiple response
functions.
That's why this form of writing is
much more compact and effective.
It basically says, no matter what combination of elements
you choose, you are at the bottom of the potential.
You should not be able to go away from it.
AUDIENCE: So the second statement
is necessary and sufficient condition?
PROFESSOR: This statement is necessary but not sufficient.
This statement is equivalent to the meaning
of what is a positive definite matrix.
It is sufficient.
Yes?
AUDIENCE: Is there any physical interpretation
to diagonalizing that matrix?
PROFESSOR: It depends on the kind of the formations
that you make.
Like if I was really thinking about a gas
where I allow some exchange with the surface or something
else so that delta N is not equal to 0,
the general form of this matrix in the case where
I have pressure volume would be something like this.
I would have minus dP by dV.
I would have d mu by dN.
I would have dP by dN.
I would have d mu by dV.
And I would have to make sure that this matrix is positive
definite, so I know certainly that the compressibility
by itself has to be positive.
d mu by dN also has to be positive.
But if I allow changes both in volume
and the number of particles, then because
of these off-diagonal terms, I have additional constraints.
So for appropriate changes, it is
sufficient to look at the diagonal terms.
But for the entire set of possible transformations,
it's not.
Yes?
AUDIENCE: If you look at different ways to do the work
on the system-- so in this case, change volume,
change number of particles, if you're saying forces have
different dimensions, and if diagonalization of these
metrics corresponds to inventing some new ways of acting
on the system which are a linear combination of old ways like
change volume a little and change particles--
PROFESSOR: Change the number of particles, yes.
AUDIENCE: Those are pretty weird manipulations,
in terms of how to physically explain
what is diagonalization matrix.
PROFESSOR: OK.
There is a question of dimensions,
and there's a question of amount.
So maybe what you are worried about
is that if I make one them volume and the other pressure,
one is extensive, one is intensive,
and the response functions would have
to carry something that is proportional to size.
But then there are other conditions,
let's say, where all of the variables are intensive.
And then it's moving around in the space that
is characterized, let's say, by pressure and chemical potential
from one point to another point.
And there is some generalized response that you would have,
depending on which direction you go to.
So essentially, I could certainly
characterize various thermodynamic functions
in this space P mu.
And what I know is that I cannot write any function in this
two-parameter space that would be consistent with
the thermodynamic principles of stability.
I have to have some kind of convexity.
So certainly, what typically-- let's say again,
going back to my favorite example here-- what people do
and what we will do later on is we
specify that you have some particular pressure
and volume of an interacting gas.
And there is no way that we can calculate
the thermodynamic properties of that system exactly.
We try to construct a free energy
as a function of these variables.
And that free energy will be approximate based
on some ideas.
And then we will see that that free energy that I construct
in some places will violate the conditions of stability.
And then I know that that free energy that I constructed,
let's say on the assumption that the box that I have
is uniformly covered by some material,
that assumption is violated.
And maybe what has happened is that part of the volume
is a liquid and part of the volume is gas.
So we will do precisely that exercise later on.
We will choose a pair of variables.
We will try to construct a free energy function out.
We make the assumption that the box is uniformly occupied
with the gas at some density.
Based on that, we estimate its energy, entropy, et cetera.
We construct a free energy function out.
And then we will see that that energy function violates
the compressibility condition at some point.
So we know that, oh, we made a mistake.
The assumption that it is uniformly occupying this box
is wrong.
All right, any other questions?
Yes?
AUDIENCE: One of the things that confuses me
about thermodynamics in general is
the limits of its applicability.
We think of it as being very general,
applying to many different systems.
For example, if we had two bar magnets
and we had the north pole close to the south pole
of the other one, and I thought of the magnetic field
in between, is it conceivable to think
of that as a thermodynamic system?
PROFESSOR: You can think of the magnet
as a thermodynamic system.
And you can certainly look at two of them,
including the electromagnetic field that surrounds them.
And there would be some conditions,
like if you were to jiggle these things with respect
to each other-- and you would maintain the entire thing
in a box that prevented escape of electromagnetic waves
or heat to the outside-- that eventually
this jiggling of the two magnets will lead to their coming
to equilibrium but at a higher temperature.
So this is what I'm saying.
So imagine that you have-- let's for simplicity think of, again,
one big magnet that you don't move,
and one small magnet here that you externally jiggle around.
And then you remove the source of work.
You wait for a while.
And then you find that after a while,
this thing comes to an equilibrium.
You ask, where did that kinetic energy go to?
You had some kinetic energy that you had set into this,
and the kinetic energy disappeared.
And I assume that there are conditions
that the kinetic energy could not have escaped from this box.
So the only thing that could have happened
is that that kinetic energy heated up
the air of the box, the magnet, or whatever you have.
So thermodynamics is applicable in that sense to your system.
Now, if you ask me, is it applicable to having
a single electron spin that is polarized because
of the magnetic field and some condition
that I set it to oscillate, then it's not applicable,
because then I know some quantum mechanical rules that
govern this.
There may be in some circumstances
some conservation law that says this with oscillate forever.
If it doesn't oscillate forever, there's
presumably a transition matrix that
says it will admit a particular photon of a particular energy.
And I have to ask what happened to that photon.
So what we will see later on is that thermodynamics
is very powerful, but it's not a fundamental theory
in this sense that the quantum mechanics
that governs the spins.
And the reason that this works for the case of the magnet
is because the magnet has billions of spins in it--
billions and billions of spins, or whatever.
So it is the of large numbers that ultimately
allows you to translate the rules of motion
into a well-defined theory that is
consistent with thermodynamics.
AUDIENCE: I know that my question seemed off-topic.
But the reason why I asked about this particular system
is because I would think that it doesn't
have a positive compressibility.
Like if you pull the two magnets apart very slowly,
the force pulling them together decreases
instead of increasing like with the rubber band.
PROFESSOR: OK.
It may be that you have to add additional agents in order
to make the whole system work.
So it may be that the picture that I had in mind
was that you have the magnet that
is being pushed up and down.
And I really do put a number of springs
that make sure there exists an equilibrium position.
If there is no mechanical equilibrium position,
then the story is not whether or not to thermodynamic applies.
It's a question of whether or not there is an equilibrium.
So in order for me to talk about thermodynamics
I need to know that there is an eventual equilibrium.
and that's why I actually recast the answer to your question,
eventually getting rid of the kinetic energy of this
so that I have a system that is in equilibrium.
If there are things that are not in equilibrium,
all bets are off.
Thermodynamics is the science of things that are in equilibrium.
Other questions?
All right, so there is only one minor part of our description--
rapid survey of thermodynamics-- that is left.
And that's the third law of thermodynamics
and that has to do something with the behavior of things
as you go towards the limit of the zero
of thermodynamic temperature.
And I think we will spend maybe 10, 15 minutes next lecture
talking about the third law of thermodynamics.
Probably from my perspective, that
is enough time with respect towards the time
we spend on the other laws of thermodynamics,
because as I will emphasize to you also,
the third law of thermodynamics is in some sense
less valid than the others.
It's certainly correct and it is valid.
It is just that its validity rests on other things than what
I was emphasizing right now, which
is the large number of degrees of freedom.
You could have a large number of degrees
of freedom in a world that is governed
by classic laws of physics, and the third law of thermodynamics
would be violated.
So it is a condition that we live
in a world that is governed by quantum mechanics that tells us
about the third law, as we will discuss next time.