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\author{Gabriel C. Drummond-Cole}
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\title{Math Physics\\January 29, 2006}
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All right.  So what did we do?  Let me remind you briefly what we've seen so far.  I gave a brief introduction to thermodynamics.  Today I'll be more precise.  I introduced ``heat'' and said, despite being vague, that it's a one-form $Q$ on $M,$ the macroscopic state space.  Now though, because I started it last week, I want to focus on $Q.$

In particular I brought up the second law of thermodynamics, and there was the theorem of Carath\'eodory:\\Let $\alpha$ be a nowhere zero $1$-form such that for each $p\in M$ there exist points arbitrarily close to $p$ which cannot be joined to $p$ by a null curve of $\alpha.$

If $\alpha$ is such a one-form, then locally there exist functions $f$ and $g$ such that $\alpha=fdg.$

We started proving this last time. So far we considered particular cases.  There is the case $\alpha_{\text{odd}}=dz+\sum^k x^i dy^i, k\ge 1$ and $\sum^k \alpha_{\text{even}}=x^idy^i, k\ge 2.$

Then $\alpha$ did not satisfy the inaccessibility condition.  So let me point some things out about these particular one-forms.  Notice that $\alpha_{\text{odd}}$ satisfies $(\alpha_{\text{odd}}\wedge (d\alpha_{\text{odd}})^k,$ if the coordinates are nowhere zero in the region, is nonzero in the region.  But $(d\alpha_{\text{odd}})^{k+1}=0$ identically.  Simiarly, $(d\alpha_{\text{even}})^k\ne 0$ but $\alpha_{\text{even}}\wedge (d\alpha_{\text{even}})^k=0$ identically.  You call these rank $2k+1$ and rank $2k.$

Now, when you have one-forms of certain ranks, you have this theorem, I'll quote it and the proof can be in the homework.
\begin{thm}Darboux $(2k+1)$\\
If $\alpha$ is a rank $2k+1$ one-form, then there exist local coordinates $z, x^i,y^i$ such that $\alpha$ locally looks like $dz+x^i dy^i.$  
\end{thm}
There is a corresponding situation for $2k.$
\begin{thm}Darboux $(2k)$\\
If $\alpha$ is a rank $2k$ one-form, then there exist local coordinates $x^i,y^i$ such that $\alpha$ locally looks like $x^i dy^i.$  
\end{thm}
You prove something similar for $2$-forms and then show it implies this for $1$-forms. 

Now, for Caratheodory's theorem, we can use the inaccessibility condition, along with Darboux's theorem, to show that $\alpha\wedge d\alpha=0.$  Just for now, I'll assume that my macroscopic state space is odd dimensional, say $2n+1.$
Step one is that either $\alpha\wedge (d\alpha)^n$ is or is not zero everywhere.  If it's not zero then there exists $p\in M$ such that this form is nowhere zero in a neighborhood.  Then by Darboux, it is rank $2n+1$ on that neighborhood, therefore it locally looks like something accessible.

So what about a lower rank?  Now we know that $\alpha\wedge (d\alpha)^n =0.$  Then either $(d\alpha)^n=0$ or it does.  If it is nonzero, then it is accessible.  We can iterate through this until we get to $\alpha\wedge d\alpha=0.$  You guys have taken some topology, you have seen the Frobenius theorem:
\begin{thm}
Let $\Ii$ be a differential ideal (an ideal closed under the differential) locally generated by $\dim\ M - p$ one-forms. (we'll be interested in this being 1.)  Then for $x\in M$ there exists a unique connected maximal integral submanifold of $\Ii$ (meaning restricting the form to the manifold gives zero of dimension $p.$
\end{thm}
A submanifold is integral for $\Ii$ when $i^*\omega=0$ for all $\omega\in \Ii.$

If $\alpha$ is nowhere vanishing then it generates an ideal.  Then $d\alpha\in \Ii(\alpha)$ if and only if $\alpha\wedge d\alpha=0.$

Thus for any $x\in M$ we can take the unique connected maximal submanifold $N$ in $M.$  In particular there exist local coordinates $g, x^1,\ldots, x^{\dim\ M-1}$ such that $i^*g=0,\ N=g^{-1}(0).$   So $\alpha$ along $N$ give you null directions, so that with respect to these coordinates $\alpha=fdg$ for some function $f.$  In particular $f$ needs to be nowhere vanishing.

That was a lot of math.  Let's get back to physics.  I'm going to return to a description of the macroscopic state space $M.$  The systems we're considering are large particle number systems so what we do, as $n$ goes off to $10^24$ is we make certain assumptions, based on experience or empirical evidence.  One stupid assumption is
\begin{enumerate}\item We can isolate a portion of the universe from the rest of the universe.  That portion is what we study, and we call that a system $\$$  The system $\$$ has ``states.'' In particular, in thermodynamics, it no longer makes sense to model this as particles.  Instead what's often done is, systems are modeled not by manifolds of large finite dimension but by infinite dimensional manifolds.  Let me give you an example.  A gas occupying a region $X$ of the space we live in might be modeled by a time dependent density $\rho:X\times \rR_{t_1,t_2}\to \rR^{>0}.$  The state space is the space of such $\rho$ functions.  From now on I'll denote a space by $\rho.$

So our isolated systems, sometimes we unisolate them in a controlled way and they interact with other isolated systems or the universe in general.  So we make certain assumptions about the interaction.
\item There is a distinguished finite dimensional submanifold $M$ of states called equilibrium states.  If you have an isolated system without outside interference.  Eventually, given time, these are states that are stable, fixed points or equilibria sets within which one can move freely.
\item  One of the things we actually study, we focus on certain types of interactions, and in particular one of the most important is called \begin{enumerate}\item Two systems 
can have diathermal contact.  This means there is no macroscopic motion nor exchange of material.  I could take a hot solid box.  I could touch it to a cold solid box.  This will be all heat energy and no work energy.  We'll pause here for a while and talk about diathermal contact.  I have two systems $\$_1,\$_2$ in diathermal contact, it's natural to consider their sum as a third system $\$_3.$  If $M_i$ is the submanifold of equilibria for the states $\$_i$ the $M_3\subset M_1\times M_2.$  You bring them together and then, if the two systems aren't in contact, it would be exactly the cartesian product.

So if $\rho_i\in M_i$ are such that $(\rho_1,\rho_2)\in M^3$ then we say that $\rho_i$ are in thermal equilibrium.

So this is the 0th law of thermodynamics.  This isn't really a law, it's one of our assumptions.  The notion of thermal equilibrium as I've described it is an equivalence relation.

Now we have the equivalence classes $A_n$ of systems in thermal equilibrium.  These are called an abstract temperature.

\begin{defi} A thermometer is a fixed system with an injective $\rR$-function $\theta$ on equivalence classes of states for the system.  It's a function on the systems which is injective on equivalence classes.
\end{defi}
$\theta$ is called an empirical temperature.

\begin{itemize}\item Part of the zeroth law is the assumption, the existence of a thermometer and smoothness of $\theta,$ such that $d\theta$ is nowhere zero.  \item If $\theta'$ is any other such empirical temperature, then $\theta'=F\circ \theta$ for some smooth monotonic $F.$
\item In practice, the equilibrium submanifold $M$ usually has physically relevant functions $x^1,\ldots x^n,$ configurational variables.  For instance, the volume of gas, total mass, total charge, pressure.  So together with $\theta,\ x^1,\ldots, x^n,$ these are local coordinates on $M.$  Usually, because we're usually working with gasses, the volume $V$ is included as $x^1,$ and so we're using $\theta, V, \ldots$
\end{enumerate}
\end{enumerate}

We were talking about certain sorts of interactions for thermal systems.  We were talking about diathermal contact.  We also have
\begin{itemize}
\item[3b.] Adiabatic interaction.  I've talked about this a little bit.  Adiabatic interactions are where the equilibrium is not disturbed unless configurational variables are changed.

We have our equilibrium.  In the big picture these should be the end points of some flow.  We change the state a little bit, and what happens is you flow to possibly a different equilibrium state.  Such a process is adiabatic if moving in temperature direction.  In a diathermal process, volume doesn't change or something like that, and in an adiabatic process, those curves are the null curves of $Q.$  Diathermal is all heat, adiabatic is all work.

Now, more definitions.  If you have a piston, and you don't allow the heat to be exchanged at all, that's an adiabatic process.  This doesn't allow heat to flow in or out.  If for some period of time a system $\$$ only experiences adiabatic interactions, then we say $\$$ is in an adiabatic enclosure.  As far as heat goes it's completely enclosed.
\item [3c.] Reversible curves are paths $\gamma:\rR\to \text{States}$ (almost) if $\gamma(t)$ is (sufficiently close to) an equilibrium state.  You want to be close enough to reasonably project onto the equilibrium states.
\begin{itemize}
\item Work is the change of internal energy.  There is a function $U:\text{states}\to \rR_+^{\text{units}}$ called energy.  Then work is the change of internal energy along adiabatic variables.  If a path is reversible then we can write $W=\nu_1dx^1+\ldots+\nu_n dx^n$ for functions $\nu_i:M\to \rR$ with $\nu_i$ called a potential.

For example, often one of the variables is $x^1=V$ the volume.  In this case the potential is $\nu_1=P,$ pressure.  $dV=Adx$ and $Fdx=W$ so $P=F/A$ is the natural choice.

\item Remember that the way we defined heat was the unaccounted change in energy for general interaction, possibly not adiabatic, is heat.  So $dU=W+Q.$  So we can define adiabatic with respect to $Q$ that along an adiabatic curve $\gamma,$ we have that $\gamma^* Q=0.$
\end{itemize}\end{itemize}}
If I take a bottle of fluid and shake it so it gets to equilibrium, I have done work an it will never come back again, this will be the second law again.  If I have an adiabatic enclosure I can do work on it but I can't get the work back along any reversible adiabatic curve.  Near any equilibrium state $\rho,$ there exist arbitrarily close equilibrium states that cannot be joined to $\rho$ by a reversible adiabatic curve.

Anyway, given this statement of the second law, $Q$ is locally $\lambda d\phi.$  There are no preferred functions for the $\lambda$ and $\phi,$ but there are physical reasons to pick preferred functions.  Let's call it a day.

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