# How do we know that heat is a differential form?

In thermodynamics, the first law can be written in differential form as $$dU = \delta Q - \delta W$$ Here, $dU$ is the differential $1$-form of the internal energy but $\delta Q$ and $\delta W$ are inexact differentials, which is emphasized with the replacement of $d$ with $\delta$.

My question is why we regard heat (or work) as differential forms? Suppose our system can be described with state variables $(x_1,\ \cdots ,\ x_n)$. To my understanding, a general $1$-form is written as $$df = f_1\ dx_1 + f_2\ dx_2 + \cdots + f_n\ dx_n$$ In particular, differential forms are linear functionals of our state variables. Is there any good reason to presuppose that $\delta Q$ is linear in the state variables?

In other words, if the infinitesimal heat transferred when moving from state $\mathbf{x}$ to $\mathbf{x} + d\mathbf{x}_1$ is $\delta Q_1$ and from $\mathbf{x}$ to $\mathbf{x} + d\mathbf{x}_2$ is $\delta Q_2$, is there any physical reason why the heat transferred from $\mathbf{x}$ to $\mathbf{x} + d\mathbf{x}_1 + d\mathbf{x}_2$ should be $\delta Q_1 + \delta Q_2$?

I apologize if the question is a bit unclear, I am still trying to formulate some of these ideas in my head.

P.S. I know that for quasi-static processes, we have $\delta Q = T\ dS$ (and $\delta W = p\ dV$) so have shown $\delta Q$ is a differential form in this case. I guess my question is about non-quasi-static processes in general.

We want to show that "infinitesimal" changes in heat along a given path in thermodynamic state space can be modeled via a differential 1-form conventionally called $\delta Q$.

The strategy.

1. We introduce a certain kind mathematical object called a cochain.

2. We argue that in thermodynamics, heat can naturally be modeled by a cochain.

3. We note a mathematical theorem which says that to every sufficiently well-behaved cochain, there corresponds exactly one differential form, and in fact that the cochain is given by integration of that differential form.

4. We argue that the differential form from step 3 is precisely what we usually call $\delta Q$ and has the interpretation of modeling "infinitesimal" changes in heat.

Some math.

In order to introduce cochains which we will argue should model heat, we need to introduce some other objects, namely singular cubes and chains. I know there is a lot of formalism in what follows, but bear with me because I think that understanding this stuff pays off in the end.

Cubes and chains.

Let the state space of the thermodynamic system be $\mathbb R^n$ for some positive integer $n$. A singular $k$-cube in $\mathbb R^n$ is a continuous function $c:[0,1]^k\to \mathbb R^n$. In particular, a singular 1-cube is simply a continuous curve segment in $\mathbb R^n$. Let $S_k$ be the set of all singular $k$-cubes in $\mathbb R^n$, and let $C_k$ denote the set of all functions $f:S_k\to\mathbb Z$ such that $f(c) = 0$ for all but finitely many $c\in S_k$. Each such function is called a $k$-chain.

The set of chains is a module.

It turns ut that the set of $k$-chains can be made into a vector space in the following simple way. For each $f,g\in C_k$, we define their sum $f+g$ as $(f+g)(c) = f(c) + g(c)$, and for each $a\in \mathbb Z$, we define the scalar multiple $af$ as $(af)(c) = af(c)$. I'll leave it to you to show that $f+g$ and $af$ are $k$-chains if $f$ and $g$ are. These operations turn the set $C_k$ into a module over the ring of integers $\mathbb Z$, the module of $k$-chains!

Ok so what the heck is the meaning of these chains? Well, if for each singular $k$-cube $c\in S_k$ we abuse notation a bit and let it also denote a corresponding $k$-chain $f$ defined by $f(c) = 1$ and $f(c') = 0$ for all $c'\neq c$, then one can show that every singular $k$-chain can be written as a finite linear combination of singular $k$-cubes: \begin{align} a_1c_1 + a_2c_2 + \cdots + a_Nc_N \end{align} For $k=1$, namely if we consider 1-chains, then it is relatively easy to visualize what these guys are. Recall that each singular 1-cube $c_i$ in the chain is just a curve segment. We can think of each scalar multiple $a_i$ of a given cube $c_i$ in the chain as an assignment of some number, a sort of signed magnitude, to that cube in the chain. We then think of adding the different cubes in the chain as gluing the different cubes (segments) of the chain together. We are left with an object that is just a piecewise-continuous curve in $\mathbb R^n$ such that each curve segment that makes up the curve is assigned a signed magnitude.

Now here's where we get to the cool stuff. Recall that the set $C_k$ of all $k$-chains is a module. It follows that we can consider the set of all linear functionals $F:C_k\to \mathbb R$, namely the dual module of $C_k$. This dual module is often denoted $C^k$. Every element of $C^k$ is then called a $k$-cochain (the "co" here being reminiscent of "covector" which is usually used synonymously with the term "dual vector). In summary, cochains are linear functionals on the module of chains.

Heat as a $1$-cochain.

I'd now like to argue that heat can naturally be thought of as a $1$-cochain, namely a linear functional on $1$-chains? We do so in steps.

1. For each piecewise-continuous path $c$ (aka a $1$-chain) in thermodynamic state space, there is a certain amount of heat that is transferred to a system when it undergoes a quasistatic process along that path. Mathematically, then, it makes sense to model heat as a functional $Q:C_k\to\mathbb R$ that associates a real number to each path that physically represents how much heat is transferred to the system when it moves along the path.

2. If $c_1+c_2$ is a $1$-chain with two segments, then the heat transferred to the system as it travels along this chain should be the sum of the heat transfers as it travels along $c_1$ and $c_2$ individually; \begin{align} Q[c_1+c_2] = Q[c_1] + Q[c_2] \end{align} In other words, the heat functional $Q$ should be additive.

3. If we reverse the orientation of a chain, which physically corresponds to traveling along a path in state space in the reverse direction, then the heat transferred to the system along this reversed path should have the opposite sign; \begin{align} Q[-c] = -Q[c] \end{align}

4. If we combine steps 2 and 3, we find that $Q$ is a linear functional on chains; it is a cochain! To see why this is so, let a chain $a_1c_1 + a_2c_2$ be given. Since $a_1$ and $a_2$ are integers, we can rewrite this chain as \begin{align} a_1c_1 + a_2c_2 = \mathrm{sgn}(a_1) \underbrace{(c_1 + \cdots + c_1)}_{\text{$|a_1|$ terms}} +\mathrm{sgn}(a_2) \underbrace{(c_2 + \cdots + c_2)}_{\text{$|a_2|$ terms}} \end{align} and we can therefore compute: \begin{align} Q[a_1c_1 + a_2c_2] &= Q[\mathrm{sgn}(a_1) (c_1 + \cdots + c_1) +\mathrm{sgn}(a_2) (c_2 + \cdots + c_2)] \\ &= \mathrm{sgn}(a_1)|a_1|Q[c_1] + \mathrm{sgn}(a_2)|a_2|Q[c_2] \\ &= a_1Q[c_1] + a_2 Q[c_2] \end{align} In summary, by thinking about heat as a functional on paths, and by imposing physically reasonable constraints on that functional, we have argued that heat is a $1$-cochain.

From cochains to differential forms.

Now that we have argued that heat can be thought of as a $1$-cochain, let's show how this leads to modeling "infinitesimal" changes in heat with a differential $1$-form.

This is where things get really mathematically interesting. We first recall the definition of a differential $k$-form over a $k$-chain. If $\omega$ is a $k$-form, and $c = a_1c_1 + \cdots a_Nc_N$ is a $k$-chain, then we define the integral of $\omega$ over $c$ as follows: \begin{align} \int_c\omega = a_1\int_{c_1} \omega + \cdots + a_N\int_{c_N}\omega. \end{align} In other words, we integrate $\omega$ over each $k$-cube $c_i$ in the chain multiplied by the appropriate signed magnitude $a_i$ associated with that cube, and then we add up all of the results to get the integral over the chain as a whole. For example, if $k=1$ then we have an integral of a $1$-form over a $1$-chain which is usually just called a line integral.

Now notice that given any $k$-form $\omega$, there exists a corresponding cochain, which we'll call $F_\omega$, defined by \begin{align} F_\omega[c] = \int_c\omega \end{align} for any $k$-chain $c$. In other words, integration of a form over a chain can simply be thought of as applying a particular linear functional to that chain.

But here's the really cool thing. The construction we just exhibited shows that to every differential form, there corresponds a cochain $F_\omega$ given by integration of $\omega$. A natural question then arises: is there a mapping that goes the other way? Namely, if $F$ is a given $k$-cochain, is there a corresponding $k$-form $\omega_F$ such that $F$ can simply be written as integration over $\omega$? The answer is yes! (provided we make suitable technical assumptions). In fact, there is a mathematical theorem which basically says that

Given a sufficiently smooth $k$-cochain $F$, there is a unique differential form $\omega_F$ such that \begin{align} F[c] = \int_c\omega_F \end{align} for all suitably non-pathological chains $c$.

If we apply this result to the heat $1$-cochain $Q$, then we find that there exists a unique corresponding differential $1$-form $\omega_Q$ such that for any reasonable chain $c$, we have \begin{align} Q[c] = \int_c \omega_Q \end{align} This is precisely what we want. If we identify $\omega_Q$ as $\delta Q$, then we have shown that

The heat transferred to a system that moves along a given path ($1$-chain) in thermodynamic state space is given by the integral of a differential $1$-form $\delta Q$ along the path.

This is a precise formulation of the statement that $\delta Q$ is a one-form that represents "infinitesimal" heat transfers.

Note. This is a new, totally revamped version of the answer that actually answers the OP's question instead of just reformulating it mathematically. Most of the earlier comments pertain to older versions.

Acknowledgement.

I did not figure this all out on my own. In the original form of the answer, I reformulated the question in a mathematical form, and I essentially posted this mathematical question on math.SE:

https://math.stackexchange.com/questions/658214/when-can-a-functional-be-written-as-the-integral-of-a-1-form

That question was answered by user studiosus who found that the theorem on cochains and forms to which I refer was proven by Hassler Whitney roughly 60 years ago in his good Geometric Integration Theory. In attempting to understand the theorem, and especially the concept of cochains, I found the paper "Isomorphisms of Differential Forms and Cochains" written by Jenny Harrison to be very illuminating. In particular, her discussion of theorem on forms and chains to which I refer above is nice.

Differential forms are not linear functions of state variables in any sense. What you probably wanted to say is that the differential form $df$ is a linear function of $dx_i$ but $dx_i$ isn't a state variable; it is a differential (infinitesimally small change) of the state variable.

One may view your notation for the differential form involving the $dx_i$ factors to be just one possible notation among many. Alternatively, we may consider a differential form to be nothing else than the collection of functions that you call $(f_1,f_2,\dots ,f_n)$. Each component $f_i$ is a function of the state variables. By definition, the space of possible configurations of these functions forms a linear space – the space of linear forms (the same thing as differential 1-forms), the dual space to the tangent space at a given point of the phase space. The particular values of all the components at a given point in the configuration/phase space are, by definition, forming a linear space so there is a well-defined addition.

The same addition is relevant when adding heat or internal energy $U$ that is composed of several terms. For the internal energy, your rule that $U(x+dx_A+dx_B) = U(x)+ dU_A+dU_B$ follows from the linear rule of the "derivative of the sum". The derivative of the sum is simply equal to the sum of derivatives. This rule requires that all three terms in that equation exists but in physics, we want to deal with continuous or smooth enough functions for which the derivatives exist.

I erased the last paragraph suggesting the possible nonlinearity for $\delta Q$'s because I decided it was misleading. $\delta Q$ is still a well-defined, unique linear differential form on the state space, albeit not an exact one, so it still obeys the linearity condition. It's just not terribly useful for particular processes that only probe the evolution along a particular curve rather than the evolution in all possible directions from each point of a curve.

• Thank you for the correction. I had meant to say that differentials are linear at each point in the state space, i.e. linear on the tangent space. I have always heard $\delta Q$ being called an inexact differential. That seems to imply that $$\delta Q(d\mathbf{x}_1 + d\mathbf{x}_2) = \delta Q(d\mathbf{x}_1) + \delta Q(d\mathbf{x}_2)$$ at some fixed point, just by the definition of "differential form". If I've understood you correctly, you're saying that this doesn't necessarily hold? Is it inappropriate to call $\delta Q$ a differential form then? – EuYu Jan 31 '14 at 8:20
• No, what you wrote here holds perfectly. It assumes that there is a particular (inexact but) differential form $\delta Q$ defined on the state space and the identity holds by linearity - the differential forms are, by definition, linear functions of the tangent vector coordinates $dx_i$. – Luboš Motl Jan 31 '14 at 8:29
• What I disagreed with was your physical interpretation of this identity: "... if the infinitesimal heat transferred when moving from state $x$ to $x+dx_1$ is $\delta Q_1$ and from $x$ to $x+dx_2$ is $\delta Q_2$". In that sentence, you are comparing two (well, three) different processes. Each of them is linked with some differential form $\delta Q$, like $\delta Q_1,\delta Q_2,\delta Q$. There's no reason why there should be a sum rule for these three completely unrelated differential forms. $Q$ is not a function of the state variables so the linearity of derivative isn't a valid proof. – Luboš Motl Jan 31 '14 at 8:32
• The analogous sentence for $dU$ would hold exactly because $dU$ is an exact differential, so you may talk about "the" change of the internal energy while going from $x$ to $x+dx_1$, and so on. But in the case of $\delta Q$, there is no "the" change you get when moving from one point of the phase space to another. How much heat gets transferred to the system depends on what the external objects are doing. – Luboš Motl Jan 31 '14 at 8:34
• I apologize for the confusion. I think my question got a bit side tracked somewhere along the way. I had always assumed that there exists some universal $\delta Q$. Each process is simply some curve $\gamma$ in the state space which links some initial state $\mathbf{x}_1$ and some final state $\mathbf{x}_2$ and the net heat transfer $Q$ is the integral $\int_\gamma \delta Q$. That is to say, while the net heat $Q$ depends on the process, the $\delta Q$ differential itself does not. You seem to be saying that each process is linked with a separate differential form. – EuYu Jan 31 '14 at 8:57

I) OP specifically asks about non-quasi-static processes, but then the thermodynamical state variables, such as, temperature $T$, pressure $p$, etc, might be ill-defined. And then for each instant $t$, it might not be possible to assign the system to a single point $x$ in phase space. See also comments by Lubos Motl.

So let us from now on only consider quasi-static processes. Furthermore, let us also only consider systems where the pertinent functions are differentiable. (Differentiability is of course not God-given.)

II) As pointed out by Christoph in his answer, heat $Q$ and work $W$ are not state functions. But are they well-defined objects $Q[\gamma]$ and $W[\gamma]$ in path space (of the phase space)? Here $\gamma$ denotes a path/curve. The opposite path $-\gamma$ is realizable for a reversible process.

For work $W[\gamma]$, the answer is Yes (assuming pressure $p$ and volume $V$ are part of phase space). And we have $W[-\gamma]=-W[\gamma]$.

For heat $Q[\gamma]$, the answer is Yes (assuming temperature $T$ and entropy $S$ are part of phase space).

Needless to say that if our phase space is missing some state variables, it is possible to render the function $Q[\gamma]$ ill-defined (as a function of $\gamma$) simply because the system may depend on external circumstances from the environment not recorded in our description of the system.

III) So to summarize, there are at least three reasons why one might not be able to write an infinitesimal heat transfer $\delta Q$ as a first order variation$^1$

$$\tag{*} \delta Q ~=~ \sum_i f_i(x)~\delta x^i$$

for some function $f_i(x)$ where $x$ denotes a point in phase space, and $\delta x$ denotes an infinitesimal change. Here it is implicitly understood that $\delta Q$ should be linear in $\delta x$ but not necessarily in $x$. The three reasons are:

1. The process could be non-quasi-static.

2. The relevant functions could be not differentiable.

3. Our phase space description lacks some state variables.

--

$^1$ We use the word first order variation rather than differential form not to get possibly sidetracked by (what we perceive as inessential) mathematical semantics.

• Thank you for your answer. Just a small question of mine. I had always assumed (and been told) that the only restriction on $\delta Q = T\ dS$ was that the process needs to be quasi-static, regardless of whether its reversible or irreversible. Can you give me an example of an irreversible quasi-static process for which $\delta Q = T\ dS$ does not hold? – EuYu Feb 1 '14 at 23:15
• the BH hysteresis loop of a ferromagnet is dissipative and irreversible at any rate so $\delta Q < TdS$ – hyportnex Feb 2 '14 at 16:01
• – Qmechanic Feb 2 '14 at 16:21
• A 4th reason [for breakdown of (*)] is possible Hysteresis. – Qmechanic Feb 3 '14 at 12:26

The point is that neither heat $Q$ nor work $W$ are functions (linear or otherwise) of the state variables, whereas thermodynamic potentials like $U$ are.

If your systems evolves from the state $p_1$ to $p_2$, you'll know the difference in internal energy $\Delta U = U(p_2)-U(p_1)$ just from that, but you won't know how that difference splits into $Q, W$ as these quantities depend on the path $\gamma$ your system took.

But that's exactly what a differential 1-form is: A thing you can integrate along a path.

Symbolically: $$\Delta U(p_1, p_2) = \int_\gamma dU = \int_\gamma \delta Q - \int_\gamma \delta W = Q[\gamma] - W[\gamma]$$

Now to the second, rather important part of your question that I originally missed: What about processes that aren't quasi-static?

Then, the intensive variables are generally not well-defined globally as the system did not have time to relax towards a single value. Even though we can still split $\Delta U = Q - W$ -- after all, $U$ is a thermodynamic potential and thus well defined at the initial and final equilibrium states, and $Q$ and $W$ are in principle measurable any time during the whole process -- there's no path $\gamma$ to integrate over and the whole formalism breaks down.

In fact, even if the processes is quasi-static and we do have a well-defined path $\gamma$, just integrating our 1-form $\delta Q$ should give the wrong result if the process is not reversible. Note that I deliberately chose the word should, as I'd have to go look at the literature or start thinking about pistons with friction or some other messy real-world nonsense before making a final judgement ;)

• I agree that this is the right intuition, but I think a stronger argument needs to be made. I'd be curious to know if you have any comments on the formulation given in my answer as I'm stuck in attempting to answer a mathematical question that I can't answer. – joshphysics Jan 31 '14 at 8:42
• I have little experience with thermodynamics of non-quasi-static processes, but it seems from your addendum that the differential formulation of the first law $dU = \delta Q - \delta W$ is practically limited to quasi-static thermodynamics. This has been a deeper question than I had anticipated. I appreciate everyone's help with this. – EuYu Jan 31 '14 at 11:10
• I'm wondering if the claim that no path could give the correct answer is actully true in general - it seemed obvious at first, but I think we all know how that can go ;) – Christoph Jan 31 '14 at 11:26
• Correct me if I am wrong but this question seems vacuous to me. A non-quasi-static process is out-of-equilibrium. So most of the time, the temperature is ill-defined. The temperature is one of the state variables $x$, so the position in the $x$-space is ill-defined as well and we can't talk about the right functions and forms at all. – Luboš Motl Jan 31 '14 at 11:34
• Well, I think you can't. You may return from one point to the same one so that $\Delta U=0$. You either do nothing which is reversible; or you may get there by lots of irreversible processes that just happen to cancel both $\Delta Q$ and $\Delta W$ as well, can't you? – Luboš Motl Jan 31 '14 at 12:44