One-forms in thermodynamics In the Lecture notes of David Tong on GR, the following is written on page $90$:
$$dE =\delta Q + \delta W$$

This is much more natural in the language of forms. All of the terms in the first law are one-forms. But the transfer of heat, $\delta Q$, and the work $\delta W$ are not exact one-forms and so can't be written as $d(something)$. In contrast, $dE$ is an exact one-form. That's what the $\delta$ notation is really telling us: it's the way of denoting non-exact one-forms before we had a notion of differential geometry.

This raised a couple of questions:

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*How is $dE$ a one-form while $\delta Q$ and $\delta W$ are not exact one-forms?

*What does it mean that $\delta Q$ and $\delta W$ are not exact one-forms?

Possible answer for the latter: it simply means there is no $Q$ and $W$ which can be written as $Q= d\psi \to dQ=0$ and $W= d\phi \to dW=0$. But if it so, why?
 A: The underlying assumption of thermostatics is that the internal energy is a differentiable function of mechanical (also electrical, magnetic, etc.,) parameters of the material plus "temperature' that represents one thermal variable that can be measured by some mechanical displacement, a thermometer. That is we have a function $E=E(V,m,T, x,y,z,...)$, and of course we can always write a total differential of it as $dE=\sum_k y_k dx_k$ where $x_k:\{V,m,T,x,y,z,....\}$.
If you split this sum $dE$ into two pieces say $\tilde a_1 = y_1dx_1+y_2dx_2$ and the the rest as $\tilde a_2=\sum_{k>2} y_kdx_k$ then these parts $a_1$ and $a_2$ cannot be written as total differential individually.
 As it happens from the 1st law  we do have a natural decomposition of the internal energy differential as $dE=\tilde q+ \tilde w$. The reason why that is interesting because the heat flux and work are separately measurable quantities but they depend on the manner (path) of change and thus are not the derivatives of any function of state parameters. Instead from the 2nd law we know that $\tilde \sigma= \frac{\tilde q}{T}$ $is$ a total differential, ie., there is a function $S$ of the state parameters $S=S(V,m,T,x,y,z,...)$ such that $dS = \tilde \sigma$

Note: I do not know if there is one or disprovable that there can be no another $useful$ decomposition of $dE$ such as $dE = \tilde a + \tilde b + ... $ where some of the various terms, say $\tilde a$ is $integrable$ the way $\tilde q$ is but with different functions of the parameters. May be this is worth a separate question.
A: To formalize this properly, we need to adopt the viewpoint of differential geometry.  So as to not be overly general, consider our system to be an ideal gas.  We will call the space of thermodynamic states $\mathcal M$, and (for the moment) we will choose coordinates $(S,V)$ for $\mathcal M$ where we interpret $S$ and $V$ as the entropy and volume of a state.

A state quantity is a function $f:\mathcal M\rightarrow \mathbb R$.  Examples of state quantities include the volume function, $\mathcal V: (S,V) \mapsto  V$ and the internal energy function $E:(S,V)\mapsto E(S,V)$ (which is somewhat tedious to write out).

A tangent vector to $\mathcal M$ can be thought of as a directional derivative, which roughly eats a state function $f$ and spits out the rate at which $f$ changes along some direction.  In coordinate form, a tangent vector $\vec X$ acts on $f$ as follows:
$$\vec X(f) = X^1 \frac{\partial f}{\partial S} + X^2 \frac{\partial f}{\partial V}$$
where $X^1$ and $X^2$ are called the components of $\vec X$ (in this basis). Accordingly, we can represent $\vec X$ all by itself as
$$\vec X = X^1 \frac{\partial}{\partial S} + X^2 \frac{\partial }{\partial V}$$
In this sense, the partial derivatives $\frac{\partial}{\partial S}$ and $\frac{\partial}{\partial V}$ are the basis vectors which span this space.

A one-form is an object which eats a tangent vector and spits out a number.  Given a choice of basis $\left\{\frac{\partial}{\partial S},\frac{\partial}{\partial V}\right\}$ for the tangent vectors, we can define a corresponding basis $\{\mathrm dS,\mathrm dV\}$ for the one-forms which has the property that
$$\mathrm dS\left(\frac{\partial}{\partial S}\right) = 1 \qquad \mathrm dS\left(\frac{\partial}{\partial V}\right) = 0$$
$$\mathrm dV\left(\frac{\partial}{\partial S}\right) = 0 \qquad \mathrm dV\left(\frac{\partial}{\partial V}\right) = 1$$
and so any one-form $\omega$ can be expressed in component form as
$$\omega = \omega_1 \mathrm dS + \omega_2 \mathrm dV$$


What does it mean that $\delta Q$ and $\delta W$ are not exact one-forms?

Given any state function $f$, we can define a one-form $\mathrm df$ which eats a tangent vector $\vec X$ and spits out $\mathrm df(\vec X):= \vec X(f)$.  In component notation, $\mathrm df$ takes the form
$$\mathrm df = \frac{\partial f}{\partial S} \mathrm dS + \frac{\partial f}{\partial V}\mathrm dV$$
We call $\mathrm df$ the gradient of $f$. Any one-form which is the gradient of a state function is called exact.  It's reasonable to ask whether all one-forms are exact, and the answer is no.  As an example, consider the one-form
$$\omega = V \mathrm dS$$
Comparing this with the general form for a gradient $\mathrm df$, we see that $\frac{\partial f}{\partial S}=V \implies f = SV + g(V)$ where $g$ is some function of $V$ alone.  But we must also have that $\frac{\partial f}{\partial V} = S + g'(V) = 0$, which implies that $g'(V)=-S$.  But since $g$ (and by extension, $g'$) is not a function of $S$, this cannot be satisfied.  We conclude that there is no function $f$ such that $\omega = \mathrm df$, which means that this $\omega$ is not exact.
So, the answer to this question is that $\delta Q$ and $\delta W$ are one-forms - that is, maps which eat tangent vectors and spit out numbers.  Physically, we can associate tangent vectors to infinitesimal paths in $\mathcal M$; in this sense, $\delta Q$ and $\delta W$ are objects which eat infinitesimal paths and spit out numbers.
However, they cannot be written as the gradients of some state functions $Q$ and $W$.  This means that $\delta Q$ and $\delta W$ are determined by the infinitesimal paths they act on, and not merely by the endpoints of those paths.  In thermodynamic language, knowing which two states are connected by a process is not enough to know how much heat was added (or work was done) to get from one to the other.


How is $\mathrm dE$ a one-form while $\delta Q$ and $\delta W$ are not exact one-forms?

Consider the function $f = SV$.  From the definition above, $\mathrm df = V\mathrm dS + S\mathrm dV$.  If we define $A = V\mathrm dS$ and $B = S \mathrm dV$, we observe that neither $A$ nor $B$ are exact one-forms, but their sum is.
The first law of thermodynamics can be understood as the physical statement that during an infinitesimal thermodynamic process, the change in the state function $E$ can be attributed to some small amount of heat $\delta Q$ and some small amount of work $\delta W$.  The latter two quantities are not exact one-forms because there is no "heat function" or "work function" which are well-defined for a given thermodynamic state; nevertheless, the difference $\delta Q - \delta W = \mathrm dE$ is exact.
The fact that no such functions exist can be understood by the following example.  I can take an ideal gas and adiabatically compress it to a smaller volume, thereby raising its temperature.  I have added no heat in this process.  On the other hand, I could take the same gas, isothermally compress it to the smaller volume, and then add heat to raise its temperature.  During this process I have added heat.
In both examples, I started and ended with the same thermodynamic state, but I added heat during the second process and no heat during the first.  If there were a state function $Q$, then it would increase during the second process but not during the first - but both processes end in the same state! Therefore, heat is a quantity which can be associated to a process but not to a state.
A: Being an exact one-form implies that integrals are independent of the path they are integrated along.
Such a path in the thermodynamic state space could lead from a state with $(T_1, V_1)$ (temperature and volume) to $(T_2,V_2)$. But there are in principle many such paths. There is little reason why either heat or work should only depend on the endpoints. In fact, you can think of going from one state to the other through for example either an isobaric or an isentropic process. Work and heat along either path are going to be different in general. They are not what is called a state function. Think of a vector field with non-vanishing curl.
Energy on the other hand should be a state function. This implies that its differential $dE$ is an exact form. Think of a conservative vector field. Thermodynamics axiomatizes this:

There is a state function called energy.

Although it is more common to axiomatize the existence of a state function called entropy.
The real justification comes from statistical mechanics though, namely by associating the entropy (a function!) with the number of microstates as Boltzmann did. The same goes for energy. In the statistical description, energy is obtained as an expectation value over the Boltzmann distribution. Why the Boltzmann distribution should be the correct description of thermodynamic equilibrium is another question.
Now how come $\delta Q$ and $\delta W$ "conspire" and add up to an exact differential?
It's rather the other way around. Say $E(T,V)$ is a function, then $dE = \partial_T E\, dT + \partial_V E\, dV$ by definition of the derivative. Now identify terms.
A: 

*

*How is $dE$ a one-form while $\delta Q$ and $\delta W$ are not exact one-forms?




*

*What does it mean that $\delta Q$ and $\delta W$ are not exact one-forms?


The inexact differential ($\delta$) for heat and work is meant to describe the amount of an incremental transfer of energy in the form of heat or work between a system and surroundings. A system does not "contain" heat or work.  So there is no "change" in the heat and work of a system. Unlike internal energy, heat and work are not properties of a system. The amount of heat and work done depends on the process (path) between two states.
The exact differential ($d$) used for internal energy means a differential change in the amount of internal energy contained in a system. The change in internal energy between two states is independent of the process (path) between the states. So it only has one value between states, making it an exact differential.
Hope this helps
