Rule for manipulating differential 2-forms in a physics derivation

Question

This question comes from a more physics perspective, particularly thermodynamics and includes the usage of differential forms.

We assume the existence of smooth functions with continuous derivatives up to the second order, for which the associated differential forms are exact.

$$\{ f \in C^2(\mathbb{R}^n) \mid d(df) = 0 \}$$

So I was looking at a fast way of deriving the Maxwell relations from the Fundamental relation with using as least physics possible and depending on just the fundamental rules of differential forms.

We start by the Fundamental relation: $$dU=TdS-pdV$$

and use the exterior derivative operator so

$$d(dU) = dT \wedge dS - dp \wedge dV$$

since one of the fundamental properties of the exterior derivative is that

$$d(df) = 0$$

we obtain:

$$dT \wedge dS=dp\wedge dV \tag{1}.$$

Now we can "divide" by $dp$ and $dS$ to get:

$$\frac {\partial T}{\partial p}= \frac {\partial V}{\partial S}$$

and also since differential forms don't commute we get

$$dT \wedge dS = -dV \wedge dp$$

and now if we "divide" by $dV$ and $dS$ we get

$$\frac {\partial T}{\partial V}= -\frac {\partial p}{\partial S}.$$

These are the two Maxwell relations I wanted to obtain.

My question is why can't we derive:

$$\frac {\partial T}{\partial V}= \frac {\partial p}{\partial S}$$

from the equation $(1)$?

I know that it is wrong, because immidiately we get a contradiction, i.e.:

$$\frac {\partial p}{\partial S}=-\frac {\partial p}{\partial S}$$

(also it doesn't make physical sense) but what is the rule when manipulating differential 2-forms in the situation like this

$$dx \wedge dy=dz \wedge dw$$ when wanting to obtain partial derivatives relations I showed above?

It seems, that we have to "divide" the most left differential form on the LHS with the most left one on the RHS and the most right on the RHS with the most right one on the LHS and all other combinations are not allowed, but that is not rigorous.

I recognize that using the term ‘divide’ in the context of differential forms is not mathematically rigorous, as it implies an operation that isn’t defined for wedge products. I understand that the asymmetry in the partial derivatives has to be a consequence of the antisymmetry property of differential forms, but I am unable to see the rule behind this.

Could someone provide a more formal rule for handling equations involving wedge products of differential forms to avoid the non-rigorous explanation I’ve used above?

Answer 1

Firstly, you can’t ‘divide’ differential forms, as you have rightly observed. The correct way to obtain the identities is to:

• write out what the condition $d^2f=0$ entails in your situation, using basic definitions/properties of exterior derivatives along the way.
• fix a coordinate system $x=(x^1,\dots, x^n)$. Note then that the collection $\{dx^{i_1}\wedge\dots\wedge dx^{i_k}\,:1\leq i_1<\dots <i_k\leq n\}$ forms a basis for all $k$-forms (on the domain $\Omega$ where these things are defined). In particular, $\{dx^1,\dots, dx^n\}$ is a basis for 1-forms, $\{dx^i\wedge dx^j\,:\,1\leq i<j\leq n\}$ is a basis for $2$-forms, etc.
• Then, for each differential form appearing in your equation, express them as a linear combination of the basis $k$-forms. For example, for a function $f$, and a coordinate system $x=(x^1,\dots, x^n)$, we have $df=\sum_{i=1}^n\frac{\partial f}{\partial x^i}\,dx^i$ which expresses $df$ as a linear combination of the basis 1-forms $dx^i$.
• Comparing the coefficients of the basis forms gives the desired equality of ‘scalar equations’.

Anyway, most of this is just (exterior/multi)linear algebra, the only place where differential forms really come in is in the definition of $df$ and $d^2f=0$.

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Let’s carry this out in your case.

• $dU=T\,dS-P\,dV$ and the fact $d^2U=0$ imply that $0=dT\wedge dS-dP\wedge dV$.
• Next, the coordinate system we shall use is $(S,V)$.
• We now have to express all our differential 1-forms as linear combinations of $dS$ and $dV$. So, we plug in $dT=\frac{\partial T}{\partial S}\,dS+\frac{\partial T}{\partial V}\,dV$ and $dP=\frac{\partial P}{\partial S}\,dS+\frac{\partial P}{\partial V}\,dV $ into the equation from the first bullet point to get (using the alternating property of wedge products) \begin{align} 0&=\left(\frac{\partial T}{\partial S}\,dS+\frac{\partial T}{\partial V}\,dV\right)\wedge dS-\left(\frac{\partial P}{\partial S}\,dS+\frac{\partial P}{\partial V}\,dV\right)\wedge dV\\ &=0+\frac{\partial T}{\partial V}\,dV\wedge dS-\frac{\partial P}{\partial S}\,dS\wedge dV-0\\ &=-\left(\frac{\partial T}{\partial V}+\frac{\partial P}{\partial S}\right)\,dS\wedge dV \end{align}
• Since we have a 2-dimensional coordinate system $(S,V)$, it follows that $\{dS\wedge dV\}$ forms a basis for 2-forms, so the above being equal to $0$ implies that the coefficient of $dS\wedge dV$ itself must be $0$, and thus $\frac{\partial T}{\partial V}=-\frac{\partial P}{\partial S}$.

The last bullet point is crucial in order to go from an equation about differential forms to a ‘scalar equation’ about partial derivatives.

Lastly, I emphasize that I am working throughout in the $(S,V)$ coordinates so $\frac{\partial T}{\partial V}$ means the derivative of $T$ with respect to $V$ while keeping $S$ constant, etc (btw, just to beat a dead horse regarding the importance of specifying the coordinate system: see this answer of mine).

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You can play a similar game with the other thermodynamic potentials. For example, when dealing with the Helmholtz free energy $F=U-TS$, you get equation $dF=-S\,dT-P\,dV$, which suggests using $(T,V)$ as a local coordinate system. Then, writing out the condition $0=-d^2F=dS\,dT+dP\wedge dV=\left(-\frac{\partial S}{\partial V}+\frac{\partial P}{\partial T}\right)\,dT\wedge dV$ then implies $\frac{\partial S}{\partial V}=\frac{\partial P}{\partial T}$ (again, relative to the coordinate system $(T,V)$), and so on with the remaining two potentials $H,G$.

Answer 2

Maybe you know this already and are just looking for an alternative, but...

One of the Maxwell relations you require comes from the definition${}^*$ of the differential of $U(S,V)$ (from standard multivariable calculus): $$ dU = T\,dS - p \, dV = \frac{\partial U(S, V)}{\partial S} \, dS + \frac{\partial U(S, V)}{\partial V} \, dV\,. $$ So, taking the partial derivative of $T$ with respect to $V$ and $p$ with respect to $S$ yields two equivalent ways to write the same thing: $$ \frac{\partial T(S, V)}{\partial V} = \frac{\partial^2 U(S, V)}{\partial V \partial S} = \frac{\partial^2 U(S, V)}{\partial S \partial V} = - \frac{\partial p(S, V)}{\partial S} $$ And the second can be found from the differential of the enthalpy, $H = U + p V$, \begin{align} dH &= dU + p dV + V dp = T dS + V dp \\ & \rightarrow \quad H = H(S, p) \quad \text{and} \quad dH = T(S,p) \, dS + V(S,p) \, dp \end{align} and the same type of manipulation using the definition of the differential of $H(S,p)$.

${}^*$ - For a definition of the differential of a function, see, e.g., Stewart's Calculus (chapter on Partial Derivatives):