# Physical meaning of the exterior derivative of the first law of thermodynamics

We know, $$dU = d \overline{q} - d \overline{W}.$$ suppose we took the exterior derivative on both sides, then:

$$0= d( d \overline{q}) - d( d \overline{W})$$

This means, $$d^2 \overline{q} = d^2 \overline{w} \tag{1}$$

However, do not be misled, the above expression is not equal to zero as followed from $$d^2 (\text{anything})=0$$, the quantity $$d \overline{q}$$ is an inexact differential.

What does the above equation (1) mean? How can we interpret the action of the exterior derivative onto a quantity containing both inexact and exact differentials?

• Jul 9 at 13:23

The first part of your question starts from a misunderstanding. It is better to forget about inexact differentials. What you write as $$dq$$ and $$dw$$ has nothing to do with differentials. In general, they are not even functions of the state variables, and there are no exterior derivatives one could take.

Working on $$dU$$ alone is a different story. $$U$$ is a state function depending on some thermodynamic quantities like $$S,V$$, and $$N$$. The fact that $$dU$$ is closed ($$d^2U=0$$) implies the vanishing of the coefficients of the resulting 2-form, i.e., the equality of the second mixed derivatives. This result is known in thermodynamics as the so-called Maxwell relations. One example is the following: $$-\left.\frac{\partial{P}}{\partial{S}}\right|_{V,N} = \frac{\partial^2{U}}{\partial{S}\partial{V}} = \frac{\partial^2{U}}{\partial{V}\partial{S}} = \left.\frac{\partial{T}}{\partial{V}}\right|_{S,N}.$$ Other Maxwell relations can be obtained either using the partial derivatives with respect to $$N$$, or by exploiting the closeness of other thermodynamic potentials.

• >The first part of your question starts from a misunderstanding. It is better to forget about inexact differentials. What you write as dq and dw has nothing to do with differentials. In general, they are not even functions of the state variables, and there are no exterior derivatives one could take. -- Hmm, let's consider the subset of cases where they can be parameterized by some state variables. In that case, the inexact differential, to my understanding, is basically a one form. Jul 9 at 16:19
• @Buraian The cases where $dq$ and $dw$ can be parameterized by some state variables is precisely the case where one is writing the explicit 1-form $dU$. That is the case I have explicitly addressed. Jul 9 at 17:03

If we write the two differential forms: $$\delta Q=C_vdT+ldV$$ and $$\delta W =-PdV$$ Then $$d\delta Q=\frac{\partial C_V} {\partial V} dV\land dT+\frac{\partial l}{\partial T}dT\land dV=\left(\frac{\partial C_V}{\partial V}-\frac{\partial l}{\partial T}\right)dV\land dT$$

Because $$dT \land dT = 0$$ , $$dV \land dV = 0$$ and $$dT \land dV = -dV \land dT$$

Also : $$d\delta W=-\left(\frac{\partial P}{\partial T}\right)dT\land dV=\left(\frac{\partial P}{\partial T}\right)dV\land dT$$

So, we find : $$\left(\frac{\partial C_V}{\partial V}-\frac{\partial l}{\partial T}\right)=-\frac{\partial P}{\partial T}$$

This is usually found with $$dU=C_vdT+\left(l-P\right)dV$$ and equaling the two cross derivatives. In usual practice, I have never found the differential forms to be very useful, but I am not an expert and maybe with more pratice.... !