0

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?

Qmechanic
  • 220,844
Clemens Bartholdy
  • 8,309

2 Answers2

6

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.

GiorgioP-DoomsdayClockIsAt-89
  • 39,465
  • 9
  • 53
  • 119
1

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.... !

Vincent Fraticelli
  • 4,960