Proof of Caratheodory's theorem

Caratheodory's formulation of second law of thermodynamics, also referred to as Caratheodory's principle states

In any neighbourhood of any thermodynamic state $$P$$ there exist states which are adiabatically inaccessible from $$P$$,

where adiabatic accessibility means that states can be connected by paths satisfying $$DQ = 0$$. Caratheodory's theorem states then that this is a sufficient condition for an integrating factor to exist, so that

$$DQ = \tau \,\text{d}\sigma$$

or in other words that $$DQ=0$$ describes a foliation (planar arrangement) of states' space.

I am looking for an intuitive proof/argument for this implication (Caratheodory's principle $$\Rightarrow$$ foliation). In the literature, I was only able to find rather technical proofs, not helping with intuition, or arguments that seem wrong (but it may very well be that I am missing something). For example, Chandrasekhar in his Introduction to the Study of Stellar Structure writes that in a given neighbourhood of $$P$$ points, adiabatically accessible from $$P$$ must form a hyperplane, since they cannot form anything of smaller dimension as $$DQ$$ is already an infinitesimal hyperplane element, nor can they form anything of higher dimension as it would contradict Cartheodory's principle. How do we however exclude the possibility of those points forming two surfaces tangential at $$P$$?

• Sep 21, 2022 at 17:04

Write the conservation of energy in differential form as $$\delta Q = dU - \delta W$$ and $$\delta W = \sum_k y_kdx_k$$, where $$U$$ is the internal energy and $$x_k$$ are the extensive parameters forming the configuration space of the system. Let the starting state of a reversible adiabatic process be denoted by $$\mathcal A^0: \{U^{0}, x_k^0\}$$ and during the process the system is moved from $$\mathcal A^0$$ to another state $$\mathcal A^1: \{U^{1}, x_k^1\}$$ after absorbing $$W=U^1-U^0$$ work.
This state $$\mathcal A^1$$ is the only state whose "configuration coordinates" are $$\{x_k^1\}$$ that the system can reach reversibly and adiabatically from $$\mathcal A^0$$. This we can see from Kelvin's principle and energy conservation because if there was another such state, say, $$\mathcal A^2: U^2, \{x_k^2\}$$, such that $$\{x_k^1\}=\{x_k^2\}$$ adiabatically accessible from $$\mathcal A^0$$ then the energy difference $$\Delta U= U^2-U^1$$ between the two states must be supplied via heat transfer $$\Delta Q = \Delta U$$ since the configuration coordinates are the same, and thus the work done must also be the same in the two processes $$0 \to 1$$ and $$0 \to 2$$. But according to Kelvin this would be impossible since the path $$2 \to 0 \to 1$$ would then convert the heat absorbed in $$1 \to 2$$ to an equivalent amount of adiabatic work.
Now you can see that for any other configuration point again there can be only one adiabatic path from $$\mathcal A^0$$, consequently (warning: heavy duty handwaving, but you have asked for intuition) the locus of these accessible points has the same dimension as that of the configuration space $$\{x_k\}$$, hence it is a hypersurface in the thermostatic space of $$U,\{x_k\}$$.