# Thermodynamic inaccessibility and its consequences

Caratheodory's principle, namely, that in any neighborhood of an arbitrary equilibrium state there are states not accessible via adiabatic reversible path, leads to the existence of an integrating factor $$T$$ for the $$\delta Q= dU-\delta W$$, that is $$dS=\frac{\delta Q}{T}=\frac{dU-\delta W}{T}\tag{1}\label{1}$$ is an exact differential, so that $$dU=TdS+\delta W \tag{2}\label{2},$$ where $$\delta W$$ is the reversible work.

Knowing that $$\eqref{2}$$ holds, the adiabatic inaccessibility just means that in a process with fixed $$S$$, that is $$dS=0$$, and $$dU=\delta W$$, an adiabatic process, one cannot move from one $$S$$ value to another; well, of course, we have just assumed that $$S$$ is unchanging.

We can turn this $$\eqref{2}$$ around by considering the Helmholtz free energy, $$F=U-TS$$, $$dF=dU-TdS-SdT$$, or $$dF=-SdT+\delta W \tag{3}\label{3}.$$ Now using a similar verbiage to that of Caratheodory, we can say that in an arbitrary neighborhood of any state there are states that cannot be reached via an isothermal process, that is one for which $$T$$ is a constant, $$dT=0$$, and in general there is an integrating factor, called entropy, for the difference of free energy and work, such that $$dT=-\frac{dF-\delta W}{S}\tag{4}\label{4}$$ is an exact differential.

Both $$\eqref{1}$$ and $$\eqref{4}$$ have an immediate geometric interpretation, namely that the thermodynamic, or better said thermostatic space of equilibrium states form a pair of families of "surfaces" covering the full space, the first family being the isentropic surfaces, the other the isothermal surfaces. No two surfaces in the same family have common points, in other words one can assign to each surface within the family a unique and monotonically increasing set of numbers.

One may visualize this as an "egg-crate" for a two-form as in Misner-Thorne-Wheeler

Looking at it this way one may surmise that any state can be reached from any state both via a pair of isothermal-isentropic process or via an isentropic-isothermal process.

My question: how could this be proven directly from either Kelvin's or Clausius's principle?

Note added to clarify the above. Here is a quote from Landsberg that actually gave me idea to ask my question.

THIS communication gives a simple deduction of Caratheodory's principle from Kelvin's principle that it is impossible to convert an amount of heat completely into work in a cyclic process without at the same time producing other changes. There has always been some difficulty in motivating Caratheodory's principle convincingly. The present argument demonstrates for the first time that this principle can be obtained directly and simply from the impossibility of a perpetuum mobile of the second kind, and some reappraisal of the relationship between these two methods of developing thermodynamics appears, therefore, to be required.

Equilibrium states are all represented in a thermodynamic phase space. Suppose this space contains a point $$\mathcal A$$ which possesses a neighbourhood $$\mathcal N$$, all points of which can be reached adiabatically from $$\mathcal A$$. Let the co-ordinates be the internal energy $$U$$, and deformation co-ordinates (volumes, magnetic fields, etc.) $$v_1,v_2,_....$$ Keeping $$v_1,v_2,...$$ fixed, take the system from a point $$\mathcal B$$, $$(U_B in $$\mathcal N$$ to the point $$\mathcal A$$. With the $$v_i$$ constant, no work is done. The increase in internal energy is due to heat $$Q > 0$$ having been supplied. Because $$\mathcal B$$ is in $$\mathcal N$$, the system can be returned from $$\mathcal A$$ to $$\mathcal B$$ by an adiabatic process, using appropriate changes of the deformation coordinates. The drop in internal energy is now entirely due to mechanical work $$W$$ which has been done by the system. Conservation of energy during the cycle $$\mathcal {B \to A \to B}$$ demands that $$W = Q$$, so that heat $$Q$$ has been completely converted into work, contrary to Kelvin's principle. It follows that points such as $$\mathcal A$$ cannot exist. Hence, in every neighbourhood of every point $$\mathcal C$$ in thermodynamic phase space there are points adiabatically inaccessible from $$\mathcal C$$. The same argument can be used if, instead of Kelvin's principle, one assumes the impossibility of a perpetuum mobile of the second kind.

The foregoing argument shows that the conventional and the geometrical methods of developing thermodynamics, which have been regarded as very different, are in fact intimately related. It seems desirable to use the best of each in an exposition of thermodynamics, namely, to start with the motivation of the conventional approach and then to switch by the foregoing argument to the more precise concepts of the geometrical method.

Landsberg: "A Deduction of Caratheodory's Principle from Kelvin's Principle", Nature, February 1, 1964, pp485-486.

• So basically you’re asking how to link any two states with a Carnot cycle?
– LPZ
Jul 30, 2023 at 8:56
• @LPZ not how to link with a Carnot cycle but rather for the proof that there is always a Carnot cycle between any two states for which the states are the "corners" of the cycle. What I wrote above proves that there always is such a cycle but it needs the proof of the simultaneous existence of both entropy and free energy surfaces. Instead I would like to know how to get the existence of such a cycle directly from saying, eg., "one cannot extract positive work from any two-stage cycle that connects two states so that one stage is isothermal and the other is an adiabatic one (Kelvin)." Jul 30, 2023 at 14:19
• We can go from "Kelvin" $\Rightarrow$ "Caratheodory" $\Rightarrow$ isentropic surfaces $\Rightarrow$ free energy surfaces. This seems to be a rather indirect way and I feel that both Kelvin's postulate and this general accessibility (isothermal + isentropic or isentropic + isothermal) are equally more primitive than what Caratheodory does with it. Jul 30, 2023 at 14:31

I rephrase the question as I understand it from the clarifications in the comments:

Prove that the statement

"S1: one cannot extract positive work from any two-stage cycle that connects two states so that one stage is isothermal and the other is an adiabatic one (Kelvin)"

is equivalent to

"S2: there is always a Carnot cycle between any two states for which the states are the "corners" of the cycle".

S1 is equivalent to the existence of entropy, $$S$$, a state function whose properties and relationship to other thermodynamic properties are established as usual (most textbooks follow this path to the definition of entropy). Among these relationships we have $$T = 1\Big/\left(\frac{\partial S}{\partial U}\right)_{V,N}$$ and $$S = \int\frac{dQ_\text{rev}}{T}$$ The first equation implies that $$S(U,V,N)$$ is differentiable, i.e., there is a $$T$$ for each $$(U,V,N)$$. The second equation implies that $$S$$ is continuous along an isothermal path. $$T$$ is also continuous, as required by thermal equilibrium.

Suppose two states, $$(S_1,T_1,N)$$ and $$(S_2,T_2,N)$$ cannot be connected via an isentropic/isothermal (or isothermal/isentropic) path. Since there is always a neighborhood around each state where $$S$$ and $$T$$ both vary continuously with $$(U,V,N)$$, the fact that the two states cannot be connected means that $$S$$ and $$T$$ can vary discontinuously. But this would mean that $$S$$ is discontinuous and non differentiable for some $$(U,N,V)$$, which contradicts the properties of entropy.

In this answer the Kelvin statement is only used insofar as it defines entropy. The accessibility of any two states o the $$TS$$ graph has to do with the geometrical properties of entropy: any "holes" in the $$TS$$ graph would mean that $$S$$ is discontinuous function of $$(U,V,N)$$. This assumes that $$(U,V,N)$$ can be varied continuously, which at constant $$N$$ and in the absence of quantum effects is true.

• It seems to me that if you start with $S$ be differentiable a phase transition involving a process from state 1 to state 2 is excluded but even then Kelvin's principle must hold along with a Carnot cycle. Anyhow, maybe it is impossible, but I am hoping for a more direct proof for $K \to C$ that does not involve a side tour that first would prove the existence of entropy, something similar the way Landsberg proves adiabatic inaccessibility from Kelvin's principle. Is that possible? Aug 4, 2023 at 19:55
• $S$ is differentiable at the phase boundary: the temperature of both phases are the same, we can't have phase equilibrium otherwise (think of $S$ touching the tie line smoothly). I suspected you are looking for a more direct proof of $K\to C$, hence the disclaimer in my last paragraph, but the way you phrased the question leaves room for proofs that are not direct. Aug 4, 2023 at 20:04
• Ah, yes, of course, my bad. But regarding the question do you think an argument a la Landsberg is hopeless? I mean, just using the concept of empirical temperature without which nothing can happen anyhow, it seems to be obvious that isothermal surfaces must exist, after all, one cannot move from one temp to another and having the same temp all the way, hence isothermal inaccessibility. (I don't need this but then Caratheodory's math would also give us the integrability of $dF-\delta W$ and the existence of $S$) But something is still missing. Does my thinking make any sense to you? Aug 4, 2023 at 20:22
• My reading of Landberg/Caratheodory is that the condition “reversible adiabatic”, I.e., “isentropic” defines a line, not area. Similarly with “isothermal”. If we can show that we cannot have isentropic and isothermal simultaneously, then it seems to me that the two families of lines must transverse the thermodynamic plane. This argument is however too qualitative and does not sound convincing enough. Aug 4, 2023 at 22:40
• I agree with your interpretation of Landsberg that he talks about isentropic "lines" not "surfaces" (those will come "after" Landsberg from Caratheodory), but as I said above I think the existence of isothermal surfaces are even more primitive and can come from the 0th law and empirical temperature ($\theta=f(v_1,v_2,...)$, and then with your phrasing all we need is to show that any temperature (or such defined isothermal surface) can be reached isentropically from any state of a different temperature, and that would just need to be a "line". Aug 5, 2023 at 8:24