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Let $A$ be a thermodynamic system and let $\Sigma$ be its state space. By the state postulate, if $A$ has $n$ two-way work channels (i.e. ways in which work can be done both positively and negatively by the system), then $\Sigma$ is $d = n+1$ dimensional.

Now starting from a given point $P\in\Sigma$, quasi-static adiabatic/work-only processes can get you anywhere in a certain subset of $\Sigma$ but not all of $\Sigma$. If we allow both quasi-static and non-quasi-static adiabatic/work-only processes, my understanding is that we can get anywhere in a larger subset $K_{P}$ of $\Sigma$, but still not all of $\Sigma$ by Caratheodory's formulation of second law of thermodynamics (Caratheodory's principle).

I am wondering, if we include both the set of states $P'$ for which $P\rightarrow_{\text{work-only}} P'$ is possible and the set of states $P'$ for which $P'\rightarrow_{\text{work-only}} P$ is possible, do we obtain all of $\Sigma$? If so, is there a proof of this? If not, is there a good counterexample?

To put it another way, let $K_{P}$ be the set of all states accessible starting from $P$ by quasi-static or non-quasi-static adiabatic/work-only processes. Let $L_{P}$ be the set of all states from which you can access $P$ by quasi-static or non-quasi-static adiabatic/work-only processes. Do we have $\Sigma = K_{P}\cup L_{P}$? I'd be interested to know how one can think about this question with some level of rigor.

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    $\begingroup$ Are you aware that a quasi-static and not quasi-static adiabatic process cannot connect the same two equilibrium states? $\endgroup$
    – Bob D
    Commented Apr 15 at 21:56
  • $\begingroup$ @BobD Honestly no, so if there is a proof it would be preferrable for it to include a proof of this claim. However, if this is too much to ask, then the details don't have to be spelled out; I can do my own research if directed the right path. $\endgroup$
    – Maximal Ideal
    Commented Apr 15 at 22:00
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    $\begingroup$ Maximal Ideal, of course, you "know" what @Bob_D is talking because a reversible adiabatic process conserves entropy, that is isentropic, while an irreversible adiabatic process increases entropy at every stage of the process while irreversible, so the two starting from the same equilibrium state, say $\mathcal P_0$, can never end in the same equilibrium state, say $\mathcal P_1$. Unfortunately, this does not help you answering the question you have posed. $\endgroup$
    – hyportnex
    Commented Apr 15 at 22:36
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    $\begingroup$ The proof is as follows: An non quasi-static adiabatic process between two states is an irreversible process. Any irreversible process between two states generates entropy. A quasi-static adiabatic process not involving friction is a reversible process. There is no change in entropy for a reversible adiabatic process between two states. Since entropy is a state function the change in entropy between two states is the same regardless of the process. Ergo a reversible and irreversible adiabatic process cannot connect the same two equilibrium states. $\endgroup$
    – Bob D
    Commented Apr 15 at 22:37
  • $\begingroup$ [...] I actually do not believe (note the verb "believe" is not the same as "know") that it is true that any two of its states of an arbitrary thermodynamic system can be connected with some adiabatic irreversible process. Instead, whether it is true or not is an assumption underlying classical thermodynamic thinking. When it may fail, as some thermodynamicist might argue, are cases that even the concept of "state" may not be classically definable. For example, what would be the appropriate definition of the thermodynamic state of a ferromagnetic material exhibiting hysteresis? $\endgroup$
    – hyportnex
    Commented Apr 15 at 22:45

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