One common way of motivating the existence of Entropy as a state function is the following. Let us take the Clausius/Kelvin-Planck statement of the second law, from which we can deduce Clausius' theorem $$\oint \frac{\delta Q}{T} \le 0,$$ where equality holds if and only if the cyclic process is reversible.

This of course means that the quantity $$\int_C \frac{\delta Q}{T} $$ is path independent for reversible paths $C$, and so it defines a function of state which we call Entropy.

But this only seems to hold on the presumption that all states within our state space is mutually accessible through reversible processes, i.e. given any two states $A$ and $B$ in our state space, there exists some reversible process $A\rightarrow B$ and some reversible process $B\rightarrow A$. I don't see why this is necessarily true. Is this taken to be an additional (and apparently implicit) assumption? Or is this assumption provable? Or is it not actually needed to define entropy this way?

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    $\begingroup$ I think you're probably right in that you need to assume that the set of all macrostates is connected by reversible paths; this seems to be implicit in the classic Carnot/Clausius argument for the universality of reversible engine efficiency/ well-definedness of $S$ as a function of state through $S=\int_c\frac{\delta\,Q}{T}$. Once you have this, the function of state still holds regardless of how you get between states: it's defined as the reversible path integral and the change in $S$ if the same change were made reversibly, regardless of whether or not the actual transition were reversible. $\endgroup$ Jun 22, 2016 at 7:42
  • $\begingroup$ The general accessibility of any state from another one is not needed in classical thermodynamics, instead either a state A is accessible from B or B is accessible from A with a reversible path is needed. I say "classical" thermodynamics because even this assumption fails in certain systems, for example hysteretic ferromagnets or plastic deformation for which no state can be reached reversibly from any other state. Bridgman has shown how to assign entropy functions to these kind of systems. $\endgroup$
    – hyportnex
    Jun 22, 2016 at 13:21

2 Answers 2


Is this taken to be an additional (and apparently implicit) assumption?

You are correct.

Take two arbitrary points $A,B$ on the $PV$ (or any other) plane, and draw an arbitrary curve connecting them: you have just defined a reversible transformation connecting $A$ and $B$.

This is because every point in the $PV$ (or any other) plane represents an equilibrium state, so every continuous set of points (such as the curve you drew) represents a reversible transformation.

The existence of reversible processes is one of the postulates of thermodynamics. Of course, a reversible process is an idealization, because it would require that the system is in equilibrium at every instant during the process, which is clearly absurd, because if the state variables are changing then clearly there is no equilibrium. This is why we talk of "quasi-static" transformations, in which an infinite number of infinitesimal steps is performed in such a way that the system is always in equilibrium.

Regarding your last question, it is actually possible to define entropy in another way in statistical mechanics. Between 1872 and 1875 Boltzmann formulated the equation

$$S=k \log(\Omega)$$

where $k$ is a constant with dimensions of $J/K$ and $\Omega$ is the number of microstates corresponding to a the macrostate of the system. This definition is in some way more fundamental than the thermodynamic one, as it gives the connection between the microscopic and the macroscopic description of Nature.

  • $\begingroup$ So, in essence, the OP's intuition that existence reversible transitions between every pair of states is an implicit assumption is correct, no? $\endgroup$ Jun 22, 2016 at 7:56
  • $\begingroup$ @WetSavannaAnimalakaRodVance Yes. If it is not clear enough, I can edit my answer to highlight this better. I also wanted to stress that it is a (useful) idealization. $\endgroup$
    – valerio
    Jun 22, 2016 at 8:11

So, you want to prove that between any arbitrary two states of a system, it exists at least one reversible path. You can prove this if you accept continuity of properties of substances. I.e. for example, if we have an ideal gas in equilibrium at initial state $(P_i,T_i)$ and final state $(P_f,T_f)$; then certainly there are infinite equilibrium states between initial and final states. enter image description here In diagram above $(P_1,T_1)$ is an equilibrium state so that $P_1=P_i+\Delta P$ and $T_1=T_i+\Delta T$.

If, $(\Delta P,\Delta T)\to (0,0)$ then $(P_1,T_1)\to(P_i,T_i)$ So, we can reach from $(P_1,T_1)$ to $(P_2,T_2)$ through infinite number of equilibrium states so that for each two adjacent states $(\Delta P,\Delta T)\to(0,0)$ and if we do this quasi-static process by insulating the system; then that process will be an isentropic process and we know that an isentropic process is a reversible process.

  • $\begingroup$ This shows the existence of a quasistatic path between any two states, which I agree exists. But quasistatic does not mean reversible. For example, consider a piston with friction. You can compress the piston arbitrarily slowly, but the process will not be reversible. $\endgroup$
    – EuYu
    Jun 22, 2016 at 4:12
  • $\begingroup$ You want to prove there is at least one reversible path between two states. So, if we have an isentropic reversible path between them, then we have at least one reversible path between them. $\endgroup$
    – lucas
    Jun 22, 2016 at 4:20
  • $\begingroup$ @EuYu "This shows the existence of a quasistatic path between any two states, which I agree exists" If you agree with that, you should accept that if we do a quasi-static process through an adiabatic path, we will have a reversible path. $\endgroup$
    – lucas
    Jun 22, 2016 at 4:29
  • $\begingroup$ There does not exist an isentropic path between any two states. By definition, isentropic paths only exist between states of constant entropy. Reversible only requires net entropy be constant, not system entropy. $\endgroup$
    – EuYu
    Jun 22, 2016 at 4:29
  • $\begingroup$ Necessary condition for $\oint \left(\frac{\delta Q}T\right)_{\textrm{sys}}\le0$ is that cycle is internally reversible. $\endgroup$
    – lucas
    Jun 22, 2016 at 4:41

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