I am a bit confused with the description of the thermodynamic equilibrium from my course. By definition the system is in equilibrium if all state parameters are constant in time and there are no flows in the system (otherwise it will be a stationary process). But I cannot understand why thermodynamic equilibrium of a system is equivalent to existence of an equation $F(a_1,a_2,...,a_p)=0$, where $a_1,a_2,...,a_p$ are state parameters.

This was used to formulate the 0th law of thermodynamic, that if for systems A, B, C pair of them (A,B) is in equilibrium and (B,C) as well, so one can write

$F_1(a_1,a_2,...,a_p;b_1,b_2,...,b_q)=0,$ $F_2(b_1,b_2,...,b_q;c_1,c_2,...,c_r)=0$

there exists function $F_3$ such that $F_3(a_1,a_2,...,a_p;c_1,c_2,...,c_r)=0$, what means that the pair of systems (A,C) is in equilibrium.

My thinking is that one can rather easily find such functions, for example, $a_1(t)$ and $a_2(t)$ that $F(a_1(t),a_2(t),...,a_p)=0$ however the system will not be in equilibrium.


4 Answers 4


Consider the time derivative

$$ \frac{dF}{dt}(a_1,...,a_p) = \frac{\partial F}{\partial a_1}\frac{da_1}{dt} + ... + \frac{\partial F}{\partial a_p}\frac{da_p}{dt} + \frac{\partial F}{\partial t} $$

In this case, $t$ is not an explicit parameter so $\partial F/\partial t=0$.

If $da_i/dt=0$ for all $i$ (system is in equilibrium) then $dF/dt=0$ for any $F$ with $\partial F/\partial t=0$.

If $dF/dt\equiv 0$ (i.e. zero for any $\{a_i\}$) then either $\partial F/\partial a_i=0$ for all $i$ (in which case you don't have a function of all of the $a_i$), or $da_i/dt=0$ for all $i$ (i.e. system is in equilibrium).

  • $\begingroup$ But there can be also a situation in which $\frac{∂ F}{∂ a_1} \frac{d a_1}{dt}=-\frac{∂ F}{∂ a_2} \frac{d a_2}{dt}$ and $\frac{d a_i}{dt}=0$ for $i\neq 1, 2$. So considering an ideal gas (pV-nRT=0) I can vary p and V in time in such a manner that pV=const. The equation of state would hold but I would have parameters which change in time. $\endgroup$ Commented Apr 17, 2017 at 14:51
  • $\begingroup$ If the process is quasistatic, then the equation of state holds at all points of time because a quasistatic process is considered to take place through infinite steps where at each step, the system is in equilibrium. $\endgroup$
    – Yashas
    Commented Apr 17, 2017 at 15:04
  • $\begingroup$ Does it mean that existence of equation F=0 is equivalent to existence of a quasistatic process not just one state? $\endgroup$ Commented Apr 17, 2017 at 15:30

The values of the state functions can be computed when the system is in equilibrium only.

For example, consider a situation where an ideal gas isn't in equilibrium such that the pressures at different points in the container is different. How would you determine the pressure of the gas? You can't; the pressure isn't defined for the system.

The existence of a function $F$ which takes state parameters as inputs implies that the values of the state functions are well defined and can be related.

Consider the equation of state for an ideal gas:

$$PV = nRT$$

This equation of state is only applicable for a system at equilibrium.

In our previous example, the ideal gas isn't in equilibrium and the pressure of the system wasn't well defined. Therefore, the equation of state for an ideal gas is not applicable; or in other words, there isn't a value of P which can be substituted in the equation of state.

  • 1
    $\begingroup$ I think this is the best answer. Strictly speaking you could argue that the "true system" is a function of a massive number of variables and thermodynamic equilibrium is equivalent to a slightly less massive system of nonlinear equations in these variables. But this is antithetical to the philosophy of classical thermodynamics. $\endgroup$
    – Ian
    Commented Apr 17, 2017 at 19:03

I just rethought my question and found an answer.

The function $F(a_1,...,a_p)$ has its zeros at such $a_1,...,a_p$ that the system is in thermodynamic equilibrium, by definition.

So even if I can find $a_1(t)$ and $a_2(t)$ such that $F(a_1(t),a_2(t),...,a_p)=0$ it just means that $\{(a_1(t),a_2(t),...,a_p)\,|\quad t\in \mathbb{R_+}\}$ is a set of points in phase space at which system is in equilibrium. In fact this set is a path of a quasistatic process as @Yashas pointed out. So even if there is a state with inconstant parameters for which equation F=0 holds in fact it is a process composed of succeeding equilibrium states.

So the 0th law says now: If systems $(A,B)$, $(B,C)$ and $(A,C)$ are described by functions $F_1(a_1,...,a_p;b_1,...,b_q)$, $F_2(b_1,...,b_q;c_1,...,c_r)$ and $F_3(a_1,...,a_p;c_1,...,c_r)$ respectively and $\{(a_1^*,...,a_p^*;b_1^*,...,b_q^*)\}$, $\{(b_1^*,...,b_q^*;c_1^*,...,c_r^*)\}$ are sets of zeros of $F_1$ and $F_2$ respectively, then $\{(a_1^*,...,a_p^*;c_1^*,...,c_r^*)\}$ is a set of zeros of $F_3$.

I would by glad if someone could audit this answer.


Your question may be largely related to the Coarse graining method, which is "aim at simulating the behavior of complex systems using their coarse-grained (simplified) representation".

In my view, in nonequilibrium conditions, the definition of some parameters is no longer suitable in the coarse-grained model. Consider a pot of heated water with a distribution of temperature, which means in different positions the temperature varies. So how to define a temperature of the system of that pot of water?

EDIT: I try to make it more clear:

  1. "A macroscopic property of a system is, in general, obtained from the exact state after a coarse graining. Properties such as the internal energy, temperature, pressure etc. are examples of such macroscopic properties and they all have to be understood as (derived from) coarse grained variables."[1]

    In general, we prefer using coarse-grained variables in the study of a thermal system mainly for two reasons:

    • simplicity: instead of describing the microstate with 6N variables ($\vec{x_i}$,$\vec{p_i}$), one can describe its macrostate physically using the coarse-grained variables.
    • measurement: any predicting result should be compared that of a well-designed experiment, in which coarse-grained variables provide bridges over these gaps.
  2. I agree with your idea that one can find coarse-grained variables $\alpha_1(t)$ and $\alpha_2{t}$ which satisfy the equation: $$ F(\alpha_1(t), \alpha_2(t),...,\alpha_p) = 0$$ Yet it is more like an evolving equation, rather than a state equation.

    What's more, take care of these coarse-grained variables $\alpha_1(t)$ and $\alpha_2(t)$ you defined. Whether the introduction of these variables simplifies your research? Or can they be measured in our experiment nowadays, or in the future?

  • $\begingroup$ I have upvoted to cancel someone's downvote. Please elaborate your answer. $\endgroup$
    – Deep
    Commented Apr 18, 2017 at 5:23
  • $\begingroup$ I just edited. Is it more clear? $\endgroup$
    – Yang
    Commented Apr 18, 2017 at 11:56
  • $\begingroup$ Your edit contains good information, but I am not sure if it answers OP's question. $\endgroup$
    – Deep
    Commented Apr 19, 2017 at 5:13

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