A thermodynamic system being in thermodynamic equilibrium is characterized by the property that for every thermodynamic potential $F$ which describes the system, its differential $dF$ is zero. Let consider for example the internal energy $U(S, V, N_i)$ for now. If the system is in thermodynamic equilibrium then

$$dU= \frac{\partial U}{\partial S}dS + \frac{\partial U}{\partial V}dV + \sum_i^n \frac{\partial U}{\partial N_i}dN_i = TdS + pdV + \sum_i^n \mu_i dN_i = 0 $$

Note that $\frac{\partial U}{\partial S} =T, \frac{\partial U}{\partial V} =p, \frac{\partial U}{\partial N_i} =\mu_i$.

Question: How this condition $dU$ helps "in practice" when one works with concrete systems and want to find out in which $(S_0, V_0, (N_i)_0)$ the system has its "equilibrium"?

When I try to apply it I obtain something nonsensical and I want to understand which mistake I make here. Back to our condition $dU=0$ implies that $\frac{\partial U}{\partial S} =T, \frac{\partial U}{\partial V} =p, \frac{\partial U}{\partial N_i} =\mu_i$ should be all zero, because $S, V $ and $N_i$ are independent variables, therefore the differentials $dS, dV$ and $dN_i$ as well. So I obtain $n+2$ conditions $\frac{\partial U}{\partial S} =0, \frac{\partial U}{\partial V} =0, \frac{\partial U}{\partial N_i} =0$

But this not make any sense to me simply because this would imply that if the state is equilibrium state, then always its $T, p $ and $\mu$ are all zeroes. But certainly there are thermodynamical systems which are in equilibrium but their $T, p, \mu_i$ are not zero.

I'm confused now, what I'm doing wrong? could anybody explain to me how to "read" and "work" with the condition $dU=0$ correctly? sorry, if my question is too easy for people with elementary knowledge on this topic but also after long search I nowhere found an answer.


You have been misled by ab incomplete statement of the equilibrium condition. The precise statement is that for a system at thermodynamic equilibrium under the condition of a fixed set of thermodynamic variables, the corresponding thermodynamic potential is minimum with respect to any additional variable representing a possible internal constraint.

In practice, this means that if you have an isolated system characterized by fixed values of entropy, volume, and number of molecules, its internal energy $U(S,V,N)$ at equilibrium is minimum with respect to any other variable different from those determining the thermodynamic state. For example, if the system is in a container and a fixed, impenetrable and insulating wall is separating subsystem $1$ from subsystem $2$, this is equivalent to have two separate subsystems with energies $U_1(S_1,V_1,N_1)$ and $U_2(S_2,V_2,N_2)$. If the constraint on thermal insulation is relaxed, and heat can flow between the two subsystems varying $S_1$ and $S_2$ but without entropy production, $S_1+S_2=S$, then only one additional independent variable, say $S_1$, represents the constraint. The vanishing of the first order variation of the total energy with respect to $S_1$ $$ \frac{\partial{U(S,V,N,S_1)}}{\partial{S_1}}=0, $$ at fixed $S,V,N$, provides the condition for thermal equilibrium after removal of the constraint, i.e. the equality of the temperatures of the two subsystems.

  • $\begingroup$ Hello Giorgio, thank you for your answer. I thought a while about it and came to following question: Is that a sufficient or only a neccessary statement on thermodynamical equilibrium? Namely can this equilibrium condition be used to decide if our given system in an equilibrium state? What I'm principaly interested in is if this criterion can be applied on following thought experiment: $\endgroup$ – Eddy Wa Feb 13 at 2:36
  • $\begingroup$ Assume we have an isolated system about which we only know that it has fixed macroparameters $S, V ,N$ and we can always measure $U$ (I purposly not wrote $U(S, V,N)$ because as far as I know the internal energy can be only expressed as $U(S, V,N)$ if the system already is in td equilibrium, but that's what we are going here to check; see my comment on it below in $**$). Before we start we measure $U$. What we do now is we try to vary artificially all possible physical parameters of the system but with respect only one rule: everything what we do should not change $S, V, N$. $\endgroup$ – Eddy Wa Feb 13 at 2:37
  • $\begingroup$ That is we can try to change the system in every way we want as far as the rule is not violated. And when we do it we continuously measure $U$. Assume we are able somehow (that's a thought experiment) to can out all possible manipulations of this system which respect to rule not to change $S, V, N$. If the minimal value of $U$ coinsides with $U$ at the beginning then our system was in equilibrium state. Does this approach make sense? $\endgroup$ – Eddy Wa Feb 13 at 2:37
  • $\begingroup$ Motivation: by the procedure in the thought experiment to change the system in all possible ways as far as $S, V, N$ not change I intended to imitate the mathematical procedure to vary any variable in the system different from $S, V, N$ $\endgroup$ – Eddy Wa Feb 13 at 2:37
  • $\begingroup$ Note on $**$: As far as I understand the ussue the function $U$ (or any other td potential) exists always but the only case when it depends only on three parameters (as here $S, V, N$) is only the case when the system is in equilibrium. In case of non equilib $U$ in general should exist but is too complicated (maybe depend on infinity many parameters) to work with, right? $\endgroup$ – Eddy Wa Feb 13 at 2:39

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