When we speak of 'Equations of State' I believe we are most often referring to a steady state equilibrium relationship between particular state variables.

Am I wrong?

For example, the famous ideal gas law (aka 'state equation') $$PV=nRT$$ While I see it perfectly legitimate to take derivatives of certain variables with respect to others (for example $\frac{dP}{dV}$ ) , it's not so clear to me that I can properly introduce time into the equation where certain variables may be functions of time, and others not, and then take the derivative of some variable with respect to time.

Mathematically it can all work, but is there anything physical that makes it wrong - or constrains what can be done?

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    $\begingroup$ I don't think you can, in principle, take derivatives of those variables with respect to time, can you? Equations of state describe statistically stationary states. To include time you would need to deal with non-equilibrium thermodynamics, master equations and etc. $\endgroup$ – QuantumBrick Jun 30 '16 at 16:31
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    $\begingroup$ If you have an ideal-gas problem, and you can define two of the three variables as functions of time (assuming that "n" is constant), there shouldn't be any reason that you can't take time derivatives. $\endgroup$ – David White Jun 30 '16 at 16:36
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    $\begingroup$ You can do it mathematically, but you'll be just moving letters around. $\endgroup$ – knzhou Jun 30 '16 at 16:38
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    $\begingroup$ @DavidWhite: So what does thermodynamics have to say about time derivatives? It says "You shall not ask me about non-thermal states because I am clueless!". In terms of statistical mechanics, if you take an ideal gas with a Boltzmann distribution of velocities and you apply a time dependent force on it (e.g. by moving a wall quickly), you end up with a non-Boltzmann distribution, hence temperature is ill-defined. $\endgroup$ – CuriousOne Jun 30 '16 at 16:44
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    $\begingroup$ Yes, it is utterly wrong to do that. Take air in thermodynamic equilibrium and apply slow time dependent motion to it, what do you get? You won't get the equations of thermodynamics but the equations of fluid mechanics. If you apply much faster motion to it, things heat up, now you are dealing with fluid mechanics and thermodynamics at the same time (welcome to aerodynamics of supersonic flight). Make things faster and it becomes plasma physics (hypersonic re-entry boundary layer physics). $\endgroup$ – CuriousOne Jun 30 '16 at 16:47

For quasi-stationary processes you can take the time derivative, no problem.

Apart from the "state equation", there is a "process equation", for example, for an adiabatic expansion $PV^{\gamma}=\rm{const}$ or so. It is also quasi-stationary.

What you can loose in taking the time derivatives are quick processes like sound or shock wave propagation and their dissipation.

Consider, for example, the Newton equation for a particle under some force $F$ and with friction force $-F_{\rm{fr}}$: $$\frac{d^2x}{dt^2}=F-F_{\rm{fr}}.$$ There is some (maybe quick) dynamics to achieve a quasi-stationary motion; the latter being described with a simplified quasi-stationary equation: $$F-F_{\rm{fr}}=0.$$ As an example consider the current behavior: $j(t)=\sigma E(t)$. What you may miss is an exponentially decaying addendum to it describing acceleration of electrons from zero velocity to the stationary velocity if the electric field $E(t)$ is switched on at $t=0$.

For quasi-stationary equation to be applicable, there must some inequalities to be fulfilled, something like $\tau_1\ll T$, where $\tau_1$ is the transition time of the quick process and $T$ is the characteristic time of the force quasi-stationary change. In other words, the full dynamics is reduced to a quasi-stationary dynamics if this inequality holds. (Physics is full of inequalities!)

Generally, quasi-stationary solutions are much easier to obtain, and generally they are called "closing relationships" in simplified physical descriptions.

  • $\begingroup$ I think variable that we know as time, is not defined (physically) for a quasi-stationary process. Quasi-stationary process is an (ideal) mathematically assumption that being independent of time is inside that. $\endgroup$ – lucas Jun 30 '16 at 17:05
  • $\begingroup$ That's an ad-hoc treatment of the problem. Quantumbrick already mentioned the theoretically correct description with the master equation. $\endgroup$ – CuriousOne Jun 30 '16 at 17:14
  • $\begingroup$ @lucas: Yes, it is well defined for any processes. $\endgroup$ – Vladimir Kalitvianski Jun 30 '16 at 17:14
  • $\begingroup$ @CuriousOne: A Master Equation is reduced to the quasi-stationary equation after some transient period. Quasi-stationary solutions differ only slightly from exact solutions, as I explained in my answer. Just build a toy model and see it. $\endgroup$ – Vladimir Kalitvianski Jun 30 '16 at 17:23
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    $\begingroup$ what would be the meaning of a reversible process if time derivatives were meaningless? $\endgroup$ – Wolphram jonny Jun 30 '16 at 17:53

This question is quite profound, but I didn't think it would go that far. I'll post my views for feedback, and since I'll write some lines it is best to leave it as an answer.

As I mentioned briefly in my first comment, time derivatives won't make sense in thermodynamics. I'll try to develop this comment a little bit more.

The entities we are talking about (pressure, temperature, volume) are given as the thermodynamical limit of a partition function obtained through entropy (if the ensemble is canonical, microcanonical, etc, doesn't matter). To obtain those quantities it is a must that we ensure the ergodic hypothesis (or, for those more mathematically inclined, the central limit theorem) is applicable: we must be able to exchange time averages for space averages. This is the foundation of statistical mechanics, and it is also the first time the word time has appeared naturally. Now, if you start allowing your state variables to depend on time, then what actually depends on time is your entropy. But well, if your entropy depends on time, then what actually depends on time is your phase space volume of accessible states. This is a problem, since if you have an integral the depends on time, then it is obvious that you won't be able to exchange a time integral by a space integral: the ergodic hypothesis is out of the table and thermodynamics is no longer an option.

There are two famous situations where an analogous problem arises: open quantum systems and dissipative classical mechanics. You need Lindblad operators; you need van der Pool-like cycles. To make the story short, you need master equations. Of course there is the possibility to approximate your system as if its non-ergodicity were of perturbative nature. You can also approximate a tenth order polynomial by a straight line. The question is: what the heck do you want? The comments by CuriousOne are quite precise, I believe: there exist a very small number of problems where phase space volume changes so "adiabatically" that you can still apply the ergodic hypothesis, but in general this will lead you to very abnormal results. Generally the conclusion you'll arrive at is that either your system has such a poor behaviour that it doesn't even make sense to try and define state variables in the first place, or you'll have a very good model that describes really well a system that practically doesn't change in time. So the question remais: what is your system? What do you want?

  • $\begingroup$ Good thoughts. I especially like your last comment (or rather posed question). There is nature and there are models of nature. I believe many forget or never realize the difference. Physicists or engineers deal with models to suit our needs. What do we want? We always strive for perfection, but wind up settling for less. In the end we work with the practical. $\endgroup$ – docscience Jul 1 '16 at 14:41
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    $\begingroup$ In the end we always deal with approximations. Theorists usually forget the stuff they create have a very definite purpose: to predict nature. Many are used to working in such a high level of abstraction that they forget what physics is about. In the end, what you choose to be "perfect" really only depends on the degree of accuracy you look for: quantum mechanics is "perfect", but it won't predict different masses fall with the same acceleration. Perfection is a human definition: nature doesn't care. Existing is enough. $\endgroup$ – QuantumBrick Jul 1 '16 at 15:09

Equilibrium thermodynamics should in principle only deal with equilibrium states. In an equilibrium state, the thermodynamic variables (for example $P,V,T$) are by definition constants: this is why the time variable $t$ never appears in thermodynamics.

However, when we use an equation of state $f(P,V,T)=0$ such as the ideal gas equation of state, $PV=nRT$, we expect that, if the transformation is slow enough, the equation should hold at every time during the transformation. For example, we expect that during a transformation at constant temperature of a rarefied gas, $P(V)$ should be an hyperbola.

We can state this formally by saying that during a quasistatic process we expect the relation

$$P(t) V(t) = n R T(t)$$

to hold $\forall t$. Of course, a quasistatic process is nothing but an idealization. Nevertheless, if the transformation is slow enough, we will observe that the equation of state is approximately verified.

For example, in numerical weather prediction, a set of six primitive equations in the variables $u,v,w,P,\rho,T$ is used ($u,v,w$ are the components of the velocity field vector, $P$ is the pressure, $\rho$ is the density and $T$ the temperature).

One of this equations is the equation of state

$$P = \rho R T$$

Without this equation, the system would not be solvable because we would have 6 unknowns and only 5 equations.

So we use the state equation in a situation in which the thermodynamic variables are clearly time-dependent. We do this because it is assumed that on atmospheric scales the change of the thermodynamic variables is slow enough for the equation of state to hold at every time.


If a fluid system is experiencing a transient irreversible change, the temperature and pressure within the system will typically be non-uniform spatially. This is the result of two factors not present in (slow) reversible changes: the inertia (mass) of the fluid (which allows non-uniformities in velocity and density) and the viscous nature of the fluid (which results in forces that depend not just on the magnitude of the deformation, but also on the time rate of change of the deformation). At a given spatial location in the (rapidly) deforming fluid, the force per unit area exerted on surfaces and on adjacent fluid parcels is not just determined by the local thermodynamic pressure (based on the local temperature and density) but by viscous stresses as well. This is what @CuriousOne was alluding to when he referred to the need for applying the equations of fluid mechanics. If the temperature T and pressure P are not even uniform throughout V, you certainly can't apply the ideal gas law (which assumes that the temperature and pressure are uniform throughout V).

  • $\begingroup$ It's all bla-bla. You have to compare the quasi-stationary solution to the exact solution and if the difference is small for your purposes, then you can use the approximate solutions. $\endgroup$ – Vladimir Kalitvianski Jul 1 '16 at 11:10
  • $\begingroup$ For the problems we typically encounter in engineering, involving fluid mechanics and heat transfer, we never find it necessary to make such a comparison, and we've actually been able to design and operate both mechanical and chemical processing equipment for hundreds of years with great success. And, throughout these many years, there has been huge amounts of money riding on our ability to accurately do this. So, if you want to live in an ivory tower and throw up your hands saying that it can't be done, that's fine. But what are you going to do if you actually have to design something. $\endgroup$ – Chet Miller Jul 1 '16 at 11:32
  • $\begingroup$ You change the subject. And you put your words in my mouth ("it can't be done"). $\endgroup$ – Vladimir Kalitvianski Jul 1 '16 at 12:09
  • $\begingroup$ I thought your response of "It's all bla-bla" was kind of rude. Maybe we should get the moderators involved? $\endgroup$ – Chet Miller Jul 1 '16 at 13:49
  • $\begingroup$ I did not mean to offend you. I was speaking of necessity of comparison, of importance of inequalities. $\endgroup$ – Vladimir Kalitvianski Jul 1 '16 at 15:38

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