So we know the first law of thermodynamics holds for isolated systems

$$\Delta E = W+Q\tag{1}$$

Now assuming we are in a thermally isolated system such that $(Q=0)$ we obtain

$$\Delta E= W\tag{2}$$

Like most of classical thermodynamics (which is sadly often not clearly said) we imply that the initial and final state when performing work on a system have to be in equilibrium.

But by reading I found that for the first law of thermodynamics to hold the system can be in non-equilibrium during the process. All that matters is apparently that the system returns to equilibrium for the first law of thermodynamics to hold.

But now I am studying the following paper: C. Jarzynski, "Nonequilibrium Equality for Free Energy Differences", Phys. Rev. Lett. 78 (1997) 2690, arXiv:cond-mat/9610209.

As far as I understand its goal is to find a generalization of the second law of thermodynamics far away from equilibrium.

In the paper, a system which is initially in equilibrium is brought out of equilibrium by performing a certain amount of work W on it.

This work is then called 'non-equilibrium work'.

The main relation of this paper is

$$\langle e^{-\beta W}\rangle = e^{-\beta \Delta F}\tag{3}$$

where W is the work that is performed on the system to take it out of equilibrium. $\Delta F$ on the other hand is the equilibrium difference of free energies. It can be calculated by taking the free energy of the equilibrium initial state and a theoretical equilibrium final state to which the final state (after W has been performed) would equilibrate if we would wait.

The author and people in general are happy about the fact that from eq. (3) directly follows a known relation which is just another way to state the second law of thermodynamics

$$\langle W \rangle \geq\Delta F\tag{4}$$

And as such we have found a relation that holds for systems far from equilibrium, which gives us the second law of thermodynamics.

This is the way I understand this.

Now in order to find eq. (4) I looked up the definition of Helmholtz-free energy here from which I derived

$$\Delta F = \Delta E - T\Delta S\tag{5}$$

so yes, when we have an irreversible process, where $\Delta S \geq 0$ we obtain

$$\Delta E \geq \Delta F\tag{6}$$

if we are additionally in a thermally isolated system we obtain through eq.(2).

$$W \geq \Delta F\tag{7}$$

But as I just said in order to obtain this result we used the first thermodynamic law as well.

Now the Jarzynski paper which comes to the same result (for systems far away from equilibrium, eq.(4)) never clearly mentioned whether the first law still holds in the scenario described in the paper.

Am I missing something? Does the first law hold for the scenario in the paper as well? If it does things would make more sense for me.


1 Answer 1


The first law of thermodynamics, namely that the total change in the energy of the system is the sum of the work done on it and the heat supplied to it, $\Delta E=W+Q$, is certainly obeyed in the scenario discussed by Jarzynski. It amounts to conservation of energy.

In many derivations of the Jarzynski equation, particularly those based on a time-dependent hamiltonian, it is actually possible to identify, separately, the heat term and the work term, at the microscopic level: that is, in each nonequilibrium trajectory. (Subsequently, an average over trajectories is carried out). This is explained well in Mark Tuckerman's textbook, Statistical mechanics: theory and molecular simulation, section 8.4, eqn (8.4.5), which I recommend you have a look at. If you can't immediately get hold of Tuckerman's book, you can see the same equation in his online lecture notes.

You may need to do a bit more background reading to follow the details of the derivation. The paper by Jarzynski that you are referencing is an early one in the field, dating from 1996. Since then, plenty of other papers and reviews have been published, which you can track down. But Tuckerman's book is a good starting place.

  • $\begingroup$ Great! Yes the whole implications of this equation are quite advanced but I'll have a look at the book your proposed. The only thing here that strikes me a little bit is your mention of heat. The word 'heat' is not mentioned once in the Jarzynski-paper but in contrast the work is defined as $W=\Delta H$ implying that this is the only mode of energy transfer Jarzynski is considering. Of course I assume that the Jarzynski eq. can be extended to other szenarios which also take heat into consideration but just to be clear: In the scenario from the paper we assume that $\Delta E = W$, correct? $\endgroup$
    – CatoMaths
    Commented Jan 24, 2019 at 15:24
  • $\begingroup$ Actually the word "heat" is mentioned several times. The first sentence is "Consider a finite classical system in contact with a heat reservoir" and later, before eqn (3), "the system evolves, in contact with a heat reservoir,". Then after eqn (5) "it is instructive to first consider what happens when there is no heat reservoir during the switching process". In that last case he states "the work done on an isolated Hamiltonian system is equal to the change in its energy", as you say. Shortly afterwards he reverts to "the system is coupled to a reservoir". $\endgroup$
    – user197851
    Commented Jan 24, 2019 at 17:20
  • $\begingroup$ .... For the system coupled to a reservoir he states "We assume that the system of interest and the reservoir together constitute a larger, isolated Hamiltonian system.". So there, you have conservation of total energy. Also, later: "So far, we have implicitly assumed that our system is coupled to a physical heat reservoir." In later papers, it is possible to discuss the definition of "heat" $Q$ in more detail (depending on the model), but here it is just $\Delta E-W$. $\endgroup$
    – user197851
    Commented Jan 24, 2019 at 17:25
  • $\begingroup$ Ah yes! I missed the later parts. Also you wrote $\Delta E - W$ (at the very end) which confused me here for a bit but I assume it means $\Delta E =W$. Now everything is clear, thank you $\endgroup$
    – CatoMaths
    Commented Jan 24, 2019 at 17:37
  • $\begingroup$ Just to be clear: When I wrote $\Delta E-W$ I meant that this is $Q$. In other words, $\Delta E=Q+W \, \Rightarrow \, Q=\Delta E-W$. $\endgroup$
    – user197851
    Commented Jan 24, 2019 at 17:39

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