11
$\begingroup$

Detailed balance is an important property of many classes of physical systems. It can be written as $$ \frac{p_{i \to j}}{p_{j \to i}} = e^{\frac{\Delta G}{k_B T}},\tag{1} $$ where $i$ and $j$ represent microscopic states of the entire system; $p$ represents the probability for the system to transition from one state to another in a particular finite time period; and $\Delta G$ represents the difference in free energy between the two states. (Whether to use the Gibbs or Helmholtz free energy or some other potential depends on the ensemble.)

Many systems obey detailed balance, but not all do. The Earth cannot fluctuate backwards in its orbit around the Sun, because this would violate conservation of angular momentum. An RLC circuit does not obey detailed balance because the fluctuations have a distinctive ringing time. For these systems the correct formula is $$ \frac{p_{i \to j}}{p_{j' \to i'}} = e^{\frac{\Delta G}{k_B T}},\tag{2} $$ where $i'$ and $j'$ represent states identical to $i$ and $j$, except that all velocities and magnetic fields have been reversed. (In quantum mechanics, they represent something like the complex conjugates of states $i$ and $j$.)

Both of these formulae guarantee that the system will obey the second law (on average), but $(1)$ is substantially stronger, because it guarantees that not only will the system tend toward equilibrium in the thermodynamic limit, but that it will not oscillate as it approaches the equilibrium. (It can still oscillate far away from equilibrium, however.)

My question is about chemical kinetics. Here we universally assume equation $(1)$ and not $(2)$. This puts strong constraints on the reaction rates, and leads to the well-known result that near-equilibrium oscillations are impossible in chemical systems. I've recently been discussing this topic with a very experienced researcher in nonlinear dynamics, and I found myself unable to convince him that $(1)$ rather than $(2)$ is a good assumption in the case of chemistry.

So I thought I'd ask here and see if anyone can help me out: what is the argument that leads us to assume the 'strong' form of detailed balance in chemical systems, rather than the weaker form in equation $(2)$?

$\endgroup$
  • $\begingroup$ I think what you are looking for is loosely related to the answer I wrote at https://physics.stackexchange.com/a/340874/59023, specifically the 3rd principle to which I elude therein. Imposing constraint (1), I think, implies that the temporal ensemble average exists which is another way of saying the system is ergodic. I think that is the direction you are going, but I wasn't sure so thus the comment instead of answer. $\endgroup$ – honeste_vivere Jul 29 '17 at 18:59
1
$\begingroup$

For physicists, the term "detailed balance" is used in one of the following contexts:

1) A closed system A is observed at a coarse-grained level (e.g. A is a vessel with a macroscopic amount of chemicals and one only tracks/observes the global concentrations of the chemicals).

2) An observable system A (e.g. a mixture of chemicals) is weakly in contact with an external "bath" B. The latter is in thermodynamic equilibrium.

Microscopically then, the contents of A resp. A+B must obey deterministic, time-reversible equations of motion. Careful: when we hit the rewind-button for the motion of a particle in time, its instantaneous position $x$ is not changed, but its velocity $v$ of course flips to $-v$. The former variables are called "odd" and the latter are "even" under time-exchange. Now, when the global system ($A$ resp. $A+B$) is in thermal equilibrium, every microscopic trajectory has the same likelihood as the time-reversed trajectory. So microscopically the probability to see a certain transition-event from a micro-state $\alpha$ to $\beta$ is equal to the probability to see a transition from $\beta$ to $\alpha$. For our macroscopic observations (concerning $P_{i \to j}$, it remains simply to count the number $W(i)$ of micro-states $\alpha$ that correspond to a certain macrostate $i$. The $\log$ of this microstate-multiplicity of the macrostate $i$ is precisely the entropy/ free energy (depending on the context) $G(i)$. Also there's the multiplication of the log by $k_B T$. We then obtain $$1=\frac{W(i)P_{i \to j}}{W(j')P_{j' \to i'}}=\frac{W(i)P_{i \to j}}{W(j)P_{j' \to i'}}.$$ So $$\frac{P_{i \to j}}{P_{j' \to i'}}=\frac{W(j)}{W(i)}=e^{\frac{G(j)-G(i)}{k_BT}}.$$ To proceed finally to your question why in the context of chemical kinetics, the time-reversed state $i'$ of $i$ is simply $i$: You presumably keep track of the macroscopic concentrations of your chemicals and those data are the content of your states {i}. Now, since the microscopic positions of particles are even under time-exchange, also concentrations of those particles in a fixed region of space are even variables. So indeed, upon applying time-reversal, an array of concentrations $i$ remains the same array of concentrations $i$.

$\endgroup$
  • $\begingroup$ I understand all this very well, and what you say in your last paragraph is true: in chemical kinetics we do indeed take the concentrations of the chemical species (and only that) as the macroscopic state. The question, though, is what makes that a good macroscopic state to use in the case of chemistry. As an example to see what I mean, imagining modelling the Earth's orbit around the sun using only its angular position as the macroscopic state, ignoring its velocity. Then the macroscopic state would be even under time reversal, and we would conclude that, in this model, ... $\endgroup$ – Nathaniel Nov 17 '17 at 2:36
  • $\begingroup$ ...the Earth obeys my equation (1) rather than (2). The problem is that this would be a terrible model, since the motion of the Earth in fact does strongly depend on its velocity, which is odd under time reversal, and by removing it from our macrostate we lost a lot of predictive power. Now from the long experience of chemists, we can ignore everything but the concentrations and still get good predictions. So it seems that velocities and other things that are odd under time reversal are not so relevant for predicting changes in concentrations. The question is, why? $\endgroup$ – Nathaniel Nov 17 '17 at 4:23
  • $\begingroup$ Indeed, in the pursuit of choosing "good" macroscopic observables, one often desires the motion of the macroscopic system to obey the Markov-property. That property is broken if you track the earth's position but forget about its velocity. However, in your example of the earth there's the more serious problem that the earth's center-of-mass motion is not in thermal equilibrium (at least on a human time-scale) with respect to any bath or internally. $\endgroup$ – Thibaut Demaerel Nov 17 '17 at 9:49
  • $\begingroup$ E.g. for my above argument to work for the earth, the probability to see it move clockwise vs. anti-clockwise around the sun should be equal on the time-scale of interest. That prerequisite is clearly not satisfied within one human life. $\endgroup$ – Thibaut Demaerel Nov 17 '17 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.