I'm trying to understand the form of the 'force' which drives chemical reactions, i.e. the difference in chemical potential, also sometimes called the 'affinity'.

$$\Delta \mu = - kT \ln{\frac{J_+}{J_-}} \tag{1}$$

where $J_+$ and $J_-$ are the forward and backward fluxes. At equilibrium, of course, $J_+ = J_-$ and the affinity is zero. But in a non-equilibrium steady state it is non-zero, and gives rise to an entropy production.

I've seen explanations that use the Arrhenius rate law (exponential of the difference in chemical energy) but these feel too empirical. I want an intuitive, theoretical explanation for this mathematical form.

Specifically, I'm interested in entropy production, which seems to always take the form 'flux times force':

$$\sigma = - J (kT)\ln{\frac{J_+}{J_-}} \tag{2}$$

where $J = J_+ - J_-$.

From a master equation approach (with probabilities $p_i$ and transition coefficients $q_{ij}$ from state $i$ to $j$), I can differentiate the Shannon entropy, and this should be $0$ at steady state. So, the entropy rate of change can be broken up, and is usually done into two pieces: the internal entropy production

$$\sigma = \frac{1}{2} \sum_{i,j} (p_i q_{ij} - p_j q_{ji} ) \ln{\frac{p_i q_ {ij}}{p_j q_{ji}}} \tag{3}$$

and the heat exchange

$$h = \frac{1}{2} \sum_{i,j} (p_i q_{ij} - p_j q_{ji} ) \ln{\frac{q_ {ji}}{ q_{ij}}} \tag{4}$$

which, when added, give the rate of change of entropy

$$\frac{dS}{dt} = \frac{1}{2} \sum_{i,j} (p_i q_{ij} - p_j q_{ji} ) \ln{\frac{p_i}{q_{ji}}} \tag{5}$$

For simplicity I set $kT = 1$. Of course, for equilibrium, these are both zero, and for non-equilibrium steady states, they are equal and opposite. And $p_i q_{ij}$ has the physical meaning of a flux, so the logarithm of the ratio has the same form as the chemical potential difference above.

The question is, intuitively/mathematically, what justifies this breaking up of the rate of change of the entropy, i.e. the form of the entropy production? And, furthermore, what is the justification of the form "entropy production = flux times force", which has been used heavily since at least Prigogine? Is it simply that it has units of power, and power ultimately equates to dissipation?

This would be okay at steady state, where all entropy produced is dissipated, but not in a non-equilibrium transient state, for which the change in entropy can be positive or negative in a far from equilibrium regime, as it approaches the steady state.


The difference in chemical potentials is NOT defined in terms of the forward and backward fluxes: rather it the chemical potential of each species is $$\mu_i=\partial F/\partial N_i$$ where $F$ is the (say) Helmholtz free energy and $N_i$ is the number of particles of type $i$ and the derivative is at constant volume, temperature and numbers of other particles.

As for relating the entropy change fluxes and forces, this does {\it not} depend on any assumptions about the system (your discussion is a close to ideal gas or ideal solution model). Rather we have the thermodynamic relationship $$dS=T^{-1}(dU-\sum_j X_j dx_j)$$ where $U$ is the internal energy and $x_j$ is a displacement and $X_j$ is its conjugate force. Taking $dx_j/dt=f_j$ to be the flux of associated with the displacement $x_j$, we immediately find that the entropy production is proportional to fluxes times displacements.

This result is not only a steady state result, and it is quite old.

  • $\begingroup$ this is nice. I'd think we need to use the product rule to differentiate, so there'd be another term. But this still doesn't answer why the force has the form it does (logarithm of ratio of fluxes), which is what I was mostly interested in, and why I can't mark this as "the answer" $\endgroup$ – Ethan Jun 8 '14 at 5:17

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