I've come across the Onsager reciprocal principle. It's almost clear, except for thermodynamic conjugate quantities - what's that, physical meaning (except the formal definitions: $X_i = -\frac{1}{k}\frac{\partial S}{\partial x_i}$, which isn't clear) and why:

\begin{equation} \langle X_i\cdot x_k\rangle = \delta_{ik} \end{equation}

The Wikipedia is lack for references in this article.


2 Answers 2


The fundamental quantity in thermodynamics is entropy, which is a function of $n$-variables $S=S(x_1, x_2,...,x_n)$. For instance, for a simple mono-component system $S=S(U,V,N)$ where $U$ is internal energy, $V$ is volume, and $N$ composition.

Taking the differential

$$\mathrm{d}S = \sum_i \left( \frac{\partial S}{\partial x_i}\right)_{j \neq i} \mathrm{d}x_i = \sum_i F_i \mathrm{d}x_i$$

The quantities $F_i \equiv ({\partial S}/{\partial x_i})_{j \neq i} $ are intensive entropic parameters and measure the change in entropy when variables change. For instance the intensive entropic parameter $(1/T)$ gives the change on entropy due to a change in the energy $U$.

Using the thermodynamic theory of fluctuations it can be shown that

$$F_i = k \left( \frac{\partial \ln P}{\partial x_i}\right)$$

where $P$ is the probability of a fluctuation in the variables near an equilibrium state.

Using the definition of average

$$\langle A \rangle = \int A P \mathrm{d}x_1 \mathrm{d}x_2 \cdots \mathrm{d}x_n$$

the demonstration of the central result of linear nonequilibrium thermodynamics

$$\langle F_i \cdot x_j \rangle = -k \delta_{ij} $$

is direct although it needs first the use of $(\partial \ln P / \partial x_i)P = \partial P / \partial x_i$ in the integrand and next integration by parts.

What Wikipedia makes is to rewrite this central result using the new quantities $X_i \equiv - F_i / k$ but the physically important quantities are the $F_i$ often also called in this context thermodynamic forces or affinities.


Thermodynamic conjugate variables are pairs of variables $x_i$, $X_i$ where the product $X_i\cdot dx_i$ has the dimension of energy and actually appears as a term in the infinitesimal variation of energy, free energy, or work such as $dE$.

In the pair, one quantity is intensive and the other is extensive.

The best example is pressure and volume because $p\,dV$ is a term in $dE$. Analogously, temperature and entropy because of $T\,dS$, chemical potential and particle number because of $\mu \,dN$, and many electric, magnetic, gravitational, and surface tension examples exist.

If you have $$ dE = \dots + p\,dV, $$ it's easy to see that $p$ is the partial derivative of $E$ with respect to $V$. This role of $p,V$ may be reverted if you consider $H=E+pV$ instead of $E$ i.e. substitute $E = H-pV$. Then you will get $$dH = \dots - V\,dp.$$ This "Lagrange duality" may be applied to any pair.

  • $\begingroup$ And what about $\langle X_i\cdot x_k\rangle = \delta_{ik}$? As I understood this implies only in the approximation of infinitesimal fluctuations. I'm right? $\endgroup$
    – m0nhawk
    Oct 29, 2012 at 9:34
  • $\begingroup$ Usually nowadays the convention is the opposite sign for the work term: $-pdV$. $\endgroup$
    – perplexity
    Oct 29, 2012 at 10:31
  • 1
    $\begingroup$ Apart from some technical issues of your answer (which I will ignore here), your variables $X$ and $x$ do not correspond to those in @m0nhawk question. Evidently he is not using the energetic representation. $\endgroup$
    – juanrga
    Oct 30, 2012 at 18:34
  • $\begingroup$ Lubos--- you should fix the answer--- his conjugate quantities are your conjugate quantities divided by T. $\endgroup$
    – Ron Maimon
    Oct 31, 2012 at 6:09
  • $\begingroup$ @RonMaimon. The relation between both quantities is not a mere division per T even if we ignore the Boltzmann constant... $\endgroup$
    – juanrga
    Oct 31, 2012 at 12:33

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