A thermodynamic argument for the reciprocal relation....

A thermodynamic argument for the reciprocal relation. Consider a situation of steady-state flow between two reservoirs separated by a thin membrane of thickness $\delta x$. The membrane is not perfectly conducting but does allow heat and particles, etc. to pass. We assume that the flow is slow enough for each reservoir to be characterized by a well-defined temperature, pressure etc., even though it is slow brane is in steady state, so a Page 433 i 464 other of the reservoirs. (i) Show that the rate of entropy generation in the whole system, per unit volume of the membrane, is $$ \frac{\mathrm{d} s}{\mathrm{~d} t}=\sum_i\left(\nabla \phi_i\right) \cdot \mathbf{j}_i, $$ where the potentials $\phi_i$ and currents $\mathbf{j}_i$ are as described in (28.31) and (28.32). This implies that $\dot{s}$ can be regarded as a function of the gradients and the currents: $$ \dot{s}=\dot{s}\left(\boldsymbol{\phi}_1^{\prime}, \boldsymbol{\phi}_2^{\prime}, \ldots, \mathbf{j}_1, \mathbf{j}_2, \ldots\right), $$ where $\phi_i^{\prime} \equiv \nabla \phi_i$. (ii) From (28.36), show that $\mathbf{j}_i=\left(\partial \dot{s} / \partial \phi_i^{\prime}\right)$ and hence (using (28.32)) obtain $$ L_{i k}=\frac{\partial^2 \dot{s}}{\partial \phi_k^{\prime} \partial \phi_i^{\prime}} . $$ Similarly, $$ L_{k i}=\frac{\partial^2 \dot{s}}{\partial \phi_i^{\prime} \partial \phi_k^{\prime}} . $$ Since $\dot{s}$ is a well-behaved analytic function, it follows that $L_{k i}=L_{i k}$. Thus we appear to have obtained Onsager's reciprocal relation without the need to consider thermal fluctuations, and without the need to invoke the regression hypothesis or the principle of microscopic reversibility. (iii) The above argument is plausible. However, when Thomson and others first discussed this subject, they were careful to raise questions over whether this type of argument is valid, because equilibrium concepts are being applied to out of equilibrium processes. ${ }^5$ Does the presence of the flow invalidate the use of the concepts being used to describe it? Consider the case of not one thin layer, but many adjacent thin layers. Then temperature, and other intensive properties, are being ascribed to a system with a non-negligible temperature gradient, through which energy and other quantities are flowing, and within which entropy is being generated. Is (28.36) correct, and does it involve any hidden assumptions?

A thermodynamic argument for the reciprocal relation. Consider a situation of steady-state flow between two reservoirs separated by a thin membrane of thickness $\delta x$. The membrane is not perfectly conducting but does allow heat and particles, etc. to pass. We assume that the flow is slow enough for each reservoir to be characterized by a well-defined temperature, pressure etc., even though it is slow brane is in steady state, so a
Page 433 i 464 other of the reservoirs. (i) Show that the rate of entropy generation in the whole system, per unit volume of the membrane, is
$$
\frac{\mathrm{d} s}{\mathrm{~d} t}=\sum_i\left(\nabla \phi_i\right) \cdot \mathbf{j}_i,
$$
where the potentials $\phi_i$ and currents $\mathbf{j}_i$ are as described in (28.31) and (28.32). This implies that $\dot{s}$ can be regarded as a function of the gradients and the currents:
$$
\dot{s}=\dot{s}\left(\boldsymbol{\phi}_1^{\prime}, \boldsymbol{\phi}_2^{\prime}, \ldots, \mathbf{j}_1, \mathbf{j}_2, \ldots\right),
$$
where $\phi_i^{\prime} \equiv \nabla \phi_i$.
(ii) From (28.36), show that $\mathbf{j}_i=\left(\partial \dot{s} / \partial \phi_i^{\prime}\right)$ and hence (using (28.32)) obtain
$$
L_{i k}=\frac{\partial^2 \dot{s}}{\partial \phi_k^{\prime} \partial \phi_i^{\prime}} .
$$
Similarly,
$$
L_{k i}=\frac{\partial^2 \dot{s}}{\partial \phi_i^{\prime} \partial \phi_k^{\prime}} .
$$
Since $\dot{s}$ is a well-behaved analytic function, it follows that $L_{k i}=L_{i k}$. Thus we appear to have obtained Onsager's reciprocal relation without the need to consider thermal fluctuations, and without the need to invoke the regression hypothesis or the principle of microscopic reversibility.
(iii) The above argument is plausible. However, when Thomson and others first discussed this subject, they were careful to raise questions over whether this type of argument is valid, because equilibrium concepts are being applied to out of equilibrium processes. ${ }^5$ Does the presence of the flow invalidate the use of the concepts being used to describe it? Consider the case of not one thin layer, but many adjacent thin layers. Then temperature, and other intensive properties, are being ascribed to a system with a non-negligible temperature gradient, through which energy and other quantities are flowing, and within which entropy is being generated. Is (28.36) correct, and does it involve any hidden assumptions?

Key Concepts

Entropy Generation

Entropy generation quantifies the irreversible production of entropy in a system undergoing non-equilibrium processes. In this context, it is expressed as the product of thermodynamic forces (gradients of intensive quantities) and the corresponding fluxes (currents). This concept is central to understanding how energy dissipation occurs and is used to derive relations between the forces and fluxes in the system.

Thermodynamic Forces and Fluxes

Thermodynamic forces are gradients of intensive parameters like temperature or chemical potential, which drive the flow of corresponding extensive quantities, or fluxes. In the framework of non-equilibrium thermodynamics, these forces and fluxes are related linearly in the near-equilibrium regime, where the fluxes are proportional to the gradients, thereby allowing the formulation of phenomenological laws.

Linear Irreversible Thermodynamics

Linear irreversible thermodynamics is the theoretical framework used to describe systems slightly out of equilibrium, where the relationships between forces and fluxes are linear. This approach facilitates the expression of entropy production as a bilinear form and leads to the definition of phenomenological coefficients, which characterize the transport processes occurring in the system.

Onsager’s Reciprocal Relations

Onsager’s reciprocal relations state that the cross-coefficients linking different pairs of forces and fluxes are equal, provided that the system is near equilibrium and obeys microscopic reversibility. This symmetry, originally derived using arguments based on microscopic fluctuations, can also be derived from the analytic properties of the entropy production function, where the equality of mixed partial derivatives enforces this reciprocity.

Analyticity of Entropy Production

The assumption that the entropy production is an analytic function of the thermodynamic forces allows the use of mathematical properties such as the equality of mixed partial derivatives (Clairaut’s theorem). This analyticity is pivotal to deriving symmetric relations between the phenomenological coefficients, thereby underpinning Onsager’s reciprocal relations in the formalism of non-equilibrium thermodynamics.

Local Equilibrium Assumption

The local equilibrium assumption posits that even though the overall system is not in thermodynamic equilibrium, each small volume element (or layer) can be characterized by equilibrium thermodynamic variables. This assumption justifies applying equilibrium concepts, such as well-defined temperature or pressure, to incremental regions within a system experiencing gradients and fluxes, thereby enabling the analysis of steady-state flows using equilibrium thermodynamic potentials.

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