# Question on (the linear, phenomenological equations of) nonequilibrium thermodynamics

1. Apr 14, 2012

### nonequilibrium

Hello,

So one presumes that $J_i = \sum_j L_{ij} X_j$ where J_i is the current for a certain variable x_i and X_j are the different kind of thermodynamical "forces". In the case of heat conduction, the "force" is temperature difference $\nabla T$ and the current J_i is a current of heat (Fourier's law! $J = - \kappa \nabla T$).

One also presumes that for the entropy production $\sigma = \frac{\mathrm d}{\mathrm dt} S = \sum_i X_i J_i$ (*), so that together with the previous one gets that $\sigma = \sum_{i,j} L_{ij} X_i X_j$.

But anyway, how can (*) possibly be correct? After all, we also know that by the chain rule $\frac{\mathrm d}{\mathrm dt} S = \sum_i \frac{\partial S}{\partial x_i} \frac{\mathrm d x_i}{\mathrm dt}$ or by definition of current $\frac{\mathrm d}{\mathrm dt} S = \sum_i \frac{\partial S}{\partial x_i} J_i$ (**).

Comparing (*) and (**) would suggest that $\boxed{ X_i = \frac{\partial S}{\partial x_i} }$.

However, take again the case of heat conduction: $x_i = E$ (energy), hence $\frac{\partial S}{\partial x_i} = \frac{1}{T}$, but on the other hand, as claimed in my first paragraph: $X_i = \nabla T$
But obviously $\boxed{ \frac{\partial S}{\partial x_i} = \frac{1}{T} \neq \nabla T = X_i }$

Where is my misunderstanding?

Last edited: Apr 14, 2012