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

  1. Apr 14, 2012 #1
    Hello,

    So one presumes that [itex]J_i = \sum_j L_{ij} X_j[/itex] 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 [itex]\nabla T[/itex] and the current J_i is a current of heat (Fourier's law! [itex]J = - \kappa \nabla T[/itex]).

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

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

    Comparing (*) and (**) would suggest that [itex]\boxed{ X_i = \frac{\partial S}{\partial x_i} }[/itex].

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

    Where is my misunderstanding?
     
    Last edited: Apr 14, 2012
  2. jcsd
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