# Is entropy maximum in equilibrium?

Usually, in the context of non-equilibrium thermodynamics, it is said that entropy achieves a maximum in equilibrium, so the Taylor series expansion of entropy around the equilibrium state as a function of fluctuations $$\xi_i$$ of internal extensive variables does not have degree 1 (linear) terms, just the contributions of the Hessian matrix (which is symmetric negative definite). \begin{align*} S\left(\boldsymbol{\xi}\right) &\approx S\left(\mathbf{0}\right) + \boldsymbol{\xi}^{\top} \nabla S\left(\mathbf{0}\right) + \frac{1}{2 \, !} \boldsymbol{\xi}^{\top} \mathbf{H} \left(S\left(\mathbf{0}\right)\right) \boldsymbol{\xi} \\ &= S_0 + \frac{1}{2} \boldsymbol{\xi}^{\top} \mathbf{H} \left(S\left(\mathbf{0}\right)\right) \boldsymbol{\xi} \end{align*} This is necessary to derive Onsager relations. But isn't there a flaw in this reasoning applied to open and closed systems? The entropy of the system should achieve max value in equilibrium only in isolated systems, shouldn't it?

• The main point is that, if you assume that the entropy belongs to the universe (isolated system), then $\nabla S(0)=0$ only at equilibrium is a valid statement. While if the entropy belongs to any other kind of system, then $\nabla S(0)=0$ is still valid, and just tells you that you have a set of local variables in thermodynamic space "$\mathbf{0}$" that form a local extremum point for the entropy, and nothing else Jul 4, 2020 at 14:40