# "$\delta^2 S$" confusion regarding "second variations" in stability conditions

As far as I am aware, for some function of $$n$$ variables $$f$$, $$\delta^2 f$$ represents the third term in its Taylor expansion.

So, I've encoutered the following expression in my thermodynamics book:

$$\sum_{a=1,2} \delta^2 S_a = - \dfrac{1}{2T} \sum_{a=1,2} \delta \mu_a \delta N_a$$ $$= - \dfrac{1}{2T} \sum_a \dfrac{\partial \mu_a}{\partial N_a}|_T \ \ \delta N^2_a$$

where we consider the entropy $$S$$ of an isolated system divided into two subsystems $$a=1,2$$ and where the number of the particles in each subsystem $$N_1 , N_2$$ is free to vary but restricted by $$N_1 + N_2 = N$$ (the systems have chemical potentials $$\mu_a$$ , volumes $$V_a$$ , temperatures $$T_a$$ ...) The book also says "$$\delta T = 0$$ , $$\delta V=0$$" for the system ,which I take to mean that the temperatures, volumes of the subsystems are fixed.

In summary: we take a isolated system , divide it in two subsystems, the volumes and temperatures of the two subsystems are assumed fixed and equal, and the number of particles of the two subsystems are free to vary but must be equal to the total number of particles in the entire system.

Question: How do we get from $$\sum \delta^2 S$$ to the second expression (and from the second to the third also!)? What does $$\delta \mu \delta N$$ mean?

How would one rewrite these stability conditions in the more familiar notation of gradients and Hessian matrices (i.e. the multivariable Taylor expansion)?

• Why don't you specify the book, so people can understand the context, assumptions, and strategy? (I'll never understand why people ask about information they read but decline to name the source.) Commented Jul 7, 2023 at 15:48
• @Chemomechanics because I cannot (these are my prof.'s notes/syllabus). Anyways, I maybe should have mentioned that there is no relevant context, strategy, assumption besides the one I have given . (I'll never understand why some people ask about information about the source but fail to see that it is irrelevant.) Commented Jul 7, 2023 at 16:05
• Let me help your understanding: People writing questions sometimes make mistakes in writing equations, defining variables, and interpreting the implications. Sources can also contain errors. Your comment about irrelevancy would be true if these factors didn't exist, but they do. You can find ample evidence from the many questions on this site asking "Why A?" or "Since A, then why B?" when A is a shaky premise. Anyway, thank you for clarifying that the source won't be revealed and that no additional information should be assumed. Commented Jul 7, 2023 at 17:01
• As an example, if we reasonably assume that $S$ is entropy and $\mu$ is the chemical potential, your first and third terms have units of J/K and J, respectively, and therefore can't be equivalent. Commented Jul 7, 2023 at 17:24
• oh first and third terms... yes ! there is a mistake there Commented Jul 7, 2023 at 17:48

It is just as easy to do the more general case. Start with the Gibbs equation $$U=TS+\sum_kY_k X_k$$ and energy conservation represented by $$dU=TdS +\sum_kY_kdX_k$$, and assume that the extensive parameters $$X_k$$, $$k=1,2,..,M$$ are conserved; they may be mass, chemical species moles, electric charge, etc., and note that $$Y_k=\frac{\partial U}{\partial X_k}$$ are the conjugate intensive parameters; for example, if $$X_k=N_k$$ then $$Y_k=\mu_k$$.
Now take two systems that are in contact with internal energies $$U$$ and $$U'$$ but together they are isolated from the environment, then their total energy is $$U+U'$$ and must be a constant, $$dU+dU'=0$$, and if with each other they exchange conserved extensive quantities $$dX_k$$ and $$dX_k'=0$$ then $$dX_k+dX_k'=0$$. Since entropy is only conserved in a reversible process, one can only state that $$dS +dS' = \sigma \ge 0$$, where the $$2^{nd}$$ order quantity $$\sigma$$ is the irreversibly generated entropy, but if you assume that the two systems are in equilibrium then any spontaneous fluctuation between them is reversible, $$\sigma = 0$$, from which it automatically follows the equality of the intensives in equilibrium.
So we have $$\sigma = \frac{1}{T}dU - \sum_k \frac{Y_k}{T}dX_k + \frac{1}{T'}dU' - \sum_k \frac{Y_k'}{T'}dX_k' \tag{1}$$ With $$dU=-dU'$$ and $$dX_k=-dX_k'$$ Eq. 1 can be rewritten as $$\sigma = \left(\frac{1}{T}-\frac{1}{T'}\right)dU - \sum_k \left(\frac{Y_k}{T}-\frac{Y_k'}{T'}\right)dX_k \tag{2}$$ or noting that $$\left(\frac{1}{T'}-\frac{1}{T}\right)=\delta \frac{1}{T} = -\frac{1}{T^2}\delta T$$ and similarly $$\frac{Y_k'}{T'}-\frac{Y_k}{T}=\delta \frac{Y_k}{T}=\frac{1}{T}\delta Y_k-\frac{Y_k}{T^2} \delta T$$ we get $$\sigma = \frac{1}{T^2}\delta TdU + \sum_k \left(\frac{1}{T}\delta Y_k-\frac{Y_k}{T^2} \delta T\right)dX_k \tag{3}.$$
Now multiply both sides with T and rearrange: $$T\sigma = \frac{1}{T}\delta T dU - \sum_k \left(\frac{Y_k}{T} \delta T\right)dX_k + \sum_k \delta Y_kdX_k \tag{4}$$ The sum of the first two terms on the right is zero: $$T\sigma = \sum_k \delta Y_kdX_k \ge 0 \tag{5}$$ The left side of Eq. 5 is the dissipation that is never negative. Your question was to derive Eq. 6 taht you get by dividing both sides by $$T$$ $$\sigma = \frac{1}{T}\sum_k \delta Y_kdX_k \ge 0 \tag{6}$$ Stability is assured if for all virtual displacements $$\sum_k \delta Y_kdX_k < 0 \tag{7}$$ The beauty of Eq.7 is that you can arbitrarily choose for the independent variables a set of indices $$K:=\{k_1,k_2,...\}$$ and $$K':=\{k'_1,k'_2,...\}$$, so that $$K\cup K' =1,2,...M$$ and the independent variables are $$X_{k_1}, X_{k_2},...,Y_{k'_1},Y_{k'_2},...$$ with dependent variables $$X_{k'_1}, X_{k'_2},...,Y_{k_1},Y_{k_2},...$$.