I am reading "A Modern Course in Statistical Mechanics" by Linda E. Reichl. Where i encountered this notation:

$$\Delta S = \bar g : \vec \alpha \vec \alpha$$

Here $\bar g$ is $$ g_{i,j}=-{ \partial^2 S \over \partial A_i \partial A_j}\bigg|_{A_i = A_i^0, A_j = A_j^0 } $$ a matrix of second derivatives of the entropy S derives after thermodynamic variables $A_i$ around their equilibrium $A_i^0$. And the vectors $ \vec \alpha $ are defined as deviations of the thermodynamic variables from their equilibrium: $\alpha_i= A_i-A_i^0$.

From the equation in question this $$ -\left({ \partial \Delta S \over \partial \alpha}\right)= \bar g \alpha$$ follows immedeatley. Maybe that helps anyone?

I have not yet found an explanation of $\Delta S = \bar g : \vec \alpha \vec \alpha$ anywhere and i would be really glad if anyone can explain this. If you need more information i am of course willing to provide.

  • $\begingroup$ Looks like a double contraction of tensors $\endgroup$ May 8, 2015 at 9:47

1 Answer 1


I think what they could mean is that $\vec{\alpha}\vec{\alpha}$ is a second rank tensor that is contracted with $\overline{g}$. I saw this notation being used in the context of electrodynamics before. It is used to get a simple notation for multi-dimensional Taylor series. So we get

$$\Delta S=\overline{g}:\vec{\alpha}\vec{\alpha} \equiv \sum_{ij} -\frac{1}{2!}g_{ij} \alpha_i \alpha_j$$

The minus takes into account that $g_{ij}$ contains the negative derivative in contrast to the ordinary taylor expansion. From this would follow that

$$\frac{\partial\Delta S}{\partial\vec{\alpha}} =(\partial_{\alpha_i}\Delta S)\ \vec{e}_i = -\sum_{kl} \frac{1}{2!}g_{kl} \ \partial_{\alpha_i}(\alpha_k \alpha_l)\ \vec{e}_i =-\sum_{kl}\frac{1}{2!}g_{kl} (\delta_{ik} \alpha_{l} + \delta_{il} \alpha_k)\ \vec{e}_i $$

And therefore

$$\frac{\partial\Delta S}{\partial\vec{\alpha}} = -\sum_{k} \frac{1}{2!}(g_{ik} \alpha_{k} + g_{ki} \alpha_k)\ \vec{e}_i = -\frac{1}{2!}(\overline{g}\vec{\alpha}+\overline{g}^T\vec{\alpha}) = -\overline{g}\vec{\alpha} $$

Which is the desired result. The last step follows from the symmetry of $\overline{g}$.

I finally consulted my notes on electrodynamics where I found the multivariate taylor expansion of a scalar field $s(\vec{x})$ around $\vec{x}_0$ written as

$$s(\vec{x}) = \sum_{n=0}^\infty \frac{1}{n!} (\vec{x}-\vec{x}_0)^n \vdots \nabla^n s(\vec{x}_0)$$


$$\nabla^n s(\vec{x}_0) \equiv (\partial_{i_1} \ldots \partial_{i_n}s(\vec{x}_0))e^{(i_1)}\ldots e^{(i_n)}$$


$$(\vec{x})^n\equiv x^{i_1}\ldots x^{i_n} e_{(i_1)}\ldots e_{(i_n)}$$

Note that here it was distinguished between co- and contravariant components and the sum-convention has been used. So that $\vdots$ represents the n-fold contraction of the two tensors.

  • $\begingroup$ I have not yet read through the details entirely because i am insanely tired. But to my untrained eyes this looks like the only sensible answer. I really don't get why they used this in several editions without any explanation. One question though: In retrospect judging by the Wikipedia definition of a dierectional derivative $\nabla_{\vec{v}}{f}(\vec {x}) = \nabla f(\vec{x}) \cdot \vec {v}$ this is the equivalent thing to $-({ \partial \Delta S \over \partial \alpha})= \bar g \alpha$ right? $\endgroup$
    – Kuhlambo
    May 8, 2015 at 10:40
  • $\begingroup$ by the way this kind of stuff is why i love this site^^ $\endgroup$
    – Kuhlambo
    May 8, 2015 at 10:42
  • $\begingroup$ I would say that is similar, yes. But you see that the ordinary directional derivative contains derivatives of first order and $\Delta S$ does not, which makes sense since you are concerned with a situation in thermodynamic equilibrium. Applying the directional derivative to S would therefore give 0. So if you want to learn something about the change in $S$ you need to consult the properties of the next-order term $\endgroup$ May 8, 2015 at 10:48
  • $\begingroup$ Oh sure i was referring just to the expression $\frac{\partial\Delta S}{\partial\vec{\alpha}}$ where we take the first derivative. I just noticed that this is just the definition of the normalized directional derivative $\nabla_{\vec{v}}{f}(\vec{x}) = \nabla f(\vec{x}) \cdot \frac{\vec{v}}{|\vec{v}|}=\partial_{x_i} f(\vec x) \vec e_i $ witch is seems to be equivalent to your definition $\frac{\partial\Delta S}{\partial\vec{\alpha}} =(\partial_{\alpha_i}\Delta S)\ \vec{e}_i$ right? $\endgroup$
    – Kuhlambo
    May 8, 2015 at 11:12
  • $\begingroup$ Aah ok. I see. Yes. It is the same indeed! $\endgroup$ May 8, 2015 at 11:16

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