Strange vector matrix operation (in "A Modern Course in Statistical Physics" by Reichl)

Question

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.

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

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

where

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

and

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

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.