# What does it mean to Taylor expand free energy density “in gradients”?

I am self-studying Statistical Mechanics from J. Sethna's book Entropy, Order parameters, complexity. In one of the exercises (page 206), the Landau theory for the Ising Model is derived.

Starting with a general, local free energy density of this form:

$$F^{Ising}\{m,T\} = F(\textbf{x},m,\partial _j m, \partial _j \partial _k m, ...) .$$

we are supposed to "Taylor expand in gradients", keeping terms with up to two gradients of m, to get:

$$F^{Ising}\{m,T\} = A(m,T) + V_i(m,T) \partial _i m + B_{ij}(m,T) \partial _i \partial_j m + C_{ij}(m,T)( \partial_i m )(\partial_j m).$$

I do not understand what "Taylor expanding in gradients" means, nor how to get to the second equation (what do $A$, $V$, $B$ and $C$ represent?). It is not explained in the book, and I could not find anything related to this online.

Note that I am not a physicist by training and I am unfamiliar with much of the jargon (I do know what gradients and Taylor expansions are, though).

Suppose we have an expression for the free energy $F(\vec{x},T,m,\partial_j m,\partial_i \partial_j m,...)$, but we also assume that $m$ is approximately constant. This means that higher-order gradients (and large products of lower-order gradients) of $m$ are very small, so we can approximate $F$ as an expression involving only up to, say, two gradients of $m$ in any given term.

What are the possible forms an expression can take under these circumstances? Well, first of all, it could be independent of any gradient, and just be a function of $T$ and $m$ itself. Assuming that term is present in our expression for $F$, we'll call it $A(m,T)$. An expression could also involve just one gradient of $m$, and it might affect different directions differently; assuming all of these possible expressions are present, we'll label them $V_i(m,T)\partial_i m$. Similarly, we'll call those expressions which involve a second-order gradient $B_{ij}(m,T)\partial_i\partial_j m$, and those expressions with a product of two gradients will be labeled $C_{ij}(m,T)(\partial_i m)(\partial_j m)$. These are all of the possible expressions involving two or fewer instances of $\partial$. Adding all four possible terms together, we get the expression you have above.

If it seems like we didn't really do anything in the last paragraph, you're half-right: this is just setting up a general expression that will be elaborated upon in the rest of the book. Applying this expression to a physical situation will impose constraints on the values of $A,V_i,B_{ij},$ and $C_{ij}$; maybe $A(m,T)=0$ in some situations, or $V_i(m,T)=Ke^{-m/T}$, where $K$ is a constant dependent on the lattice spacing or something. The point is, the values for the parameters in your expression will be determined by the physical situation you're considering.

• Indeed, later in the problem it does exactly what I said: various symmetries are imposed to better determine the form of the functions above. – probably_someone Jun 19 '17 at 18:40
• Thank you, this very much clarifies what is done in the book. I still don't see how this is a Taylor expansion though. – enricosandro Jun 20 '17 at 18:32
• It's a generalization of what you're used to seeing. This might help: math.ubc.ca/~feldman/m200/taylor2dSlides.pdf – probably_someone Jun 20 '17 at 18:41