datomerki8a5yj

2022-05-18

I am trying to get the relation between developing a Ginzburg-Landau theory, let's say for a ferromagnet with magnetization field $\stackrel{\to }{m}=\stackrel{\to }{m}\left(\stackrel{\to }{r}\right)$, and the formal expansion of the free energy density $\mathcal{F}=\mathcal{F}\left(\stackrel{\to }{m}\right)$ in terms of a Taylor series.
Considering an isotropic ferromagnet, the lowest-order terms in our Ginzburg-Landau theory should be given by
$\mathcal{F}=\frac{r}{2}{\stackrel{\to }{m}}^{2}+\frac{U}{4}{\left({\stackrel{\to }{m}}^{2}\right)}^{2}+\frac{J}{2}\left[{\left({\mathrm{\partial }}_{x}\stackrel{\to }{m}\right)}^{2}+{\left({\mathrm{\partial }}_{y}\stackrel{\to }{m}\right)}^{2}+{\left({\mathrm{\partial }}_{z}\stackrel{\to }{m}\right)}^{2}\right]$
with r<0 and U,J>0.
However, when I think of a Taylor expansion of $\mathcal{F}\left(\stackrel{\to }{m}\right)$ around the origin
$\mathcal{F}={\mathcal{F}}_{0}+{\stackrel{\to }{m}}^{T}\cdot D\mathcal{F}+\frac{1}{2}{\stackrel{\to }{m}}^{T}\cdot {D}^{2}\mathcal{F}\cdot \stackrel{\to }{m}+\dots$
this is giving me terms of all the individual powers in $\stackrel{\to }{m}$ , which are either zero or identified with the r- and U-term, but no gradient terms for the J-term? How to motivate these through a Taylor expansion?

Cortez Hughes

Note that this is rather an opinion than a fully rigorous statement:
For vanishing gradients, i.e. for uniform systems (the case originally considered by Landau), the expansion of the free energy is indeed a Taylor expansion (in even powers of m) near the transition. However the addition of gradient terms in the case of non-uniform systems removes this interpretation as is rather a phenomenological modification. However, you could still see this an expression for a classical field theory.

Raphael Mccullough

It looks to me like you are Taylor expanding $\mathcal{F}$ as a function of $\stackrel{\to }{m}$ without considering that $\stackrel{\to }{m}$ is a function of space. In particular, your $D$ and ${D}^{2}$ involve things like $\frac{\mathrm{\partial }}{\mathrm{\partial }{m}_{x}}$ but not $\frac{\mathrm{\partial }}{\mathrm{\partial }x}$. You are implicitly assuming that $\mathcal{F}\left(\stackrel{\to }{x}\right)$ only depends on $\stackrel{\to }{m}\left(\stackrel{\to }{x}\right)$ and not also for example $\stackrel{\to }{m}\left(\stackrel{\to }{x}+d\stackrel{\to }{x}\right)$ for small $d\stackrel{\to }{x}$