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authorJaron Kent-Dobias <jaron@kent-dobias.com>2019-06-27 00:11:50 -0400
committerJaron Kent-Dobias <jaron@kent-dobias.com>2019-06-27 00:11:50 -0400
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lots of progress on writing the theory sections
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diff --git a/hidden-order.tex b/hidden-order.tex
index 364602e..b326d0d 100644
--- a/hidden-order.tex
+++ b/hidden-order.tex
@@ -69,7 +69,8 @@ Consider a generic order parameter $\eta$. To write down its free energy, we in
\[
f_{\X}=\tfrac12b^{(i)}\epsilon_{\X}^{(i)}\cdot\eta
+\tfrac12e^{(i)}\epsilon_{\Aog}^{(i)}\eta^2
-V\]
+ +\tfrac12h^{(ij)}\epsilon_{\Aog}^{(i)}\epsilon_{\X}^{(j)}\eta
+\]
The total free energy is
\[
F=\int d^3x\,(f_{\mathrm e}+f_{\mathrm o}+f_{\X})
@@ -90,12 +91,12 @@ The most general quartic free energy density (discounting total derivatives) is
independent of the symmetry of $\eta$. In principle we could have $D_\parallel\neq D_\perp$, but this does not affect the physics at hand. This is the free energy for a Lifshitz point, and so we expect to see that phenomenology in $\eta$.
Before doing anything, we can minimize the free energy with respect to strain alone to find the strain in terms of $\eta$ exactly. We have
\[
- 0=\frac{\delta F}{\delta\epsilon_{\mathrm X}^{(1)}(x)}=\lambda_{\mathrm X}^{(11)}\epsilon_{\mathrm X}^{(1)}(x)+\frac12b^{(1)}\eta(x)
+ 0=\frac{\delta F}{\delta\epsilon_{\mathrm X}^{(1)}(x)}=\lambda_{\mathrm X}^{(11)}\epsilon_{\mathrm X}^{(1)}(x)+\frac12b^{(1)}\eta(x)+\frac12h^{(i)}\epsilon_{\Aog}^{(i)}\eta
\]
whence we immediately have $\epsilon_{\mathrm X}^{(1)}=-\frac{b^{(1)}}{2\lambda_{\mathrm X}^{(11)}}\eta(x)$. We also have
\[
0=\frac{\delta F}{\delta\epsilon_{\Aog}^{(i)}(x)}
- =\lambda_{\Aog}^{(ij)}\epsilon_{\Aog}^{(j)}(x)+\frac12 e^{(i)}\eta^2(x)
+ =\lambda_{\Aog}^{(ij)}\epsilon_{\Aog}^{(j)}(x)+\frac12 e^{(i)}\eta^2(x)+\frac12h^{(i)}\epsilon_\X\eta
\]
which is a linear system whose solutions are
\begin{align*}