clarifed what we are doing with the strain trace

1 files changed, 20 insertions, 13 deletions

diff --git a/main.tex b/main.tex
index 9be54af..d20c7f4 100644
--- a/main.tex
+++ b/main.tex
@@ -186,7 +186,6 @@ components present in \eqref{eq:strain-components}, this rules out all of the
 \emph{u}-reps (which are odd under inversion) and the $\Atg$ irrep.
 
 If the \op\ transforms like $\Aog$ (e.g. a fluctuation in valence number), odd terms are allowed in its free energy and any transition will be first order and not continuous without fine-tuning. Since the \ho\ phase transition is second-order \brad{cite something}, we will henceforth rule out $\Aog$ \op s as well. 
-
 For the \op\ representation $\X$ as any of $\Bog$, $\Btg$, or $\Eg$, the most general
 quadratic free energy density is
 \begin{equation}
@@ -197,7 +196,8 @@ quadratic free energy density is
   \end{aligned}
   \label{eq:fo}
 \end{equation}
-where $\nabla_\parallel=\{\partial_1,\partial_2\}$ transforms like $\Eu$, and $\nabla_\perp=\partial_3$ transforms like $\Atu$. Other quartic terms are
+where $\nabla_\parallel=\{\partial_1,\partial_2\}$ transforms like $\Eu$, and
+$\nabla_\perp=\partial_3$ transforms like $\Atu$. Other quartic terms are
 allowed---especially many for an $\Eg$ \op---but we have included only those
 terms necessary for stability when either $r$ or $c_\perp$ become negative. The
 full free energy functional of $\eta$ and $\epsilon$ is
@@ -207,20 +207,27 @@ full free energy functional of $\eta$ and $\epsilon$ is
       &=F_\op[\eta]+F_\e[\epsilon]+F_\i[\eta,\epsilon] \\
       &=\int dx\,(f_\op+f_\e+f_\i).
   \end{aligned}
+  \label{eq:free_energy}
 \end{equation}
-The only strain relevant to the \op\ at linear coupling is $\epsilon_\X$, which can be traced out
-of the problem exactly in mean field theory. Extremizing with respect to
-$\epsilon_\X$,
+
+Rather than analyze this two-argument functional directly, we begin by tracing
+out the strain and studying the behavior of \op\ alone, assuming the
+strain is equilibrated. Later we will invert this procedure and trace out the
+\op when we compute the effective elastic moduli. The only strain relevant to
+the \op\ at linear coupling is $\epsilon_\X$, which can be traced out of the
+problem exactly in mean field theory. Extremizing the functional
+\eqref{eq:free_energy} with respect to $\epsilon_\X$ gives
 \begin{equation}
-  0=\frac{\delta F[\eta,\epsilon]}{\delta\epsilon_\X(x)}\bigg|_{\epsilon=\epsilon_\star}=C^0_\X\epsilon^\star_\X(x)
-    -b\eta(x)
+  0
+    =\frac{\delta F[\eta,\epsilon]}{\delta\epsilon_\X(x)}\bigg|_{\epsilon=\epsilon_\star}
+    =C^0_\X\epsilon^\star_\X(x)-b\eta(x),
 \end{equation}
-\textbf{talk more about the functional-ness of these parameters!, also, why are we tracinig out strain?}
-gives the optimized strain conditional on the \op\ as
-$\epsilon_\X^\star[\eta](x)=(b/C^0_\X)\eta(x)$ and $\epsilon_\Y^\star[\eta]=0$
-for all other $\Y$. Upon substitution into the free energy, the resulting
-effective free energy $F[\eta,\epsilon_\star[\eta]]$ has a density identical to
-$f_\op$ with $r\to\tilde r=r-b^2/2C^0_\X$. 
+which in turn gives the strain field conditioned on the state of the \op\ field
+as $\epsilon_\X^\star[\eta](x)=(b/C^0_\X)\eta(x)$ at all spatial coordinates
+$x$, and $\epsilon_\Y^\star[\eta]=0$ for all other irreps $\Y\neq\X$. Upon
+substitution into the free energy, the resulting single-argument free energy
+functional $F[\eta,\epsilon_\star[\eta]]$ has a density identical to $f_\op$
+with $r\to\tilde r=r-b^2/2C^0_\X$. 
 
 \begin{figure}[htpb]
   \includegraphics[width=\columnwidth]{phase_diagram_experiments}