leave nothing undefined! better free energy notation

1 files changed, 9 insertions, 5 deletions

diff --git a/main.tex b/main.tex
index a458623..3c299ea 100644
--- a/main.tex
+++ b/main.tex
@@ -208,7 +208,11 @@ where $\nabla_\parallel=\{\partial_1,\partial_2\}$ transforms like $\Eu$ and
 $\nabla_\perp=\partial_3$ transforms like $\Atu$. We'll take $D_\parallel=0$
 since this does not affect the physics at hand. The full free energy functional of $\eta$ and $\epsilon$ is then
 \begin{equation}
-  F[\eta,\epsilon]=\int dx\,(f_\o+f_\e+f_\i)
+  \begin{aligned}
+    F[\eta,\epsilon]
+      &=F_\o[\eta]+F_\e[\epsilon]+F_\i[\eta,\epsilon] \\
+      &=\int dx\,(f_\o+f_\e+f_\i)
+  \end{aligned}
 \end{equation}
 Neglecting interaction terms
 higher than quadratic order, the only strain relevant to the problem is
@@ -218,7 +222,7 @@ $\epsilon_\X$, and this can be traced out of the problem exactly, since
     +\frac12b\eta_i(x)
 \end{equation}
 gives $\epsilon_\X[\eta]=-(b/2C_\X)\eta$. Upon substitution into the free
-energy, the resulting effective free energy $F_\o[\eta]=F[\eta,\epsilon[\eta]]$ has a density identical to $f_\o$
+energy, the resulting effective free energy $F[\eta,\epsilon[\eta]]$ has a density identical to $f_\o$
 with $r\to\tilde r=r-b^2/4C_\X$.
 
 With the strain traced out \eqref{eq:fo} describes the theory of a Lifshitz
@@ -270,7 +274,7 @@ The susceptibility of the order parameter to a field linearly coupled to it is g
 \begin{equation}
   \begin{aligned}
     &\chi^\recip(x,x')
-      =\frac{\delta^2F_\o[\eta]}{\delta\eta(x)\delta\eta(x')}
+    =\frac{\delta^2F[\eta,\epsilon[\eta]]}{\delta\eta(x)\delta\eta(x')}
       =\big(\tilde r-c_\parallel\nabla_\parallel^2-c_\perp\nabla_\perp^2 \\
     &\qquad\qquad+D_\perp\nabla_\perp^4+12u\eta^2(x)\big)
     \delta(x-x'),
@@ -309,7 +313,7 @@ energy functional of $\epsilon$. Extremizing over $\eta$ yields
     +\frac12b\epsilon_\X(x),
   \label{eq:implicit.eta}
 \end{equation}
-which implicitly gives $\eta[\epsilon]$ and $F_\e[\epsilon]=F[\eta[\epsilon],\epsilon]$. Since $\eta$ is a functional of $\epsilon_\X$ alone, only the stiffness $\lambda_\X$ is modified from its bare value $C_\X$. Though this
+which implicitly gives $\eta[\epsilon]$. Since $\eta$ is a functional of $\epsilon_\X$ alone, only the stiffness $\lambda_\X$ is modified from its bare value $C_\X$. Though this
 cannot be solved explicitly, we can make use of the inverse function theorem.
 First, denote by $\eta^{-1}[\eta]$ the inverse functional of $\eta$ implied by
 \eqref{eq:implicit.eta}, which gives the function $\epsilon_\X$ corresponding
@@ -331,7 +335,7 @@ Finally, \eqref{eq:implicit.eta} and \eqref{eq:inv.func} can be used in concert
 \begin{equation}
   \begin{aligned}
     \lambda_\X(x,x')
-    &=\frac{\delta^2F_\e[\epsilon]}{\delta\epsilon_\X(x)\delta\epsilon_\X(x')} \\
+    &=\frac{\delta^2F[\eta[\epsilon],\epsilon]}{\delta\epsilon_\X(x)\delta\epsilon_\X(x')} \\
     &=C_\X\delta(x-x')+
     b\frac{\delta\eta(x)}{\delta\epsilon_\X(x')}
     +\frac12b\int dx''\,\epsilon_{\X k}(x'')\frac{\delta^2\eta_k(x)}{\delta\epsilon_\X(x')\delta\epsilon_\X(x'')} \\