From 55217a41b7e35e193c06bf965036331d1a089e6a Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Mon, 24 Jun 2019 10:15:43 -0400
Subject: better way of getting results, some cleaning

---
 hidden-order.tex | 298 ++++++++++++++++++++++++++++++++++++++-----------------
 1 file changed, 206 insertions(+), 92 deletions(-)

(limited to 'hidden-order.tex')

diff --git a/hidden-order.tex b/hidden-order.tex
index 57ec702..364602e 100644
--- a/hidden-order.tex
+++ b/hidden-order.tex
@@ -44,7 +44,7 @@ The most general quadratic free energy density is built from combinations of str
     f_{\mathrm e}&=\frac12\sum_{\X}\lambda_{\X}^{(ij)}\epsilon_{\X}^{(i)}\epsilon_{\X}^{(j)}\\
     &=\frac12\Big[\lambda^{(11)}_{\Aog}(\epsilon_{\Aog}^{(1)})^2+
       \lambda_{\Aog}^{(22)}(\epsilon_{\Aog}^{(2)})^2+
-      \lambda_{\Aog}^{(12)}\epsilon_{\Aog}^{(1)}\epsilon_{\Aog}^{(2)}+
+      2\lambda_{\Aog}^{(12)}\epsilon_{\Aog}^{(1)}\epsilon_{\Aog}^{(2)}+
       \lambda_{\Bog}^{(11)}(\epsilon_{\Bog}^{(1)})^2+
       \lambda_{\Btg}^{(11)}(\epsilon_{\Btg}^{(1)})^2+
     \lambda_{\Eg}^{(11)}\epsilon_{\Eg}^{(1)}\cdot\epsilon_{\Eg}^{(1)}\Big]
@@ -53,144 +53,185 @@ The most general quadratic free energy density is built from combinations of str
 where the sum is over the irreducible representations of the point group.
 We can now identify
 \begin{align*}
-  \lambda_{\Aog}^{(1)}=\tfrac12(\lambda_{1111}+\lambda_{1122})
+  \lambda_{\Aog}^{(11)}=\tfrac12(\lambda_{1111}+\lambda_{1122})
   &&
-  \lambda_{\Aog}^{(2)}=\lambda_{3333}
+  \lambda_{\Aog}^{(22)}=\lambda_{3333}
   &&
-  \lambda_{\Aog}^{(3)}=2\lambda_{1133}
+  \lambda_{\Aog}^{(12)}=\lambda_{1133}
   \\
-  \lambda_{\Bog}^{(1)}=\tfrac12(\lambda_{1111}-\lambda_{1122})
+  \lambda_{\Bog}^{(11)}=\tfrac12(\lambda_{1111}-\lambda_{1122})
   &&
-  \lambda_{\Btg}^{(1)}=4\lambda_{1212}
+  \lambda_{\Btg}^{(11)}=4\lambda_{1212}
   &&
-  \lambda_{\Eg}^{(1)}=4\lambda_{1313}
+  \lambda_{\Eg}^{(11)}=4\lambda_{1313}
 \end{align*}
-Consider a generic order parameter $\eta$. To write down its free energy, we in principle must know both how it transforms under symmetry operations of the lattice and how derivative operators transform. For derivative operators, the two independent combinations are $\nabla_\parallel=\{\partial_x,\partial_y\}$ which transforms like $\Eg$, and $\nabla_\perp=\partial_3$ which transform like $\Aog$. The most general quartic free energy density (discounting total derivatives) is
-\[
-  f_{\mathrm o}=\frac12\Big[r\eta^2+
-                c_\parallel(\nabla_\parallel\eta)^2+
-                c_\perp(\partial_3\eta)^2+
-                D\big[(\nabla_\parallel^2+\partial_3^2)\eta\big]^2\Big]
-                +u\eta^4
-\]
-independent of the symmetry of $\eta$. This is the free energy for a Lifshitz point, and so we expect to see that phenomenology in $\eta$. The irreducible symmetry that $\eta$ transforms like determines its coupling to strain. To quadratic order we have
+Consider a generic order parameter $\eta$. To write down its free energy, we in principle must know both how it transforms under symmetry operations of the lattice and how derivative operators transform. For derivative operators, the two independent combinations are $\nabla_\parallel=\{\partial_x,\partial_y\}$ which transforms like $\Eg$, and $\nabla_\perp=\partial_3$ which transform like $\Aog$. The irreducible symmetry that $\eta$ transforms like determines its coupling to strain. To quadratic order we have
 \[
   f_{\X}=\tfrac12b^{(i)}\epsilon_{\X}^{(i)}\cdot\eta
   +\tfrac12e^{(i)}\epsilon_{\Aog}^{(i)}\eta^2
-\]
+V\]
 The total  free energy is 
 \[
   F=\int d^3x\,(f_{\mathrm e}+f_{\mathrm o}+f_{\X})
 \]
-Replacing $\eta$ with its inverse Fourier transform, we have
+There are three distinct possibilities here: $\eta$ transforms like a one-component irreducible representation of the point group that is not $\Aog$ ($\Bog$ or $\Btg$), $\eta$ transforms like $\Aog$, and $\eta$ is a two-component vector that transforms like $\Eg$.
+
+\section{$\Bog$ or $\Btg$ order parameter}
+
+We will first tackle the case of a non-$\Aog$, one-component order parameter.
+The most general quartic free energy density (discounting total derivatives) is
 \[
-  F_{\mathrm e}=\frac V2\sum_q\sum_X\lambda_X^{(ij)}\tilde\epsilon_X^{(i)}(q)\tilde\epsilon_X^{(j)}(-q)
+  f_{\mathrm o}=\frac12\Big[r\eta^2+
+                c_\parallel(\nabla_\parallel\eta)^2+
+                c_\perp(\nabla_\perp\eta)^2+
+                D(\nabla^2\eta)^2\Big]
+                +u\eta^4
 \]
+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
 \[
-  F_{\mathrm o}=\frac V2\sum_q\Big[r|\tilde\eta_q|^2+c_\parallel q_{\parallel}^2|\tilde\eta_q|^2
-  +c_\perp q_{\perp}^2|\tilde\eta_q|^2+Dq^4|\tilde\eta_q|^2\Big]
-  +Vu\sum_q\sum_{q'}\sum_{q''}\tilde\eta_q\tilde\eta_{q'}\tilde\eta_{q''}\tilde\eta_{-(q+q'+q'')}
+  0=\frac{\delta F}{\delta\epsilon_{\mathrm X}^{(1)}(x)}=\lambda_{\mathrm X}^{(11)}\epsilon_{\mathrm X}^{(1)}(x)+\frac12b^{(1)}\eta(x)
 \]
+whence we immediately have $\epsilon_{\mathrm X}^{(1)}=-\frac{b^{(1)}}{2\lambda_{\mathrm X}^{(11)}}\eta(x)$. We also have
 \[
-  F_{\X}=\frac V2\sum_q\Big[b^{(i)}\tilde\epsilon^{(i)}_{\X}(-q)\tilde\eta_q+\sum_qe^{(i)}\tilde\epsilon_{\Aog}^{(i)}(-(q+q'))\tilde\eta_q\tilde\eta_{q'}\Big]
+  0=\frac{\delta F}{\delta\epsilon_{\Aog}^{(i)}(x)}
+  =\lambda_{\Aog}^{(ij)}\epsilon_{\Aog}^{(j)}(x)+\frac12 e^{(i)}\eta^2(x)
 \]
-There are three distinct possibilities here: $\eta$ transforms like a one-component irreducible representation of the point group that is not $\Aog$ ($\Bog$ or $\Btg$), $\eta$ transforms like $\Aog$, and $\eta$ is a two-component vector that transforms like $\Eg$.
-
-We will first tackle the case of a non-$\Aog$, one-component order parameter. We will simply write $\epsilon=\epsilon_{\X}^{(1)}$, $\lambda=\lambda_{\X}^{(11)}$, and $\alpha^{(i)}=\epsilon_{\Aog}^{(i)}$ for the duration of this section.
-We will assume that our system orders at some specific $q_\perp=q^*$. This ansatz is equivalent to
+which is a linear system whose solutions are
 \begin{align*}
-  \tilde\eta_q=\tfrac12\eta_*\big[\delta_{q_\perp,q_*}+\delta_{q_\perp,-q_*}\big]\delta_{q_\parallel,0}
+  \epsilon_{\Aog}^{(1)}(x)=\frac12\frac{e^{(1)}\lambda_{\Aog}^{(22)}-e^{(2)}\lambda_{\Aog}^{(12)}}{((\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}}\eta^2(x)
   &&
-  \tilde\epsilon_q=\tfrac12\epsilon_*\big[\delta_{q_\perp,q_*}+\delta_{q_\perp,-q_*}\big]\delta_{q_\parallel,0}
+  \epsilon_{\Aog}^{(2)}(x)=\frac12\frac{e^{(2)}\lambda_{\Aog}^{(11)}-e^{(1)}\lambda_{\Aog}^{(12)}}{((\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}}\eta^2(x)
+\end{align*}
+All other strain fields are extremized for $\epsilon_{\mathrm Y}^{(i)}(x)=0$. These solutions may be substituted into the free energy to arrive at one that is only a functional of $\eta(x)$ whose free energy density is identical to $f_{\mathrm o}$ but with
+\begin{align*}
+  r\to \tilde r=r-\frac{(b^{(1)})^2}{4\lambda_{\X}^{(11)}}
   &&
-  \tilde\alpha_q^{(i)}=\delta_{q,0}
+  u\to \tilde u=u+\frac18\frac{(e^{(1)})^2\lambda_{\Aog}^{(22)}+(e^{(2)})^2\lambda_{\Aog}^{(11)}-2e^{(1)}e^{(2)}\lambda_{\Aog}^{(12)}}{(\lambda_{(12)}^{\Aog})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}}
 \end{align*}
-where because of the coupling of $\alpha$ to $\eta$ we don't expect it to produce modulated order (this can also be confirmed directly by minimizing over a modulated $\alpha$).
-For $q_*\neq0$, we have
+so that the total free energy is
 \[
-  F=\frac V2\bigg[\frac12(r+c_\perp q_*^2+Dq_*^4)\eta_*^2+\frac34u\eta_*^4+\frac12\lambda\epsilon_*^2+\lambda^{(ij)}\alpha^{(i)}_*\alpha^{(j)}_*+\frac12b\epsilon_*\eta_*+\frac12e^{(i)}\alpha_*^{(i)}\eta_*^2\bigg]
+  F=\int dx\bigg(\frac12\Big[\tilde r\eta^2+c_\parallel(\nabla_\parallel\eta)^2+c_\perp(\nabla_\perp\eta)^2+D(\nabla^2\eta)^2\Big]+\tilde u\eta^4\bigg)
 \]
-while for $q^*=0$ we have
+Replacing $\eta$ with its inverse Fourier transform, the free energy is
 \[
-  F=\frac V2\big(r\eta_*^2+2u\eta_*^4+\lambda\epsilon_*^2+\lambda^{(ij)}\alpha^{(i)}_*\alpha^{(j)}_*+b\epsilon_*\eta_*+e^{(i)}\alpha_*^{(i)}\eta_*^2\big)
+  F=\frac V2\sum_q\Big[\tilde r|\tilde\eta_q|^2+c_\parallel q_{\parallel}^2|\tilde\eta_q|^2
+  +c_\perp q_{\perp}^2|\tilde\eta_q|^2+Dq^4|\tilde\eta_q|^2\Big]
+  +V\tilde u\sum_q\sum_{q'}\sum_{q''}\tilde\eta_q\tilde\eta_{q'}\tilde\eta_{q''}\tilde\eta_{-(q+q'+q'')}
 \]
-For $r>b^2/4\lambda$ and $c_\perp>0$ or $r>b^2/4\lambda+c_\perp^2/4D$ and $c_\perp<0$, there is a free energy minimizer with all fields zero. For $r<b^2/4\lambda$ there is a local minimum with $q_*=0$ and 
-\begin{align*}
-  \eta_*^2=\frac12|r-r_c|\bigg(2u-\frac{(e^{(1)})^2\lambda_{\Aog}^{(22)}+(e^{(2)})^2\lambda_{\Aog}^{(11)}-e^{(1)}e^{(2)}\lambda_{\Aog}^{(12)}}{4\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}-(\lambda_{\Aog}^{(12)})^2}\bigg)^{-1}
-\end{align*}
-\begin{align*}
-  \epsilon_*=-\frac b{2\lambda}\eta_*
-  &&
-  \epsilon^{(1)}_*=\frac{e^{(2)}\lambda_{\Aog}^{(12)}-2e^{(1)}\lambda_{\Aog}^{(22)}}{4\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}-(\lambda_{\Aog}^{(12)})^2}\eta_*^2
-  &&
-  \epsilon^{(2)}_*=\frac{e^{(1)}\lambda_{\Aog}^{(12)}-2e^{(2)}\lambda_{\Aog}^{(11)}}{4\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}-(\lambda_{\Aog}^{(12)})^2}\eta_*^2
-\end{align*}
-with $r_c=b^2/4\lambda$. For $c_\perp<0$ and $r<b^2/4\lambda+c_\perp^2/4D$ there is a local minimum with $q_*=\sqrt{-c_\perp/2D}$ and
-\begin{align*}
-  \eta_*^2=|r-r_c|\bigg(3u-\frac{(e^{(1)})^2\lambda_{\Aog}^{(22)}+(e^{(2)})^2\lambda_{\Aog}^{(11)}-e^{(1)}e^{(2)}\lambda_{\Aog}^{(12)}}{4\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}-(\lambda_{\Aog}^{(12)})^2}\bigg)^{-1}
-\end{align*}
-\begin{align*}
-  \epsilon_*=-\frac b{2\lambda}\eta_*
-  &&
-  \epsilon^{(1)}_*=\frac12\frac{e^{(2)}\lambda_{\Aog}^{(12)}-2e^{(1)}\lambda_{\Aog}^{(22)}}{4\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}-(\lambda_{\Aog}^{(12)})^2}\eta_*^2
-  &&
-  \epsilon^{(2)}_*=\frac12\frac{e^{(1)}\lambda_{\Aog}^{(12)}-2e^{(2)}\lambda_{\Aog}^{(11)}}{4\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}-(\lambda_{\Aog}^{(12)})^2}\eta_*^2
-\end{align*}
-now with $r_c=b^2/4\lambda+c_\perp^2/4D$. Between the disordered and either of these phases is a continuous phase transition, and between the nontrivial phases is an abrupt transition.
-
-We are interested in the response of the system near the critical lines. The susceptibility can be found by adding an additional modulated field $h$ to the free energy linearly coupled to $\eta$ and computing
+The susceptibility of this system is
 \[
-  \chi(q)=\frac{\delta^2F}{\delta\tilde h(q)^2}\bigg|_{\tilde h(q)=0}
+  \begin{aligned}
+    \chi^{-1}(x,x')
+    &=\frac{\delta^2F}{\delta\eta(x)\delta\eta(x')}
+    =\frac{\delta}{\delta\eta(x')}\Big[\tilde r\eta(x)-c_\parallel\nabla_\parallel^2\eta(x)-c_\perp\nabla_\perp^2\eta(x)+D\nabla^4\eta(x)+4\tilde u\eta^3(x)\Big] \\
+    &=\Big[\tilde r-c_\parallel\nabla_\parallel^2-c_\perp\nabla_\perp^2+D\nabla^4+12\tilde u\eta^2(x)\Big]\delta(x-x')
+  \end{aligned}
 \]
-The susceptibility in the disordered (trivial) phase is
+or in Fourier space,
 \[
-  \chi(q)=\frac12\big(r-b^2/4\lambda+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4\big)^{-1}
+  \chi^{-1}(q)
+  =\tilde r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4+12\tilde u\sum_{q'}\tilde\eta(q')\tilde\eta(-q')
 \]
-The susceptibility in the in ordered (uniform) phase is
+To get the elastic susceptibility we must now extremize over $\eta$ and $\epsilon_{\Aog}^{(i)}$. The latter is done exactly as before, but the former givs
 \[
-  \chi(q)=\frac12\big(b^2/2\lambda-2r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4\big)^{-1}
+  0=\frac{\delta F}{\delta\eta}=\frac{\delta F_\eta}{\delta\eta}+\frac12b^{(1)}\epsilon_\X^{(1)}
 \]
-The susceptibility in the modulated phase is
+whose solutions give $\eta[\epsilon_\X^{(1)}]$ implicitly as
 \[
-  \chi(q)=\frac12\big(b^2/4\lambda+c_\perp^2/2D-r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4\big)^{-1}
+  \eta^{-1}[\eta(x)]=-\frac2{b^{(1)}}\frac{\delta F_\eta}{\delta\eta(x)}
 \]
-This makes the susceptibility near the disordered--ordered transition
+The inverse function theorem gives
 \[
-  \chi(q)=\frac12\begin{cases}\big[Dq^4+c_\parallel q_\parallel^2+c_\perp q_\perp^2+|\Delta r|\big]^{-1}&r>r_c\\\big[Dq^4+c_\parallel q_\parallel^2+c_\perp q_\perp^2+2|\Delta r|\big]^{-1}&r<r_c\end{cases}=\frac1{2c_\perp}\frac{\xi_\perp^2}{1+\xi_\parallel^2q_\parallel^2+\xi_\perp^2q_\perp^2+\xi_\perp^2(D/c_\perp)q^4}
+  \begin{aligned}
+    \bigg(\frac{\delta\eta(x)}{\delta\epsilon_\X^{(1)}(x')}\bigg)^{-1}
+    =\frac{\eta^{-1}[\eta(x)]}{\eta(x')}
+    =-\frac2{b^{(1)}}\frac{\delta^2F_\eta}{\delta\eta(x)\delta\eta(x')}
+    =-\frac2{b^{(1)}}\bigg[\chi^{-1}(x,x')+\frac{(b^{(1)})^2}{4\lambda_\X^{(11)}}\delta(x-x')\bigg]
+  \end{aligned}
 \]
-for
+The elastic susceptibility then follows as
 \begin{align*}
-  \xi_\perp=\begin{cases}(|\Delta r|/c_\perp)^{-1/2}&r<r_c\\(2|\Delta r|/c_\perp)^{-1/2}&r>r_c\end{cases}
-  &&
-  \xi_\parallel=\begin{cases}(|\Delta r|/c_\parallel)^{-1/2}&r<r_c\\(2|\Delta r|/c_\parallel)^{-1/2}&r>r_c\end{cases}
+  (\chi_{\X}^{(11)}(x,x'))^{-1}
+  &=\frac{\delta^2F}{\delta\epsilon_{\X}^{(1)}(x)\delta\epsilon_{\X}^{(1)}(x')} \\
+  &=\lambda_\X^{(11)}\delta(x-x')+
+  b^{(1)}\frac{\delta\eta(x)}{\delta\epsilon_\X^{(1)}(x')}+\frac12b^{(1)}\int dx''\,\epsilon(x'')\frac{\delta^2\eta(x)}{\delta\epsilon_\X^{(1)}(x')\delta\epsilon_\X^{(1)}(x'')} \\
+  &\qquad+\int dx''\,dx'''\,\frac{\delta^2F_\eta}{\delta\eta(x'')\delta\eta(x''')}\frac{\delta\eta(x'')}{\delta\epsilon_\X^{(1)}(x)}\frac{\delta\eta(x''')}{\delta\epsilon_\X^{(1)}(x')} \\ 
+  &=\lambda_\X^{(11)}\delta(x-x')+
+  b^{(1)}\frac{\delta\eta(x)}{\delta\epsilon_\X^{(1)}(x')}
+  -\frac12b^{(1)}\int dx''\,dx'''\,\bigg(\frac{\partial\eta(x'')}{\partial\epsilon_\X^{(1)}(x''')}\bigg)^{-1}\frac{\delta\eta(x'')}{\delta\epsilon_\X^{(1)}(x)}\frac{\delta\eta(x''')}{\delta\epsilon_\X^{(1)}(x')} \\ 
+  &=\lambda_\X^{(11)}\delta(x-x')+
+  b^{(1)}\frac{\delta\eta(x)}{\delta\epsilon_\X^{(1)}(x')}
+  -\frac12b^{(1)}\int dx''\,\delta(x-x'')\frac{\delta\eta(x'')}{\delta\epsilon_\X^{(1)}(x')} \\
+  &=\lambda_\X^{(11)}\delta(x-x')+
+  \frac12b^{(1)}\frac{\delta\eta(x)}{\delta\epsilon_\X^{(1)}(x')}
+\end{align*}
+whence
+\[
+  \chi_\X^{(11)}(x,x')
+  =\bigg(\lambda\delta(x-x')+\frac12b^{(1)}\frac{\delta\eta(x)}{\delta\epsilon_\X^{(1)}(x')}\bigg)^{-1}
+  =\frac1{\lambda_\X^{(11)}}\bigg(\delta(x-x')+\frac{(b^{(1)})^2}{4\lambda_\X^{(11)}}\chi(x,x')\bigg)
+\]
+or
+\[
+  \chi_\X^{(11)}(q)=\frac1{\lambda_\X^{(11)}}\bigg(1+\frac{(b^{(1)})^2}{4\lambda_\X^{(11)}}\chi(q)\bigg)
+\]
+The same process can be used to get the susceptibility for the $\Aog$ components. Now extremizing over $\epsilon_{\X}^{(1)}$ and $\eta$, we have
+\[
+  0=\frac{\delta F}{\delta\eta}=\frac{\delta F_\eta}{\delta\eta}+e^{(i)}\epsilon_{\Aog}^{(i)}\eta
+\]
+Unfortunately we cannot use the inverse function theorem as before. I'm still figuring out how to do this cleanly.
+
+We will assume that our system orders at some specific $q_\perp=q^*$. This ansatz is equivalent to
+\begin{align*}
+  \tilde\eta(q)=\tfrac12\eta_*\big[\delta_{q_\perp,q_*}+\delta_{q_\perp,-q_*}\big]\delta_{q_\parallel,0}
 \end{align*}
-and the susceptibility near the disordered--modulated transition is
+For $q_*\neq0$, we have
+\[
+  F=\frac V2\bigg[\frac12(\tilde r+c_\perp q_*^2+Dq_*^4)\eta_*^2+\frac34\tilde u\eta_*^4\bigg]
+\]
+while for $q^*=0$ we have
+\[
+  F=\frac V2\big(\tilde r\eta_*^2+2\tilde u\eta_*^4\big)
+\]
+For $\tilde r>0$ and $c_\perp>0$ or $r>c_\perp^2/4D$ and $c_\perp<0$, there is a free energy minimizer with $\eta_*=0$. For $r<0$ there is a local minimum with $q_*=0$ and $\eta_*^2=-\tilde r/4\tilde u$. For $c_\perp<0$ and $r<c_\perp^2/4D$ there is a local minimum with $q_*^2=-c_\perp/2D$ and
 \[
-  \chi(q)=\frac12\big[c_\parallel q_\parallel^2+D(q_\parallel^4+2q_\parallel^2q_\perp^2)+D(q_*^2-q_\perp^2)^2+|\Delta r|\big]^{-1}=\frac1{2D}\frac{\xi_\perp^4}{1+\xi_\parallel^2q_\parallel^2+\xi_\perp^4(q_\parallel^4+2q_\parallel^2q_\perp^2)+\xi_\perp^{4}(q_*^2-q_\perp^2)^2}
+  \eta_*^2=\frac{c_\perp^2-4D\tilde r}{12D\tilde u}=-\frac{\tilde r-\tilde r_c}{3\tilde u}
 \]
-where $\xi_\perp=(|\Delta r|/D)^{-1/4}$ and $\xi_\parallel=(|\Delta r|/c_\parallel)^{-1/2}$. We're also interested in the elastic response, defined by
+with $\tilde r_c=c_\perp^2/4D$. Between the disordered and either of these phases is a continuous phase transition, and between the nontrivial phases is an abrupt transition.
+
+In the disordered phase, $\sum_{q}\tilde\tilde\eta(q)\tilde\eta(-q)=0$ and the susceptibility is
 \[
-  \frac12\tilde\lambda_{\X}^{(ij)}(q)=\frac{\delta^2F}{\delta\epsilon_{\mathrm X}^{(i)}(q)\delta\epsilon_{\mathrm X}^{(j)}(q)}\bigg|_{\epsilon=\epsilon_*}
+  \chi^{-1}(q)
+  =\tilde r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4
 \]
-The effective elastic susceptibility for the symmetry component of our order parameter at the disordered--ordered transition has a response of the form
+In the ordered phase, $\sum_q\tilde\eta(q)\tilde\eta(-q)=\eta_*^2=-\tilde r/4\tilde u$, and
 \[
-  \frac{\lambda}{\tilde\lambda(q)}=1+\frac{b^2}{4\lambda}\begin{cases}\big[Dq^4+c_\perp q^2+c_\parallel q_\parallel^2+|\Delta r|\big]^{-1}&r>r_c\\
-    \big[Dq^4+c_\perp q_\perp^2+c_\parallel q_\parallel^2+2|\Delta r|\big]^{-1}&r<r_c
-  \end{cases}
-  =1+\frac{b^2}{2\lambda}\chi(q)
+  \chi^{-1}(q)
+  =\tilde r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4-3\tilde r
+  =-2\tilde r+c_\parallel q_\parallel^2+c_\perp q_\perp^2
 \]
-and the effective susceptibility for $\Aog$ has a response like
+In the modulated phase, $\sum_q\tilde\eta(q)\tilde\eta(-q)=\frac12\eta_*^2=(c_\perp^2-4D\tilde r)/24D\tilde u$ and
 \[
-  \tilde\lambda_{\Aog}^{(ij)}(q)-\lambda_{\Aog}^{(ij)}=\delta_{q,0}\begin{cases}0&r>r_c\\-e^{(i)}e^{(j)}/4u&r<r_c\end{cases}
+  \chi^{-1}(q)
+  =\tilde r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4+c_\perp^2/2D-2\tilde r
+  =-(\tilde r-\tilde r_c)+c_\parallel q_\parallel^2+D(q_\parallel^4+2q_\parallel^2q_\perp^2)+D(q_*^2-q_\perp^2)^2
 \]
-At the disordered--modulated transition the responses are of the form
+At the disordered--ordered transition with $\tilde r_c=0$, this yields
 \[
-  \frac{\lambda}{\tilde\lambda(q)}=1+\frac{b^2}{4\lambda}\big[c_\parallel q_\parallel^2+D(q_\parallel^4+q_\parallel^2q_\perp^2)+D(q_*^2-q_\perp^2)^2+|\Delta r|\big]^{-1}=1+\frac{b^2}{2\lambda}\chi(q)
+  \chi(q)=\begin{cases}\big[Dq^4+c_\parallel q_\parallel^2+c_\perp q_\perp^2+|\Delta\tilde r|\big]^{-1}&\Delta\tilde r>0\\\big[Dq^4+c_\parallel q_\parallel^2+c_\perp q_\perp^2+2|\Delta\tilde r|\big]^{-1}&\Delta\tilde r<0\end{cases}=\frac1{c_\perp}\frac{\xi_\perp^2}{1+\xi_\parallel^2q_\parallel^2+\xi_\perp^2q_\perp^2+\xi_\perp^2(D/c_\perp)q^4}
 \]
-and
+where the correlation lengths are
+\begin{align*}
+  \xi_\perp=\begin{cases}(|\Delta\tilde r|/c_\perp)^{-1/2}&\tilde r<\tilde r_c\\(2|\Delta \tilde r|/c_\perp)^{-1/2}&\tilde r>\tilde r_c\end{cases}
+  &&
+  \xi_\parallel=\begin{cases}(|\Delta\tilde r|/c_\parallel)^{-1/2}&\tilde r<\tilde r_c\\(2|\Delta \tilde r|/c_\parallel)^{-1/2}&\tilde r>\tilde r_c\end{cases}
+\end{align*}
+At the disordered--modulated transition with $\tilde r_c=c_\perp^2/4D$, this yields
 \[
-  \tilde\lambda_{\Aog}^{(ij)}(q)-\lambda_{\Aog}^{(ij)}=\delta_{q,0}\begin{cases}0&r>r_c\\-e^{(i)}e^{(j)}/6u&r<r_c\end{cases}
+  \chi(q)=\big[c_\parallel q_\parallel^2+D(q_\parallel^4+2q_\parallel^2q_\perp^2)+D(q_*^2-q_\perp^2)^2+|\Delta\tilde r|\big]^{-1}=\frac1{D}\frac{\xi_\perp^4}{1+\xi_\parallel^2q_\parallel^2+\xi_\perp^4(q_\parallel^4+2q_\parallel^2q_\perp^2)+\xi_\perp^{4}(q_*^2-q_\perp^2)^2}
 \]
+where $\xi_\perp=(|\Delta\tilde r|/D)^{-1/4}$ and $\xi_\parallel=(|\Delta \tilde r|/c_\parallel)^{-1/2}$. 
 
 The Ginzburg criterion gives the proximity $t_G$ of the critical point at which mean field theory is no longer self consistent, with
 \begin{align*}
@@ -214,13 +255,86 @@ and therefore $t_G\sim1.2\times10^{-6}$, which means one would need to get withi
   \label{fig:B1g.fit}
 \end{figure}
 
+\section{$\Aog$ order parameter}
+
+For the case where $\eta$ transforms like $\Aog$, the most general free energy is
+\[
+  f_{\mathrm o}=\frac12\Big[r\eta^2+
+                c_\parallel(\nabla_\parallel\eta)^2+
+                c_\perp(\partial_3\eta)^2+
+                g(\partial_3\eta)^3+
+                D\big[(\nabla_\parallel^2+\partial_3^2)\eta\big]^2\Big]
+                +v\eta^3
+                +u\eta^4
+\]
+Extremizing over $\epsilon$ requires
+\[
+  0=\frac{\delta F}{\delta\epsilon_{\Aog}^{(i)}(x)}
+  =\lambda_{\Aog}^{(ij)}\epsilon_{\Aog}^{(j)}(x)+\frac12 b^{(i)}\eta(x)+\frac12 e^{(i)}\eta^2(x)
+\]
+which is a system of equations with solution
+\begin{align*}
+  \epsilon_{\Aog}^{(1)}(x)&=-\frac1{2\big[(\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}\big]}\bigg[
+      \Big(b^{(2)}\lambda_{\Aog}^{(12)}-b^{(1)}\lambda_{\Aog}^{(22)}\Big)\eta(x)+\Big(e^{(2)}\lambda_{\Aog}^{(12)}-e^{(1)}\lambda_{\Aog}^{(22)}\Big)\eta^2(x)
+    \bigg]\\
+    \epsilon_{\Aog}^{(2)}(x)&=-\frac1{2\big[(\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}\big]}\bigg[
+      \Big(b^{(1)}\lambda_{\Aog}^{(12)}-b^{(2)}\lambda_{\Aog}^{(22)}\Big)\eta(x)+\Big(e^{(1)}\lambda_{\Aog}^{(12)}-e^{(2)}\lambda_{\Aog}^{(11)}\Big)\eta^2(x)
+    \bigg]
+\end{align*}
+All other strains are zero. Like the previous case, substitution of this solution into the free energy results in a functional of $\eta$ alone. As before, the result is a free energy density with the form of $f_{\mathrm o}$ but with
+\begin{align*}
+  r&\to \tilde r=r+\frac18\frac{(b^{(1)})^2\lambda_{\Aog}^{(22)}+(b^{(2)})^2\lambda_{\Aog}^{(11)}-2b^{(1)}b^{(2)}\lambda^{(12)}_{\Aog}}{(\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}} \\
+  u&\to \tilde u=u+\frac18\frac{(e^{(1)})^2\lambda_{\Aog}^{(22)}+(e^{(2)})^2\lambda_{\Aog}^{(11)}-2e^{(1)}e^{(2)}\lambda^{(12)}_{\Aog}}{(\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}}
+\end{align*}
+Unlike before, there is now an additional term in $f$ proportional to $\eta^3(x)$ with coefficient
+\[
+  v\to\tilde v=v+\frac14\frac{b^{(1)}e^{(1)}\lambda_{\Aog}^{(22)}+b^{(2)}e^{(2)}\lambda_{\Aog}^{(11)}-\lambda_{\Aog}^{(12)}(b^{(1)}e^{(2)}+b^{(2)}e^{(1)})}{(\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}}
+\]
+This implies that, unless the model is tuned to a place where this term vanishes, all transitions will be abrupt. We won't study this phenomenology any further.
+
+\section{$\Eg$ order parameter}
+
+For a $\Eg$ order parameter, the most general quartic free energy density is
+\[
+  f_{\mathrm o}=\frac12\Big[r\eta^2+c_\parallel(\nabla_\parallel\cdot\eta)^2+c_\perp(\nabla_\perp\eta)^2+D(\nabla^2\eta)^2\Big]+u\eta^4
+\]
+Minimizing the strain in $\Aog$ proceeds exactly as before, while $\Eg$ gives
+\[
+  0=\frac{\delta F}{\delta\epsilon_{\Eg i}^{(1)}(x)}=\lambda_{\Eg}^{(11)}\epsilon_{\Eg i}^{(1)}+\frac12b^{(1)}\eta_i(x)
+\]
+whence $\epsilon_{\Eg i}^{(1)}(x)=-b^{(1)}/2\lambda_{\Eg}^{(11)}\eta_i(x)$. This leads to the same shifts in $r$ and $u$ as before, and we will use $\tilde r$ and $\tilde u$ to take the same meanings as they did for the $\Bog$ case. We now have
+\[
+  \begin{aligned}
+    \chi^{-1}_{ij}(x,x')
+    &=\frac{\delta^2F}{\delta\eta_i(x)\delta\eta_j(x')}
+    =\frac\delta{\delta\eta_j(x')}\Big[\tilde r\eta_i(x)-c_\parallel\partial_i^2\eta_i(x)-c_\perp\nabla_\perp^2\eta_i(x)+D\nabla^4\eta_i(x)+4\tilde u\eta_k(x)\eta_k(x)\eta_i(x)\Big] \\
+    &=\Big\{\Big[\tilde r-c_\parallel\partial_i^2-c_\perp\nabla_\perp^2+D\nabla^4+4\tilde u\eta^2(x)\Big]\delta_{ij}+8\tilde u\eta_i(x)\eta_j(x)\Big\}\delta(x-x')
+  \end{aligned}
+\]
+The Fourier transformed version is
+\[
+  \chi^{-1}_{ij}(q)=\Big[\tilde r+c_\parallel q_i^2+c_\perp q_\perp^2+Dq^4+4\tilde u\sum_{q'}\tilde\eta_i(q')\tilde\eta_i(-q')\Big]\delta_{ij}+8\tilde u\sum_{q'}\tilde\eta_i(q')\tilde\eta_j(-q')
+\]
 
-The conditions for $\eta$ being a stationary function of $F$ is
+We will assume the system orders at some $q_\perp=q_*$ as before, with
+\begin{align*}
+  \tilde\eta_1(q)=\tfrac12\eta_*(\delta_{q_\perp,q_*}+\delta_{q_\perp,-q_*})\delta_{q_\parallel,0}
+  &&
+  \tilde\eta_2(q)=\tfrac12\eta_*(e^{-i\phi_*}\delta_{q_\perp,q_*}+e^{i\phi_*}\delta_{q_\perp,-q_*})\delta_{q_\parallel,0}
+\end{align*}
+where we have introduced a relative phase $\phi_*$ between the two components. For $q_*\neq0$ we have
+\[
+  F=\frac V2\big[\tilde r+c_\perp q_*^2+Dq_*^4\big]\eta_*^2+\frac V4\tilde u\big[5+\cos(2\phi)\big]\eta_*^4
+\]
+while for $q_*=0$ we have
+\[
+  F=\frac14\tilde r\big[3+\cos(2\phi)\big]\eta_*^2+\frac34\tilde u\big[3+\cos(2\phi)\big]^2\eta_*^4
+\]
+There is as before an $\eta^*=0$ solution corresponding to a trivial state, with the same boundaries as the $\Bog$ case. There is an ordered state with $\eta_*^2=-\tilde r/4\tilde u$ and $\phi_*=\pm\pi/2$. Finally, there is a modulated state with $q_*^2=-c_\perp/2D$, $\phi^*=\pm\pi/2$, and
 \[
-  0=\frac{\delta F}{\delta\eta}=\frac{\partial f}{\partial\eta}-\partial_i\frac{\partial f}{\partial(\partial_i\eta)}+\partial_i^2\frac{\partial f}{\partial(\partial_i^2\eta)}
-  =r\eta-c_\parallel\nabla_\parallel^2\eta-c_\perp\partial_3^2\eta+D\nabla^4\eta+2u\eta^3
+  \eta_*^2=\frac{c_\perp^2-4D\tilde r}{16D\tilde u}
 \]
-We can estimate the error in our ansatz by how much the total variation in $F$ differs from zero.
+Now, however, the ordered state becomes unstable when $c_\perp<0$, and there is a continuous transition between the ordered and metastable states along the $c_\perp=0$, $r<0$ line.
 
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 \bibliography{hidden_order.bib}
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