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\title{Elastic Lifshitz point}
\author{Jaron Kent-Dobias}

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\def\Aog{\mathrm A_{1\mathrm g}}
\def\Bog{\mathrm B_{1\mathrm g}}
\def\Btg{\mathrm B_{2\mathrm g}}
\def\Eg{\mathrm E_{\mathrm g}}
\def\X{\mathrm X}

\maketitle

The free energy density of our model has three terms, respectively corresponding to the strain, the order parameter, and the strain--order parameter coupling. The free energy density for the strain, taken to quadratic order under the assumption of small strains, is generically of the form $f_{\mathrm e}=\frac12\lambda_{ijk\ell}\epsilon_{ij}\epsilon_{k\ell}$. Symmetries of the strain tensor, which is symmetric, and those of the lattice restrict the form of $\lambda$. The most general form for a system with a tetragonal point group is
\[
  f_{\mathrm e}=\frac12\Big[\lambda_{1111}(\epsilon_{11}^2+\epsilon_{22}^2)+
    \lambda_{3333}\epsilon_{33}^2+
      2\lambda_{1133}(\epsilon_{11}+\epsilon_{22})\epsilon_{33}+
      2\lambda_{1122}\epsilon_{11}\epsilon_{22}+
      4\lambda_{1212}\epsilon_{12}^2+
      4\lambda_{1313}(\epsilon_{13}^2+\epsilon_{23}^2)\Big]
\]
A very convenient way to come by this form is to use knowledge of the point group of the lattice and the way that strain transforms under it. The following linear combinations of strains transform like a particular irreducible representation of the point group:
\begin{align*}
  \epsilon_{\Aog}^{(1)}=\epsilon_{11}+\epsilon_{22}
  &&
  \epsilon_{\Aog}^{(2)}=\epsilon_{33}
  \\
  \epsilon_{\Bog}^{(1)}=\epsilon_{11}-\epsilon_{22}
  &&
  \epsilon_{\Btg}^{(1)}=\epsilon_{12}
  &&
  \epsilon_{\mathrm E_{\mathrm G}}^{(1)}=\{\epsilon_{11},\epsilon_{22}\}
\end{align*}
The most general quadratic free energy density is built from combinations of strains that transform like scalars, or $\Aog$, yielding
\[
  \begin{aligned}
    f_{\mathrm e}&=\frac12\sum_{\X}\lambda_{\X}^{(ij)}\epsilon_{\X}^{(i)}\epsilon_{\X}^{(j)}\\
    &=\frac12\Big[\lambda^{(11)}_{\Aog}(\epsilon_{\Aog}^{(1)})^2+
      \lambda_{\Aog}^{(22)}(\epsilon_{\Aog}^{(2)})^2+
      \lambda_{\Aog}^{(12)}\epsilon_{\Aog}^{(1)}\epsilon_{\Aog}^{(2)}+
      \lambda_{\Bog}^{(11)}(\epsilon_{\Bog}^{(1)})^2+
      \lambda_{\Btg}^{(11)}(\epsilon_{\Btg}^{(1)})^2+
    \lambda_{\Eg}^{(11)}\epsilon_{\Eg}^{(1)}\cdot\epsilon_{\Eg}^{(1)}\Big]
  \end{aligned}
\]
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}^{(2)}=\lambda_{3333}
  &&
  \lambda_{\Aog}^{(3)}=2\lambda_{1133}
  \\
  \lambda_{\Bog}^{(1)}=\tfrac12(\lambda_{1111}-\lambda_{1122})
  &&
  \lambda_{\Btg}^{(1)}=4\lambda_{1212}
  &&
  \lambda_{\Eg}^{(1)}=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
\[
  f_{\X}=\tfrac12b^{(i)}\epsilon_{\X}^{(i)}\cdot\eta
  +\tfrac12e^{(i)}\epsilon_{\Aog}^{(i)}\eta^2
\]
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
\[
  F_{\mathrm e}=\frac V2\sum_q\sum_X\lambda_X^{(ij)}\tilde\epsilon_X^{(i)}(q)\tilde\epsilon_X^{(j)}(-q)
\]
\[
  F_{\mathrm o}=\frac V2\sum_q\Big[r|\tilde\eta_q|^2+c_\parallel q_{\parallel}^2|\tilde\eta_q|^2
  +c_\perp q_{\perp}^2|\tilde\eta_q|^2+Dq^4|\tilde\eta_q|^2\Big]
  +Vu\sum_q\sum_{q'}\sum_{q''}\tilde\eta_q\tilde\eta_{q'}\tilde\eta_{q''}\tilde\eta_{-(q+q'+q'')}
\]
\[
  F_{\X}=\frac V2\sum_q\Big[b^{(i)}\tilde\epsilon^{(i)}_{\X}(-q)\tilde\eta_q+\sum_qe^{(i)}\tilde\epsilon_{\Aog}^{(i)}(-(q+q'))\tilde\eta_q\tilde\eta_{q'}\Big]
\]
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
\begin{align*}
  \tilde\eta_q=\tfrac12\eta_*\big[\delta_{q_\perp,q_*}+\delta_{q_\perp,-q_*}\big]\delta_{q_\parallel,0}
  &&
  \tilde\epsilon_q=\tfrac12\epsilon_*\big[\delta_{q_\perp,q_*}+\delta_{q_\perp,-q_*}\big]\delta_{q_\parallel,0}
  &&
  \tilde\alpha_q^{(i)}=\delta_{q,0}
\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
\[
  F=\frac V2\bigg[\frac12(r+c_\perp q_*^2+Dq_*^4)\eta_*^2+\frac34u\eta_*^4+\frac12\lambda\epsilon_*^2+\lambda^{(ij)}\alpha^{(i)}_*\alpha^{(j)}_*+\frac12b\epsilon_*\eta_*+\frac12e^{(i)}\alpha_*^{(i)}\eta_*^2\bigg]
\]
while for $q^*=0$ we have
\[
  F=\frac V2\big(r\eta_*^2+2u\eta_*^4+\lambda\epsilon_*^2+\lambda^{(ij)}\alpha^{(i)}_*\alpha^{(j)}_*+b\epsilon_*\eta_*+e^{(i)}\alpha_*^{(i)}\eta_*^2\big)
\]
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)}_*=\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*}
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*}
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
\[
  \chi(q)=\frac{\delta^2F}{\delta\tilde h(q)^2}\bigg|_{\tilde h(q)=0}
\]
The susceptibility in the disordered (trivial) phase is
\[
  \chi(q)=\frac12\big(r-b^2/4\lambda+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4\big)^{-1}
\]
The susceptibility in the in ordered (uniform) phase is
\[
  \chi(q)=\frac12\big(b^2/2\lambda-2r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4\big)^{-1}
\]
The susceptibility in the modulated phase is
\[
  \chi(q)=\frac12\big(b^2/4\lambda+c_\perp^2/2D-r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4\big)^{-1}
\]
This makes the susceptibility near the disordered--ordered transition
\[
  \chi(q)=\frac12\begin{cases}\big[Dq^4+c_\parallel q_\parallel^2+c_\perp q_\perp^2+|\Delta r|\big]^{-1}&r>r_c\\\big[Dq^4+c_\parallel q_\parallel^2+c_\perp q_\perp^2+2|\Delta r|\big]^{-1}&r<r_c\end{cases}=\frac1{2c_\perp}\frac{\xi_\perp^2}{1+\xi_\parallel^2q_\parallel^2+\xi_\perp^2q_\perp^2+\xi_\perp^2(D/c_\perp)q^4}
\]
for
\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}
\end{align*}
and the susceptibility near the disordered--modulated transition is
\[
  \chi(q)=\frac12\big[c_\parallel q_\parallel^2+D(q_0^2-q_\perp^2)^2+|\Delta r|\big]^{-1}=\frac1{2D}\frac{\xi_\perp^4}{1+\xi_\parallel^2q_\parallel^2+\xi_\perp^{4}(q_0^2-q^2)^2}
\]
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
\[
  \frac12\tilde\lambda_{\X}^{(ij)}(q)=\frac{\delta^2F}{\delta\epsilon_{\mathrm X}^{(i)}(q)\delta\epsilon_{\mathrm X}^{(j)}(q)}\bigg|_{\epsilon=\epsilon_*}
\]
The effective elastic susceptibility for the symmetry component of our order parameter at the disordered--ordered transition has a response of the form
\[
  \frac{\lambda}{\tilde\lambda(q)}=1+\frac{b^2}{4\lambda}\begin{cases}\big[Dq^4+c_\perp q^2+c_\parallel q_\parallel^2+|\Delta r|\big]^{-1}&r>r_c\\
    \big[Dq^4+c_\perp q_\perp^2+c_\parallel q_\parallel^2+2|\Delta r|\big]^{-1}&r<r_c
  \end{cases}
  =1+\frac{b^2}{2\lambda}\chi(q)
\]
and the effective susceptibility for $\Aog$ has a response like
\[
  \tilde\lambda_{\Aog}^{(ij)}(q)-\lambda_{\Aog}^{(ij)}=\delta_{q,0}\begin{cases}0&r>r_c\\-e^{(i)}e^{(j)}/4u&r<r_c\end{cases}
\]
At the disordered--modulated transition the responses are of the form
\[
  \frac{\lambda}{\tilde\lambda(q)}=1+\frac{b^2}{4\lambda}\big[c_\parallel q_\parallel^2+D(q_0^2-q_\perp^2)^2+|\Delta r|\big]^{-1}=1+\frac{b^2}{2\lambda}\chi(q)
\]
and
\[
  \tilde\lambda_{\Aog}^{(ij)}(q)-\lambda_{\Aog}^{(ij)}=\delta_{q,0}\begin{cases}0&r>r_c\\-e^{(i)}e^{(j)}/6u&r<r_c\end{cases}
\]

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*}
  t_G=\frac{k_B^2}{32\pi^2(\Delta c_V)^2\xi_0^6}
\end{align*}
Experiments give $\Delta c_V\sim1\times10^5\,\mathrm J\,\mathrm m^{-3}\,\mathrm K^{-1}$ \cite{fisher_specific_1990} and $T_c\sim17.5\,\mathrm K$. A fit of $\tilde\lambda$ to experimental data (Fig.~\ref{fig:B1g.fit}) yields
\begin{align*}
  \lambda=71\,\mathrm{GPa}-(0.10\,\mathrm{GPa}\,\mathrm{K}^{-1})T &&
  \frac{b^2}{4\lambda Dq_0^4}=0.084&&\frac a{Dq_0^4}=0.0038\,\mathrm K^{-1}
\end{align*}
with $|r-r_c|=a|T-T_c|$. We suspect that the modulation at the transition at very low pressure is on the order of the lattice spacing, which would give $q_0^{-1}\sim9.568\,\text{\r A}$. Combined with our fit, this gives
\[
  \xi_0=(aT_c/D)^{-1/4}=\big[T_c(a/Dq_0^4)q_0^4\big]^{-1/4}\sim19\,\text{\r A}
\]
and therefore $t_G\sim1.2\times10^{-6}$, which means one would need to get within about $\Delta T=T_ct_G\sim20\,\mu\mathrm K$ of the critical point to see mean field theory break down.

\begin{figure}
  \centering
  \includegraphics[width=0.5\textwidth]{cusp}
  \caption{Experimental data for the $\Bog$ effective elastic constant $\tilde\lambda_{\Bog}$ (blue dots), along with a fit of the above theory (orange line).}
  \label{fig:B1g.fit}
\end{figure}


The conditions for $\eta$ being a stationary function of $F$ is
\[
  0=\frac{\delta F}{\delta\eta}=\frac{\partial f}{\partial\eta}-\partial_i\frac{\partial f}{\partial(\partial_i\eta)}+\partial_i^2\frac{\partial f}{\partial(\partial_i^2\eta)}
  =r\eta-c_\parallel\nabla_\parallel^2\eta-c_\perp\partial_3^2\eta+D\nabla^4\eta+2u\eta^3
\]
We can estimate the error in our ansatz by how much the total variation in $F$ differs from zero.

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