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@@ -1,5 +1,5 @@ -\documentclass[aps,prb,reprint,longbibliography,floatfix]{revtex4-1} +\documentclass[aps,prb,reprint,longbibliography,floatfix,fleqn]{revtex4-1} \usepackage[utf8]{inputenc} \usepackage{amsmath,graphicx,upgreek,amssymb,xcolor} \usepackage[colorlinks=true,urlcolor=purple,citecolor=purple,filecolor=purple,linkcolor=purple]{hyperref} @@ -111,9 +111,10 @@ broken symmetry remains unknown. This state, known as \emph{hidden order} lower temperatures. At sufficiently large hydrostatic pressures, both superconductivity and \ho\ give way to local moment antiferromagnetism (\afm).\cite{Hassinger_2008} Modern theories~\cite{Kambe_2018, Haule_2009, - Kusunose_2011, Kung_2015, Cricchio_2009, Ohkawa_1999, Santini_1994, -Kiss_2005, Harima_2010, Thalmeier_2011, Tonegawa_2012, Rau_2012, Riggs_2015, -Hoshino_2013, Ikeda_1998, Chandra_2013a, 1902.06588v2, Ikeda_2012} propose +Kusunose_2011_On, Kung_2015, Cricchio_2009, Ohkawa_1999, Santini_1994, +Kiss_2005, Harima_2010, Thalmeier_2011, Tonegawa_2012_Cyclotron, +Rau_2012_Hidden, Riggs_2015_Evidence, Hoshino_2013_Resolution, +Ikeda_1998_Theory, Chandra_2013_Hastatic, 1902.06588v2, Ikeda_2012} propose associating any of a variety of broken symmetries with \ho. This work analyzes a family of phenomenological models with order parameters of general symmetry that couple linearly to strain. Of these, only one is compatible with two @@ -128,9 +129,10 @@ examine the thermodynamic discontinuities in the elastic moduli at $T_{\text{\ho}}$.\cite{1903.00552v1} The observation of discontinues only in compressional, or $\Aog$, elastic moduli requires that the point-group representation of \ho\ be one-dimensional. This rules out many order parameter -candidates~\cite{Thalmeier_2011, Tonegawa_2012, Rau_2012, Riggs_2015, -Hoshino_2013, Ikeda_2012, Chandra_2013b} in a model-independent way, but -doesn't differentiate between those that remain. +candidates~\cite{Thalmeier_2011, Tonegawa_2012_Cyclotron, Rau_2012_Hidden, +Riggs_2015_Evidence, Hoshino_2013_Resolution, Ikeda_2012, Chandra_2013_Origin} +in a model-independent way, but doesn't differentiate between those that +remain. Recent x-ray experiments discovered rotational symmetry breaking in \urusi\ under pressure.\cite{Choi_2018} Above 0.13--0.5 $\GPa$ (depending on @@ -462,33 +464,34 @@ signatures of a continuous transition by locating thermodynamic singularities at nonzero $q=q_*$. The remaining clue at $q=0$ is a particular kink in the corresponding modulus. -\begin{figure}[htpb] +\section{Comparison to experiment} + +\begin{figure*}[htpb] \centering - \includegraphics[width=\columnwidth]{fig-stiffnesses} + \includegraphics{fig-stiffnesses} \caption{ - \Rus\ measurements of the elastic moduli of \urusi\ at ambient pressure as a - function of temperature from recent experiments\cite{1903.00552v1} (blue, - solid) alongside fits to theory (magenta, dashed). The solid yellow region - shows the location of the \ho\ phase. (a) $\Btg$ modulus data and a fit to - the standard form.\cite{Varshni_1970} (b) $\Bog$ modulus data and a fit to - \eqref{eq:static_modulus}. The fit gives - $C^0_\Bog\simeq\big[71-(0.010\,\K^{-1})T\big]\,\GPa$, $D_\perp - q_*^4/b^2\simeq0.16\,\GPa^{-1}$, and - $a/b^2\simeq6.1\times10^{-4}\,\GPa^{-1}\,\K^{-1}$. Addition of a quadratic - term in $C^0_\Bog$ was here not needed for the fit.\cite{Varshni_1970} (c) - $\Bog$ modulus data and the fit of the \emph{bare} $\Bog$ modulus. (d) - $\Bog$ modulus data and the fit transformed by - $[C^0_\Bog(C^0_\Bog/C_\Bog-1)]]^{-1}$, which is predicted from - \eqref{eq:static_modulus} to equal $D_\perp q_*^4/b^2+a/b^2|T-T_c|$, e.g., - an absolute value function. The failure of the Ginzburg--Landau prediction - below the transition is expected on the grounds that the \op\ is too large - for the free energy expansion to be valid by the time the Ginzburg - temperature is reached. + \Rus\ measurements of the elastic moduli of \urusi\ at ambient pressure as a + function of temperature from recent experiments\cite{1903.00552v1} (blue, + solid) alongside fits to theory (magenta, dashed). The solid yellow region + shows the location of the \ho\ phase. (a) $\Btg$ modulus data and a fit to + the standard form.\cite{Varshni_1970} (b) $\Bog$ modulus data and a fit to + \eqref{eq:static_modulus}. The fit gives + $C^0_\Bog\simeq\big[71-(0.010\,\K^{-1})T\big]\,\GPa$, $D_\perp + q_*^4/b^2\simeq0.16\,\GPa^{-1}$, and + $a/b^2\simeq6.1\times10^{-4}\,\GPa^{-1}\,\K^{-1}$. Addition of a quadratic + term in $C^0_\Bog$ was here not needed for the fit.\cite{Varshni_1970} (c) + $\Bog$ modulus data and the fit of the \emph{bare} $\Bog$ modulus. (d) + $\Bog$ modulus data and the fit transformed by + $[C^0_\Bog(C^0_\Bog/C_\Bog-1)]]^{-1}$, which is predicted from + \eqref{eq:static_modulus} to equal $D_\perp q_*^4/b^2+a/b^2|T-T_c|$, e.g., + an absolute value function. The failure of the Ginzburg--Landau prediction + below the transition is expected on the grounds that the \op\ is too large + for the free energy expansion to be valid by the time the Ginzburg + temperature is reached. } \label{fig:data} -\end{figure} +\end{figure*} -\section{Comparison to experiment} \Rus\ experiments~\cite{1903.00552v1} yield the individual elastic moduli broken into irreps; data for the $\Bog$ and $\Btg$ components defined in \eqref{eq:strain-components} are shown in Figures \ref{fig:data}(a--b). The @@ -497,18 +500,27 @@ the presence of the transition, exhibiting the expected linear stiffening upon cooling from room temperature, with a low-temperature cutoff at some fraction of the Debye temperature.\cite{Varshni_1970} The $\Bog$ modulus Fig.~\ref{fig:data}(b) has a dramatic response, softening over the course of -roughly $100\,\K$ and then cusping at the \ho\ transition. While the -low-temperature response is not as dramatic as the theory predicts, mean field -theory---which is based on a small-$\eta$ expansion---will not work -quantitatively far below the transition where $\eta$ has a large nonzero value -and higher powers in the free energy become important. The data in the +roughly $100\,\K$ and then cusping at the \ho\ transition. The data in the high-temperature phase can be fit to the theory \eqref{eq:static_modulus}, with a linear background modulus $C^0_\Bog$ and $\tilde r-\tilde r_c=a(T-T_c)$, and -the result is shown in Figure \ref{fig:data}(b). The data and theory appear -quantitatively consistent in the high temperature phase, suggesting that \ho\ -can be described as a $\Bog$-nematic phase that is modulated at finite $q$ -along the $c-$axis. The predicted softening appears over hundreds of Kelvin; -Figures \ref{fig:data}(c--d) show the background modulus $C_\Bog^0$ and the +the result is shown in Figure \ref{fig:data}(b). + +The behavior of the modulus below the transition does not match +\eqref{eq:static_modulus} well, but this is because of the truncation of the +free energy expansion used above. Higher order terms like $\eta^2\epsilon^2$ +contribute to the modulus starting at order $\eta_*^2$, and therefore while +they do not affect the behavior above the transition, they change the behavior +below it. To demonstrate this, in Appendix~\ref{sec:higher-order} we compute +the modulus in a theory where the interaction free energy is truncated after +fourth order with new term $\frac12g\eta^2\epsilon^2$. The thin solid black +line in Fig.~\ref{fig:data} shows the fit of the \rus\ data to \eqref{eq:C0} +and shows that high-order corrections can account for the low-temperature +behavior. + +The data and theory appear quantitatively consistent, suggesting that \ho\ can +be described as a $\Bog$-nematic phase that is modulated at finite $q$ along +the $c-$axis. The predicted softening appears over hundreds of Kelvin; Figures +\ref{fig:data}(c--d) show the background modulus $C_\Bog^0$ and the \op--induced response isolated from each other. We have seen that the mean-field theory of a $\Bog$ \op\ recreates the topology @@ -624,6 +636,203 @@ such as ultrasound, that could further support or falsify this idea. Elena Hassinger. We thank Sayak Ghosh for \rus\ data. \end{acknowledgements} +\appendix + +\section{Adding a higher-order interaction} +\label{sec:higher-order} + +In this appendix, we compute the $\Bog$ modulus for a theory with a high-order +interaction truncation to better match the low-temperature behavior. Consider +the free energy density $f=f_\e+f_\i+f_\op$ with +\begin{equation} + \begin{aligned} + f_\e&=\frac12C_0\epsilon^2 \\ + f_\i&=-b\epsilon\eta+\frac12g\epsilon^2\eta^2 \\ + f_\op&=\frac12\big[r\eta^2+c_\parallel(\nabla_\parallel\eta)^2+c_\perp(\nabla_\perp\eta)^2+D(\nabla_\perp^2\eta)^2\big]+u\eta^4. + \end{aligned} + \label{eq:new_free_energy} +\end{equation} +The mean-field stain conditioned on the order parameter is found from +\begin{equation} + \begin{aligned} + 0 + &=\frac{\delta F[\eta,\epsilon]}{\delta\epsilon(x)}\bigg|_{\epsilon=\epsilon_\star[\eta]} \\ + &=C_0\epsilon_\star[\eta](x)-b\eta(x)+g\epsilon_\star[\eta](x)\eta(x)^2, + \end{aligned} +\end{equation} +which yields +\begin{equation} + \epsilon_\star[\eta](x)=\frac{b\eta(x)}{C_0+g\eta(x)^2}. + \label{eq:epsilon_star} +\end{equation} +Upon substitution into \eqref{eq:new_free_energy} and expanded to fourth order +in $\eta$, $F[\eta,\epsilon_\star[\eta]]$ can be written in the form +$F_\op[\eta]$ alone with $r\to\tilde r=r-b^2/C_0$ and $u\to\tilde +u=u+b^2g/2C_0^2$. The phase diagram in $\eta$ follows as before with the +shifted coefficients, and namely $\langle\eta(x)\rangle=\eta_*\cos(q_*x_3)$ for +$\tilde r<c_\perp^2/4D=\tilde r_c$ with $q_*^2=-c_\perp/2D$ and +\begin{equation} + \eta_*^2=\frac{c_\perp^2-4D\tilde r}{12D\tilde u} + =\frac{|\Delta\tilde r|}{3\tilde u}. +\end{equation} +We would like to calculate the $q$-dependant modulus +\begin{equation} + C(q) + =\frac1V\int dx\,dx'\,C(x,x')e^{-iq(x-x')}, +\end{equation} +where +\begin{widetext} +\begin{equation} + C(x,x') + =\frac{\delta^2F[\eta_\star[\epsilon],\epsilon]}{\delta\epsilon(x)\delta\epsilon(x')}\bigg|_{\epsilon=\langle\epsilon\rangle} + =\frac{\delta^2F_\e[\eta_\star[\epsilon],\epsilon]}{\delta\epsilon(x)\delta\epsilon(x')}+ + \frac{\delta^2F_\i[\eta_\star[\epsilon],\epsilon]}{\delta\epsilon(x)\delta\epsilon(x')}+ + \frac{\delta^2F_\op[\eta_\star[\epsilon],\epsilon]}{\delta\epsilon(x)\delta\epsilon(x')} + \bigg|_{\epsilon=\langle\epsilon\rangle} +\end{equation} +and $\eta_\star$ is the mean-field order parameter conditioned on the strain defined implicitly by +\begin{equation} + 0=\frac{\delta F[\eta,\epsilon]}{\delta\eta(x)}\bigg|_{\eta=\eta_\star[\epsilon]} + =-b\epsilon(x)+g\epsilon(x)^2\eta_\star[\epsilon](x)+\frac{\delta F_\op[\eta]}{\delta\eta(x)}\bigg|_{\eta=\eta_\star[\epsilon]}. + \label{eq:eta_star} +\end{equation} +We will work this out term by term. The elastic term is the most straightforward, giving +\begin{equation} + \frac{\delta^2F_\e[\epsilon]}{\delta\epsilon(x)\delta\epsilon(x')} + =\frac12C_0\frac{\delta^2}{\delta\epsilon(x)\delta\epsilon(x')}\int dx''\,\epsilon(x'')^2 + =C_0\delta(x-x'). +\end{equation} +The interaction term gives +\begin{equation} + \begin{aligned} + \frac{\delta^2F_\i[\eta_\star[\epsilon],\epsilon]}{\delta\epsilon(x)\delta\epsilon(x')} + &=-b\frac{\delta^2}{\delta\epsilon(x)\delta\epsilon(x')}\int dx''\,\epsilon(x'')\eta_\star[\epsilon](x'') + +\frac12g\frac{\delta^2}{\delta\epsilon(x)\delta\epsilon(x')}\int dx''\,\epsilon(x'')^2\eta_\star[\epsilon](x'')^2 \\ + &=-b\frac{\delta\eta_\star[\epsilon](x')}{\delta\epsilon(x)} + -b\frac{\delta}{\delta\epsilon(x)}\int dx''\,\epsilon(x'')\frac{\delta\eta_\star[\epsilon](x'')}{\delta\epsilon(x')} + +g\frac{\delta}{\delta\epsilon(x)}\big[\epsilon(x')\eta_\star[\epsilon](x')^2\big] \\ + &\qquad+g\frac{\delta}{\delta\epsilon(x)}\int dx''\,\epsilon(x'')^2\eta_\star[\epsilon](x'')\frac{\delta\eta_\star[\epsilon](x'')}{\delta\epsilon(x')} \\ + &=-2(b-2g\epsilon(x)\eta_\star[\epsilon](x))\frac{\delta\eta_\star[\epsilon](x)}{\delta\epsilon(x')}-b\int dx''\,\epsilon(x'')\frac{\delta^2\eta_\star[\epsilon](x'')}{\delta\epsilon(x)\delta\epsilon(x')} + +g\eta_\star[\epsilon](x)^2\delta(x-x') \\ + &\qquad+g\int dx''\,\epsilon(x'')^2\frac{\delta\eta_\star[\epsilon](x'')}{\delta\epsilon(x)}\frac{\delta\eta_\star[\epsilon](x'')}{\delta\epsilon(x')} + +g\int dx''\,\epsilon(x'')^2\eta_\star[\epsilon](x'')\frac{\delta^2\eta_\star[\epsilon](x'')}{\delta\epsilon(x)\delta\epsilon(x')}. + \end{aligned} +\end{equation} +The order parameter term relies on some other identities. First, \eqref{eq:eta_star} implies +\begin{equation} + \frac{\delta F_\op[\eta]}{\delta\eta(x)}\bigg|_{\eta=\eta_\star[\epsilon]} + =b\epsilon(x)-g\epsilon(x)^2\eta_\star[\epsilon](x), + \label{eq:dFodeta} +\end{equation} +and therefore that the functional inverse $\eta_\star^{-1}[\eta]$ is +\begin{equation} + \eta_\star^{-1}[\eta](x)=\frac b{2g\eta(x)}\Bigg(1-\sqrt{1-\frac{4g\eta(x)}{b^2}\frac{\delta F_\op[\eta]}{\delta\eta(x)}}\Bigg). +\end{equation} +The inverse function theorem further implies (with substitution of \eqref{eq:dFodeta} after the derivative is evaluated) that +\begin{equation} + \bigg(\frac{\delta\eta_\star[\epsilon](x)}{\delta\epsilon(x')}\bigg)^{\{-1\}} + =\frac{\delta\eta_\star^{-1}[\eta](x)}{\delta\eta(x')}\bigg|_{\eta=\eta_\star[\epsilon]} + =\frac{g\epsilon(x)^2\delta(x-x')+\frac{\delta^2F_\op[\eta]}{\delta\eta(x)\delta\eta(x')}\big|_{\eta=\eta_\star[\epsilon]}}{b-2g\epsilon(x)\eta_\star[\epsilon](x)} +\end{equation} +and therefore that +\begin{equation} + \frac{\delta^2F_\op[\eta]}{\delta\eta(x)\delta\eta(x')}\bigg|_{\eta=\eta_\star[\epsilon]} + =(b-2g\epsilon(x)\eta_\star[\epsilon](x))\bigg(\frac{\delta\eta_\star[\epsilon](x)}{\delta\epsilon(x')}\bigg)^{\{-1\}} + -g\epsilon(x)^2\delta(x-x'). + \label{eq:d2Fodetadeta} +\end{equation} +Finally, we evaluate the order parameter term, using \eqref{eq:dFodeta} and \eqref{eq:d2Fodetadeta} which give +\begin{equation} + \begin{aligned} + \frac{\delta^2F_\op[\eta_\star[\epsilon]]}{\delta\epsilon(x)\delta\epsilon(x')} + &=\frac{\delta}{\delta\epsilon(x)}\int dx''\,\frac{\delta\eta_\star[\epsilon](x'')}{\delta\epsilon(x')}\frac{\delta F_\op[\eta]}{\delta\eta(x'')}\bigg|_{\eta=\eta_\star[\epsilon]} \\ + &=\int dx''\,\frac{\delta^2\eta_\star[\epsilon](x'')}{\delta\epsilon(x)\delta\epsilon(x')}\frac{\delta F_\op[\eta]}{\delta\eta(x'')}\bigg|_{\eta=\eta_\star[\epsilon]} + +\int dx''dx'''\frac{\delta\eta_\star[\epsilon](x'')}{\delta\epsilon(x)}\frac{\delta\eta_\star[\epsilon](x''')}{\delta\epsilon(x')}\frac{\delta^2F_\op[\eta]}{\delta\eta(x'')\delta\eta(x''')}\bigg|_{\eta=\eta_\star[\epsilon]} \\ + &=\int dx''\,\frac{\delta^2\eta_\star[\epsilon](x'')}{\delta\epsilon(x)\delta\epsilon(x')}\big(b\epsilon(x)-g\epsilon(x)^2\eta_\star[\epsilon](x)\big) + +(b-2g\epsilon(x)\eta_\star[\epsilon](x))\frac{\delta\eta_\star[\epsilon](x)}{\delta\epsilon(x')} \\ + &\qquad-g\int dx''\,\epsilon(x'')^2\frac{\delta\eta_\star[\epsilon](x'')}{\delta\epsilon(x)}\frac{\delta\eta_\star[\epsilon](x'')}{\delta\epsilon(x')}. + \end{aligned} +\end{equation} +Summing all three terms, we see a great deal of cancellation, with +\[ + \frac{\delta^2F[\eta_\star[\epsilon],\epsilon]}{\delta\epsilon(x)\delta\epsilon(x')}=C_0\delta(x-x')+g\eta_\star[\epsilon](x)^2\delta(x-x')-(b-2g\epsilon(x)\eta_\star[\epsilon](x))\frac{\delta\eta_\star[\epsilon](x)}{\delta\epsilon(x')}. +\] +We new need to evaluate this at $\langle\epsilon\rangle$. First, $\eta_\star[\langle\epsilon\rangle]=\langle\eta\rangle$, and +\[ + \frac{\delta^2F[\eta_\star[\epsilon],\epsilon]}{\delta\epsilon(x)\delta\epsilon(x')}\bigg|_{\epsilon=\langle\epsilon\rangle}=C_0\delta(x-x')+g\langle\eta(x)\rangle^2\delta(x-x')-(b-2g\langle\epsilon(x)\rangle\langle\eta(x)\rangle)\frac{\delta\eta_\star[\epsilon](x)}{\delta\epsilon(x')}\bigg|_{\epsilon=\langle\epsilon\rangle}. +\] +Computing the final functional derivative is the most challenging part. We will +first compute its functional inverse, take the Fourier transform of that, and +then use the basic relationship between Fourier functional inverses to find the +form of the non-inverse. First, we note +\begin{equation} + \frac{\delta^2F_\op[\eta]}{\delta\eta(x)\delta\eta(x')}\bigg|_{\eta=\langle\eta\rangle} + =\big[r-c_\perp\nabla_\perp^2-c_\parallel\nabla_\parallel^2+D\nabla_\perp^4+12u\langle\eta(x)\rangle^2\big]\delta(x-x'), +\end{equation} +which gives +\begin{equation} + \begin{aligned} + \bigg(\frac{\delta\eta_\star[\epsilon](x)}{\delta\epsilon(x')}\bigg)^{\{-1\}}\bigg|_{\epsilon=\langle\epsilon\rangle} + &=\frac1{b-2g\langle\epsilon(x)\rangle\langle\eta(x)\rangle}\bigg[g\langle\epsilon(x)\rangle^2\delta(x-x')+\frac{\delta^2F_\op[\eta]}{\delta\eta(x)\delta\eta(x')}\bigg]_{\eta=\langle\eta\rangle} \\ + &=\frac1{b-2g\langle\epsilon(x)\rangle\langle\eta(x)\rangle}\Big[ + g\langle\epsilon(x)\rangle^2+r-c_\perp\nabla_\perp^2-c_\parallel\nabla_\parallel^2+D\nabla_\perp^4+12u\langle\eta(x)\rangle^2 + \Big]\delta(x-x'). + \end{aligned} +\end{equation} +Upon substitution of \eqref{eq:epsilon_star} and expansion to quadratic order it $\langle\eta(x)\rangle$, we find +\begin{equation} + \begin{aligned} + \bigg(\frac{\delta\eta_\star[\epsilon](x)}{\delta\epsilon(x')}\bigg)^{\{-1\}}\bigg|_{\epsilon=\langle\epsilon\rangle} + &=\frac1b\Bigg\{r-c_\perp\nabla_\perp^2-c_\parallel\nabla_\parallel^2+D\nabla_\perp^4\\ + &\qquad\qquad+\langle\eta(x)\rangle^2\bigg[12u+\frac{b^2g}{C_0^2}+\frac{2g}{C_0}(r-c_\perp\nabla_\perp^2-c_\parallel\nabla_\parallel^2+D\nabla_\perp^4)\bigg]+O(\langle\eta\rangle^4)\Bigg\}\delta(x-x'). + \end{aligned} +\end{equation} +Defining $\widehat{\langle\eta\rangle^2}=\int dq'\,\langle\hat\eta(q')\rangle\langle\hat\eta(-q')\rangle$, its Fourier transform is then +\begin{equation} + \begin{aligned} + G(q) + &=\frac1V\int dx\,dx'\,e^{-iq(x-x')}\bigg(\frac{\delta\eta_\star[\epsilon](x)}{\delta\epsilon(x')}\bigg)^{\{-1\}}\bigg|_{\epsilon=\langle\epsilon\rangle} \\ + &=\frac1b\Bigg\{r+c_\perp q_\perp^2+c_\parallel q_\parallel^2+Dq_\perp^4+\widehat{\langle\eta\rangle^2}\bigg[12u+\frac{b^2g}{C_0^2}+\frac{2g}{C_0}(r+c_\perp q_\perp^2+c_\parallel q_\parallel^2+Dq_\perp^4)\bigg]+O(\langle\hat\eta\rangle^4)\Bigg\}. + \end{aligned} +\end{equation} +We can now compute $C(q)$ by taking its Fourier transform, using the convolution theorem for the second term: +\begin{equation} + \begin{aligned} + C(q) + &=C_0+g\widehat{\langle\eta\rangle^2}-\int dq''\bigg(b\delta(q'')-\frac{gb}{C_0}\int dq'\langle\hat\eta_{q'}\rangle\langle\hat\eta_{q''-q'}\rangle\bigg)/G(q-q'') \\ + &=C_0+g\widehat{\langle\eta\rangle^2}-b^2\bigg(\frac1{r+c_\perp q_\perp^2+c_\parallel q_\parallel^2+Dq_\perp^4}-\widehat{\langle\eta\rangle^2}\frac{12u+b^2g/C_0^2+\frac{2g}{C_0}(r+c_\perp q^2+c_\parallel q_\parallel^2+Dq_\perp^4)}{(r+c_\perp q_\perp^2+c_\parallel q_\parallel^2+Dq_\perp^4)^2}\bigg)\\ + &\qquad+\frac{gb^2}{C_0}\int dq'\,dq''\frac{\langle\hat\eta_{q'}\rangle\langle\hat\eta_{q''-q'}\rangle}{r+c_\perp(q_\perp-q_\perp'')^2+c_\parallel(q_\parallel-q_\parallel'')^2+D(q_\perp-q_\perp'')^4}+O(\langle\hat\eta\rangle^4). + \end{aligned} +\end{equation} +Upon substitution of $\langle\hat\eta_q\rangle=\frac12\eta_*\big[\delta(q_\perp-q_*)+\delta(q_\perp+q_*)\big]\delta(q_\parallel)$, we have +\begin{equation} + \begin{aligned} + C(q) + &=C_0+\frac14g\eta_*^2-b^2\bigg(\frac1{r+c_\perp q_\perp^2+c_\parallel q_\parallel^2+Dq_\perp^4}-\frac{\eta_*^2}4\frac{12u+b^2g/C_0^2+\frac{2g}{C_0}(r+c_\perp q^2+c_\parallel q_\parallel^2+Dq_\perp^4)}{(r+c_\perp q_\perp^2+c_\parallel q_\parallel^2+Dq_\perp^4)^2}\bigg)\\ + &\qquad+\frac{gb^2\eta_*^2}{4C_0}\bigg(\frac2{r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq_\perp^4}+\frac1{r+c_\parallel q_\parallel^2+c_\perp(q_\perp-2q_*)^2+D(q_\perp-2q_*)^4} \\ + &\qquad\qquad+\frac1{r+c_\parallel q_\parallel^2+c_\perp(q_\perp+2q_*)^2+D(q_\perp+2q_*)^4}\bigg)+O(\eta_*^4). + \end{aligned} +\end{equation} +Evaluating at $q=0$, we have +\begin{equation} + \begin{aligned} + C(0) + &=C_0-\frac{b^2}r+\frac{\eta_*^2}4\bigg(g+\frac{b^2}{r^2}(12u+b^2g/C_0^2)+\frac{2gb^2}{C_0r}\frac{16Dq_*^4+3r}{8Dq_*^4+r}\bigg) + \end{aligned} + \label{eq:C0} +\end{equation} +Above the transition this has exactly the form of \eqref{eq:static_modulus} for +any $g$; below the transition it has the same form at $g=0$ to order +$\eta_*^2$. With $r=a\Delta T+c^2/4D+b^2/C_0$, $u=\tilde u-b^2g/2C_0^2$, and +\begin{equation} + \eta_*^2=\begin{cases} + 0 & \Delta T > 0 \\ + -a\Delta T/3\tilde u & \Delta T \leq 0, + \end{cases} +\end{equation} +we can fit the ratios $b^2/a=1665\,\mathrm{GPa}\,\mathrm K$, $b^2/Dq_*^4=6.28\,\mathrm{GPa}$, and $b\sqrt{-g/\tilde u}=14.58\,\mathrm{GPa}$ with $C_0=(71.14-(0.010426\times T)/\mathrm K)\,\mathrm{GPa}$. The resulting fit the thin solid black line in Fig.~\ref{fig:data}. +\end{widetext} + \bibliographystyle{apsrev4-1} \bibliography{hidden_order} |