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Diffstat (limited to 'main.tex')
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1 files changed, 20 insertions, 13 deletions
@@ -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} |