From d1158d2e35d690ddf2ce58c34cc3c71da1d83d0e Mon Sep 17 00:00:00 2001 From: Jaron Kent-Dobias Date: Fri, 23 Aug 2019 17:42:32 -0400 Subject: leave nothing undefined! better free energy notation --- main.tex | 14 +++++++++----- 1 file changed, 9 insertions(+), 5 deletions(-) (limited to 'main.tex') 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'')} \\ -- cgit v1.2.3-70-g09d2