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author | Jaron Kent-Dobias <jaron@kent-dobias.com> | 2020-04-15 14:13:51 -0400 |
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committer | Jaron Kent-Dobias <jaron@kent-dobias.com> | 2020-04-15 14:13:51 -0400 |
commit | 1537dc5cdc78359399f271b9619d45e9b8c01fe0 (patch) | |
tree | 476899387d52682a86775cc7f6957cd493f579ad /main.tex | |
parent | 02187d60a03d2f7bed87813fe4c15646c93b5eb5 (diff) | |
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Nudged the end of widetext.
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-rw-r--r-- | main.tex | 2 |
1 files changed, 1 insertions, 1 deletions
@@ -821,6 +821,7 @@ Evaluating at $q=0$, we have \end{aligned} \label{eq:C0} \end{equation} +\end{widetext} 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 @@ -831,7 +832,6 @@ $\eta_*^2$. With $r=a\Delta T+c^2/4D+b^2/C_0$, $u=\tilde u-b^2g/2C_0^2$, and \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} |