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\title{Identifying the downturn issue with clean case}

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\section{The problem}

As discussed earlier there was an issue with a sharp drop in $T_1(z)$ as temperature approached $T_c$.
The dirty case looks fine in \cref{fig:exdirtynodownturn}, but the clean system in \cref{fig:excleandownturn} shows the problem.
Both of those are at $z$ around $20 \lambda_F$ ($\lambda_F$ is $\approx \SI{0.4}{\nm}$.)

\begin{figure}[htp]
	\centering
	\includegraphics[width=\linewidth]{downturnExampleDirty-no-downturn}
	\caption{$T_1(z)$, everything looks appropriately good for a dirty system.} \label{fig:exdirtynodownturn}
\end{figure}

\begin{figure}[htp]
	\centering
	\includegraphics[width=\linewidth]{downturnExampleClean}
	\caption{$T_1(z)$, for a clean system we have the downturn.} \label{fig:excleandownturn}
\end{figure}

The obvious question is whether this matters, because it's $T$ so close to $T_c$ that cutting it off is probably not an issue.
But the effect matters because in the intermediate values of $\tau$ both the turning up to the normal state and the problematic downturn are present, and that's not great.
\Cref{fig:downturnintermediate} shows what that looks like.
I \emph{think} that that's still probably explainable as having a rise up to the normal state for $T \rightarrow T_c$, masked by the downturn because it only rises close to $T_c$.
But it's not really a convincing graph, so I proceed under the assumption that it's worth investigation.

\begin{figure}[htp]
	\centering
	\includegraphics[width=\linewidth]{downturnIntermediate}
	\caption{$T_1(z)$, for a clean system we have the downturn.} \label{fig:downturnintermediate}
\end{figure}

Interestingly, this problem is dependent on the specific $z$.
$T_1$ plotted for different zs in \cref{fig:varioust1vsT}.
The smaller $z$ doesn't have that effect.
But that's the direction that's less helpful, because it's already a very small $z$.

\begin{figure}[htp]
	\centering
	\includegraphics[width=\linewidth]{t1vsTexamples}
	\caption{$T_1(T)$, for different $z$, some showing the downturn others not.} \label{fig:varioust1vsT}
\end{figure}

I plotted $T_1$ as a function of $z$ at very high temperatures ($T = 0.99999 T_c$), in the downturn, and again for $T = 0.99 T_c$, which is above the downturn.
In \cref{fig:t1vszdirty}, the dirty case, everything looks sensible.
However, in \cref{fig:t1vszclean} we can see that for $z > \lambda_F$ the $T_1(z)$ curve has a very different flat shape at the very high temperature, which ultimately leads to the downturn.

\begin{figure}[htp]
	\centering
	\includegraphics[width=\linewidth]{t1VsZDirtyCase-no-problems}
	\caption{$T_1(z)$, at two different temperatures.} \label{fig:t1vszdirty}
\end{figure}

\begin{figure}[htp]
	\centering
	\includegraphics[width=\linewidth]{t1VsZcleanCase-bad}
	\caption{$T_1(z)$, at two different temperatures.} \label{fig:t1vszclean}
\end{figure}

I looked at where that might come from.
We have that $T_1 \propto \frac{1}{\chi}$, and 
\begin{equation}
	\chi_{B,zz} \propto \int_0^\infty \dd{u} u^2 e^{-2 u z} \Im r_s(u),
\end{equation}
in a regime where $u \gg \frac{1}{\lambda}$ for vacuum wavelength $\lambda$.
This is always the regime we care about.
So $\chi(z) \approx \Im r_s(\frac{1}{z})$ or so.
I plotted $\Im r_s(u)$ in \cref{fig:imrs}, and it doesn't really have any of the unusual features we might exepect.
That figure is pretty ugly, but the takeaway is that for $u > 0.2 \lambda_F$, all that matters is the temperature.
However, for $u < 0.2 \lambda_F$ the dirty and clean lines start to split, and for small enough $u$ the difference between the higher and lower temperature cases is less important than the clean-dirty difference.
That makes sense, and doesn't at all explain why $T_1$ is problematic.

\begin{figure}[htp]
	\centering
	\includegraphics[width=\linewidth]{Im rs-no-weirdness}
	\caption{$\Im r_s$, showing no weirdness.} \label{fig:imrs}
\end{figure}

So that leaves the integral as a problem.
I haven't been able to excise the issue yet though, either by increasing precision or rewriting the integral different ways to pull $z$ out.
Using something like 
\begin{equation}
	\chi_{B,zz} \propto \frac{1}{z^3} \int_0^\infty \dd{u'} u'^2 e^{-2 u'} \Im r_s(\frac{u'}{z})
\end{equation}
should be numerically nicer, but I get about the same answers (with Mathematica's integral handler internally doing something similar to that transform anyway?).
The problem is definitely still apparent in the $\chi$ calculation, which is plotted in \cref{fig:chi}.
So I'm currently confused on this.

\begin{figure}[htp]
	\centering
	\includegraphics[width=\linewidth]{chiVsZCleanCaseWithProblem}
	\caption{$\chi(z)$, showing the same issue as $T_1$ (as expected).} \label{fig:chi}
\end{figure}

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