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\begin{document}

Daniel:

In trying to derive the expression for the shift in critical temperature
for the two-cycle model, I think I am missing some physics.

I start with (5.18):
\begin{align*}
	\rho_c^{(\alpha)}
	&= \rho_c^{(0)} - \frac{1-e^{-2\alpha}}{(8\pi\beta)^{3/2}}.
\end{align*}
Expanding to first order in $\alpha$:
\begin{align*}
	\rho_c^{(\alpha)}
	&\approx \rho_c^{(0)} - \frac{2\alpha}{(8\pi\beta)^{3/2}}.
\end{align*}
Using (5.7), namely,
$\alpha=(8/\pi\beta)^{1/2} a$:
\begin{align*}
	\rho_c^{(a)}
	&\approx \rho_c^{(0)} - \frac
	{2 \cdot 8^{1/2} \cdot a}
	{\pi^{1/2}\cdot \beta^{1/2}\cdot 8^{3/2}\cdot \pi^{3/2}\cdot \beta^{3/2}}
	\\
%
	&= \rho_c^{(0)} - \frac{a}{4 \pi^2\beta^2}.
\end{align*}
Using $\beta=1/T_c^{(0)}$:
\begin{align*}
	\rho_c^{(a)}
	&= \rho_c^{(0)} - \frac{{T_c^{(0)}}^2 a}{4 \pi^2}.
\end{align*}
Using the near-critical approximation $\rho \sim T^{3/2}$ from the old version
of your 2007 paper with Betz:
\begin{align*}
	{T_c^{(a)}}^{3/2} &= {T_c^{(0)}}^{3/2} - \frac{{T_c^{(0)}}^2 a}{4 \pi^2} \\
	\left(\frac{T_c^{(a)}}{T_c^{(0)}} \right)
		&= 1 - \frac{{T_c^{(0)}}^{1/2} a}{4 \pi^2} \\
	\frac{T_c^{(a)}}{T_c^{(0)}}
		&= \left( 1 - \frac{{T_c^{(0)}}^{1/2} a}{4 \pi^2} \right)^{2/3}.
\end{align*}
Taylor-expanding the right-hand side in the small parameter $a$:
\begin{align*}
	\frac{T_c^{(a)}}{T_c^{(0)}} &= 1 - \frac{{T_c^{(0)}}^{1/2} a}{6 \pi^2} \\
	\frac{T_c^{(a)}}{T_c^{(0)}} -1 &= - \frac{{T_c^{(0)}}^{1/2} a}{6 \pi^2} \\
	\frac{T_c^{(a)}-T_c^{(0)}}{T_c^{(0)}}
		&= - \frac{{T_c^{(0)}}^{1/2} a}{6 \pi^2}.
\end{align*}
Again using the near-critical approximation $T^{1/2} \sim \rho^{1/3}$:
\begin{align*}
	\frac{T_c^{(a)}-T_c^{(0)}}{T_c^{(0)}}
		&= - \frac{\rho^{1/3} a}{6 \pi^2}.
\end{align*}
The minus sign does not belong there, but I don't see the error in my work.

Questions:
\begin{itemize}
\item Why is it true that $\rho \sim T^{3/2}$ near $T_c$?
	(The original statement in your 2007 paper with Betz was that
	$\beta \sim \rho^{-2/3}$ near $T_c{(0)}$.)
\item What is the scale factor connecting $\rho$ and $T^{3/2}$:  is it 1?
\item Even though $\rho \sim T^{3/2}$, can one conclude that necessarily
	$\rho_c \sim T_c^{3/2}$?
\item I cannot reproduce the factor of 0.74 in (5.19).  If in fact
	$\rho \sim K T^{3/2}$ for some scale factor $K$ other than 1, I get
	$$
	\frac{T_c^{(a)}-T_c^{(0)}}{T_c^{(0)}}
		= - \frac{\rho^{1/3} a}{6 K^2 \pi^2}
	$$
	so I would hope that $1/6K^2\pi^2 \approx 0.74$.
\end{itemize}

Thank you!

\end{document}
