--- /dev/null
+\linenumbers
+\chapter{Temperature dependent unfolding}
+
+\section{Energy landscape roughness}
+
+I'm skeptical about \HTeq{8} to \HTeq{9}, so I'll rework as much of
+their math as I am capable of\ldots
+
+\begin{align}
+ \fs &= \frac{\kT}{\dx} \left[ \logp{ \frac{\r \dx}{\kexp \kT} }
+ + \logp{1 + \fs\frac{\dx'}{\dx} - \frac{\FO'}{\dx} + \frac{\vD'}{\vD}\cdot\frac{\kT}{\dx}}
+ + \logp{\avg{e^{\bt F_1}}}^2 \right]
+ & \HTeq{8} \nonumber
+\end{align}
+
+We simplify by dropping the 2\nd term
+(``In obtaining Eq.\ \textbf{9}, we have assumed that the second term in Eq.\ \textbf{8} is small.''),
+and defining $\alpha \equiv \kT$,
+ $\rho \equiv \logp{ \frac{\r \dx}{\kexp \kT} }$, and
+ $e^{\bt \ep} \equiv \avg{e^{\bt F_1}}$, yielding
+\begin{equation}
+ \fs = \frac{\alpha}{\dx} \left( \rho + \frac{\ep^2}{\alpha^2} \right)
+\end{equation}
+
+We obtain our version of \HTeq{9} by taking two measurements of equal mode force
+\begin{align}
+ 0 &= \fs_1 - \fs_2 \\
+ &= \frac{1}{\dx} \left( \alpha_1\rho_1 + \frac{\ep^2}{\alpha_1}
+ -\alpha_2\rho_2 - \frac{\ep^2}{\alpha_2} \right) \\
+ \ep^2\left(\frac{1}{\alpha_2} - \frac{1}{\alpha_1}\right) &= \alpha_1\rho_1 - \alpha_2\rho_2 \\
+ \ep^2 \cdot \frac{\alpha_1 - \alpha_2}{\alpha_1\alpha_2} &= \\
+ \ep^2 &= \frac{\alpha_1\alpha_2}{\alpha_1 - \alpha_2} \left( \alpha_1\rho_1 - \alpha_2\rho_2 \right)\\
+ \ep^2 &= \frac{\kT_1\kT_2}{\kT_1 - \kT_2} \left[ \kT_1\logp{\frac{\rs1\dxs1}{\kexps1 \kT_1}}
+ - \kT_2\logp{\frac{\rs2\dxs2}{\kexps2 \kT_2}} \right]
+\end{align}
+
+Which is different from \HTeq{9} by the sign in the prefactor, and the replacement $\vD \rightarrow \kf$.
+\begin{align}
+ \ep^2 &= \frac{\kT_1\kT_2}{\kT_2 - \kT_1} \left[ \kT_1\logp{\frac{\rs1\dxs1}{\vDs1 \kT_1}}
+ - \kT_2\logp{\frac{\rs2\dxs2}{\vDs2 \kT_2}} \right]
+ & \HTeq{9} \nonumber
+\end{align}
+
+Alternatively, noting that \dx can vary as a function of temperature, we follow Nevo et al.\ in keeping it in.
+Using $\delta \equiv \dx$
+\begin{align}
+ 0 &= \fs_1 - \fs_2 \\
+ &= \frac{\alpha_1\rho_1}{\delta_1} + \frac{\ep^2}{\delta_1\alpha_1}
+ -\frac{\alpha_2\rho_2}{\delta_2} - \frac{\ep^2}{\delta_2\alpha_2} \\
+ \ep^2\left(\frac{1}{\delta_2\alpha_2} - \frac{1}{\delta_1\alpha_1}\right)
+ &= \frac{\alpha_1\rho_1}{\delta_1} - \frac{\alpha_2\rho_2}{\delta_2} \\
+ \ep^2 \cdot \frac{\delta_1\alpha_1 - \delta_2\alpha_2}{\delta_1\delta_2\alpha_1\alpha_2}
+ &= \frac{\delta_2\alpha_1\rho_1 - \delta_1\alpha_2\rho_2}{\delta_1\delta_2} \\
+ \ep^2 &= \frac{\alpha_1\alpha_2}{\delta_1\alpha_1 - \delta_2\alpha_2}
+ \left( \delta_2\alpha_1\rho_1 - \delta_1\alpha_2\rho_2 \right)\\
+ \ep^2 &= \frac{\kT_1\kT_2}{\dxs1\kT_1 - \dxs2\kT_2}
+ \left[ \dxs2\kT_1\logp{\frac{\rs1\dxs1}{\kfs1 \kT_1}}
+ - \dxs1\kT_2\logp{\frac{\rs2\dxs2}{\kfs2 \kT_2}} \right]
+\end{align}
\newcommand{\ie}{\emph{i.e.}} % "id est" or "in other words"
\newcommand{\sawsim}{\texttt{sawsim}}
+
+\newcommand{\ensuretext}[1]{\ensuremath{\text{#1}}}
+% Hyeon and Thirumalai equation number #1
+\newcommand{\HTeq}[1]{\ensuretext{\emph{H\&T eq.\ {#1}}}}
+\newcommand{\kT}{\ensuremath{k_B T}}
+\newcommand{\bt}{\ensuremath{\beta}}
+\newcommand{\fs}{\ensuremath{f^*}}
+\newcommand{\dx}{\ensuremath{\Delta x(\fs)}}
+\newcommand{\dxs}[1]{\ensuremath{\Delta x_{#1}(\fs)}} % for subscripting
+\newcommand{\FO}{\ensuremath{\Delta F_0^\ddagger(\fs)}}
+\newcommand{\vD}{\ensuremath{\nu_D(\fs)}}
+\newcommand{\vDs}[1]{\ensuremath{\nu_{D{#1}}(\fs)}}
+\newcommand{\kexp}{\ensuremath{\vD e^{-\bt \FO}}}
+\newcommand{\kexps}[1]{\ensuremath{\vDs{#1} e^{-\bt_{#1} \FO_{#1}}}}
+\newcommand{\kf}{\ensuremath{k(\fs)}}
+\newcommand{\kfs}[1]{\ensuremath{k_{#1}(\fs)}}
+%\newcommand{\avg}[1]{\ensuremath{\left\langle {#1} \right\rangle}}
+\newcommand{\logp}[1]{\ensuremath{\log\!\!\!\left( {#1} \right)}}
+% \! is a negative thin space to get the paren closer to the log
+\renewcommand{\r}{\ensuremath{r_f}}
+\newcommand{\rs}[1]{\ensuremath{r_{f{#1}}}}% to avoid double-subscripting
+\newcommand{\ep}{\varepsilon}