\section{Single-domain proteins under constant loading}

Let $x$ be the end to end distance of the protein, $t$ be the time since loading began, $F$ be tension applied to the protein, $P$ be the surviving population of folded proteins.
Make the definitions
\begin{align}
  v &\equiv \deriv{t}{x} && \text{the pulling velocity} \\
  k &\equiv \deriv{x}{F} && \text{the loading spring constant} \\
  P_0 &\equiv P(t=0) && \text{the initial number of folded proteins} \\
  D &\equiv P_0 - P && \text{the number of dead (unfolded) proteins} \\
  \kappa &\equiv -\frac{1}{P} \deriv{t}{P} && \text{the unfolding rate}
\end{align}
The proteins are under constant loading because
\begin{equation}
  \deriv{t}{F} = \deriv{x}{F}\deriv{t}{x} = kv\;,
\end{equation}
a constant, since both $k$ and $v$ are constant (\citet{evans97} in the text on the first page, \citet{dudko06} in the text just before Eqn.~4).

The instantaneous likelyhood of a protein unfolding is given by $\deriv{F}{D}$, and the unfolding histogram is merely this function discretized over a bin of width $W$(This is similar to \citet{dudko06} Eqn.~2, remembering that $\dot{F}=kv$, that their probability density is not a histogram ($W=1$), and that their pdf is normalized to $N=1$).
\begin{equation}
  h(F) \equiv \deriv{\text{bin}}{F}
    = \deriv{F}{D} \cdot \deriv{\text{bin}}{F}
    = W \deriv{F}{D}
    = -W \deriv{F}{P}
    = -W \deriv{t}{P} \deriv{F}{t}
    = \frac{W}{vk} P\kappa \label{eq.hist}
\end{equation}
Solving for theoretical histograms is merely a question of taking your chosen $\kappa$, solving for $P(f)$, and plugging into Eqn. \ref{eq.hist}.
We can also make a bit of progress solving for $P$ in terms of $\kappa$ as follows:
\begin{align}
  \kappa &\equiv -\frac{1}{P} \deriv{t}{P} \\
  -\kappa \dd t \cdot \deriv{t}{F} &= \frac{\dd P}{P} \\
  \frac{-1}{kv} \int \kappa \dd F &= \ln(P) + c \\
  P &= C\exp{\p({\frac{-1}{kv}\integ{F}{\kappa}})} \;, \label{eq.P}
\end{align}
where $c \equiv \ln(C)$ is a constant of integration scaling $P$.

\subsection{Constant unfolding rate}

In the extremely weak tension regime, the proteins' unfolding rate is independent of tension, we have
\begin{align}
  P &= C\exp{\p({\frac{-1}{kv}\integ{F}{\kappa}})}
     = C\exp{\p({\frac{-1}{kv}\kappa F})}
     = C\exp{\p({\frac{-\kappa F}{kv}})} \\
  P(0) &\equiv P_0 = C\exp(0) = C \\
  h(F) &= \frac{W}{vk} P \kappa
     = \frac{W\kappa P_0}{vk} \exp{\p({\frac{-\kappa F}{kv}})}
\end{align}
Suprise!  A constant unfolding-rate/hazard-function gives exponential decay.
Not the most earth shattering result, but it's a comforting first step, and it does show explicitly the dependence in terms of the various unfolding-specific parameters.

\subsection{Bell model}

Stepping up the intensity a bit, we come to Bell's model for unfolding
(\citet{hummer03} Eqn.~1 and the first paragraph of \citet{dudko06} and \citet{dudko07}).
\begin{equation}
  \kappa = \kappa_0 \cdot \exp\p({\frac{F \dd x}{k_B T}})
         = \kappa_0 \cdot \exp(a F) \;,
\end{equation}
where we've defined $a \equiv \dd x/k_B T$ to bundle some constants together.
The unfolding histogram is then given by
\begin{align}
  P &= C\exp\p({\frac{-1}{kv}\integ{F}{\kappa}})
     = C\exp\p[{\frac{-1}{kv} \frac{\kappa_0}{a} \exp(a F)}]
     = C\exp\p[{\frac{-\kappa_0}{akv}\exp(a F)}] \\
  P(0) &\equiv P_0 = C\exp\p({\frac{-\kappa_0}{akv}}) \\
  C &= P_0 \exp\p({\frac{\kappa_0}{akv}}) \\
  P &= P_0 \exp\p\{{\frac{\kappa_0}{akv}[1-\exp(a F)]}\} \\
  h(F) &= \frac{W}{vk} P \kappa
     = \frac{W}{vk} P_0 \exp\p\{{\frac{\kappa_0}{akv}[1-\exp(a F)]}\} \kappa_0 \exp(a F)
     = \frac{W\kappa_0 P_0}{vk} \exp\p\{{a F + \frac{\kappa_0}{akv}[1-\exp(a F)]}\} \label{eq.bell_pdf}\;.
\end{align}
The $F$ dependent behavior reduces to
\begin{equation}
  h(F) \propto \exp\p[{a F - b\exp(a F)}] \;,
\end{equation}
where $b \equiv \kappa_0/akv \equiv \kappa_0 k_B T / k v \dd x$ is
another constant rephrasing.

This looks an awful lot like the the Gompertz/Gumbel/Fisher-Tippett
distribution, where
\begin{align}
  p(x) &\propto z\exp(-z) \\
  z &\equiv \exp\p({-\frac{x-\mu}{\beta}}) \;,
\end{align}
but we have
\begin{equation}
  p(x) \propto z\exp(-bz) \;.
\end{equation}
Strangely, the Gumbel distribution is supposed to derive from an
exponentially increasing hazard function, which is where we started
for our derivation.  I haven't been able to find a good explaination
of this discrepancy yet, but I have found a source that echos my
result (\citet{wu04} Eqn.~1).

Oh wait, we can do this:
\begin{equation}
  p(x) \propto z\exp(-bz) = \frac{1}{b} z'\exp(-z')\propto z'\exp(-z') \;,
\end{equation}
with $z'\equiv bz$.  I feel silly...  From
\href{Wolfram}{http://mathworld.wolfram.com/GumbelDistribution.html},
the mean of the Gumbel probability density
\begin{equation}
  P(x) = \frac{1}{\beta} \exp\p[{\frac{x-\alpha}{\beta}
                                 -\exp\p({\frac{x-\alpha}{\beta}})
                                 }]
\end{equation}
is given by $\mu=\alpha-\gamma\beta$, and the variance is
$\sigma^2=\frac{1}{6}\pi^2\beta^2$, where $\gamma=0.57721566\ldots$ is
the Euler-Mascheroni constant.  Selecting $\beta=1/a=k_BT/\dd x$,
$\alpha=-\beta\ln(\kappa\beta/kv)$, and $F=x$ we have
\begin{align}
  P(F)
    &= \frac{1}{\beta} \exp\p[{\frac{F+\beta\ln(\kappa\beta/kv)}{\beta}
                               -\exp\p({\frac{F+\beta\ln(\kappa\beta/kv)}
                                             {\beta}})
                             }] \\
    &= \frac{1}{\beta} \exp(F/\beta)\exp[\ln(\kappa\beta/kv)]
                       \exp\p\{{-\exp(F/\beta)\exp[\ln(\kappa\beta/kv)]}\} \\
    &= \frac{1}{\beta} \frac{\kappa\beta}{kv} \exp(F/\beta)
                       \exp\p[{-\kappa\beta/kv\exp(F/\beta)}] \\
    &= \frac{\kappa}{kv} \exp(F/\beta)\exp[-\kappa\beta/kv\exp(F/\beta)] \\
    &= \frac{\kappa}{kv} \exp(F/\beta - \kappa\beta/kv\exp(F/\beta)] \\
    &= \frac{\kappa}{kv} \exp(aF - \kappa/akv\exp(aF)] \\
    &= \frac{\kappa}{kv} \exp(aF - b\exp(aF)]
    \propto h(F) \;.
\end{align}
So our unfolding force histogram for a single Bell domain under
constant loading does indeed follow the Gumbel distribution.

\subsection{Saddle-point Kramers model}

For the saddle-point approximation for Kramers' model for unfolding
(\citet{evans97} Eqn.~3, \citet{hanngi90} Eqn. 4.56c, \citet{vanKampen07} Eqn. XIII.2.2).
\begin{equation}
  \kappa = \frac{D}{l_b l_{ts}} \cdot \exp\p({\frac{-E_b(F)}{k_B T}}) \;,
\end{equation}
where $E_b(F)$ is the barrier height under an external force $F$,
$D$ is the diffusion constant of the protein conformation along the reaction coordinate,
$l_b$ is the characteristic length of the bound state $l_b \equiv 1/\rho_b$,
$\rho_b$ is the density of states in the bound state, and
$l_{ts}$ is the characteristic length of the transition state
\begin{equation}
  l_{ts} = TODO
\end{equation} 

\citet{evans97} solved this unfolding rate for both inverse power law potentials and cusp potentials.

\subsubsection{Inverse power law potentials}

\begin{equation}
  E(x) = \frac{-A}{x^n}
\end{equation}
(e.g. $n=6$ for a van der Waals interaction, see \citet{evans97} in
the text on page 1544, in the first paragraph of the section
\emph{Dissociation under force from an inverse power law attraction}).
Evans then gets funky with diffusion constants that depend on the
protein's end to end distance, and I haven't worked out the math
yet\ldots


\subsubsection{Cusp potentials}

\begin{equation}
  E(x) = \frac{1}{2}\kappa_a \p({\frac{x}{x_a}})^2
\end{equation}
(see \citet{evans97} in the text on page 1545, in the first paragraph
of the section \emph{Dissociation under force from a deep harmonic well}).
