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Commit 0a844497 authored by Benjamin Cumming's avatar Benjamin Cumming
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Merge branch 'mechanisms' of github.com:eth-cscs/cell_algorithms into mechanisms

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......@@ -72,6 +72,9 @@
\newcommand{\dder}[2]{\frac{\deriv{#1}}{\deriv{#2}}}
\newcommand{\vv}[1]{\bm{#1}\xspace}
\newcommand{\unit}[1]{\left[{#1}\right]}
\newcommand{\txtunit}[1]{$\left[{#1}\right]$}
%----------------------------------------------------------------------------------------
% ARTICLE INFORMATION
%----------------------------------------------------------------------------------------
......
%-------------------------------------------------------------------------------
%\subsubsection{Balancing Units}
%-------------------------------------------------------------------------------
Ensuring that units are balanced and correct requires care.
Take the description of the nonlinear system of ODEs that arises from the finite volume discretisation
\begin{equation}
\label{eq:linsys_FV}
V_i^{k+1} + \sum_{j\in\mathcal{N}_i} {\frac{\Delta t \alpha_{ij}}{\sigma_i} (V_i^{k+1}-V_j^{k+1})}
= V_i^k - \frac{\Delta t}{c_m}(i_m^{k} - i_e).
\end{equation}
The choice of units for a parameter, e.g. $\mu m^2$ or $m^2$ for the area $\sigma_{ij}$, introduces a constant of proportionality wherever it is used ($10^{-12}$ in the case of $\mu m^2 \rightarrow m^2$).
Wherever terms are added in \eq{eq:linsys_FV} the units must be checked, and constants of proportionality balanced.
First, appropriate units for each of the parameters and variables are chosen in~\tbl{tbl:units}.
We try to use the same units as NEURON, except for the specific membrane capacitance $c_m$, for which $F\cdot m^{-2}$ is used in place of $nF\cdot mm^{-2}$.
In \eq{eq:linsys_FV} we choose units of $mV \equiv 10^{-3}V$ for each term because of the $V_i$ terms on either side of the equation.
\begin{table}[hp!]
\begin{tabular}{lllr}
\hline
term & units & normalized units & NEURON \\
\hline
$t$ & $ms$ & $10^{-3} \cdot s$ & yes \\
$V$ & $mV$ & $10^{-3} \cdot V$ & yes \\
$a,~\Delta x$ & $\mu m$ & $10^{-6} \cdot m$ & yes \\
$\sigma_{i},~\sigma_{ij}$ & $\mu m^2$ & $10^{-12} \cdot m^2$ & yes \\
$c_m$ & $F\cdot m^{-2}$ & $s\cdot A\cdot V^{-1}\cdot m^{-2}$ & no \\
$r_L$ & $\Omega\cdot cm$ & $ 10^{-2} \cdot A^{-1}\cdot V\cdot m$ & yes \\
$\overline{g}$ & $S\cdot cm^{-2}$ & $10^{4} \cdot A\cdot V^{-1}\cdot m^{-2}$ & yes \\
$I_e$ & $nA$ & $10^{-9} \cdot A$ & yes \\
\hline
\end{tabular}
\caption{The units chosen for parameters and variables in NEST MC. The NEURON column indicates whether the same units have been used as NEURON.}
\label{tbl:units}
\end{table}
%------------------------------------------
\subsubsection{current terms}
%------------------------------------------
Membrane current is calculated as follows $i_m = \overline{g}(E-V)$, with units
\begin{align}
\unit{ i_m } &= \unit{ \overline{g} } \unit{ V } \nonumber \\
&= 10^{4} \cdot A\cdot V^{-1}\cdot m^{-2} \cdot 10^{-3} \cdot V \nonumber \\
&= 10 \cdot A \cdot m^{-2}. \label{eq:im_unit}
\end{align}
The injected current $I_e$ has units $nA$, which has to be expressed in terms of current per unit area $i_e=I_e / \sigma_i$ with units
\begin{align}
\unit{ i_e } &= \unit{ I_e } \unit{ \sigma_i }^{-1} \nonumber \\
&= 10^{-9}\cdot A \cdot 10^{12} \cdot m^{-2} \nonumber \\
&= 10^{3} \cdot A \cdot m ^{-2}, \label{eq:ie_unit}
\end{align}
which must be scaled by $10^2$ to match $i_m$ in \eq{eq:im_unit}.
The units for the flux coefficent can be calculated as follows:
\begin{align}
\unit{ \frac{\Delta t}{c_m} } &= 10^{-3} \cdot s \cdot s^{-1}\cdot A^{-1}\cdot V\cdot m^2 \nonumber \\
&= 10^{-3} \cdot A^{-1} \cdot V\cdot m^2. \label{eq:dtcm_unit}
\end{align}
From \eq{eq:im_unit} and \eq{eq:dtcm_unit} that the units of the full current term are
\begin{align}
\unit{ \frac{\Delta t}{c_m}\left(i_m - i_e\right) }
&= 10^{-3} \cdot A^{-1} \cdot V\cdot m^2 \cdot 10 \cdot A \cdot m^{-2} \nonumber \\
&= 10^{-2} \cdot V,
\end{align}
which must be scaled by $10$ to match the units of $mV\equiv10^{-3}V$.
%------------------------------------------
\subsubsection{flux terms}
%------------------------------------------
The coefficients in the linear system have the units
\begin{equation}
\unit{ \frac{\Delta t\alpha_{ij}}{\sigma_i} }
=
\unit{ \frac{\Delta t \sigma_{ij} } {c_m r_L \Delta x_{ij} \sigma_i} }
=
\unit{ \frac{\Delta t } {c_m r_L \Delta x_{ij} } },
\end{equation}
where we we simplify by noting that $\unit{\sigma_{ij}}=\unit{\sigma_i}$.
The units of the term $c_m r_L$ on the denominator are calculated as follows
\begin{align}
\unit{c_m r_L}
&= s \cdot A \cdot V^{-1} \cdot m^{-2} \cdot 10^{-2} \cdot A^{-1} \cdot V \cdot m \nonumber \\
&= 10^{-2} \cdot s \cdot m^{-1},
\end{align}
so the units of the denominator are
\begin{align}
\unit{c_m r_L \Delta x_{ij}}
&= 10^{-2} \cdot s \cdot m^{-1} \cdot 10^{-6} \cdot m \nonumber \\
&= 10^{-8} \cdot s,
\end{align}
and hence
\begin{align}
\unit{\frac{\Delta t } {c_m r_L \Delta x_{ij} }}
&= 10^{8} \cdot s^{-1} \cdot 10^{-3} \cdot s \nonumber \\
&= 10^{5}.
\end{align}
So, the terms with $\alpha_{ij}$ must be scaled by $10^5$ to match the units of $mV$.
%------------------------------------------
\subsubsection{discretization with scaling}
%------------------------------------------
Here is something that I wish the NEURON documentation had provided:
\begin{align}
& V_i^{k+1} + \sum_{j\in\mathcal{N}_i} {10^5 \cdot \frac{\Delta t \alpha_{ij}}{\sigma_i} (V_i^{k+1}-V_j^{k+1})} \nonumber \\
& = V_i^k - 10\cdot \frac{\Delta t}{c_m}(i_m^{k} - 10^2\cdot I_e/\sigma_i).
\end{align}
%------------------------------------------
\subsection{Supplementary Unit Information}
%------------------------------------------
Here is some information about units scraped from Wikipedia for convenience.
\begin{table*}[htp!]
\begin{center}
......@@ -15,6 +125,8 @@
\hline
\end{tabular}
\vspace{20pt}
\begin{tabular}{llll}
\hline
symbol & unit & equivalents & SI base \\
......@@ -44,27 +156,4 @@
\end{center}
\caption{Symbols and quantities.}
\end{table*}
%-------------------------------------------------------------------------------
\subsubsection{Units}
%-------------------------------------------------------------------------------
Reffering to the cable equation first defined in~\eq{eq:cable}
\begin{equation*}
c_m \pder{V}{t} = \frac{1}{2\pi a r_{L}} \pder{}{x} \left( a^2 \pder{V}{x} \right) - i_m + i_e,
\end{equation*}
If the units are taken to be
\begin{itemize}
\item $c_m = F\cdot cm^{-2}$
\item $V = V$
\item $a = cm$
\item $r_L = \Omega\cdot cm$
\end{itemize}
Then the units of each term in equation are $A\cdot cm^{-2}$.
In practice, the units above are not used, for example distances are usually measured in $\mu m$ and areas in $cm^2$.
But if we work term-by-term, scaling for these factors is manageable.
A useful identity to use when performing the dimensional analysis relates capacitance and resistance
\begin{equation*}
1~F = 1~\Omega^{-1} \cdot s
\end{equation*}
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