Nernst Potential
The Nernst Potential is a fundamental concept for the understanding of the electrical properties of biological membranes. It stands for the potential difference caused by different ionic concentrations inside and outside a membrane. Some neuronal models use the Nernst Potential to model the dynamics equilibrium of the flow of several species of ions through the membrane [1].
In the case of a single species of charge $q$, the potential difference across the membrane $\Delta V$ at temperature $T$ and particle densities $n_1$ and $n_2$ respectively inside and outside the cell is given by:
\[\Delta V = k_B\frac{T}{q}\log\left(\frac{n_1}{n_2}\right)\]Our goal is to reach this expression starting from first principles of statistical mechanics.
Model Assumptions
Our model will assume that the membrane separate a system of $N$ particles of charge $q$ that are kept under constant temperature $T$ in a heat bath. The exact shape of this boundary is assumed to be unimportant. The number of particles is kept constant.
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| The model assumes that the membrane separates the system into two regions $\Omega$ and $\Sigma$ with different concentration of particles. |
The Classical Hamiltonian for a single particle $j$ inside the membrane is given by $H_j$:
\[H_j = \frac{p_j^2}{2m} + q\sum_{i\neq j}^{N}V_{ji}(\left|\vec{x}_j - \vec{x}_i\right|)\]Where $p_j$ is the momentum of the particle, $m$ is its mass and $V_{ji}$ is the potential energy of the particle at position $x_j$ caused by all other $N-1$ particles around it. Our next approximation will be to assume that our system have an enormous amount of particles $N»1$, and that the temperature is high enough so that their motion is akin to a fluid. This will allow us to use the Continuum Approximation, setting potential energy of a particle $j$ at position $\vec{x}_j$ by the average potential energy of all particles around it:
\[\sum_{i \neq j}^{N}V_{ji}(\left|\vec{x}_j - \vec{x}_i\right|) \approx V(\vec{x}_j)\]This will heavily simplify our calculations as we will now be able to treat this system as a gas of non-interacting particles, even though they do interact by an electric field! The Hamiltonian for a single particle $j$ is now given by:
\[H_j = \frac{p_j^2}{2m} + qV(\vec{x}_j)\]Probability and Partition Function
The assumptions allow us to model the problem by treating its possible microstates being drawn from a canonical ensemble. The probability distribution in this case is given by the Boltzmann distribution:
\[P(\vec{x}, \vec{p}) = \frac{1}{Z}e^{-\beta H(\vec{x}, \vec{p})}\]Where $\beta = \frac{1}{k_BT}$ and $Z$ is the partition function:
\[Z = \sum_{\left\{\vec{x}, \vec{p}\right\}}e^{-\beta H(\vec{x}, \vec{p})}\]In this notation, the vectors $\vec{x}$ and $\vec{p}$ represent the collection of all $N$ positions and momenta of the system, that is:
\[\vec{x} := \left(\vec{x}_1,...,\vec{x}_N\right)\] \[\vec{p} := \left(\vec{p}_1,...,\vec{p}_N\right)\]The total Hamiltonian is just the sum of all individual Hamiltonians:
\[H(\vec{x}, \vec{p}) = \sum_{j=1}^{N}H_j(\vec{x}_j, \vec{p}_j) = \sum_{j=1}^{N}\frac{p_j^2}{2m} + q\sum_{j=1}^{N}V(\vec{x}_j)\]Plugging this into the expression for probability allows us to expand the sum into a product of exponentials:
\[P(\vec{x}, \vec{p}) = \frac{1}{Z}\prod_{j=1}^{N}e^{-\beta\frac{p_j^2}{2m}}e^{-\beta qV(\vec{x}_j)}\]Since we’re assuming to deal with non-interacting particles and looking at only the averaging behavior of the system, we can express the total probability as “copies” of the probability of a single particle, meaning that
\[P(\vec{x}, \vec{p}) = P_1(\vec{x}_1, \vec{p}_1)^N\]Where $P_1$ is the probability of a single particle (chosen with $j=1$ for convenience):
\[P_1(\vec{x}_1, \vec{p}_1) := \frac{1}{Z_1}e^{-\beta\frac{p_1^2}{2m}}e^{-\beta qV(\vec{x}_1)}\] \[Z_1 = \sum_{\left\{\vec{x_1}, \vec{p_1}\right\}}e^{-\beta\frac{p_1^2}{2m}}e^{-\beta qV(\vec{x}_1)}\]Derivation
Defining by $n(\vec{x}’)$ the expected particle density at a position $\vec{x}’$ (a position vector with 3 components), we can compute this quantity in the following manner:
\[n(\vec{x}') = \left\langle\sum_{j=1}^{N}\delta(\vec{x}_j-\vec{x}')\right\rangle\]The expected value of the dirac sum is calculated with the probability distribution of the system:
\[n(\vec{x}') = \left\langle\sum_{j=1}^{N}\delta(\vec{x}_j-\vec{x}')\right\rangle = \sum_{\left\{\vec{x},\vec{p}\right\}}\sum_{j=1}^{N}\delta(\vec{x}_j-\vec{x}')P(\vec{x}, \vec{p})\]The delta function can now be computed by expanding the sum over $\left\{\vec{x},\vec{p}\right\}$ as an integral. To be pedantic, this would yield a $\frac{1}{\Delta \vec{x} \Delta \vec{p}}$ factor related to the uncertainty of the measure, which is replaced by Planck’s constant as $\frac{1}{h}$; but since this would also show up on the integral of the partition function, they would cancel out. The expression then becomes:
\[= \int\prod_{j=1}^{N}\mathrm{d}\vec{x}_j\mathrm{d}\vec{p}_j\sum_{k=1}^{N}\delta(\vec{x}_k-\vec{x}')P_j(\vec{x}_j, \vec{p}_j)\] \[= \sum_{k=1}^{N}\int\prod_{j=1}^{N}\mathrm{d}\vec{x}_j\mathrm{d}\vec{p}_j\delta(\vec{x}_k-\vec{x}')P_j(\vec{x}_j, \vec{p}_j)\]By identifying $\int\mathrm{d}\vec{p}_j P_j(\vec{x_j},\vec{p_j})$ as just the probability $P_j(\vec{x_j})$, we can further simplify this as
\[= \sum_{k=1}^{N}\int\prod_{j = 1}^{N}\mathrm{d}\vec{x}_j\delta(\vec{x}_k-\vec{x}')P_j(\vec{x}_j)\]Since the delta function acts only on $\vec{x}_k$, we can integrate over it and get
\[= \sum_{k=1}^{N}\int\prod_{j\neq k}^{N}\mathrm{d}\vec{x}_j P_k(\vec{x}')P_j(\vec{x}_j)\]The remaining terms over the variables $\vec{x_j}$ with $j \neq k$ are just individual probabilities for each particle, whose sum must be unity. Thus, we get
\[n(\vec{x}') = \sum_{k=1}^{N} P_k(\vec{x}')\]Just as before, we can replace the $N$ probabilities $P_j$ by $N$ copies of a single one $P_1$. From our previous definition, that is a factor proportional to $e^{-\beta q V(\vec{x}_1)}$ only since the momentum part is cancelled out by an identical term in the partition function $Z_1$. Then:
\[n(\vec{x}') = N\frac{e^{-\beta q V(\vec{x}')}}{\int\mathrm{d}\vec{x_1}e^{-\beta q V(\vec{x}'_1)}}\] \[= N\alpha e^{-\beta q V(\vec{x}')}\]Where the integral in the denominator got replaced by $\alpha$ for being just a scalar. If the system in question had a boundary dividing it into two regions with concentrations $n_1$ and $n_2$, we could then write the ratio between the two as:
\[\frac{n_1}{n_2} = e^{-\beta q (V_1 - V_2)}\]Calling the difference in potential of the two regions $\Delta V$ and solving for it, we get:
\[\Delta V = \frac{1}{\beta q}\ln\left(\frac{n_1}{n_2}\right)\]Which is the Nernst Potential with $\beta = \frac{1}{k_B T}$.
- [1]W. M. K. Wulfram Gerstner, Spiking neuron models: single neurons, populations, plasticity, 1st ed. Cambridge University Press, 2002.