Engineering Deriving the Cable Equation (neuroscience) from Fundamental Physics Laws

[Icaro Lorran]

Homework Statement

I want to derive the cable equation 12 on this link: https://en.wikipedia.org/wiki/Cable_theory

The way wikipedia derives wasn't very convincing to me, so I wanted to start from Maxwell's equations and some basic principles instead.

Relevant Equations

$\nabla \cdot D = \rho$
$\frac{\partial \rho}{\partial t} + \nabla \cdot J = 0$
$J = \sigma E$

> Note: I am using SageMath to do the manipulations, I will attach it with the post

I modeled the problem as a cylinder of height $\Delta z$ and anisotropic conductivity: the conductivity along the axis is different from the one along the radius. Using $J = \sigma E$, where $\sigma$ is a tensor encoding both directions, I can write the total current as

$$J = \sigma_z (1 - H(r-a)) E_z \hat{z} + \sigma_r (1 - H(r-a)) E_r \hat{r} $$
.

The reason I added the heaviside function is that the conductivity should be zero outside the tube. With a similar reasoning, I also said that the potential must vanish outside as well

$$V(t, r, z) = u(t,z)(1 - H(r-a))$$

I don't include $\theta$ due to rotational symmetry.

The first Maxwell equation for a linear medium is

$$\nabla \cdot D = \rho$$

With $D = \epsilon E$, where $\epsilon$ is also a tensor:

$$\epsilon = \epsilon_0 \hat{z} \otimes \tilde{z} + \epsilon(1 - H(r-a)) \hat{r} \otimes \tilde{r} + \epsilon_0 H(r-a) \hat{r} \otimes \tilde{r}$$

Plugging that on $D$, it becomes

$$D = (\epsilon + (\epsilon_0 - \epsilon) H(r-a))E_r \hat{r} + \epsilon_0 E_z \hat{z}$$
.

Now, my initial plan was to join the first Maxwell equation with the equation for conservation of charge (consider all these integrals to be closed):

$$Q = \iint D \cdot \mathrm{d}^2 \vec{r} $$

and
$$\iint J \cdot \mathrm{d}^2 \vec{r} = -\frac{\mathrm{d}}{\mathrm{d} t} Q$$

$$\iint J \cdot \mathrm{d}^2 \vec{r} = -\frac{\mathrm{d}}{\mathrm{d} t} \iint D \cdot \mathrm{d}^2 \vec{r}$$

$$\iint \left( J + \frac{\partial}{\partial t} D\right) \cdot \mathrm{d}^2 \vec{r} = 0$$

The part that I get stuck is how to go on from here. I tried to use a cylinder as a surface to enclose everything, but what comes out of it doesn't really resemble the desired equation. I did notice that integrating the result over $r$, does get something close, but I don't have any explanation as to why that works.

Here's the link to the notebook (I couldn't attach): https://github.com/icarosadero/cable_equation/blob/main/cable_equation.ipynb

 

[osilmag]

Cable theory involves Ohmic principles more than Maxwell. You have to incorporate the electric potential $V$ into your equations. I would try to integrate it after putting in $E$ for where there is an electric field (ex. $J=\sigma E$) in your integrals and then substitute $E=V/d$.