- #1
Icaro Lorran
- 12
- 3
- 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
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
