Calculating Angle Between E-Field and Current Vectors in Anisotropic Mat.

[Karl Karlsson]

Homework Statement

In a certain anisotropic conductive material, the relationship between the current density $\vec j$ and
the electric field $\vec E$ is given by: $\vec j = \sigma_0\vec E + \sigma_1\vec n(\vec n\cdot\vec E)$ where $\vec n$ is a constant unit vector.

i) Calculate the angle between the vectors $\vec j$ and $\vec E$ if the angle between $\vec E$ and $\vec n$ is α

ii) Now assume that $\vec n=\vec e_3$ and define a coordinate transformation ξ = x, η = y, ζ = γz where γ is a constant. For what value of γ does the conductivity tensor component take the form $\sigma_{ab} = \bar \sigmaδ_{ab}$ and what is the value of the constant $\bar\sigma$ in the new coordinate system?

Relevant Equations

$\vec j = \sigma_0\vec E + \sigma_1\vec n(\vec n\cdot\vec E)$
ξ = x, η = y, ζ = γz

In a certain anisotropic conductive material, the relationship between the current density $\vec j$ and
the electric field $\vec E$ is given by: $\vec j = \sigma_0\vec E + \sigma_1\vec n(\vec n\cdot\vec E)$ where $\vec n$ is a constant unit vector.

i) Calculate the angle between the vectors $\vec j$ and $\vec E$ if the angle between $\vec E$ and $\vec n$ is α

ii) Now assume that $\vec n=\vec e_3$ and define a coordinate transformation ξ = x, η = y, ζ = γz where γ is a constant. For what value of γ does the conductivity tensor component take the form $\sigma_{ab} = \bar \sigmaδ_{ab}$ and what is the value of the constant $\bar\sigma$ in the new coordinate system?My attempt:

I don't really know if I get it into the simplest possible form but i guess one way of solving i) would be:

$\vec E\cdot\vec j = |\vec E|\cdot|\vec j|\cdot cos(\phi)= \sigma_0\vec E^{2} + \sigma_1\vec n\cdot \vec E(\vec n\cdot\vec E) \implies \phi =arccos(\frac {\sigma_0|\vec E^{2}| + \sigma_1\cdot cos(α)\cdot|\vec E|\cdot cos(α)|\cdot|\vec E|} {|\vec E|\cdot|\vec j|})$

Is this the best way to solve this?

On ii) i am completely lost. What do the coordinate transformations mean? x, y and z are not even in the given expression $\vec j = \sigma_0\vec E + \sigma_1\vec n(\vec n\cdot\vec E)$. I have already found a matrix $\sigma$ that transforms $\vec E$ to $\vec j$. Do they want me to find eigenvectors and eigenvalues? Why?

Thanks in advance!

 

[Fred Wright]

I agree with your result for part i). For part ii), since you already found the conductivity tensor, you can tell immediately by inspection that the tensor is diagonal with $\sigma_{xx} = \sigma_{yy}=\sigma_0=\sigma$ and $\sigma_{zz}=\sigma_0 + \sigma_1$. By making the coordinate transformation you are asked to solve,
$$
\begin {pmatrix}
\sigma & 0 & 0 \\
0 & \sigma & 0 \\
0 & 0 & \sigma_0 + \sigma_1 \\
\end {pmatrix}
\begin {pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \gamma \\
\end {pmatrix}
=
\begin {pmatrix}
\sigma & 0 & 0 \\
0 & \sigma & 0 \\
0 & 0 & \sigma \\
\end {pmatrix}
$$

 

[Karl Karlsson]

I agree with your result for part i). For part ii), since you already found the conductivity tensor, you can tell immediately by inspection that the tensor is diagonal with $\sigma_{xx} = \sigma_{yy}=\sigma_0=\sigma$ and $\sigma_{zz}=\sigma_0 + \sigma_1$. By making the coordinate transformation you are asked to solve,
$$
\begin {pmatrix}
\sigma & 0 & 0 \\
0 & \sigma & 0 \\
0 & 0 & \sigma_0 + \sigma_1 \\
\end {pmatrix}
\begin {pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \gamma \\
\end {pmatrix}
=
\begin {pmatrix}
\sigma & 0 & 0 \\
0 & \sigma & 0 \\
0 & 0 & \sigma \\
\end {pmatrix}
$$

Hi Fred!
Are you sure one can't express the answer in i) in any other or simpler way because it feels like my answer for the expression of the angle between $\vec E$ and $\vec j$Ok! Then I might understand but I would really appreciate verification by someone or comment on what is not correct.

Since $\vec n$ is just $\vec e_3$ i get $$\sigma = \begin{bmatrix}\sigma_0&0&0\\0&\sigma_0&0\\0&0&\sigma_0+1\end{bmatrix}$$.$$\vec E_3(\sigma_0+1)=\sigma_0\cdot\vec E_\zeta$$. I write $\vec E_\zeta = \gamma\cdot\vec E_3$ because of the change of basis matrix
$$\begin {pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \gamma \\
\end {pmatrix}
$$
and therefore $\gamma=1+1/\sigma_0$ and $\bar\sigma = \sigma_0$. Is this correct?

 

[Fred Wright]

Are you sure one can't express the answer in i) in any other or simpler way because it feels like my answer for the expression of the angle between $\vec E$ and $\vec j$

You can simplify your expression by substituting an expression for $|j|$. I suggest,
$$
j\cdot j=(\sigma_0 \vec E + \sigma_1 \hat n (\hat n \cdot \vec E)\cdot (\sigma_0 \vec E + \sigma_1 \hat n (\hat n \cdot \vec E))=(\sigma_0|\vec E | + \sigma_1 (\hat n \cdot \vec E ))^2
$$
$$
|j|=\sigma_0|\vec E | + \sigma_1 (\hat n \cdot \vec E )
$$

Ok! Then I might understand but I would really appreciate verification by someone or comment on what is not correct.

Since $\vec n$ is just $\vec e_3$ i get $$\sigma = \begin{bmatrix}\sigma_0&0&0\\0&\sigma_0&0\\0&0&\sigma_0+1\end{bmatrix}$$.

This is wrong. It should be,
$$
\sigma = \begin {pmatrix}
\sigma_0 & 0 & 0 \\
0 & \sigma_0 & 0 \\
0 & 0 & \sigma_0 +\sigma_1 \\
\end {pmatrix}
$$
and therefore $\gamma = \frac{\sigma_0}{\sigma_0 + \sigma_1}$.