# Calculate the angle between the E-field and Current vectors in an anisotropic conductive material

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?

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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}##.