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

#### Karl Karlsson

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

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

ii) Now assume that $$\displaystyle \vec n=\vec e_3$$ 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 $$\displaystyle \sigma_{ab} = \bar \sigmaδ_{ab}$$ and what is the value of the constant $$\displaystyle \bar\sigma$$ 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:

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

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 $$\displaystyle \vec j = \sigma_0\vec E + \sigma_1\vec n(\vec n\cdot\vec E)$$. I have already found a matrix $$\displaystyle \sigma$$ that transforms $$\displaystyle \vec E$$ to $$\displaystyle \vec j$$. Do they want me to find eigenvectors and eigenvalues? Why?

#### Klaas van Aarsen

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

i) Calculate the angle between the vectors $$\displaystyle \vec j$$ and $$\displaystyle \vec E$$ if the angle between $$\displaystyle \vec E$$ and $$\displaystyle \vec n$$ is α
My attempt:
$$\displaystyle \vec E\cdot\vec j = |\vec E|\cdot|\vec j|\cdot cos(\phi)= \sigma_0\vec E^{2} + \sigma_1\vec n\cdot \vec E(\vec n\cdot\vec E) \implies \phi =arccos(\frac {\sigma_0|\vec E^{2}| + \sigma_1\cdot cos(α)\cdot|\vec E|\cdot cos(α)|\cdot|\vec E|} {|\vec E|\cdot|\vec j|})$$
That is correct.
We can still simplify the expression a bit though.

ii) Now assume that $$\displaystyle \vec n=\vec e_3$$ 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 $$\displaystyle \sigma_{ab} = \bar \sigmaδ_{ab}$$ and what is the value of the constant $$\displaystyle \bar\sigma$$ in the new coordinate system?

On ii) i am completely lost. What do the coordinate transformations mean? x, y and z are not even in the given expression $$\displaystyle \vec j = \sigma_0\vec E + \sigma_1\vec n(\vec n\cdot\vec E)$$. I have already found a matrix $$\displaystyle \sigma$$ that transforms $$\displaystyle \vec E$$ to $$\displaystyle \vec j$$. Do they want me to find eigenvectors and eigenvalues? Why?
The formula for the conductivity tensor is $j_i = \sigma_{ij} E_j$.
With $\vec n=\vec e_3=\delta_{i3}\vec e_i$ and the given equation, we get $j_i = \sigma_{ij} E_j = \sigma_0 E_i + \sigma_1 \delta_{i3}(\vec e_3 \cdot \vec E) =\sigma_0 E_i + \sigma_1 \delta_{i3}E_3$, so $\vec j = \sigma_0 E_1 \vec e_1 + \sigma_0 E_2 \vec e_2 + (\sigma_0 + \sigma_1) E_3 \vec e_3$.
In the new coordinate system, this must become $j_a^* = \bar\sigma \delta_{ab}E_b^*=\bar\sigma E_a^*$. That is $\vec{j}=\bar\sigma E_1 \vec e_1 + \bar\sigma E_2 \vec e_2 + \bar\sigma E_3 \gamma \vec e_3$.
It follows that $\bar\sigma=\sigma_0$ and $\gamma = \frac{\sigma_0}{\sigma_0+\sigma_1}$.