Calculate the angle between the given vectors. When does the conductivity tensor component take the form [MATH]\sigma_{ab} = \bar \sigmaδ_{ab}[/MATH]?

Karl Karlsson

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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?



Thanks in advance!
 
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