Eigenvalues and eigenvectors of the momentum current density dyadic

[m3mb3r]

Homework Statement

What are the eigenvalues and eigenvectors of the momentum
current density dyadic \overleftrightarrow{T} (Maxwell tensor)? Then make use of these eigenvalues in finding the determinant of \overleftrightarrow{T} and the trace of \overleftrightarrow{T}^2

Homework Equations

\overleftrightarrow{T}=\overleftrightarrow{1}U-\frac{1}{4\pi}(\overrightarrow{E}\overrightarrow{E}+\overrightarrow{B}\overrightarrow{B})
U=\frac{1}{8\pi}(\overrightarrow{|E}|^{2}+|\overrightarrow{B|^{2}})

The Attempt at a Solution

<br /> <br /> \overleftrightarrow{T}x=\lambda x

det(\overleftrightarrow{T}-\overleftrightarrow{1}\lambda)=0

<br /> det\left(\begin{array}{ccc}<br /> T_{11}-\lambda &amp; T_{12} &amp; T_{13}\\<br /> T_{21} &amp; T_{22}-\lambda &amp; T_{23}\\<br /> T31 &amp; T_{32} &amp; T_{33}-\lambda\end{array}\right)=0

Since the tensor is symmetric, we have T_{ij}=T_{ji}, (after simplifying) our equation become:

-\lambda^{3}+(T_{11}+T_{22}+T_{33})\lambda^{2}+(T_{12}^{2}+T_{23}^{2}+T_{31}^{2}-T_{11}T_{22}-T_{22}T_{33}-T_{33}T_{11})\lambda+T_{11}T_{22}T_{33}+2T_{12}T_{23}T_{31}-T_{11}T_{23}^{2}-T_{22}T_{31}^{2}-T_{33}T_{12}^{2}=0

with T_{ij}=\delta_{ij}U-\frac{1}{4\pi}(E_{i}E_{j}+B_{i}B_{j}) (from the definition of \overleftrightarrow{T})

The problem is that as I expand T_{ij} the equation become more and more complicated. Am I doing the right thing here?
And later how to find the determinant of \overleftrightarrow{T} and the trace of \overleftrightarrow{T}^2 after obtaining the eigenvalue?

Thanks

 

[m3mb3r]

Anybody got any idea?