Eigenvalues and eigenvectors of the momentum current density dyadic

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
m3mb3r
Messages
4
Reaction score
0

Homework Statement



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

Homework Equations



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


The Attempt at a Solution



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

[tex]det(\overleftrightarrow{T}-\overleftrightarrow{1}\lambda)=0[/tex]

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

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

[tex]-\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[/tex]

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

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

Thanks
 
Physics news on Phys.org