Symmetry of the permittivity tensor of lossless media

[Ngineer]

I read in various sources (such as page 8 of these notes) that the dielectric permittivity tensor of a lossless medium is always symmetric. I am wondering how this can be the case, when:

• Phase accumulation in the medium could in theory depend on direction
• Coordinate system may be rotated to begin with?

I also fail to see why this is a specific property of lossless media. As far as the math is involved, both the absorptive and reactive losses are "losses". Only difference is that one is imaginary and the other is real.

 

[bobob]

I read in various sources (such as page 8 of these notes) that the dielectric permittivity tensor of a lossless medium is always symmetric. I am wondering how this can be the case, when:

• Phase accumulation in the medium could in theory depend on direction
• Coordinate system may be rotated to begin with?

I also fail to see why this is a specific property of lossless media. As far as the math is involved, both the absorptive and reactive losses are "losses". Only difference is that one is imaginary and the other is real.

It says why in your notes. The tensor is hermitian, so that if the A_ij are real, then A_ij = A_ji* and therefor, it's symmetric. If the medium is lossless, then the A_ij are real.

 

[jasonRF]

Your notes are incorrect. Even for lossless media, the elements of the permittivity tensor are in general complex. If a medium is lossless then the permittivity is Hermitian and positive definite. A more general statement is given by the Onsanger relations, which for a medium that has a DC magnetic field ($\mathbf{B}_0$) applied,, $$\epsilon_{ij}(\omega,\mathbf{k},\mathbf{B}_0) = \epsilon_{ji}(\omega,-\mathbf{k},-\mathbf{B}_0)$$.

I am personally most familiar with magnetized plasmas which are not spatially dispersive, so for the lossless case we have $$\epsilon_{ij}(\omega,\mathbf{B}_0) = \epsilon_{ji}(\omega,-\mathbf{B}_0)= \epsilon^\ast_{ji}(\omega,\mathbf{B}_0)$$
Jason