# Symmetry of the permittivity tensor of lossless media

• A
• Ngineer
, in summary, the dielectric permittivity tensor of a lossless medium is always symmetric because the elements of the permittivity tensor are in general complex.

#### 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.

Ngineer said:
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.

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

## What is the concept of symmetry in the permittivity tensor?

The concept of symmetry in the permittivity tensor refers to the relationship between the different components of the tensor. In a lossless media, the permittivity tensor is symmetric, meaning that the off-diagonal components are equal to each other.

## Why is the symmetry of the permittivity tensor important in materials science?

The symmetry of the permittivity tensor is important in materials science because it provides information about the anisotropic properties of a material. The symmetry of the tensor can help determine the directionality of properties such as refractive index and electrical conductivity.

## How is the symmetry of the permittivity tensor related to the crystal structure of a material?

The symmetry of the permittivity tensor is closely related to the crystal structure of a material. In crystalline materials, the symmetry of the tensor is determined by the symmetry of the crystal lattice. This means that different crystal structures will have different symmetries in their permittivity tensors.

## What is the mathematical representation of symmetry in the permittivity tensor?

The mathematical representation of symmetry in the permittivity tensor is through the use of indices. In a symmetric tensor, the subscripts of the off-diagonal elements are equal to each other, while the diagonal elements have the same subscript twice.

## How does the symmetry of the permittivity tensor affect the behavior of electromagnetic waves in a material?

The symmetry of the permittivity tensor affects the behavior of electromagnetic waves in a material by determining the polarization and direction of propagation. In materials with different symmetries, the wave will interact differently with the permittivity tensor, resulting in changes in its polarization and direction of propagation.