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I have a material specified by a permittivity tensor written in Cartesian coordiantes: $$\begin{pmatrix} \epsilon_{xx} & \epsilon_{xy} & \epsilon_{xz}\\ \epsilon_{yx} &\epsilon_{yy} & \epsilon_{yz}\\ \epsilon_{zx} & \epsilon_{zy} &\epsilon_{zz} \end{pmatrix}$$

I want to write all my equations in cylindrical coordinates and so I have to write the permittivity tensor in cylindrical coordinates too. How can I do that?

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    $\begingroup$ Hmm, I'd presume by using a covariant transformation law.... $\endgroup$
    – Kyle Kanos
    Commented Jun 13, 2014 at 17:52
  • $\begingroup$ Do you merely want to write the tensor in terms of cylindrical coordinates, or do you also want to write the components in terms of cylindrical basis vectors? $\endgroup$
    – Muphrid
    Commented Jun 14, 2014 at 5:04
  • $\begingroup$ @Muphrid I want to use the transformed tensor in Maxwell's equations. For the transformation, If the components where functions of $(x,y,z)$, then it was not sufficient to just write $x,y,z$ in terms of $\rho, \theta, z$, and some more complicated transformation of elements of the tensor had to be done. $\endgroup$
    – user215721
    Commented Jun 14, 2014 at 5:07
  • $\begingroup$ Paisanco is right. Please refer to this paper which published in JOSAa <hal.archives-ouvertes.fr/hal-00831552/file/…> $\endgroup$
    – Eric Z
    Commented Oct 23, 2020 at 7:13

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In cylindrical coordinates the tensor components would look like

$ T_{r\theta z}= \left( \begin{array}{ccc} \epsilon_{rr} & \epsilon_{r\theta} &\epsilon_{rz} \\ \epsilon_{\theta r} & \epsilon_{\theta \theta} & \epsilon_{\theta z} \\ \epsilon_{zr} & \epsilon_{z\theta} & \epsilon_{zz} \end{array} \right)\ $

Since both coordinate systems are orthogonal, the transformation of your Cartesian tensor $T_{xyz}$ to a tensor $T_{r\theta z}$ would be given by

$ T_{r\theta z}= {Q^T}T_{xyz} {Q} $

with the transformation matrix $Q$ is the same matrix you would use to transform a vector from Cartesian to cylindrical, i.e.

$ Q= \left( \begin{array}{ccc} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{array} \right)\ $

and ${Q^T}$ is the transpose of $Q$.

Note that this discussion assumes that the goal is to solve the non-covariant Maxwell's equations in cylindrical coordinates, neglecting special relativity. A more general treatment of the permittivity tensor and its transformation would be necessary if solving the covariant form of the Maxwell's equations

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  • $\begingroup$ Your $Q$ goes from unit vector basis to unit vectors. That isn't what is usually done in tensor calculus. Do you think that is worth going into? $\endgroup$
    – Muphrid
    Commented Jun 14, 2014 at 14:52
  • $\begingroup$ The OP does need to compute the unit vectors in cylindrical coordinates and use the divergence, curl and Laplacian in cylindrical coordinates to solve Maxwell's equations, but the question was how to transform the tensor. $\endgroup$
    – paisanco
    Commented Jun 14, 2014 at 15:57
  • $\begingroup$ I'm aware of that, but I'm talking about your expression for $Q$: the electric field is a covector field, and so the natural basis to express it in are the basis covectors: $\hat r, \hat \theta/r, \hat z$. Your expression for $Q$ does not express the field in terms of this basis; it's not wrong, but I feel it's worthy of some comment. $\endgroup$
    – Muphrid
    Commented Jun 14, 2014 at 16:11
  • $\begingroup$ @Muphrid Your point is more general, but the OP did not say if the problem was to solve the covariant Maxwell's equations or the three dimensional "junior E&M" form. I had assumed the latter from the question context, perhaps the OP can comment? $\endgroup$
    – paisanco
    Commented Jun 14, 2014 at 16:37
  • $\begingroup$ @Muphrid Added edit to my answer to address your points that a more general treatment would be needed if solving the covariant Maxwell's equations. $\endgroup$
    – paisanco
    Commented Jun 14, 2014 at 16:53

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