14
$\begingroup$

I've read that in some materials, $\mu$ can be a tensor, not a mere scalar. (I haven't actually dealt with tensors before, but I'm assuming for my purposes here, it is synonymous with "matrix".)

I'm not sure if the same holds for $\epsilon$, but I'm assuming it might there too.

I'm fine with either of those, but then I'm wondering what the formula

$$\epsilon\,\mu = \frac{1}{c_m^2}$$

turns into when dealing with tensors? ($c_m$ being the speed of light in the medium)

$\endgroup$
0

1 Answer 1

20
$\begingroup$

If $\varepsilon$ or $\mu$ are tensors (read, matrices), then so is $c_m$: $$ \overbrace{\varepsilon}^\mathrm{matrix} \underbrace{\mu}_\mathrm{matrix}=c_m^{-2}\ \leftarrow\ \text{matrix as well} $$

In other words, if the permeability and/or permittivity are matrices, then the speed of light is a matrix as well. In this case, the $\_^{\color{red}{-1}}$ is understood in the sense of matrix inverse.

In the coordiante system where $c_m$ is diagonal, we have $$ c_m=\begin{pmatrix} c_1&\cdot&\cdot\\\cdot&c_2&\cdot\\\cdot&\cdot&c_3\end{pmatrix} $$ where $c_i$ is the speed of light along the $x_i$ axis.

So yes: the speed of light in a medium can be direction-dependent; see Birefringence.

$\endgroup$
4
  • $\begingroup$ +1 Wow, I didn't expect this one, but it makes sense! I interpret this to mean that the speed of light is different in each direction? $\endgroup$
    – user541686
    Commented Dec 26, 2016 at 19:48
  • 2
    $\begingroup$ @Mehrdad yes, that's it! see the link in the post for more details. $\endgroup$ Commented Dec 26, 2016 at 19:52
  • 1
    $\begingroup$ Still, that is only the effective speed of light, accounting for different material properties in different directions. In between collisions and interactions with the material molecules etc the speed is always the same, c. $\endgroup$
    – Bob Bee
    Commented Dec 28, 2016 at 2:14
  • $\begingroup$ @BobBee if your wavelength is larger than the structure components (atoms) then this 'in between collisions' model doesnt hold anymore (in my opinion) $\endgroup$
    – lalala
    Commented Jun 15, 2021 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.