Transformation optics

Remarks

Example: let $\Omega = \Omega' = \R^3$, let the medium be vacuum, $\varepsilon = \varepsilon_0 I$ and $\mu = \mu_0 I$, and let $F(x) = 2x$. Then $A = 2I$ and $\det A = 8$, so $\varepsilon' = \frac{4 \varepsilon_0}{8} I = \frac{\varepsilon_0}{2} I$ and $\mu' = \frac{\mu_0}{2} I$. The new medium has refractive index $\sqrt{\varepsilon' \mu' / (\varepsilon_0 \mu_0)} = 1/2$: lengths are doubled and the index is halved, so optical path lengths do not change.

Non-example: the folding map $F(x_1, x_2, x_3) = (\abs{x_1}, x_2, x_3)$ is not injective and not smooth on the plane $x_1 = 0$. So it is not a diffeomorphism, and the rule does not apply to it as stated. Applied locally on the half-space $x_1 < 0$, where $\det A = -1$, the same formula gives negative parameters; this is how complementary media are described.

The transformed medium is in general anisotropic. If $\mu$ is a constant multiple of $\varepsilon$, then $\mu'$ is the same multiple of $\varepsilon'$, so the impedance is preserved. In the language of differential geometry, the rule says that $\varepsilon$ and $\mu$ transform as tensor densities.