# Formulation of Transformation optics using a Material Manifold

### Dear Community,

recently, Transformation optics celebrates some sort of scientific revival due to its (possible) applications for cloaking, see e.g. Broadband Invisibility by Non-Euclidean Cloaking by Leonhard and Tyc.

Willie Wong explained to me in another thread how a so-called material manifold could be used to include the effects of nonlinear optics as some sort of effective metric $\tilde{g}$. Indeed, this should not only work in that case, the easiest example of such an approach should be transformation optics.

Bevore explaining what is meant with such a construction, I will formulate the question:
How can one understand transformation optics in terms of the material manifold?

### The material manifold

The following explanations are directly linked to Willie's comments in the mentioned thread. Since I will formulate them on my own, mistakes might occur, which are in turn only related to me. Willie further gives the references Relativistic and nonrelativistic elastodynamics with small shear strains by Tahvildar-Zahdeh for a mathematical introduction and Beig & Schmidt's Relativistic Elasticity which is a very clear work.

Given a spacetime $(M,g)$ and a timelike killing vector field $\xi$, $g(\xi, \xi) = 0$ and a three-dimensional Riemannian material manifold $(N,h)$, the material map $\phi{:} M\rightarrow N$ for which $d\phi{(\xi)} = 0$ is an equivalence class mapping trajectories of point particles on $M$ onto one point in $N$. One can further define a metric $\tilde{g}$ as the pull-back of $h$, $\tilde{g}_{(x)}(X,Y) = h_{(\phi{(x)})}(d\phi{(X)}, d\phi{(Y)})$.

So, my question could be recast:

How can $\tilde{g}$ be used to explain transformation optics?

Sincerely

Robert

## Special case: Geometrical optics

Geometrical optics is widely used when wave-phenomena like interference do not play a major role. In this case we approach the question from relativity. I will scetch my thoughts in the following.

Minkowsky spacetime is described by the metric $\eta = \eta_{\mu\nu}dx^\mu dx^\nu$. Fixing coordinates, we can always bring it into the form $$\eta = -c^2 dt^2 + d\mathbf{r}^2$$ where I explicitly kept the speed of light $c$ and differ between space and time.
Light rays can be described by null geodesics $v$ for which $\eta(v,v) = 0$, hence $$\left( v^t \right)^2\eta_{tt} = \frac1{c^2} v^i v^j \eta_{ij}$$

Interpreting the speed of light not as a constant but as a function, $\frac1{c^2} = \epsilon_0\mu_0 \epsilon_r(x^\mu) \mu_r(x^\mu)$ (isotropic quantities assumed here, not tensorial ones - generaliziation is obvious but would blow up formulas here), we can directly interpret $$\tilde{g} = \epsilon_0\mu_0 \epsilon_r(x^\mu) \mu_r(x^\mu)\eta_{ij}dx^i dx^j$$

where, as before, $i$ and $j$ correspond to space-coordinates only. Since $\epsilon_r$ and $\mu_r$ are functions, $\tilde{g}$ will correspond to a curved three dimensional space. Furthermore, the material manifold can be identified as $$(d\phi)_i (d\phi)_j = \epsilon_r \mu_r \eta_{ij}$$ and has (under the given assumption of isotropy) the form of a conformal transformation, also discussed in another question.

I am not sure about refraction at jumps of $\epsilon_r$ and $\mu_r$ at this point - the formulation given makes only sense for differentiable quantities.

Just to add: I have a feeling that $\tilde{g}$ should be the dielectric permittivity $\epsilon$ (maybe in some rescaling) but I am not familliar with the approach the question is referring to so I left it out in favour of a clear statement. –  Robert Filter Dec 17 '10 at 15:58
not to quibble but how is the "material manifold" notion any different from what I stated in my answer to the non-linear optics question? And yes, the various components of $\tilde g$ do correspond to the dielectric permittivity and permeability. As I also mentioned in my previous answer. I used susceptibility $\chi_e$, but $\chi_e = \epsilon_r - 1$ where $\epsilon_r$ is the relative permittivity. –  user346 Dec 17 '10 at 17:51