# 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.

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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
On its own this question is still relevant, because the answers to non-linear optics question did not address the details of the correspondence between the effective metric and the electrical properties of the substance. – user346 Dec 17 '10 at 17:54
@space_cadet: Thank you for your thoughts. Indeed, the only difference is that this metric §\tilde{g}$is not that of the spacetime itself. You might not have seen the comment in the other thread where I had to admit that your idea was perfectly fine in the first place but I did not get it :) – Robert Filter Dec 18 '10 at 10:49 A useful way to write electrodynamics is as$dF=0$and$d\star F = J$where$\star$is the Hodge star and is the only thing involving a metric. – genneth Jan 20 '12 at 10:55 ## 1 Answer I am not an expert in this field, but I have done a fair amount of reading on covariant formulations electromagnetism in curved spacetime, and I happen to find using differential forms particularly useful. As genneth mentioned, formulating Maxwell's equations using the electromagnetic field tensor definitely seems like the way to go. This is a great intro to using geometric algebra in this type of formulation, but I think this one is the winner. I personally like the fact that no isotropy is assumed. I would try to summarize it, but I really do not think I could do any better than the author in that second link. ----Pure undocumented speculation here---- My knowledge of this field is entirely from reading, so read at your own risk! I was asked how geometric algebra might help here. In geometric algebra, an basis$\gamma$spans the your space in question. I think of them as a representation of direction--so if you wedge two of them you get an orientable 2D simplex, wedge three of them, and you get an orientable 3D simplex, and so forth. The electromagnetic field is spin 1, so its basis is$\gamma_i \wedge \gamma_j$for all$i\not=j$. Then it is is a matter of sticking a number next to each of those$\gamma_i \wedge \gamma_j$to represent to electromagnetic field. In goemetric algebra, B is "imaginary"--i.e. it is a three form. I prefer to keep it a sum of two forms and redefine the electromagnetic field tensor replacing E/c==>E so the units work out. Then, given an E and B field at each point in space, you can just "tack on" a$\gamma_i \wedge \gamma_j$and get an electromagnetic field tensor for your curved spacetime. As you noted when posing your question, you can encode all the information concerning the permittivity and permeability in the "effective metric" of your spacetime. Then you can use: $$\gamma_i \cdot \gamma_j=g_{ij}$$ to define your anticommutation relations. Now you can define Maxwell's equation [sic] $$\nabla F = J$$ where J is the current 3-form and$J_0=\rho\$

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Dear @Feynman and thank you for your answer! Even though I think geometric algebra is a nice thing and should be tought in general, I don't see how one can use this formulation to understand transformation optics in terms of a material manifold. Surely this might be a nice way to show it but could you be more specific? Greets – Robert Filter Feb 29 '12 at 7:28
Well, the latter link is far more applicable to a material manifold. That one is not on geometric algebra per se but on EM propagation in arbitrary materials. They use differential forms extensively, but no mention of geometric algebra. I figured the geometric algebra formulation could--if nothing else--be useful for ease of notation. – Alex Eftimiades Feb 29 '12 at 12:53