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Let's have metrics $$ ds^{2} = f(\mathbf r)dt^{2} - h(\mathbf r )\delta_{ij}dx^{i}dx^{j}. $$ Hot to show that motion of light in spacetime with this metrics is equal to motion in continuous media with refractive index $n = \sqrt{\frac{h}{f}}$?

Is it logical to start from Maxwell equations in curved spacetime, $$ D_{\mu}D^{\mu}A^{\nu} - R^{\nu}_{\quad \sigma}A^{\sigma} = 4\pi J^{\nu}, $$ and then show that it is equal to the Maxwell's equations for continuous media (by the way, how to write them down for this case?)?

Edit.

Hmm, the task is almost solved.

One more edit.

It is solved.

First, I wrote Maxwell's equation for media in a form of $$ \partial_{\alpha}F_{\beta \gamma} + \partial_{\beta}F_{\gamma \alpha} + \partial_{\gamma}F_{\alpha \beta} = 0, \qquad (1) $$ $$ \quad \partial_{\beta}H^{\beta \alpha} = 4\pi j^{\alpha}, \qquad (2) $$ where $H^{\alpha \beta} = \varepsilon \eta^{\alpha \mu}\eta^{\beta \nu}F_{\mu \nu}$, $$ \eta^{\alpha \mu} = diag \left(\sqrt{\mu \varepsilon}, -\frac{1}{\sqrt{\mu \varepsilon}}, -\frac{1}{\sqrt{\mu \varepsilon}}, -\frac{1}{\sqrt{\mu \varepsilon}} \right). $$ This connection will obvious if write down connection $$ F(\mathbf E , \mathbf B ) \to F(\mathbf D , \mathbf H ) = H(\mathbf D , \mathbf H ). $$ Second, I wrote down the equations for curved spacetime: $$ D_{\alpha}f_{\beta \gamma} + D_{\beta}f_{\gamma \alpha} + D_{\gamma}f_{\alpha \beta} = \partial_{\alpha}f_{\beta \gamma} + \partial_{\beta}f_{\gamma \alpha} + \partial_{\gamma}f_{\alpha \beta} = 0, \qquad (3) $$ $$ D_{\beta}h^{\beta \alpha} = \frac{1}{\sqrt{-g}}\partial_{\beta}(\sqrt{-g}h^{\beta \alpha}) = 4 \pi J^{\alpha}, \qquad (4) $$ where $h^{\beta \alpha} = g^{\beta \mu}g^{\alpha \nu}f_{\mu \nu}$, and where I used the equality for covariant derivative of antisymmetric tensor when there is the convolution.

So if use substitution $$ F_{\alpha \beta} = f_{\alpha \beta}, \quad H^{\beta \alpha} = \sqrt{-g}g^{\alpha \mu}g^{\beta \nu}f_{\mu \nu}, \quad j^{\alpha} = \sqrt{-g}J^{\alpha}, $$ the equations $(3), (4)$ will be "reduced" to $(1), (2)$.

Then all that remains is to find connection between metrics and $\varepsilon , \mu$. It is not hard when metrics is diagonal: $$ D^{i} = g^{i \alpha}g^{0\beta }F_{\alpha \beta} = -g^{i0}g^{0j}F_{j0} + g^{ij}g^{0l}F_{jl} + g^{ij}g^{00}F_{j0} = g^{ii}g^{00}F_{i0} = \frac{1}{fh}E^{i} = \varepsilon E^{i}, $$ $$ F^{ij} = g^{i \alpha}g^{j \beta }F_{\alpha \beta} = -g^{0i}g^{jk}F_{k0} + g^{ik}g^{j0}F_{k0} + g^{ik}g^{jl}F_{kl} = g^{ii}g^{jj}F_{ij} = \frac{1}{h^{2}}F_{ij} \Rightarrow $$ $$ H_{m} = -\frac{1}{2}\varepsilon_{m ij}F^{ij} = \frac{B_{m}}{h^{2}} = \frac{B_{m}}{\mu}, $$ so $$ n = \sqrt{\varepsilon \mu} = \sqrt{\frac{h^{2}}{hf}} = \sqrt{\frac{h}{f}}. $$

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    $\begingroup$ You should put your solution down here. It is an interesting problem. $\endgroup$
    – Prahar
    Nov 10 '13 at 21:31
  • $\begingroup$ @Prahar : I added it. $\endgroup$ Nov 10 '13 at 21:56
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Null geodesics in curved space-time could be found using Hamilton-Jacobi equation: $$ g^{\mu\nu}\partial_\mu S \partial_\nu S = 0. $$

The metric in question is static, so we could separate the $t$ variable by writing $S = W(\mathbf{r}) - E\,t,$ with $E$ a constant. Inverse metric is diagonal: $g^{\mu\nu} = \mathrm{diag}(f^{-1},-h^{-1},-h^{-1},-h^{-1})$. After substitution we get: $$ f^{-1} E^2 - h^{-1} (\nabla W )^2 = 0, $$ where $(\nabla W)^2 = \delta_{ij} \partial _i W \partial _j W$. That equation could easily be converted into conventional form of eikonal equation governing propagation of light in continuous media. We set $W = E\, \tau (\mathbf{r}) $ and obtain $$ (\nabla \tau)^2 = \left(\sqrt{\frac h f}\right)^2 = n^2, $$ where we introduced refractive index $n(\mathbf{r })=\sqrt{\frac h f}$.

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  • $\begingroup$ If I read this correctly, this is recovering what is called the "Optical metric", for the special case of a comoving background material? $\endgroup$ Feb 2 at 22:14

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