Relating Energy to Wavelength in curved space Consider a curved space, e.g. Schwarzschild:
\begin{align*}
ds^2 = -\left(1-\frac{2M}{r}\right)dt^2+\left(1-\frac{2M}{r}\right)^{-1}dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2
\end{align*}
Now, the energy of a photon is $E = \hbar \omega$, and $|\mathbf{k}|= \frac{2\pi}{\lambda}$, but am I correct in assuming that $\omega \neq |\mathbf{k}|$?
Because if $k^\mu = (\omega,\mathbf{k})$ then $k_\mu k^\mu = 0$ implies that:
\begin{align*}
g_{tt} \omega^2 + g_{rr}(k^1)^2+g_{\theta\theta}(k^2)^2+g_{\phi\phi}(k^3)^2 =0
\end{align*}
So basically, is it correct that the relationship between $\omega$ and $|\mathbf{k}|$ will vary in curved space? (And so relationships like $E = \frac{h}{\lambda}$ no longer hold?)
 A: I would think what you are saying is correct, the dispersion relation for light in vacuum $\omega = kc$ is derived by considering plane wave solutions to Maxwell's equations $\partial_{\mu} F^{\mu \nu}=0$. For curved spacetime, you would have to replace the usual derivative by a covariant derivative. 
A: The main thing to note is that the definition of the wavenumber you cite above is dependent on the underlying function satisfying the standard wave equation, because any function that satisfies
$$\eta^{ab}\partial_{a}\partial_{b}\phi(x) = 0$$
Will have its Fourier transform, $\Phi(k) = \int d^{4}x e^{ik^{a}x_{a}}\phi(x)$ satisfy 
$$k^{a}k_{a}\Phi(k) = 0$$ 
But this is no longer true, because for the case of curved spacetime, the wave equation is 
$$g^{ab}\partial_{a}\partial_{b}\phi(x) - g^{ab}\Gamma_{ab}{}^{c}\partial_{c}\phi(x) = 0$$
And this will require some modification to the Fourier transform to work.  More physically, you have effects like gravitational lensing that cause light to interact with the gravitational field, so you don't get simple straight-line propogation of monochromatic modes.
Note, however, that it is always possible to locally transform to a coordinate system where $\Gamma_{ab}{}^{c} =0$ and $g_{ab} = \eta_{ab}$, and there, you will be able to have a well defined wavenumber and frequency which satisfies $\omega^{2} = k^{i}k_{i}$.  This just won't work outside of your local neighborhood.
