Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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?)

share|cite|improve this question

2 Answers 2

up vote 1 down vote accepted

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.

share|cite|improve this answer
$\Gamma_{ab}{}^c$ can in general only be made 0 at a point, not in a neighborhood. The latter is possible iff space is flat in that neighborhood. – Robin Ekman Jun 8 '14 at 16:20
@RobinEkman: you're right, of course. In my mind, I always imagine this as taking some tolerance for the size of the christoffel symbols, and adjusting the size of the neighborhood so that the christoffel symbols are below this tolerance, and then imagine spacetime as "approximately/essentially" flat in this region -- so that my picture of the equivalence principle is a little more physical than just talking about a single point, which is obviously not realizable physically. – Jerry Schirmer Jun 8 '14 at 16:31
but I shouldn't use my personal language like that when communicating with the public. – Jerry Schirmer Jun 8 '14 at 16:43

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.