2
$\begingroup$

Light creates gravity, and the greater the light's frequency, the greater this gravitational effect is. It stands to reason then that light of different colors would react slightly differently to gravitational fields. Namely, bluer light would bend more than red light, causing gravitational lenses to act like prisms.

$\endgroup$
2
  • 1
    $\begingroup$ Gravitational lenses act equally on all kinds of electromagnetic radiation, not just visible light, but also in non-electromagnetic radiation, like gravitational waves. quote from; en.wikipedia.org/wiki/Gravitational_lens#Description $\endgroup$ Sep 18, 2020 at 10:12
  • $\begingroup$ The optical terminology for the effect you are asking about is "chromatic aberration"; in the context of a lens this is the term for a wavelength-dependent focal length. Gravitational lenses have huge amounts of spherical aberration but no chromatic aberration in the geometric limit. $\endgroup$ Feb 22, 2021 at 11:15

1 Answer 1

7
$\begingroup$

In general relativity, any test particle's worldline (whether the particle is massive or massless) is determined entirely by its initial four-velocity $u^\mu$, and not by any other properties. This is manifested by the geodesic equation $$ \frac{d u^\mu}{d\tau} = \Gamma^{\mu} {}_{\rho \sigma} u^\rho u^\sigma. $$ From the properties of ordinary differential equations, you can see that if you know $u^\mu$ at some initial time, then it determines the value of $u^\mu$ for the future of the particle.

Intuitively, you can think of this as a consequence of the equivalence principle, extended to photons. A feather and a hammer released from rest will follow the same trajectories, though they have different masses—which is to say they have different amounts of energy (including rest energy.) If you accept that photons also follow spacetime curvature in a similar way, then it should seem plausible that the energy content of a photon shouldn't affect its trajectory either, and that two photons with the same energy and the same initial four-velocity will have identical trajectories.

It is true that the test-particle limit is an approximation (both for the geodesic equation, and for the hammer-feather equivalence principle test); if we take the gravity of the photon/light wave into account, there would probably be some minuscule difference in the scattering angles due to the effect the photon has on the body creating the gravitational field. But I would bet a sizable amount of money that such effects will not be detectable in my lifetime. The test particle approximation is an exceedingly good approximation for quite a few gravitational physics situations, including this one.

$\endgroup$
5
  • $\begingroup$ The geodesic equation however, is only valid in the test particle limit, i.e. when the gravitational field of the particle itself can be ignored. In principle there could (and should) be energy dependent correction to the scattering angle. $\endgroup$
    – TimRias
    Sep 18, 2020 at 13:22
  • 1
    $\begingroup$ @mmeent: That's true, but it's true in the same sense that a hammer takes a little less time to fall to the ground than a feather does when released from rest, because the hammer pulls the Earth upwards to it harder than the feather does. It's hard for me to imagine that such effects would ever be practically measurable; the test particle approximation is an excellent approximation for a lot gravitational physics problems, including this one. (That said, edited to reflect this.) $\endgroup$ Sep 18, 2020 at 13:32
  • 1
    $\begingroup$ Agreed, the effect should be absolutely tiny. However, there is an important distinction between "the effect should not exist at all in principle" vs "the effect exist in principle but is practically almost zero". $\endgroup$
    – TimRias
    Sep 18, 2020 at 14:18
  • 1
    $\begingroup$ I wonder what the order of magnitude estimate for the dispersion would be? I suppose that the relevant scales would be $\Delta E$ the energy difference between two test photons, maybe the Ricci scalar, and a length $L$ that the photons travel over; perhaps the deviation in length between the two photons goes something like $\delta L \sim L R \Delta E \frac{8 \pi G}{c^4}$? I probably got my units wrong, but a derivation of a relation like this would be interesting to see $\endgroup$ Feb 15, 2021 at 18:03
  • 1
    $\begingroup$ It seems to me that even a microscopic difference of angle between light of different colors would become significant after lightyears of travel? $\endgroup$ Feb 16, 2021 at 19:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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