When gravity bends light, does it effect all frequencies equally. Absolutely equally?
- nebLv 71 month agoFavourite answer
Yes. The geodesic equation - the path that something follows in spacetime free fall in a gravitational field - depends only on the local geometry of spacetime and the velocity. The velocity of light is independent of frequency and therefore all frequencies will follow the exactly the same geodesic.
This ignores any infinitesimal gravity generated by the energy of the light itself (per Einsteins general relativity) since higher frequencies of light have more energy than lower frequencies of light. This would not be measureable in practice.
- 1 month ago
Well, if the wavelength is really long, comparable to the radius of curvature of space, the wave will diffract in a direction different from shorter waves. But that's only for really long wavelength radio waves.
- TomLv 71 month ago
Equally, because it is the SPACE the light is traveling through that is being bent---Since all Light Frequencies travel in space in a straight line, we will see the entire BEAM bent the same despite it's frequency/s. Light is not really being bent only the conduit (Space) that is carrying it.
- Anonymous1 month ago
Yeahhh : same treatment
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- formengLv 61 month ago
Gravity doesn't really bend light. What happens is that a massive object bends space time and if light comes close enough, it follows the bent space-time. But, yes, it effects all frequencies equally if there are no intervening objects for it to intercept.
- VamanLv 71 month ago
No. Actually electromagnetic waves have kinetic mass. Red light has very less mass than x rays and gamma rays. Therefore, guess work is gamma rays bend more than red light.
- DylanLv 61 month ago
Light travels in a straight line, when we talk about gravity bending light it actually the space that is bending.
- Anonymous1 month ago
Within current experimental limits, all frequencies are equally affected.
(If not, you would be able to detect different amounts of gravitational lensing from the same object in different parts of the object’s spectrum – which has not been observed.)
- Chris AncorLv 71 month ago
At a guess, I would think so.