13
$\begingroup$

If the energy of light is high, does its curvature differ from that of low-energy light around the Sun? In other words, if the wavelength of the light is shorter than another wavelength of light, then does the bending of the two lights differ around the Sun?

$\endgroup$
7
  • 2
    $\begingroup$ No, the amount of bending is the same for all wavelengths. See astronomy.stackexchange.com/q/33341/16685 & physics.stackexchange.com/q/46996/123208 $\endgroup$
    – PM 2Ring
    Apr 5 at 4:43
  • 2
    $\begingroup$ What is dependent on the energy of the beam is how much it bends space (to affect other light beams or matter). $\endgroup$ Apr 5 at 14:24
  • $\begingroup$ @PM2Ring I enjoy taking things to their extreme. What if the photon was so energetic it actually had a significant gravitational pull by itself? $\endgroup$ Apr 6 at 12:20
  • 1
    $\begingroup$ @StianYttervik That's very extreme! As Ross said, in theory, a light beam does affect the spacetime curvature, but the effect is tiny that it's usually neglected, although cosmologists do include the energy density due to EM radiation in their calculations of spacetime curvature and expansion. $\endgroup$
    – PM 2Ring
    Apr 6 at 13:43
  • 1
    $\begingroup$ @RossPresser Indeed! But the effect is small. Imagine we could focus the entire light output of the Sun into a cylindrical beam of radius R. The luminosity of the Sun is L=3.828E26 watts, which is equivalent to ~4.259 billion kg/s. The density of the beam is $\rho=\frac{L}{\pi R^2c^3}$. Using the formula here for the surface gravity of an infinite cylinder, $g=2\pi G\rho R$, we get $g=\frac{2GL}{c^3R}$. Using R = 1 mm, Google Calculator says (2*(3.828E26 W)*(6.6743E-11 m^3kg^-1s^-2)/c^3)/(1 mm) is ~$1.8965×10^{-6}\,m/s^2$ $\endgroup$
    – PM 2Ring
    Apr 6 at 14:00
17
$\begingroup$

The amount of "gravitational light bending" is independent of the photon energy (light wavelength).

The reason is that the light follows a path through spacetime that is appropriate for a massless particle and this is unique for a given set of initial conditions.

That this is so is amply demonstrated by the consistent angular displacement of "stars" near the limb of the sun whether observed at optical or radio wavelengths.

As pointed out in comments - there are small effects that must be taken into account, associated with the well-understood phenomenon of refraction in the corona of the Sun. However, these do not affect observations of lensing taken well away from the solar limb - which is easily possible at radio wavelengths and now becoming possible for the same sources using Gaia data.

Further evidence comes from the wavelength-independent nature of gravitational lensing and microlensing seen outside the solar system.

$\endgroup$
9
  • 1
    $\begingroup$ "light follows a path" In particular, correct my layperson's understanding as needed, light follows a straight line in curved space. A straight line is a straight line regardless of the nature of what's following it. $\endgroup$ Apr 5 at 20:40
  • 1
    $\begingroup$ @ProfRob This is a great answer accounting for the bending of light due to gravity, but what about the much larger effect of refraction by the Solar Atmosphere? $\endgroup$
    – Connor Garcia
    Apr 5 at 22:44
  • 1
    $\begingroup$ @ProfRob oh, I though they were the same. How are they different? $\endgroup$ Apr 5 at 22:53
  • 2
    $\begingroup$ @DonBranson if you shine a laser or throw a ball, they follow different paths through spacetime. Both are geodesics and there is no force on either. $\endgroup$
    – ProfRob
    Apr 6 at 0:02
  • 2
    $\begingroup$ @DonBranson The curvature of a trajectory through space is not the same as the curvature of a worldline through spacetime. Please see Why does the speed of an object affect its path if gravity is warped spacetime? $\endgroup$
    – PM 2Ring
    Apr 6 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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