<p>The amount of &quot;gravitational light bending&quot; is independent of the photon energy (light wavelength).</p>
<p>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.</p>
<p>That this is so is amply demonstrated by the consistent angular displacement of &quot;stars&quot; near the limb of the sun whether observed at optical or radio wavelengths.</p>
<p>As pointed out in comments - there are <em>small</em> 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.</p>
<p>Further evidence comes from the wavelength-independent nature of gravitational lensing and microlensing seen outside the solar system.</p>

CC BY-SA 4.0
2021-04-05T20:12:43.757Z
I've seen the experiment of bending light from stars around the sun. But has this been confirmed in any practical experiment?
Robert Andrzejuk
CC BY-SA 4.0
2021-04-05T20:40:55.340Z
"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.
Don Branson
CC BY-SA 4.0
2021-04-05T22:04:56.380Z
@DonBranson except that objects with mass also follow "straight lines", but they are different to the "straight lines" followed by light, despite no forces acting upon them.
ProfRob
CC BY-SA 4.0
2021-04-05T22:44:45.067Z
@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?
Connor Garcia
CC BY-SA 4.0
2021-04-05T22:53:46.417Z
@ProfRob oh, I though they were the same. How are they different?
Don Branson
CC BY-SA 4.0
2021-04-06T00:02:22.830Z
@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.
ProfRob
CC BY-SA 4.0
2021-04-06T07:44:14.423Z
@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?](https://physics.stackexchange.com/q/586835/123208)
PM 2Ring
CC BY-SA 4.0
2021-04-27T19:25:08.073Z
@PM2Ring - Finally made time to read that question. Thanks, it really clears things up.
Don Branson
CC BY-SA 4.0
2021-04-05T08:46:41.347Z
2021-04-06T06:52:13.213Z
2021-04-06T06:52:13.213Z
ProfRob
<p>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?</p>

CC BY-SA 4.0
2021-04-05T04:43:29.910Z
No, the amount of bending is the same for all wavelengths. See https://astronomy.stackexchange.com/q/33341/16685 & https://physics.stackexchange.com/q/46996/123208
PM 2Ring
CC BY-SA 4.0
2021-04-05T14:24:34.973Z
What **is** dependent on the energy of the beam is how much **it** bends space (to affect other light beams or matter).
Ross Presser
CC BY-SA 4.0
2021-04-06T12:20:10.703Z
@PM2Ring I enjoy taking things to their extreme. What if the photon was so energetic it actually had a significant gravitational pull by itself?
Stian
CC BY-SA 4.0
2021-04-06T13:43:53.813Z
@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.
PM 2Ring
CC BY-SA 4.0
2021-04-06T14:00:02.117Z
@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](https://physics.stackexchange.com/a/326085) 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 ~$1.8965×10^{-6}\,m/s^2$
PM 2Ring
CC BY-SA 4.0
2021-04-06T17:43:52.920Z
@PM2Ring Interesting calculation. What about https://en.wikipedia.org/wiki/GRB_080916C?
Peter - Reinstate Monica
CC BY-SA 4.0
2021-04-06T18:12:43.950Z
@Peter Yes, that probably bent spacetime a fair bit. ;)
PM 2Ring
<the-sun><gravity><light><deimos>
Does bending of light around the Sun depend on the wavelength?
Red Bel
# Does bending of light around the Sun depend on the wavelength?

(Asked by: username_0 on 2021-04-05)

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?

**Tags:** the-sun, gravity, light, deimos

> **Comments:**
> **username_1**: No, the amount of bending is the same for all wavelengths. See https://astronomy.stackexchange.com/q/33341/16685 & https://physics.stackexchange.com/q/46996/123208  
> **username_2**: What **is** dependent on the energy of the beam is how much **it** bends space (to affect other light beams or matter).  
> **username_3**: @PM2Ring I enjoy taking things to their extreme. What if the photon was so energetic it actually had a significant gravitational pull by itself?  
> **username_1**: @username_3Yttervik 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.  
> **username_1**: @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](https://physics.stackexchange.com/a/326085) 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 ~$1.8965×10^{-6}\,m/s^2$  
> **username_4**: @PM2Ring Interesting calculation. What about https://en.wikipedia.org/wiki/GRB_080916C?  
> **username_1**: @Peter Yes, that probably bent spacetime a fair bit. ;)  

---

## 1 answers

### Answer by username_5 on 2021-04-05. Score: 18

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.

> **Comments:**
> **username_6**: I've seen the experiment of bending light from stars around the sun. But has this been confirmed in any practical experiment?  
> **username_7**: "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.  
> **username_5**: @DonBranson except that objects with mass also follow "straight lines", but they are different to the "straight lines" followed by light, despite no forces acting upon them.  
> **username_8**: @username_5 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?  
> **username_7**: @username_5 oh, I though they were the same. How are they different?  
> **username_5**: @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.  
> **username_1**: @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?](https://physics.stackexchange.com/q/586835/123208)  
> **username_7**: @PM2Ring - Finally made time to read that question. Thanks, it really clears things up.  

---

