<p>The article 'A practical relativistic model for microarcsecond astrometry in space' (<a href="https://iopscience.iop.org/article/10.1086/367593/pdf" rel="nofollow noreferrer">Klioner 2003</a>; caution, very heavy duty maths...) describes the framework which was developed to support the data processing for the ESA <em>Gaia</em> astrometric space mission. This mission has the goal to measure the position (astrometry) of stars and other objects to approx. 20 microarcseconds (one millionth of an arcsecond, which is 1/3600th of a degree). In order to do this, you need to account for all the sources of error and deflection, such as light bending near massive objects, to a level that is well below you expected accuracy (this level of systematic error was chosen to be 1 microarcsecond, hence that part of title).</p>

<p>Section 6 of the paper discusses gravitational light bending and Table 1 gives the size of the effect for various bodies in the Solar System. The column headed by <span class="math-container">$\delta_{pN}$</span> gives the size of the effect in microarcseconds (<span class="math-container">$\mu$</span>as). As expected the Sun has the largest effect at <span class="math-container">$1.75 \times10^6\,\mu$</span>as or 1.75", with Jupiter second etc. The next two columns gives the angular separation a light ray has to pass to the body to cause a <span class="math-container">$1 \mu as$</span> or <span class="math-container">$10 \mu as$</span> effect. So for example, any light ray from a distant object passing within <span class="math-container">$11.3^\circ$</span> of Jupiter will cause at least a  <span class="math-container">$10 \mu as$</span> effect. The smallest bodies that have an effect larger than <span class="math-container">$1 \mu as$</span> are the dwarf planet Ceres (mass <span class="math-container">$8.958 \times 10^{20}$</span> kg), and the moons Dione (<span class="math-container">$1.05 \times 10^{21}$</span> kg) and Umbriel (<span class="math-container">$1.27 \times 10^{21}$</span> kg) - these masses are several thousand times smaller than the Earth for example.</p>

<p>Now the level of precision needed by <em>Gaia</em> is far more precise than and the deflections are tens of thousands times smaller than you would see in an optical image with a regular telescope e.g. Eddington's confirmation of general relativity during the 1919 total solar eclipse (<a href="https://www.space.com/37018-solar-eclipse-proved-einstein-relativity-right.html" rel="nofollow noreferrer">space.com story</a>). The Klioner (2003) paper includes a formula to estimate the size of the effect of the light deflection as a function of the density but you need to know what level of deflection you care about to get an answer. The formula (number 35 on page 11 of the paper) for the radius <span class="math-container">$L$</span> of a body producing a light deflection than <span class="math-container">$\delta$</span> is:
<span class="math-container">$$
L\geq \left(\frac{\rho}{1\,\textrm{g cm}^{-3}}\right)^{-1/2}\left(\frac{\delta}{1\,\mu\textrm{as}}\right)^{1/2}\times 624\,\textrm{km}
$$</span>
where <span class="math-container">$\rho$</span> is the mean density of the body. Once you have chosen a level of deflection/lensing you want (<span class="math-container">$\delta$</span>), you can chose suitable densities (e.g. asteroids are <span class="math-container">$\sim2\,\textrm{g cm}^{-3}$</span>, Earth and other rocky planets are  <span class="math-container">$\sim5.5\,\textrm{g cm}^{-3}$</span>)</p>

CC BY-SA 4.0
2020-02-21T01:34:51.927Z
2020-02-21T01:34:51.927Z
astrosnapper
<p>I wonder whether we know at how much gravitation a gravitational lensing around a celestial body becomes observable (visible in the sense of that we see stars on the wrong places, stars behind the body etc, <a href="https://upload.wikimedia.org/wikipedia/en/1/18/Neutron_Star_gravitational_lensing.png" rel="nofollow noreferrer">similar like around this neutron star</a>, but of course by far not as strong, I'd like to know where the boundary lies). Such visible gravitational lensing occurs around neutron stars, black holes and white dwarfs, right? What about red and brown dwarfs? They're also quite dense and have a high surface gravity (though not as high as the other mentioned bodies of course). Do they curve the space around them visibly? </p>

CC BY-SA 4.0
2020-02-05T18:25:08.303Z
Depends on what wavelength you want the effect to be observable at. For pulsar timing and radio astronomy, the extra time delay (the Shapiro delay) caused by the gravitational lensing is included in the standard `TEMPO2` analysis code for everything more massive than Venus (see Table 2 of [Hobbs, Edwards and Manchester 2006](https://academic.oup.com/mnras/article/369/2/655/3101501)) as it has effects on the signals if they are passing close enough to one or more of the planets/Sun. So you may want to edit and clarify the question for what you mean by "observable"
astrosnapper
CC BY-SA 4.0
2020-02-05T19:20:32.493Z
@astrosnapper Sorry, I don't understand a word. :-) I mean "visible to our eyes", I 'll clarify my question.
user30007
CC BY-SA 4.0
2020-02-05T19:56:30.067Z
Sorry let me try again. The effects of gravitational lensing from the Sun and planets are "observable" to radio astronomers looking at the pulses of radio "light" coming from distant pulsars. The extra distance the radio pulses have to travel as they bend round the Sun, Jupiter or Saturn have to be accounted for when they look at their data. Although this is "observable" and  "light" in the broad electromagnetic spectrum sense, it's probably not what you meant so the clarification to mean "visible light" is helpful.
astrosnapper
CC BY-SA 4.0
2020-02-06T06:06:35.897Z
@astrosnapper Now I understand. You can't see the Sun bending visible light at all. I'll improve my question again.
user30007
CC BY-SA 4.0
2020-02-06T09:32:25.153Z
What do you mean? Gravitational lensing by the Sun affecting the apparent position of stars during a solar eclipse was one of the first demonstrations of the correctness of general relativity. See https://en.wikipedia.org/wiki/Eddington_experiment
PM 2Ring
CC BY-SA 4.0
2020-02-06T09:43:14.210Z
@PM2Ring I mean we can't see the Sun bending visible light, that is what astrosnapper meant: it are radio pulses. I mean the bending of space being visible.
user30007
CC BY-SA 4.0
2020-02-06T09:50:19.283Z
As that Wikipedia article says, "For light grazing the surface of the sun, the approximate angular deflection is roughly 1.75 arcseconds". Sure, you won't notice that by naked eye observation, but it is observable by careful observation with a telescope.
PM 2Ring
CC BY-SA 4.0
2020-02-06T11:40:19.713Z
@PM2Ring And at what gravity does it become visible in a telescope? Can you see it e.g. around Jupiter which has about 2.5 g or around Neptune with 1.14 g?
user30007
CC BY-SA 4.0
2020-02-06T12:18:11.943Z
@user30007  I don't want to make an answer as I'd probably blow the math, but it's not just gravity, it's also the size of the gravitational field.   Jupiter's mass is about 1/1000th that of the Suns.  It's gravity is (bad approximations ahead) 1/10th.  It's angular deflection, 1/10th the size and 1/10th the strength puts it at about 1/100th of the Suns - measurable with fine equipment.  It's not just the gravity but the size of the field, that's part of the reason why galaxies with very low gravity at the edges still work as gravitational lenses.   It's both gravity or Gs and size.
userLTK
CC BY-SA 4.0
2020-02-06T12:25:23.163Z
@userLTK Thank you.
user30007
<gravity><general-relativity><gravitational-lensing>
At what gravity does a body's gravitational lensing become observable?
user30007
# At what gravity does a body's gravitational lensing become observable?

(Asked by: username_0 on 2020-02-05)

I wonder whether we know at how much gravitation a gravitational lensing around a celestial body becomes observable (visible in the sense of that we see stars on the wrong places, stars behind the body etc, [similar like around this neutron star](https://upload.wikimedia.org/wikipedia/en/1/18/Neutron_Star_gravitational_lensing.png), but of course by far not as strong, I'd like to know where the boundary lies). Such visible gravitational lensing occurs around neutron stars, black holes and white dwarfs, right? What about red and brown dwarfs? They're also quite dense and have a high surface gravity (though not as high as the other mentioned bodies of course). Do they curve the space around them visibly?

**Tags:** gravity, general-relativity, gravitational-lensing
> **Comments:**
> **username_1**: Depends on what wavelength you want the effect to be observable at. For pulsar timing and radio astronomy, the extra time delay (the Shapiro delay) caused by the gravitational lensing is included in the standard `TEMPO2` analysis code for everything more massive than Venus (see Table 2 of [Hobbs, Edwards and Manchester 2006](https://academic.oup.com/mnras/article/369/2/655/3101501)) as it has effects on the signals if they are passing close enough to one or more of the planets/Sun. So you may want to edit and clarify the question for what you mean by "observable"  
> **username_0**: @astrosnapper Sorry, I don't understand a word. :-) I mean "visible to our eyes", I 'll clarify my question.  
> **username_1**: Sorry let me try again. The effects of gravitational lensing from the Sun and planets are "observable" to radio astronomers looking at the pulses of radio "light" coming from distant pulsars. The extra distance the radio pulses have to travel as they bend round the Sun, Jupiter or Saturn have to be accounted for when they look at their data. Although this is "observable" and  "light" in the broad electromagnetic spectrum sense, it's probably not what you meant so the clarification to mean "visible light" is helpful.  
> **username_0**: @astrosnapper Now I understand. You can't see the Sun bending visible light at all. I'll improve my question again.  
> **username_2**: What do you mean? Gravitational lensing by the Sun affecting the apparent position of stars during a solar eclipse was one of the first demonstrations of the correctness of general relativity. See https://en.wikipedia.org/wiki/Eddington_experiment  
> **username_0**: @PM2Ring I mean we can't see the Sun bending visible light, that is what astrosnapper meant: it are radio pulses. I mean the bending of space being visible.  
> **username_2**: As that Wikipedia article says, "For light grazing the surface of the sun, the approximate angular deflection is roughly 1.75 arcseconds". Sure, you won't notice that by naked eye observation, but it is observable by careful observation with a telescope.  
> **username_0**: @PM2Ring And at what gravity does it become visible in a telescope? Can you see it e.g. around Jupiter which has about 2.5 g or around Neptune with 1.14 g?  
> **username_3**: @user30007  I don't want to make an answer as I'd probably blow the math, but it's not just gravity, it's also the size of the gravitational field.   Jupiter's mass is about 1/1000th that of the Suns.  It's gravity is (bad approximations ahead) 1/10th.  It's angular deflection, 1/10th the size and 1/10th the strength puts it at about 1/100th of the Suns - measurable with fine equipment.  It's not just the gravity but the size of the field, that's part of the reason why galaxies with very low gravity at the edges still work as gravitational lenses.   It's both gravity or Gs and size.  
> **username_0**: @userLTK Thank you.  

---

## 1 answers

### Answer by username_1 on 2020-02-21. Score: 1

The article 'A practical relativistic model for microarcsecond astrometry in space' ([Klioner 2003](https://iopscience.iop.org/article/10.1086/367593/pdf); caution, very heavy duty maths...) describes the framework which was developed to support the data processing for the ESA *Gaia* astrometric space mission. This mission has the goal to measure the position (astrometry) of stars and other objects to approx. 20 microarcseconds (one millionth of an arcsecond, which is 1/3600th of a degree). In order to do this, you need to account for all the sources of error and deflection, such as light bending near massive objects, to a level that is well below you expected accuracy (this level of systematic error was chosen to be 1 microarcsecond, hence that part of title).

Section 6 of the paper discusses gravitational light bending and Table 1 gives the size of the effect for various bodies in the Solar System. The column headed by $\delta\_{pN}$ gives the size of the effect in microarcseconds ($\mu$as). As expected the Sun has the largest effect at $1.75 \times10^6\,\mu$as or 1.75", with Jupiter second etc. The next two columns gives the angular separation a light ray has to pass to the body to cause a $1 \mu as$ or $10 \mu as$ effect. So for example, any light ray from a distant object passing within $11.3^\circ$ of Jupiter will cause at least a $10 \mu as$ effect. The smallest bodies that have an effect larger than $1 \mu as$ are the dwarf planet Ceres (mass $8.958 \times 10^{20}$ kg), and the moons Dione ($1.05 \times 10^{21}$ kg) and Umbriel ($1.27 \times 10^{21}$ kg) - these masses are several thousand times smaller than the Earth for example.

Now the level of precision needed by *Gaia* is far more precise than and the deflections are tens of thousands times smaller than you would see in an optical image with a regular telescope e.g. Eddington's confirmation of general relativity during the 1919 total solar eclipse ([space.com story](https://www.space.com/37018-solar-eclipse-proved-einstein-relativity-right.html)). The Klioner (2003) paper includes a formula to estimate the size of the effect of the light deflection as a function of the density but you need to know what level of deflection you care about to get an answer. The formula (number 35 on page 11 of the paper) for the radius $L$ of a body producing a light deflection than $\delta$ is:
$$
L\geq \left(\frac{\rho}{1\,\textrm{g cm}^{-3}}\right)^{-1/2}\left(\frac{\delta}{1\,\mu\textrm{as}}\right)^{1/2}\times 624\,\textrm{km}
$$
where $\rho$ is the mean density of the body. Once you have chosen a level of deflection/lensing you want ($\delta$), you can chose suitable densities (e.g. asteroids are $\sim2\,\textrm{g cm}^{-3}$, Earth and other rocky planets are $\sim5.5\,\textrm{g cm}^{-3}$)


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