<p>The Einstein angle is given by:
<span class="math-container">$$\theta_E = \left(\frac{4GM}{c^2}\frac{D_{LS}}{D_L D_S}\right)^{1/2}$$</span>
<span class="math-container">$\theta_E$</span> is in radians.  <span class="math-container">$D_L$</span> is the distance to the lens, <span class="math-container">$D_S$</span> is the distance to the source, and <span class="math-container">$D_{LS}$</span> is the distance from the lens to the source.  For cosmological distances, these are angular distances, but within the Galaxy, we can use normal distances.</p>
<p>For a one solar mass star, we have
<span class="math-container">$4GM_\odot/c^2 = 6~$</span>km and for <span class="math-container">$D_S \gg D_L$</span> we have,</p>
<p><span class="math-container">$$\theta_E = \left(\frac{6}{3.086 \times 10^{13} (\frac{D_L}{pc})}\right)^{1/2}= 4.4 \times 10^{-7} \left(\frac{D_L}{pc}\right)^{-1/2}\rm{rad}= $$</span>
<span class="math-container">$$0.091 \left(\frac{D_L}{pc}\right)^{-1/2}\rm{arcsec},$$</span></p>
<p>where <span class="math-container">$\frac{D_L}{pc}$</span> is the distance to the lens in parsecs.  The Einstein ring around the very nearest stars would have a radius of order 0.1 arcsec and would be smaller for more distant stars.  To see a ring would require resolution sizes several times smaller than this.</p>
<p>The VLTI, consisting of four 8.5-m telescopes and four 1.8-m telescopes working as an optical interferometer, has resolved two images of a star microlensed by a dark object, probably a stellar remnant.  This work was published in
<a href="https://ui.adsabs.harvard.edu/abs/2019ApJ...871...70D/abstract" rel="nofollow noreferrer">First Resolution of Microlensed Images</a> in 2019.</p>
<p>Notice the last line in the abstract is:</p>
<blockquote>
<p>Our measurement contradicts Einstein’s prediction that “the luminous
circle [i.e., microlensed image] cannot be distinguished” from a star.</p>
</blockquote>

CC BY-SA 4.0
2024-03-09T01:47:49.077Z
Right. Not VLT but VLTI.  This is an optical interferometer.  I had forgotten about that. It did resolve a microlens, so I will add that to the answer.
eshaya
CC BY-SA 4.0
2024-03-09T02:18:22.260Z
Don't know how, but when I made the edits I got two copies of my answer to show up.
eshaya
CC BY-SA 4.0
2024-03-09T08:24:42.667Z
it's just a [glitch in the matrix](https://youtu.be/z_KmNZNT5xw) that happens from time to time, cf. [Second time I've ended up with my answer becoming double-posted. What might be happening here?](https://space.meta.stackexchange.com/q/960/12102)
uhoh
CC BY-SA 4.0
2024-03-09T00:02:14.480Z
2024-03-09T14:52:45.220Z
2024-03-09T14:52:45.220Z
eshaya
<p>We know that when two galaxies are perfectly aligned they form an Einstein Ring by gravitational lensing but can two stars form it? We know gravitational bending by stars, but can two perfectly aligned stars form an Einstein Ring?</p>

CC BY-SA 4.0
2022-08-23T18:50:25.077Z
Yup, sure! This effect is usually called [microlensing](https://en.m.wikipedia.org/wiki/Gravitational_microlensing) and it is used to detect exoplanets, among other things
Prallax
CC BY-SA 4.0
2022-08-23T22:05:45.643Z
_Resolving_ an Einstein ring from two stars is another question. [The angle](https://en.wikipedia.org/wiki/Einstein_ring#Introduction) is the square root of [typical stellar Schwarzschild radii, divided by typical distances between stars], or $\sqrt{\frac{\sim\mathrm{km}}{\sim\mathrm{pc}}}$, which is very small.
pela
CC BY-SA 4.0
2022-08-24T01:40:38.893Z
@pela Ignoring the observational challenges, is there a mass or density constraint on the closer star necessary for the ring to appear *outside* the disk of the star? In other words, for sufficiently low mass or low density nearer stars is the ring not visible because the nearer star is opaque?
uhoh
CC BY-SA 4.0
2022-08-24T14:57:00.257Z
@uhoh Good question, which I'd be happy to look into once I've met a handful of deadlines (unless you find the answer first, using the lens equation above :) ).
pela
CC BY-SA 4.0
2022-08-24T14:58:34.657Z
Being a non-native English, I'm unsure if that sounded snarky.  Definitely didn't mean it that way, but I already spent too much time on this site today 
pela
CC BY-SA 4.0
2022-08-24T22:29:40.297Z
@pela oh no, it sounded exactly like something I would write; I understand just what you mean. I also find the problem intriguing but in my case am in the middle of an experiment so if someone else wants to try in the mean time that will be great!
uhoh
CC BY-SA 4.0
2022-08-25T10:23:42.293Z
I found no observational data on Star rings! If they exist I think we would have already seen them because there are more stars than galaxies.
Sandeep Jassal
<gravity><gravitational-lensing>
Can two Stars form an Einstein Ring?
Sandeep Jassal
# Can two Stars form an Einstein Ring?

(Asked by: username_0 on 2022-08-23)

We know that when two galaxies are perfectly aligned they form an Einstein Ring by gravitational lensing but can two stars form it? We know gravitational bending by stars, but can two perfectly aligned stars form an Einstein Ring?

**Tags:** gravity, gravitational-lensing
> **Comments:**
> **username_1**: Yup, sure! This effect is usually called [microlensing](https://en.m.wikipedia.org/wiki/Gravitational_microlensing) and it is used to detect exoplanets, among other things  
> **username_2**: _Resolving_ an Einstein ring from two stars is another question. [The angle](https://en.wikipedia.org/wiki/Einstein_ring#Introduction) is the square root of [typical stellar Schwarzschild radii, divided by typical distances between stars], or $\sqrt{\frac{\sim\mathrm{km}}{\sim\mathrm{pc}}}$, which is very small.  
> **username_3**: @pela Ignoring the observational challenges, is there a mass or density constraint on the closer star necessary for the ring to appear *outside* the disk of the star? In other words, for sufficiently low mass or low density nearer stars is the ring not visible because the nearer star is opaque?  
> **username_2**: @uhoh Good question, which I'd be happy to look into once I've met a handful of deadlines (unless you find the answer first, using the lens equation above :) ).  
> **username_2**: Being a non-native English, I'm unsure if that sounded snarky.  Definitely didn't mean it that way, but I already spent too much time on this site today   
> **username_3**: @pela oh no, it sounded exactly like something I would write; I understand just what you mean. I also find the problem intriguing but in my case am in the middle of an experiment so if someone else wants to try in the mean time that will be great!  
> **username_0**: I found no observational data on Star rings! If they exist I think we would have already seen them because there are more stars than galaxies.  

---

## 1 answers

### Answer by username_4 on 2024-03-09. Score: 4

The Einstein angle is given by:
$$\theta\_E = \left(\frac{4GM}{c^2}\frac{D\_{LS}}{D\_L D\_S}\right)^{1/2}$$
$\theta\_E$ is in radians. $D\_L$ is the distance to the lens, $D\_S$ is the distance to the source, and $D\_{LS}$ is the distance from the lens to the source. For cosmological distances, these are angular distances, but within the Galaxy, we can use normal distances.

For a one solar mass star, we have
$4GM\_\odot/c^2 = 6~$km and for $D\_S \gg D\_L$ we have,

$$\theta\_E = \left(\frac{6}{3.086 \times 10^{13} (\frac{D\_L}{pc})}\right)^{1/2}= 4.4 \times 10^{-7} \left(\frac{D\_L}{pc}\right)^{-1/2}\rm{rad}= $$
$$0.091 \left(\frac{D\_L}{pc}\right)^{-1/2}\rm{arcsec},$$

where $\frac{D\_L}{pc}$ is the distance to the lens in parsecs. The Einstein ring around the very nearest stars would have a radius of order 0.1 arcsec and would be smaller for more distant stars. To see a ring would require resolution sizes several times smaller than this.

The VLTI, consisting of four 8.5-m telescopes and four 1.8-m telescopes working as an optical interferometer, has resolved two images of a star microlensed by a dark object, probably a stellar remnant. This work was published in
[First Resolution of Microlensed Images](https://ui.adsabs.harvard.edu/abs/2019ApJ...871...70D/abstract) in 2019.

Notice the last line in the abstract is:

> Our measurement contradicts Einstein’s prediction that “the luminous
> circle [i.e., microlensed image] cannot be distinguished” from a star.

> **Comments:**
> **username_4**: Right. Not VLT but VLTI.  This is an optical interferometer.  I had forgotten about that. It did resolve a microlens, so I will add that to the answer.  
> **username_4**: Don't know how, but when I made the edits I got two copies of my answer to show up.  
> **username_3**: it's just a [glitch in the matrix](https://youtu.be/z_KmNZNT5xw) that happens from time to time, cf. [Second time I've ended up with my answer becoming double-posted. What might be happening here?](https://space.meta.stackexchange.com/q/960/12102)  

---

