<p>If I interpret your question and the comments to Rob's answer correctly you are concerned that the ray trajectories would look like this (the deflection is <em>greatly</em> exaggerated for clarity):</p>
<p><a href="https://i.sstatic.net/DmK7fb4E.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/DmK7fb4E.png" alt="Rays" /></a></p>
<p>and therefore that the ray reaching the observer is not the one with the maximum deflection. If I have misunderstood you ignore this answer!</p>
<p>Anyhow, you are quite correct that this is the case for a star directly behind the Sun i.e. one on the optic axis. However if we start at the observer and work backwards we get rays like this:</p>
<p><a href="https://i.sstatic.net/U9kv1fED.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/U9kv1fED.png" alt="Rays 2" /></a></p>
<p>Assuming there is a dense enough distribution of stars there will be a ray that reaches the observer that grazes the Sun and has the maximum deflection, but the light from that ray was not initially parallel to the optic axis.</p>

CC BY-SA 4.0
2024-08-05T06:52:16.160Z
None of the stars you have drawn would be seen at the limb of the Sun and none would have positions deflected by 1.75 arcsec. Star C comes closest.
ProfRob
CC BY-SA 4.0
2024-08-05T06:55:35.900Z
@ProfRob Presumably you mean in the second diagram, and I'd be happy to accept corrections to it. Alternatively if I have pinpointed the OP's concern you could extend your own answer to include it and I'll delete mine (though we have yet to hear back from the OP whether this *is* what they are asking about).
John Rennie
CC BY-SA 4.0
2024-08-05T06:42:09.430Z
2024-08-05T06:42:09.430Z
John Rennie
<p>The <span class="math-container">$R$</span> in the formula is the closest approach of the light ray to the centre of the Sun. If a star were seen right at the limb of the Sun, then <span class="math-container">$R \simeq R_\odot$</span> and the GR deflection angle would be around 1.7 arcsec.</p>
<p>In practice stars are not viewed right at the limb of the Sun, so <span class="math-container">$R$</span> is larger and the deflection is smaller. When the deflection angle is this small, one just uses a radius corresponding to the projected separation on the sky - so <span class="math-container">$R=R_\odot$</span> corresponds to 15 arc minutes angular separation from the solar centre projected on the sky.</p>
<p>Note that the background star is very far away, so all rays reaching Earth are parallel. The ones that pass closer or further from the Sun, are not the ones that reach the observer on Earth. Basically you draw one ray from the star to the observer, going past the Sun. You are correct that you <em>could draw</em> an infinite number of rays that have differing values of <span class="math-container">$R$</span>, but only one is correct according to General Relativity.</p>
<p>In the picture you have drawn, you have massively exaggerated the deflection angle, but the value of <span class="math-container">$R$</span> would be one that is the minimum distance between the light ray and the solar centre. It looks like <span class="math-container">$R \sim 4R_\odot$</span>. The deflection angle would be 4 times less than if the star were viewed at the limb of the Sun. Possibly some pop-sci desriptions of the phenomenon are confusing. In practice, the deflections of stars at a variety of angular distances from the Sun are measured during an eclipse. One can then plot a straight-line graph of deflection versus the reciprocal of angular distance to extrapolate to what the maximum deflection would be for a <em>hypothetical star</em> at the limb of the Sun.</p>
<p>Edit: To follow up on that final point. Here are two diagrams taken from the original paper by <a href="https://royalsocietypublishing.org/doi/10.1098/rsta.1920.0009" rel="nofollow noreferrer">Dyson et al. (1920)</a>. The first shows the positions of the stars, with respect to the Sun, that were used for the deflection measurement. Note that none of them are very close to the limb. The second graph is of the deflection angle versus the angular separation from the Sun (scaled so that the reciprocal of that angle is linear along the x-axis). You can see from this plot that the largest deflections <em>actually measured</em> were just over 1 arcsecond, not 1.7 arcseconds.</p>
<p><a href="https://i.sstatic.net/jtYfgYdF.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/jtYfgYdF.png" alt="Stars near the Sun" /></a>
<em>Picture (from photographic plates) of the stars used to measure the gravitational deflection due to the Sun in the 1919 eclipse (from Dyson et al. 1920).</em></p>
<p><a href="https://i.sstatic.net/LGOaf7dr.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/LGOaf7dr.png" alt="Deflection angle versus angular distance from the Sun" /></a>
<em>Declection angle vs angular separation from the Sun for those stars. The x-axis is scaled so that the result should be a straight line. The dashed line shows the Newtonian prediction, thicker solid line is the GR prediction and the thinner solid line is the best fit.</em></p>

CC BY-SA 4.0
2024-08-05T04:53:29.363Z
I agree not all rays will reach the observer, and that's why I'm confused why it seems people are assuming the one deflected at the limb of the sun is the one that will be observed. As you said in the second paragraph "in practice stars are not viewed right at the limb of the Sun". I've added a sketch in the question since it won't let me insert image in the comment. In the picture, given the location of the source, the light ray that passes at the limb of the sun will not reach the observer. The ray that **does** reach the observer is deflected at a distance larger than the solar limb.
ABC
CC BY-SA 4.0
2024-08-05T18:14:49.470Z
" In practice, the deflections of stars at a variety of angular distances from the Sun are measured during an eclipse ... to extrapolate to what the maximum deflection would be for a hypothetical star at the limb of the Sun."
-- My goodness THANK YOU this is it!  I always wondered how could everyone just happen to observe that one star that is exactly at the limb of the sun to compare to the theoretical prediction... Little did I know they didn't; it's an extrapolation!
ABC
CC BY-SA 4.0
2024-08-04T17:00:09.660Z
2024-08-05T18:37:55.333Z
2024-08-05T18:37:55.333Z
ProfRob
<p>The Newtonian prediction and GR prediction of light deflection angles are
<span class="math-container">$\frac{2GM}{c^2 R}$</span>  vs. <span class="math-container">$\frac{4GM}{c^2 R}$</span>. One gives 0.8 arcsec and the other 1.7 arcsec for deflection around the Sun, and the observation confirmed it should be 1.7 arcsec as observed during the solar eclipse.</p>
<p>What value of &quot;R&quot; is used to get the predicted 0.8 or 1.7 arcsec? It seems solar radius is used, but why should it be? R is supposed to be distance between light ray and the sun, which could be at any distance from the Sun. Why do we assume the light is deflected right at the Sun's surface?</p>
<p><a href="https://i.sstatic.net/wjrFPZpYm.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wjrFPZpYm.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2024-08-05T04:58:43.737Z
Here's a question from a few months ago on this topic, which you may enjoy. https://astronomy.stackexchange.com/q/55786/16685
PM 2Ring
<observational-astronomy><general-relativity><solar-eclipse><gravitational-lensing>
Newtonian vs General Relativistic light deflection angle
ABC
# Newtonian vs General Relativistic light deflection angle

(Asked by: username_0 on 2024-08-04)

The Newtonian prediction and GR prediction of light deflection angles are
$\frac{2GM}{c^2 R}$ vs. $\frac{4GM}{c^2 R}$. One gives 0.8 arcsec and the other 1.7 arcsec for deflection around the Sun, and the observation confirmed it should be 1.7 arcsec as observed during the solar eclipse.

What value of "R" is used to get the predicted 0.8 or 1.7 arcsec? It seems solar radius is used, but why should it be? R is supposed to be distance between light ray and the sun, which could be at any distance from the Sun. Why do we assume the light is deflected right at the Sun's surface?

[![enter image description here](https://i.sstatic.net/wjrFPZpYm.png)](https://i.sstatic.net/wjrFPZpYm.png)

**Tags:** observational-astronomy, general-relativity, solar-eclipse, gravitational-lensing
> **Comments:**
> **username_1**: Here's a question from a few months ago on this topic, which you may enjoy. https://astronomy.stackexchange.com/q/55786/16685  

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## 2 answers

### Answer by username_3 on 2024-08-04. Score: 12 | Accepted answer

The $R$ in the formula is the closest approach of the light ray to the centre of the Sun. If a star were seen right at the limb of the Sun, then $R \simeq R\_\odot$ and the GR deflection angle would be around 1.7 arcsec.

In practice stars are not viewed right at the limb of the Sun, so $R$ is larger and the deflection is smaller. When the deflection angle is this small, one just uses a radius corresponding to the projected separation on the sky - so $R=R\_\odot$ corresponds to 15 arc minutes angular separation from the solar centre projected on the sky.

Note that the background star is very far away, so all rays reaching Earth are parallel. The ones that pass closer or further from the Sun, are not the ones that reach the observer on Earth. Basically you draw one ray from the star to the observer, going past the Sun. You are correct that you *could draw* an infinite number of rays that have differing values of $R$, but only one is correct according to General Relativity.

In the picture you have drawn, you have massively exaggerated the deflection angle, but the value of $R$ would be one that is the minimum distance between the light ray and the solar centre. It looks like $R \sim 4R\_\odot$. The deflection angle would be 4 times less than if the star were viewed at the limb of the Sun. Possibly some pop-sci desriptions of the phenomenon are confusing. In practice, the deflections of stars at a variety of angular distances from the Sun are measured during an eclipse. One can then plot a straight-line graph of deflection versus the reciprocal of angular distance to extrapolate to what the maximum deflection would be for a *hypothetical star* at the limb of the Sun.

Edit: To follow up on that final point. Here are two diagrams taken from the original paper by [Dyson et al. (1920)](https://royalsocietypublishing.org/doi/10.1098/rsta.1920.0009). The first shows the positions of the stars, with respect to the Sun, that were used for the deflection measurement. Note that none of them are very close to the limb. The second graph is of the deflection angle versus the angular separation from the Sun (scaled so that the reciprocal of that angle is linear along the x-axis). You can see from this plot that the largest deflections *actually measured* were just over 1 arcsecond, not 1.7 arcseconds.

[![Stars near the Sun](https://i.sstatic.net/jtYfgYdF.png)](https://i.sstatic.net/jtYfgYdF.png)
*Picture (from photographic plates) of the stars used to measure the gravitational deflection due to the Sun in the 1919 eclipse (from Dyson et al. 1920).*

[![Deflection angle versus angular distance from the Sun](https://i.sstatic.net/LGOaf7dr.png)](https://i.sstatic.net/LGOaf7dr.png)
*Declection angle vs angular separation from the Sun for those stars. The x-axis is scaled so that the result should be a straight line. The dashed line shows the Newtonian prediction, thicker solid line is the GR prediction and the thinner solid line is the best fit.*

> **Comments:**
> **username_0**: I agree not all rays will reach the observer, and that's why I'm confused why it seems people are assuming the one deflected at the limb of the sun is the one that will be observed. As you said in the second paragraph "in practice stars are not viewed right at the limb of the Sun". I've added a sketch in the question since it won't let me insert image in the comment. In the picture, given the location of the source, the light ray that passes at the limb of the sun will not reach the observer. The ray that **does** reach the observer is deflected at a distance larger than the solar limb.  
> **username_0**: " In practice, the deflections of stars at a variety of angular distances from the Sun are measured during an eclipse ... to extrapolate to what the maximum deflection would be for a hypothetical star at the limb of the Sun." -- My goodness THANK YOU this is it!  I always wondered how could everyone just happen to observe that one star that is exactly at the limb of the sun to compare to the theoretical prediction... Little did I know they didn't; it's an extrapolation!  

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### Answer by username_2 on 2024-08-05. Score: 4

If I interpret your question and the comments to Rob's answer correctly you are concerned that the ray trajectories would look like this (the deflection is *greatly* exaggerated for clarity):

[![Rays](https://i.sstatic.net/DmK7fb4E.png)](https://i.sstatic.net/DmK7fb4E.png)

and therefore that the ray reaching the observer is not the one with the maximum deflection. If I have misunderstood you ignore this answer!

Anyhow, you are quite correct that this is the case for a star directly behind the Sun i.e. one on the optic axis. However if we start at the observer and work backwards we get rays like this:

[![Rays 2](https://i.sstatic.net/U9kv1fED.png)](https://i.sstatic.net/U9kv1fED.png)

Assuming there is a dense enough distribution of stars there will be a ray that reaches the observer that grazes the Sun and has the maximum deflection, but the light from that ray was not initially parallel to the optic axis.

> **Comments:**
> **username_3**: None of the stars you have drawn would be seen at the limb of the Sun and none would have positions deflected by 1.75 arcsec. Star C comes closest.  
> **username_2**: @ProfRob Presumably you mean in the second diagram, and I'd be happy to accept corrections to it. Alternatively if I have pinpointed the OP's concern you could extend your own answer to include it and I'll delete mine (though we have yet to hear back from the OP whether this *is* what they are asking about).  

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