<p>Most literature will define the diameter of the Sun up to the photosphere, the layer of the solar atmosphere you would see if you were to observe the Sun in white light.</p>

<p>The base of the <a href="https://en.wikipedia.org/wiki/Photosphere" rel="noreferrer">Photosphere</a> is defined as the region where the <a href="https://en.wikipedia.org/wiki/Optical_depth" rel="noreferrer">optical depth</a> is around 2/3, or the region where the plasma becomes transparent to most optical light wavelengths. </p>

<p>Of course the true edge of the solar atmosphere could be considered as the heliopause, where the direct influence of the Sun's magnetic field and solar wind end and interstellar space begins.</p>

CC BY-SA 3.0
2014-04-03T10:24:18.403Z
2014-04-03T19:09:32.660Z
2014-04-03T19:09:32.660Z
Cadair
<p>I thought I'd contribute an answer because there's a <em>very</em> recent paper on the subject:</p>

<p><a href="http://iopscience.iop.org/0004-637X/798/1/48" rel="noreferrer">Measuring the solar radius from space during the 2012 Venus Transit</a></p>

<p>It appeared in my RSS feeds this morning!  A related writeup is online <a href="http://hmi.stanford.edu/hminuggets/?p=522" rel="noreferrer">at the HMI website</a>. </p>

<p>To answer the question, this measurement uses the transit of Venus to fit the limb-darkening law of the Sun. That is, the Sun is a bit fainter the further from the centre that you look. As you reach the optically thinner layers near the "surface", the brightness falls off rapidly, towards zero in the vacuum of space. The inflection point of the curve (as a function of distance from the centre of the disk) is a reasonable estimate of the "radius".  As pointed out elsewhere, the value changes depending on which wavelength you use, but only by a few hundred km, compared to the Sun's overall radius of about 700 000km (actually more like 695 946 km), so the uncertainty is at or below the 0.1% level. Phil Plait <a href="http://blogs.discovermagazine.com/badastronomy/2012/03/21/the-sun-is-1392684-65-km-across/#.VJP1EuDkA" rel="noreferrer">wrote</a> about a similar measurement (by the same team, I believe) that used the transits of Mercury in 2003 and 2006.</p>

<p>Finally, the team also used the limb-darkening (I think) to <a href="http://www.sciencemag.org/content/337/6102/1638.abstract" rel="noreferrer">measure how round the Sun is</a>. i.e. the diameter from top-to-bottom versus left-to-right. Answer: the Sun is <em>very very</em> round, with the radii differing by a few parts per million.</p>

CC BY-SA 3.0
2014-12-19T09:55:46.420Z
2014-12-19T09:55:46.420Z
Warrick
<p>The radius of the Sun depends highly on which wavelength you are looking (well, taking a photo). In each of them, you'll have a well defined sharp boundary as explained by Zsbán Ambrus in his answer, but it is not the same: it varies with the wavelenght.</p>

CC BY-SA 3.0
2014-04-03T15:29:26.573Z
2014-04-03T15:29:26.573Z
Envite
<p>Look at the Sun.  You shouldn't do this directly with the naked eye, but you can do so through a very dark filter, or project a suitably dark image through a pinhole.  You can even find <a href="http://apod.nasa.gov/apod/ap120605.html" rel="nofollow">photos of the Sun on the internet</a>.  </p>

<p>What you see is a disk, uniformly bright and with a sharp boundary, surrounded by a comparatively much darker sky.  The bright region is the part we consider the Sun, and that's how we get the radius.  </p>

CC BY-SA 3.0
2014-12-18T19:58:29.190Z
Except you would get a wavelength dependent answer.
ProfRob
CC BY-SA 3.0
2017-07-24T19:14:03.567Z
It's not uniformly bright at all. The limb darkening accelerates as you get close to the edge, it sort-of goes down hill. As @RobJeffries points out it's wavelength dependent limb-darkening.
uhoh
CC BY-SA 3.0
2014-03-15T12:45:43.777Z
A sensibly experimentalist answer.
Stan Liou
CC BY-SA 3.0
2013-12-07T14:02:00.747Z
2014-12-18T11:57:42.557Z
2014-12-18T11:57:42.557Z
b_jonas
<p>Fusion reactions taking place inside the core of the star produce a huge amount of energy, most of which becomes heat. These reactions are not evenly distributed through the star and so there are phenomena such as sun spots and solar flares, however the total amount of energy produced tends to be reasonably constant.</p>

<p>I would say that the edge is defined by the average point where the gravity reaches equilibrium with the pressure of the star's super-heated gases (as a result of internal fusion).</p>

<p>See the picture of the <a href="http://en.wikipedia.org/wiki/Sun" rel="noreferrer">Sun on Wikipedia</a></p>

<p>That edge/balance will change when the sun begins to run low on hydrogen. At this time, the reactions inside the star will change causing it to become become a <a href="http://en.wikipedia.org/wiki/Sun#After_core_hydrogen_exhaustion" rel="noreferrer">giant red star</a>. </p>

<p>I guess you could compare it with the surface of sea water on Earth. It's technically not still and stable, but we can calculate an average value of the sea level. And it is because it's an average value that we can rely on that to determine altitude and earth radius as well.</p>

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2014-12-18T19:56:58.777Z
The whole of the Sun is in an equilibrium between pressure (gradient) and gravity.
ProfRob
CC BY-SA 3.0
2014-12-22T06:44:21.197Z
As said above, *all* of the Sun is in hydrostatic equilibrium. The physical principle that defines the surface is really the transition from optically thick to optically thin material.
Warrick
CC BY-SA 3.0
2013-12-05T15:55:31.003Z
Thank you Donald.McLean for completing my answer with more scientific terms.
Thibault
CC BY-SA 3.0
2014-03-15T12:45:03.007Z
This is a very confusing answer, it defines the edge as the point of hydrostatic equilibrium... but stable stars tend to be at least near hydrostatic equilibrium *everywhere*, so there is no apparent reason why this definition would unambiguously pick out the "edge".
Stan Liou
CC BY-SA 3.0
2013-12-05T13:40:11.527Z
2013-12-05T15:39:08.623Z
2013-12-05T15:39:08.623Z
Thibault
<p>That is, it's twice the radius where the radius is from the centre of the sun to some edge. But what is that edge?</p>

<the-sun>
How do you define the diameter of the Sun
Bathsheba
# How do you define the diameter of the Sun

(Asked by: username_0 on 2013-12-05)

That is, it's twice the radius where the radius is from the centre of the sun to some edge. But what is that edge?

**Tags:** the-sun

---

## 5 answers

### Answer by username_8 on 2013-12-05. Score: 7 | Accepted answer

Fusion reactions taking place inside the core of the star produce a huge amount of energy, most of which becomes heat. These reactions are not evenly distributed through the star and so there are phenomena such as sun spots and solar flares, however the total amount of energy produced tends to be reasonably constant.

I would say that the edge is defined by the average point where the gravity reaches equilibrium with the pressure of the star's super-heated gases (as a result of internal fusion).

See the picture of the [Sun on Wikipedia](http://en.wikipedia.org/wiki/Sun)

That edge/balance will change when the sun begins to run low on hydrogen. At this time, the reactions inside the star will change causing it to become become a [giant red star](http://en.wikipedia.org/wiki/Sun#After_core_hydrogen_exhaustion).

I guess you could compare it with the surface of sea water on Earth. It's technically not still and stable, but we can calculate an average value of the sea level. And it is because it's an average value that we can rely on that to determine altitude and earth radius as well.

> **Comments:**
> **username_8**: Thank you Donald.McLean for completing my answer with more scientific terms.  
> **username_7**: This is a very confusing answer, it defines the edge as the point of hydrostatic equilibrium... but stable stars tend to be at least near hydrostatic equilibrium *everywhere*, so there is no apparent reason why this definition would unambiguously pick out the "edge".  
> **username_5**: The whole of the Sun is in an equilibrium between pressure (gradient) and gravity.  
> **username_2**: As said above, *all* of the Sun is in hydrostatic equilibrium. The physical principle that defines the surface is really the transition from optically thick to optically thin material.  

---

### Answer by username_1 on 2014-04-03. Score: 7

Most literature will define the diameter of the Sun up to the photosphere, the layer of the solar atmosphere you would see if you were to observe the Sun in white light.

The base of the [Photosphere](https://en.wikipedia.org/wiki/Photosphere) is defined as the region where the [optical depth](https://en.wikipedia.org/wiki/Optical_depth) is around 2/3, or the region where the plasma becomes transparent to most optical light wavelengths.

Of course the true edge of the solar atmosphere could be considered as the heliopause, where the direct influence of the Sun's magnetic field and solar wind end and interstellar space begins.


---

### Answer by username_2 on 2014-12-19. Score: 5

I thought I'd contribute an answer because there's a *very* recent paper on the subject:

[Measuring the solar radius from space during the 2012 Venus Transit](http://iopscience.iop.org/0004-637X/798/1/48)

It appeared in my RSS feeds this morning! A related writeup is online [at the HMI website](http://hmi.stanford.edu/hminuggets/?p=522).

To answer the question, this measurement uses the transit of Venus to fit the limb-darkening law of the Sun. That is, the Sun is a bit fainter the further from the centre that you look. As you reach the optically thinner layers near the "surface", the brightness falls off rapidly, towards zero in the vacuum of space. The inflection point of the curve (as a function of distance from the centre of the disk) is a reasonable estimate of the "radius". As pointed out elsewhere, the value changes depending on which wavelength you use, but only by a few hundred km, compared to the Sun's overall radius of about 700 000km (actually more like 695 946 km), so the uncertainty is at or below the 0.1% level. Phil Plait [wrote](http://blogs.discovermagazine.com/badastronomy/2012/03/21/the-sun-is-1392684-65-km-across/#.VJP1EuDkA) about a similar measurement (by the same team, I believe) that used the transits of Mercury in 2003 and 2006.

Finally, the team also used the limb-darkening (I think) to [measure how round the Sun is](http://www.sciencemag.org/content/337/6102/1638.abstract). i.e. the diameter from top-to-bottom versus left-to-right. Answer: the Sun is *very very* round, with the radii differing by a few parts per million.


---

### Answer by username_3 on 2014-04-03. Score: 3

The radius of the Sun depends highly on which wavelength you are looking (well, taking a photo). In each of them, you'll have a well defined sharp boundary as explained by Zsbán Ambrus in his answer, but it is not the same: it varies with the wavelenght.


---

### Answer by username_4 on 2013-12-07. Score: 3

Look at the Sun. You shouldn't do this directly with the naked eye, but you can do so through a very dark filter, or project a suitably dark image through a pinhole. You can even find [photos of the Sun on the internet](http://apod.nasa.gov/apod/ap120605.html).

What you see is a disk, uniformly bright and with a sharp boundary, surrounded by a comparatively much darker sky. The bright region is the part we consider the Sun, and that's how we get the radius.

> **Comments:**
> **username_7**: A sensibly experimentalist answer.  
> **username_5**: Except you would get a wavelength dependent answer.  
> **username_6**: It's not uniformly bright at all. The limb darkening accelerates as you get close to the edge, it sort-of goes down hill. As @RobJeffries points out it's wavelength dependent limb-darkening.  

---

