<p>If you have a look at the top-left panel of Fig.11 in <a href="https://www.staff.science.uu.nl/~rutte101/rrweb/rjr-edu/coursenotes/rutten_issf_notes_1993.pdf" rel="nofollow noreferrer">these lecture notes</a> by Rob Rutten, you will see that the continuum opacity at optical wavelengths at the photosphere is about <span class="math-container">$10^{-6.7}$</span> cm<span class="math-container">$^{-1}$</span>.</p>

<p>The inverse of this is the optical depth. You can see stuff through about 2-3 optical depths, so your "horizon", looking horizontally in the solar atmosphere is about 100 km.</p>

<p>However, the same plot shows that, because of the temperature and density variations with height, the photosphere becomes an order of magnitude more or less opaque at only 50km below or above the "photosphere" respectively.</p>

<p>What this means is that you can see about 100 km horizontally, but less than this (guesstimate 30km) looking downwards and more than this (guesstimate infinity - that is how the photosphere is defined!) looking upwards.</p>

<p>These numbers would all be revised downwards at the wavelengths of prominent absorption lines.</p>

<p>If you could look all around you, you would see a very asymmetric brightness/colour distribution. Overhead would be dimmer and cooler (redder); underneath, much brighter and hotter (more or less the same colour we see the photosphere).</p>

CC BY-SA 4.0
2018-11-01T09:27:25.610Z
2018-11-01T09:33:37.647Z
2018-11-01T09:33:37.647Z
ProfRob
<p>I suspect part of the problem here is that cosmological optical depth is defined vertically (so far as I can discern from the wikipedia article), while you seem to be asking for the extinction coefficient for a horizontal path at the specified star-radius.  So you'll need to find out what the <span class="math-container">$\alpha(z)$</span> is at that radius, play some games with integrated radiance due to the plasma in the horizontal line of site to estimate the noise factor, and compare that with the putative radiance from the distant object you wish to see.  </p>

CC BY-SA 4.0
2018-10-31T17:45:48.057Z
2018-10-31T17:45:48.057Z
Carl Witthoft
<p>I was trying to compose an answer to <a href="https://space.stackexchange.com/questions/22115/is-it-possible-to-touch-stars-with-our-bare-hands">this question here</a>, about touching the surface of a star, and I was going to mention that the popular depiction of the sun's surface as 'basically lava' (as depicted in <a href="https://www.youtube.com/watch?v=TBzVPXr8SNY" rel="noreferrer">Sunshine</a>, for example) is wrong.</p>

<p>The usual definition of a star's radius is where <a href="https://en.wikipedia.org/wiki/Solar_radius" rel="noreferrer">the optical depth is 2/3.</a> At this radius, the density of the sun is about 1/10th of air at sea level and can hardly be described as solid or liquid. </p>

<p>I wanted to get a sense of what this diffuse plasma would look like if you were in it and somehow impervious to the heat and radiation. Would it appear as a thin, glowing fog, through which you could still discern distant objects? Or would it appear as opaque as, say, a bonfire?</p>

<p>Rephrasing in a little more technical way, <strong>What is the extinction coefficient of solar plasma at ~1 Solar radius in the visible spectrum? 
What is the path length at which all light is effectively attenuated?</strong></p>

<p>(Since it varies by wavelength, I mean specifically the region of the visible spectrum where the plasma is at its most transparent).</p>

CC BY-SA 4.0
2018-10-31T20:21:33.643Z
It is of order a few 100 km. Do you need it more accurately than that? Then specify the wavelength.
ProfRob
CC BY-SA 4.0
2018-11-01T00:13:56.917Z
So I'm looking out my window at the mountains, and can tell they're far away by how blue-tinted they appear. What you're saying is that the view would be just as good, if not better if the earth (sans atmosphere) were within the sun's photosphere? I mean briefly, of course. 
If you were able to provide calculations for 400 nm (which I suspect of being the wavelength of least attenuation) and turn that into an answer, I'll accept it.
Ingolifs
<the-sun><plasma-physics><opacity>
How far can you see in solar plasma at just under 1 solar radius?
Ingolifs
# How far can you see in solar plasma at just under 1 solar radius?

(Asked by: username_0 on 2018-10-31)

I was trying to compose an answer to [this question here](https://space.stackexchange.com/questions/22115/is-it-possible-to-touch-stars-with-our-bare-hands), about touching the surface of a star, and I was going to mention that the popular depiction of the sun's surface as 'basically lava' (as depicted in [Sunshine](https://www.youtube.com/watch?v=TBzVPXr8SNY), for example) is wrong.

The usual definition of a star's radius is where [the optical depth is 2/3.](https://en.wikipedia.org/wiki/Solar_radius) At this radius, the density of the sun is about 1/10th of air at sea level and can hardly be described as solid or liquid.

I wanted to get a sense of what this diffuse plasma would look like if you were in it and somehow impervious to the heat and radiation. Would it appear as a thin, glowing fog, through which you could still discern distant objects? Or would it appear as opaque as, say, a bonfire?

Rephrasing in a little more technical way, **What is the extinction coefficient of solar plasma at ~1 Solar radius in the visible spectrum?
What is the path length at which all light is effectively attenuated?**

(Since it varies by wavelength, I mean specifically the region of the visible spectrum where the plasma is at its most transparent).

**Tags:** the-sun, plasma-physics, opacity

> **Comments:**
> **username_1**: It is of order a few 100 km. Do you need it more accurately than that? Then specify the wavelength.  
> **username_0**: So I'm looking out my window at the mountains, and can tell they're far away by how blue-tinted they appear. What you're saying is that the view would be just as good, if not better if the earth (sans atmosphere) were within the sun's photosphere? I mean briefly, of course.  If you were able to provide calculations for 400 nm (which I suspect of being the wavelength of least attenuation) and turn that into an answer, I'll accept it.  

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## 2 answers

### Answer by username_1 on 2018-11-01. Score: 4 | Accepted answer

If you have a look at the top-left panel of Fig.11 in [these lecture notes](https://www.staff.science.uu.nl/~rutte101/rrweb/rjr-edu/coursenotes/rutten_issf_notes_1993.pdf) by Rob Rutten, you will see that the continuum opacity at optical wavelengths at the photosphere is about $10^{-6.7}$ cm$^{-1}$.

The inverse of this is the optical depth. You can see stuff through about 2-3 optical depths, so your "horizon", looking horizontally in the solar atmosphere is about 100 km.

However, the same plot shows that, because of the temperature and density variations with height, the photosphere becomes an order of magnitude more or less opaque at only 50km below or above the "photosphere" respectively.

What this means is that you can see about 100 km horizontally, but less than this (guesstimate 30km) looking downwards and more than this (guesstimate infinity - that is how the photosphere is defined!) looking upwards.

These numbers would all be revised downwards at the wavelengths of prominent absorption lines.

If you could look all around you, you would see a very asymmetric brightness/colour distribution. Overhead would be dimmer and cooler (redder); underneath, much brighter and hotter (more or less the same colour we see the photosphere).


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### Answer by username_2 on 2018-10-31. Score: -1

I suspect part of the problem here is that cosmological optical depth is defined vertically (so far as I can discern from the wikipedia article), while you seem to be asking for the extinction coefficient for a horizontal path at the specified star-radius. So you'll need to find out what the $\alpha(z)$ is at that radius, play some games with integrated radiance due to the plasma in the horizontal line of site to estimate the noise factor, and compare that with the putative radiance from the distant object you wish to see.


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