What is the density of the Sun's surface?

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DaveC426913
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I was looking at the http://antwrp.gsfc.nasa.gov/apod/ap060710.html" and marvelled how well-defined the Sun's surface is. I know about coronae and such, but still, it looks almost solid (though I know it's entirely gaseous).

How well-defined is the surface? Does it go from transparent to opaque over a short distance?

A bit of Googling has revealed that the Sun's core density is about 150g/cm^3, but what is the density at the surface?
 
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This Wikipedia article has some values for particle density in the photosphere and atmosphere of the sun (reliability uncertain).
http://en.wikipedia.org/wiki/Structure_of_the_Sun#Photosphere

This page has mass densities for layers of the sun.
http://solar-center.stanford.edu/vitalstats.html

Gravitational Energy of the Sun explains 'Coronal Heating'
http://www.plasmaphysics.org.uk/research/sun.htm

A profile of the sun's layers - little more than half-way down the page.
http://ircamera.as.arizona.edu/astr_250/Lectures/Lecture_12.htm
 
sun modeling

for theoretical work there are several operational definitions of the suns surface depending on what qualities you are interested in. The concept of local thermodynamic equilibrium LTE comes into play a lot. If you want to calculate the spectrum from the different ionization states of Hydrogen for example, a layer of the sun should be defined as the thickness that is needed for the H atoms in that layer to collide often enough that they can reach the same temperature. If you go out far enough the layer thickness you need will be too large and you can say you've reached the edge of the sun.
 
The 'surface' is usually defined as optical depth of 1 where optical depth is given by

[tex] \tau = \int_0^s \kappa \rho ds[/tex]

where [tex]\kappa[/tex] is the Rosseland mean opacity, and [tex]\rho[/tex] is the density and s is the path inward from where the density is "zero".
 
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franznietzsche said:
The 'surface' is usually defined as optical depth of 1 where optical depth is given by

[tex] \tau = \int_0^s \kappa \rho ds[/tex]

where [tex]\kappa[/tex] is the Rosseland mean opacity
...so the answer is... :frown:
 
DaveC426913 said:
...so the answer is... :frown:


Complicated :biggrin: .

Simplest way to say is the point where the photon mean free path reaches out to infinity (not technically right, but that takes less typing that explaining in even more detail). So its the lowest point where the photons on average, escape to infinity without being scattered. Sorta.
 
franznietzsche said:
Complicated :biggrin: .

Simplest way to say is the point where the photon mean free path reaches out to infinity (not technically right, but that takes less typing that explaining in even more detail). So its the lowest point where the photons on average, escape to infinity without being scattered. Sorta.
So... the answer is...

the density a the surface of the Sun is _____
 
DaveC426913 said:
So... the answer is...

the density a the surface of the Sun is _____
The visible surface of the sun (i.e. the region of the sun from where we get most of the light is called the photosphere. This layer is very thin, only about 500 km thick as compared to the radius of the sun of 700,000 km. The density of the photosphere is very, very low, about 0.1% of the density of air at sea level on earth. We would call that kind of density a good vacuum and is much less than the density of wood. [The density of air at sea level is 0.0013 gm per cubic cm].
http://www.madsci.org/posts/archives/may2001/988762969.As.r.html