# Sun's surface density

1. Jul 12, 2006

### DaveC426913

I was looking at the http://antwrp.gsfc.nasa.gov/apod/ap060710.html" [Broken] 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?

Last edited by a moderator: May 2, 2017
2. Jul 12, 2006

### neutrino

I wonder if one can calculate the density at the surface which so much "random" activity there. SOHO's site simply says...
http://sohowww.nascom.nasa.gov/explore/sun101.html

3. Jul 12, 2006

### Astronuc

Staff Emeritus
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

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

4. Jul 13, 2006

### Allday

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.

5. Jul 13, 2006

### franznietzsche

The 'surface' is usually defined as optical depth of 1 where optical depth is given by

$$\tau = \int_0^s \kappa \rho ds$$

where $$\kappa$$ is the Rosseland mean opacity, and $$\rho$$ is the density and s is the path inward from where the density is "zero".

Last edited: Jul 13, 2006
6. Jul 13, 2006

### DaveC426913

7. Jul 13, 2006

### franznietzsche

Complicated .

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.

8. Jul 13, 2006