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How functionaly dense are nebulae? Are they so sparse they are only visible from an interstellar or intergalactic perspective or would you be unable to see your hand in one?

Do they vary widely in density, between nebulae or even within a single one?

What would it look like from the inside of one?

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They are very sparse. Typical densities are in the range of 100 to 10,000 particles per $\textrm{cm}^3$.

This is much more dense than the general interstellar medium (1 particle per $\textrm{cm}^3$), but much, much less dense than anything you are used to - air is around $10^{19}$ particles per $\textrm{cm}^3$. You would very easily see your own hand in a nebula.

Density variations can be quite sharp within the nebula; in star-forming regions, the variations are strong and the density variations appear to be organized like a fractal, produced by turbulence within the cloud.

However, most nebulae are basically the same, and there aren't huge differences between the densities of different starforming regions. Planetary nebulae and supernova remnants, of course, can have very different densities depending on their ages, since they are expanding balls of gas rather than broad molecular clouds loosely bound by gravity.

If you were within a nebula, it is hard to say what it would look like. But nebulae are so large that the optical depth of the cloud would actually probably be quite high, and I would guess that it would look like you were surrounded by glowing green and red gas in the far distance - instead of space looking black and dark, it would be colored all over. But this would only be an effect caused by the fact that you are looking through so much gas - even if your spaceship were a thousand kilometers away, it probably wouldn't look much different if you were inside a nebula versus outside of it.

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    $\begingroup$ books.google.com/… says that the mean free path in a nebula is about 5800 km. The mean free path is the distance at which things start to get "lost in the fog." $\endgroup$ – Andrew Jul 4 '11 at 21:25
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    $\begingroup$ 5800km as the mean free path for what? Depending on whether its for gas particles, photons of a particular wavelength (absorption line?) or visible photons as a whole wound make a pretty big difference to visibility. $\endgroup$ – Kyle Oman Jan 24 '13 at 16:41
  • $\begingroup$ It obviously isn't the mfp of a photon. $\endgroup$ – Rob Jeffries Aug 16 '16 at 6:42
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Nebulae is too nebulous a term to provide a specific answer since it basically includes almost all extended Galactic gas and dust structures.

The density of the interstellar medium ranges from below 1 particle $cm^{-3}$ to about $10^6$ cm$^{-3}$. The lower densities tend to be associated with hot ionised gas and this is essentially optically thin (transparent). The cold atomic and molecular gas in the denser regions is also optically thin to visible light.

But we "see" examples of dark molecular clouds because it is dust that provides the opacity and the cloud material is about 1 part in a 100 made of dust (small particles of soot, silicates and other condensed matter).

Let's take the molecular cloud Barnard 68 as an example. This "dark nebula" has a diameter of half a light year and measurements of extinction by looking at star counts in the optical (left) and infrared (right) reveal an optical extinction of 33 magnitudes at cloud centre.

Barnard 68 seen in the visible (left) and infrared (right)

The mean free path of a visible photon corresponds roughly to 1.1 magnitudes of extinction. Thus the cloud is roughly 30 photon mean free paths thick (utterly opaque). If the cloud were uniform (it isn't), the mfp would correspond to about $10^{14}$m. So if you were in such a cloud you would have no problem seeing your hand, but you wouldn't be able to see anything of the universe outside the cloud unless you looked with infrared or radio telescopes.

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