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?

  • $\begingroup$ For further comparison, the atmosphere of the moon is somewhere around 25,000~200,000 atoms per cubic cm. (It varies a lot.) $\endgroup$
    – Mr. Nichan
    Sep 6, 2020 at 5:21
  • $\begingroup$ "Planetary nebulae" (remnants of dead stars whose cores become white dwarfs) have densities between 100~10,000 atoms per cubic cm according to Wikipedia. The most rarefied laboratory vacuum I've ever heard of comes from the description on the following page and seems to be about 3,000 molecules per cubic cm, whereas the "intermediate" vacuum listed is 30,000,000,000,000. vacaero.com/information-resources/vac-aero-training/… $\endgroup$
    – Mr. Nichan
    Sep 6, 2020 at 5:34
  • $\begingroup$ Obviously nebulae collapse into stars and planets, so it must get very thick somewhere, but I guess one way to think about it is this: If you can't see you're hand, something is clearly dense enough to count as an atmosphere, even if it might be much less dense than Earth's air and just more opaque because of dust, etc., which basically means you're already in/on a planet/star, or maybe some very thick accretion disk or circumplanetary disk/ring very close to planet or star or stellar remnant. $\endgroup$
    – Mr. Nichan
    Sep 6, 2020 at 5:45

2 Answers 2


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.

  • 3
    $\begingroup$ 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, 2011 at 21:25
  • 4
    $\begingroup$ 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, 2013 at 16:41
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    $\begingroup$ It obviously isn't the mfp of a photon. $\endgroup$
    – ProfRob
    Aug 16, 2016 at 6:42
  • $\begingroup$ I imagine that what it would look like would depend on the type of nebula. If were a dark cold nebula rather than a glowing one, the sky would look even blacker than it does to us, since there would be no stars except for any stars, planets, etc. that were near you. If the optical depth were enough for you to see a star from a planet orbiting it (which I think it would often be), it would be just like the planet Krikkit from Hitchhiker's Guide to the Galaxy, which was supposed to have been in a dark nebula. $\endgroup$
    – Mr. Nichan
    Sep 6, 2020 at 5:02

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.

  • $\begingroup$ What would the mean free path of the matter particles? Is it a good continuum oris the Knudsen number tol high? The mean free path of a photon is not that much interesting. Water is transparent and still very thick. $\endgroup$ Mar 25, 2023 at 22:47
  • $\begingroup$ @VladimirFГероямслава the mean free path of a photon is the only thing of interest in determining how far you can see - which is what the question asks about. $\endgroup$
    – ProfRob
    Mar 26, 2023 at 9:41
  • $\begingroup$ Well OK, even though it can be interprettted more generally. But anyway I am still interested what the typical mean free path is. $\endgroup$ Mar 26, 2023 at 11:36

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