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According to the Wikipedia article Elliptical galaxy, elliptical galaxies have a much lower concentration of gas between the stars than spiral galaxies. I know that until the gas is below a certain concentration, its atoms don't travel very far between collisions and it follows the normal gas laws and the gas actually flows into the stars when they pass by. I know that when the interstellar gas is at such a low concentration that atoms on average travel a much larger distance than the average diameter of a star in the elliptical galaxy it's in, the gas doesn't follow the normal gas laws and gas gets mainly absorbed into stars by happening to be headed towards one and colliding with it rather than by flowing towards one and the ratio of the rate of absorption of gas into stars to the concentration of interstellar gas is lower, and a star has about the same expected time before colliding with another star as an interstellar gas atom has before colliding with a star. I also know that when the interstellar medium is at the equilibrium concentration, there are 3 possible driving forces for the reduction of interstellar gas; star formation, flow of interstellar gas into stars, and random collisions of interstellar gas atoms with a star which can only occur at very low concentration; and 2 driving forces for the increase in interstellar gas, the slow escape of gas molecules from stars and collisions between stars because collisions occur at a high enough speed to produce an explosion instead of combining into a larger star. Which driving force for the reduction of interstellar gas is the largest? Also, which driving force for the increase in interstellar gas is larger? If it's star formation, that means the interstellar gas is at a high enough concentration to follow the normal gas laws but since the amount of interstallar gas is at equilibrium, the main driving force for the increase in interstellar gas is collisions between stars. If it's the slow escape of gas from stars, the main driving force for the decrease in interstellar gas could be the flow of gas into stars or the random collisions of gas molecules with stars but not star formation. If the main driving force for the reduction in interstallr gas concentration is collisions of gas molecules with stars, then a star has about the same expected time beofre a collision with another star as an intersteller gas atom has before colliding with another star so the expected time before a star collides with another star is less than the age of the galaxy and if it's flow of gas into stars, the main driving force for the increase in interstellar gas could be either of the two driving forces but if the driving force for the increase is collisions between stars, the expected time before a star collides with another star is less than the age of the galaxy,

On the other hand, if the interstellar gas concentration has not yet reached equilibrium, it could be that not enough time has gone by for the interstellar gas to not follow the normal gas laws and star formation occurs mainly by the gas coalescing into stars. It could also be that the interstellar gas started diving exponentially until it stopped following the normal gas laws then the rate at which it divides exponentially greatly reduced so it has not yet had time to reach the equilibrium concentration.

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    $\begingroup$ This question is currently a "wall of text". It would be nice if you could break it up into paragraphs, and shorter sentences. $\endgroup$ – HDE 226868 Sep 11 '17 at 17:53
  • $\begingroup$ My eyes hurt from trying to read the question, but: The number density of atoms, even in the most dilute regions of an elliptical, is of the order $10^{-3}$–$10^{-2}\,\mathrm{cm}^{-3}$. In contrast, the number density of stars in the most dense stellar regions is at most $10^3$ stars per cubic parsec, or $10^{-54}\,\mathrm{cm}^{-3}$. So an atom is much, much, much more likely to collide with a star, or another atom, than two stars are. $\endgroup$ – pela Sep 11 '17 at 18:20
  • $\begingroup$ Your changed your question, which might have invalidated my answer. This is frowned upon on SE; I'd ask that you rollback your edit and ask a new question if need be. $\endgroup$ – HDE 226868 Sep 12 '17 at 1:24
  • $\begingroup$ You wrote an answer so soon after I wrote my question. $\endgroup$ – Timothy Sep 12 '17 at 17:25
  • $\begingroup$ I edited my question because you complained that it was a wall of text. I did not see an answer at the time I started making the edit then while I was in the middle of editing, I saw that you wrote an answer. According to my question at meta.stackexchange.com/questions/298932/…, I thought it was better to think carefully what I want to write before I write it. I couldn't do that because I did not yet see a problem with the way it was before you complained that it was a wall of text. $\endgroup$ – Timothy Sep 19 '17 at 3:06
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The collision timescale for a star in the solar neighborhood is1 $$t_c\simeq5\times10^{10}\text{ Gyr}\left(\frac{R}{R_{\odot}}\right)^{-2}\left(\frac{v}{30\text{ km s}^{-1}}\right)^{-1}\left(\frac{n}{0.1\text{ pc}^{-3}}\right)^{-1}$$ where $R$ is the radius of a star, $v$ is its speed relative to the stars around it, and $n$ is the stellar number density. That number isn't in years; it's in billions of years. You're looking at collision timescales on the order of $\sim10^{19}\text{ yrs}$, many orders of magnitude larger than the current age of the universe. Even changing $v$ and $n$ a bit can't really lower that significantly.

Yes, there are lots of stars in a galaxy. Current estimates put 100 billion or so stars in the Milky Way, give or take. So yes, it's certainly likely for a galaxy to have two of its stars collide sometime before now (and I'm ignoring mergers in binary systems, as well as globular clusters, which also see many stellar collisions). But the collision timescale for any single star is much, much greater than the current age of its parent galaxy.

You can probably see why galaxies are usually modeled as collisionless systems. The collisionless version of the Boltzmann equation is used, which simplifies computations quite a bit.


1 I'm referencing my textbook, Foundations of Astrophysics, Ryden & Peterson. Eqn 22.17, page 516.

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  • $\begingroup$ You're making sure that collision equation is talking about physical collisions right? Astronomers tend to use the term collision to include all sorts of other types such as gravitational collisions. $\endgroup$ – zephyr Sep 12 '17 at 14:24
  • $\begingroup$ @zephyr Yes, I was just referring to physical collisions. The gravitational cross-section of a star shouldn't change things by more than an order of magnitude or two; for this sort of order-of-magnitude approximation, I figured that it would work well enough. $\endgroup$ – HDE 226868 Sep 12 '17 at 15:40

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