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How does the ISM resist gravity? That's the only force acting on it, and all other particles seem to collect together to form stars. What makes the ISM so special among other particles?

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    $\begingroup$ Why do you think that the ISM resists gravity? $\endgroup$ – Walter Oct 12 '17 at 17:51
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It is not true that the particles in the interstellar medium (ISM) are only acted upon by gravity. For instance,

  1. In many cases a significant part of the ISM is ionized, in which case it interacts with magnetic field which permeates the gas and may in some cases be quite strong.
  2. In the vicinity of massive and hence luminous stars, radiation pressure may exert a strong force on the ISM. They also emit copious amounts of cosmic rays (i.e. relativistic particles) that transfer momentum to the surrounding gas.
  3. Supernova explosions create hot bubbles that expand and sweep through the ISM, resulting in shock waves and galactic outflows.

In most cases, however, what may prevent a gas cloud from collapsing is simply its temperature. Despite all the above processes, and despite gravity being the weakest force, gas clouds do sometimes collapse to form stars. The criterion for doing so is that the gas is dense enough, and that its internal pressure (or thermal energy) is weak enough. This is described by Jeans instability, which formulates the criterion for a cloud of gas to collapse through equating pressure forces, or thermal energy, to gravity. One way to express this is the Jeans mass $M_J$ (Jeans 1902) which is the critical mass of a cloud where thermal energy is exactly balanced by gravitational forces: $$ M_J = \rho \left( \frac{\pi k_B T}{4 \mu m_\mathrm{u} G \rho} \right)^{3/2} \\ \propto \frac{T^{3/2}}{\rho^{1/2}}. $$ Here, $k_B$, $G$, and $m_u$ are Boltzmannn's constant, the gravitational constant, and the atomic mass unit, while $T$, $\mu$, and $\rho$ are the temperature, the mean molecular mass, and the density of the gas.

In the second line of the equation it is emphasized that $M_J$ increases with temperature, and decreases with density. In other words, if the gas is too hot, or too dilute, the total mass needed to collapse must be higher.

In general, gas will not collapse to form stars if the temperature is above some $10^4\,\mathrm{K}$. If the temperature is higher, the particles simply move too fast. Since various processes may easily heat the ISM to millions of degrees, the gas has to cool before it can collapse. On way to do this is by cooling radiation: Fast-moving atoms collide (either with each other or, more often, with electrons). Some of the kinetic energy of the atoms is spent exciting their electrons to higher levels. When the atoms de-excite, photons are emitted which can leave the system. The net result is that thermal energy is removed from the cloud, until at some point it has cooled enough to collapse.

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    $\begingroup$ So, ...., this begs the question: "What is the temperature and density of the ISM?". $\endgroup$ – Eric Towers Oct 10 '17 at 14:43
  • $\begingroup$ @EricTowers: Temperatures in the ISM may, in principle, take any value from a few Kelvin to several (tens of) millions of Kelvin. However, various cooling processes makes the gas reach certain "plateaus" of temperatures. I previously discussed exactly that in an answer to How cold is interstellar space?. Wrt. densities ($n$), the various phases of the ISM tend to be very roughly in pressure equilibrium, such that the product $nT$ is more or less constant. $\endgroup$ – pela Oct 10 '17 at 18:51
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    $\begingroup$ That is, whereas a warm $T\sim10^4$ K cloud may have a density of $n\sim0.1$–$1$ particles per cm$^3$, a surrounding, hot $T\sim10^6$ K envelope will have $n\sim0.001$–$0.01$ cm$^{-3}$. And a small $T\sim10^2$ K molecular cloud will have densities of $n\sim10$–$10^2$ cm$^{-3}$ (and higher). $\endgroup$ – pela Oct 10 '17 at 18:55
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    $\begingroup$ @EricTowers: it doesn't beg the question, it raises, or prompts the question. Begging the question, or petitio principii, is a logical fallacy in which the writer or speaker assumes the statement under examination to be true. See grammarist.com/rhetoric/begging-the-question-fallacy $\endgroup$ – Jim421616 Oct 12 '17 at 20:23
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    $\begingroup$ @Jim421616 : "It's a translation of the Latin phrase petitio principii, and it's used to mean that someone has made a conclusion based on a premise that lacks support." "In modern vernacular usage, 'to beg the question' frequently appears to mean 'to raise the question' (as in, 'This begs the question, whether...')" As this is not the 16th century, I am not limited by the past's mistranslations and usage of the phrase. All I asked for was the support. $\endgroup$ – Eric Towers Oct 13 '17 at 1:28
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First, consider that gravity is weak.

The nearest star system to the Sun is Alpha Centauri, at a distance of about 4 light-years. Consider the acceleration due to the Sun's gravity at half that distance: $$a_S=\frac{GM_\odot}{r^2}\simeq3.7\times10^{-13}\text{ m/s}^2$$ where $M_\odot$ is the mass of the Sun. That's an incredibly small acceleration, meaning that most massive bodies have a very small gravitational influence on the ISM. Particles are attracted to massive objects . . . but very, very weakly.

That said, the ISM sometimes does collapse. The number density varies in different parts, from $\sim10^{-3}$ particles per cubic centimeter to $\sim10^6$ particles per cubic centimeter (1, 2). Above about $10^4$ particles per cubic centimeter, though, you venture into the regime of molecular clouds, which sometimes do collapse to form stars.

On a final note, gravity isn't the only force acting on the ISM. Galactic magnetic fields, for instance, can influence the ISM's dynamics in various scenarios, including preventing or enabling the collapse of molecular clouds (see Ferrier (2005)).

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How does the ISM resist gravity?

It doesn't. There are two distinct sources of gravity: internal and external. The internal or self-gravity of the ISM may in fact result in collapse and subsequent star formation, as explained in another answer.

The external gravitational attraction from any star or gas cloud on the ISM is too weak to be of relevance and can be neglected, as demonstrated in yet another answer.

However, the ISM is subjected to the combined gravitational pull from all stars, gas, and dark matter in the Galaxy, i.e.\ the gravity of the Galaxy itself. In response to this pull, the ISM orbits the Galaxy on near-circular orbits, as do most of the stars (in a disc galaxy such as ours). Thus, the ISM is not special in this respect.

Why does the ISM not fall into the inner Galaxy (where it is pulled to)? This is simply because it has too much angular momentum. The situation is exactly the same as for the Earth pulled to the Sun, yet orbiting (nearly) on a circle around it.

Finally, note that magnetic forces and radiation pressure from nearby stars are much weaker than Galactic gravity and can be neglected when considering the Galactic orbits of the ISM.

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