<p>The gas cloud will have a tendency to collapse because of gravity, but that collapse is counteracted by the pressure of the gas as you expect.</p>
<p>When something happens that causes local density fluctuations (e.g. a nearby supernova) the gas will be contracted locally. As this globule of cloud contracts, pressure will build up (increasing the total kinetic energy of the cloud) but so will gravity (increasing the potential energy). If the mass of the gas is below a specific threshold, the pressure will cause the gas to expand again as you expected, but if the mass of gas is above the threshold, the pressure increase will not be able to counteract the increase in gravity and the globule of gas will continue to contract into a protostar.</p>
<p>This contraction will continue until some other process (e.g. pressure caused by nuclear fusion or centrifugal force due to rotation) is able to counteract gravity.</p>
<p><strong>Using math and physics:</strong></p>
<p>In a stable gas cloud the virial theorem holds, which states that the total kinetic energy of the cloud is equal to half the potential energy.</p>
<p><span class="math-container">$$
K = \frac{1}{2}U
$$</span>
where <span class="math-container">$K$</span> is the kinetic energy and <span class="math-container">$U$</span> is the potential energy.</p>
<p>The kinetic energy is due to the motions of the particles in the gas cloud. This motion of particles and their interactions is what constitutes pressure. The kinetic energy is, therefore a measure of pressure. When <span class="math-container">$N$</span> is the number of particles, <span class="math-container">$T$</span> is the temperature and <span class="math-container">$k$</span> is Boltzmann's constant we have a kinetic energy equal to:</p>
<p><span class="math-container">$$
K = \frac{3}{2}NkT
$$</span></p>
<p>The potential energy is due to gravity and is equal to:</p>
<p><span class="math-container">$$
U = -\frac{3}{5} \frac{GM^2}{R}
$$</span></p>
<p>where <span class="math-container">$G$</span> is the gravitational constant, <span class="math-container">$M$</span> is the mass and <span class="math-container">$R$</span> is the size of the cloud.</p>
<p>The virial theorem then becomes:</p>
<p><span class="math-container">$$
NkT = \frac{1}{5}\frac{GM^2}{R}
$$</span></p>
<p>And because <span class="math-container">$N=M/m$</span>, where <span class="math-container">$m$</span> is the average mass of the particles, we can write:
<span class="math-container">$$
kT\frac{M}{m} = \frac{1}{5}\frac{GM^2}{R}
$$</span>
If this equality holds, the gas cloud is stable and is in virial equilibrium.</p>
<p>If, however, the right side of the equation is bigger, the cloud will collapse. The right side depends on the potential (gravitational) energy, so if gravity is stronger the cloud collapses:</p>
<p><span class="math-container">$$
kT\frac{M}{m} &lt; \frac{1}{5}\frac{GM^2}{R}
$$</span></p>
<p>If, on the other hand the kinetic energy wins out, the left side is bigger and the cloud will expand. This can be due to an increase in temperature which increases the average speed of the particles (and thereby the cloud's pressure).</p>
<p>To find the mass at which a cloud collapses given its size we can work some on the equation:</p>
<p><span class="math-container">$$
kT\frac{M}{m} &lt; \frac{1}{5}\frac{GM^2}{R}
$$</span></p>
<p><span class="math-container">$$
\frac{kT}{m} &lt; \frac{1}{5}\frac{GM}{R}
$$</span></p>
<p><span class="math-container">$$
M &gt; \frac{5kT}{Gm}R
$$</span></p>
<p>If we assume that the density <span class="math-container">$\rho$</span> is constant we can relate the radius <span class="math-container">$R$</span> to the mass <span class="math-container">$M$</span>. The mass of the cloud is equal to the volume <span class="math-container">$V = \frac{4}{3}\pi R^3$</span> times the density, which gives us:
<span class="math-container">$$
M = \frac{4}{3}\pi \rho R^3
$$</span>
which gives us an expression for the radius <span class="math-container">$R$</span>:
<span class="math-container">$$
R = \left[ \frac{3 M}{4 \pi \rho} \right]^{\frac{1}{3}}
$$</span>
Using this equation for <span class="math-container">$R$</span>, we get:
<span class="math-container">$$
M &gt; \frac{5kT}{Gm}\left[ \frac{3 M}{4 \pi \rho} \right]^{\frac{1}{3}}
$$</span>
<span class="math-container">$$
M^{\frac{2}{3}} &gt; \frac{5kT}{Gm}\left[ \frac{3}{4 \pi \rho} \right]^{\frac{1}{3}}
$$</span></p>
<p><span class="math-container">$$
M &gt; \left[ \frac{5kT}{Gm}\right]^{\frac{3}{2}}\left[ \frac{3}{4 \pi \rho} \right]^{\frac{1}{2}}
$$</span></p>
<p>This value of <span class="math-container">$M$</span> is the threshold at which a globule of gas will start to contract and is called the Jean's mass <span class="math-container">$M_J$</span>:</p>
<p><span class="math-container">$$
M_J = \left[ \frac{5kT}{Gm}\right]^{\frac{3}{2}}\left[ \frac{3}{4 \pi \rho} \right]^{\frac{1}{2}}
$$</span></p>
<ul>
<li>If <span class="math-container">$M &gt; M_J$</span>, gravity is stronger --&gt; Collapse</li>
<li>If <span class="math-container">$M &lt; M_J$</span>, kinetic energy is stronger --&gt; Expansion</li>
</ul>
<p>References:</p>
<ol>
<li><a href="https://www.christopherlovell.co.uk/blog/2016/04/26/jeans-mass.html" rel="nofollow noreferrer">Christhoper Lovell, 2016, Deriving the Jean's Mass</a></li>
<li><a href="http://burro.astr.cwru.edu/Academics/Astr221/LifeCycle/jeans.html" rel="nofollow noreferrer">Chris Mios, 2005, Gravitational Collapse of Gas Clouds</a></li>
</ol>

CC BY-SA 4.0
2022-02-09T13:02:19.123Z
I can't resist wondering how this applies to the famous  https://what-if.xkcd.com/4/
Carl Witthoft
CC BY-SA 4.0
2022-02-10T00:45:03.343Z
Pressure and gravity have different dimensions/units. They can never be equal.
ProfRob
CC BY-SA 4.0
2022-02-10T08:51:00.273Z
@ProfRob You are correct of course, pressure is the force (of gas molecules) per square metre and gravity is the force on a molecule or on a volume of space containing some mass. I will edit the answer accordingly.
Dieudonné
CC BY-SA 4.0
2022-02-10T09:00:40.537Z
I have removed the section as it was not really conducive to the argument.
Dieudonné
CC BY-SA 4.0
2022-02-09T11:22:33.933Z
2022-02-10T10:35:59.367Z
2022-02-10T10:35:59.367Z
Dieudonné
<p>Whilst the answer given by @Dieudonné addresses one of the fundamental criteria for gravitational collapse here (the 'Jeans criterion') it contains some inaccuracies and omissions (which have been adopted from the cited references).</p>
<p>First of all, the equation assumed for the gravitational potential energy in the derivation implies a sphere of constant density. In hydrostatic equilibrium however (the assumption made in the reference this answer is based on) the density in an isothermal cloud must decrease like <span class="math-container">$1/R^2$</span> (as is not too difficult to show) which results in a gravitational potential energy</p>
<p><span class="math-container">$$U=-\frac{GM^2}{R}$$</span></p>
<p>i.e. a factor <span class="math-container">$5/3$</span> larger than assumed here, with the Jeans Mass changing correspondingly.</p>
<p>For arbitrary density distributions, the potential energy is</p>
<p><span class="math-container">$$U=-\frac{kGM^2}{R}$$</span></p>
<p>where <span class="math-container">$\frac{1}{2}\le k\lt \infty$</span> (see also <a href="https://physics.stackexchange.com/questions/341065/gravitational-potential-energy-of-any-spherical-distribution">this SE post</a>)</p>
<p>More importantly, the impression is given here that star formation would proceed as long as the Jeans criterion (in whatever form) is fulfilled. This crucially ignores the fact that for a closed system the total energy must be constant, so the potential energy lost during collapse must be turned into kinetic energy i.e. the temperature must increase (given that the thermal collision times/distances for atoms are usually orders of magnitude shorter than the free fall times/ dimensions of the cloud). This would stop any collapse rather soon.
From the equations</p>
<p><span class="math-container">$$U=-\frac{GM^2}{R}$$</span>
<span class="math-container">$$E=\frac{U}{2}$$</span></p>
<p>(where <span class="math-container">$E$</span> is the total energy as per the virial theorem) we can see that</p>
<p><span class="math-container">$$R=-\frac{GM^2}{2E}$$</span></p>
<p>Since the total energy <span class="math-container">$E$</span> is constant, this means the radius <span class="math-container">$R$</span> must be (on average) constant as well.
Even if we assume that initially the kinetic energy i.e. the temperature is zero (in which case we would have <span class="math-container">$E=U$</span> instead) the cloud could collapse only to half its original size.</p>
<p>This means that for star formation to occur, they gas cloud must continuously lose energy. This can only happen through inelastic collisions between the atoms and molecules, with the lost energy radiated away in the resulting atomic decay processes. A theoretical problem is here that the initial temperature of the cloud is much too low for any electronic states of atoms to be excited from the ground state. Hence it is generally assumed that the excitation of vibrational and rotational states of certain molecules are responsible for the energy losses. It is however also possible that highly excited states ('Rydberg states') of hydrogen atoms contribute to this. Although their density will be very small, the excitation cross section of these Rydberg states is extremely large, which could result in a significant effect (I have theoretically examined the significance of highly excited Rydberg states for the <a href="https://www.researchgate.net/publication/253543274_Scattering_of_radio_waves_by_high_atomic_Rydberg_states" rel="nofollow noreferrer">scattering of radio waves</a>, which is a different context and applied to different physical conditions, but the general principles should be transferable).</p>

CC BY-SA 4.0
2022-02-12T19:17:56.797Z
2022-02-13T15:29:11.363Z
2022-02-13T15:29:11.363Z
Thomas
<p>Regions of higher density mean that there are indeed some extra atoms of gas in there relative to their surroundings, and the gravitational pull (if you will ignore GR here and go with the Newtonian models) would not balance for a particular atom situated in the center of two spheres of mismatching number of gas that exert unequal forces. By that force, all the regions with the slightest of a &quot;fluctuation&quot; relative to the other surrounding regions, will slowly but steadily accumulate more matter and will start rotating due to unbalanced forces on it too by its neighbors, and then it will spin and then collapse into itself.</p>
<p>(the heat and temperature will be enough to trigger nuclear fusion reactions and the gravitational pressure from outside is then balanced from the exothermic reaction which produces heat and causes an outward tendency to expand: forming a star when after many oscillations, the protostar settles to a definite volume)</p>
<p>To answer <em>why indeed do density fluctuations occur in the first place</em>,</p>
<ol>
<li>God made the universe with a specific set of <em>chosen (for unknown reasons)</em> like this and let it evolve thereafter: we as a species developed in those conditions as a <em>result-there might be other life developed in non-Earth conditions as well.</em></li>
</ol>
<p>2)Our observable universe maybe just one of the many with innumerable conditions-it is the way it is because it is. +chaotic boundary conditions (either the universe is spatially infinite or there are infinite universes-ours is just one of them)</p>
<ol start="3">
<li>The weak anthropic principle</li>
</ol>
<p>So, the conditions may arise due to divine inspiration or as predicted by the weak anthropic principle.</p>
<p>The fluctuations themselves arise because the universe functions according to quantum mechanics <em>and</em> relativity, and with the uncertainty principle we know that the early universe must have been uncertain with just the minimum fluctuations in the <em>positions-(possibly causing clumps)</em> and velocity. The small irregularities would then have been blown up by the rapid expansion of the universe as the inflationary models suggest.</p>
<p>All changes and people who point out errors are welcome.</p>

CC BY-SA 4.0
2022-02-09T11:06:44.463Z
Hi Aveer, mentioning divine intervention before Jeans criterion on this (mainstream physics) site is probably going to frowned upon.
pela
CC BY-SA 4.0
2022-02-09T10:15:27.877Z
2022-02-09T10:15:27.877Z
Aveer
<p>Stars form when gas cloud collapse under gravity, becoming hot and subsequently initiating nuclear fusion. I have read that the collapse is triggered by density fluctuations, where regions of high density accumulate mass and then collapse? Why is that the case? When we say regions of high density, what does it mean? That they are regions where there are more gas particles? If that is the case, it should have more pressure and since particles move from high pressure to low pressure areas, accumulation of mass shouldn't happen right? Am I missing something here ?</p>

<star><astrophysics><cosmology><stellar-evolution><star-formation>
Why does the gas cloud collapse in regions of high density?
Rian
# Why does the gas cloud collapse in regions of high density?

(Asked by: username_0 on 2022-02-09)

Stars form when gas cloud collapse under gravity, becoming hot and subsequently initiating nuclear fusion. I have read that the collapse is triggered by density fluctuations, where regions of high density accumulate mass and then collapse? Why is that the case? When we say regions of high density, what does it mean? That they are regions where there are more gas particles? If that is the case, it should have more pressure and since particles move from high pressure to low pressure areas, accumulation of mass shouldn't happen right? Am I missing something here ?

**Tags:** star, astrophysics, cosmology, stellar-evolution, star-formation


---

## 3 answers

### Answer by username_1 on 2022-02-09. Score: 2

The gas cloud will have a tendency to collapse because of gravity, but that collapse is counteracted by the pressure of the gas as you expect.

When something happens that causes local density fluctuations (e.g. a nearby supernova) the gas will be contracted locally. As this globule of cloud contracts, pressure will build up (increasing the total kinetic energy of the cloud) but so will gravity (increasing the potential energy). If the mass of the gas is below a specific threshold, the pressure will cause the gas to expand again as you expected, but if the mass of gas is above the threshold, the pressure increase will not be able to counteract the increase in gravity and the globule of gas will continue to contract into a protostar.

This contraction will continue until some other process (e.g. pressure caused by nuclear fusion or centrifugal force due to rotation) is able to counteract gravity.

**Using math and physics:**

In a stable gas cloud the virial theorem holds, which states that the total kinetic energy of the cloud is equal to half the potential energy.

$$
K = \frac{1}{2}U
$$
where $K$ is the kinetic energy and $U$ is the potential energy.

The kinetic energy is due to the motions of the particles in the gas cloud. This motion of particles and their interactions is what constitutes pressure. The kinetic energy is, therefore a measure of pressure. When $N$ is the number of particles, $T$ is the temperature and $k$ is Boltzmann's constant we have a kinetic energy equal to:

$$
K = \frac{3}{2}NkT
$$

The potential energy is due to gravity and is equal to:

$$
U = -\frac{3}{5} \frac{GM^2}{R}
$$

where $G$ is the gravitational constant, $M$ is the mass and $R$ is the size of the cloud.

The virial theorem then becomes:

$$
NkT = \frac{1}{5}\frac{GM^2}{R}
$$

And because $N=M/m$, where $m$ is the average mass of the particles, we can write:
$$
kT\frac{M}{m} = \frac{1}{5}\frac{GM^2}{R}
$$
If this equality holds, the gas cloud is stable and is in virial equilibrium.

If, however, the right side of the equation is bigger, the cloud will collapse. The right side depends on the potential (gravitational) energy, so if gravity is stronger the cloud collapses:

$$
kT\frac{M}{m} < \frac{1}{5}\frac{GM^2}{R}
$$

If, on the other hand the kinetic energy wins out, the left side is bigger and the cloud will expand. This can be due to an increase in temperature which increases the average speed of the particles (and thereby the cloud's pressure).

To find the mass at which a cloud collapses given its size we can work some on the equation:

$$
kT\frac{M}{m} < \frac{1}{5}\frac{GM^2}{R}
$$

$$
\frac{kT}{m} < \frac{1}{5}\frac{GM}{R}
$$

$$
M > \frac{5kT}{Gm}R
$$

If we assume that the density $\rho$ is constant we can relate the radius $R$ to the mass $M$. The mass of the cloud is equal to the volume $V = \frac{4}{3}\pi R^3$ times the density, which gives us:
$$
M = \frac{4}{3}\pi \rho R^3
$$
which gives us an expression for the radius $R$:
$$
R = \left[ \frac{3 M}{4 \pi \rho} \right]^{\frac{1}{3}}
$$
Using this equation for $R$, we get:
$$
M > \frac{5kT}{Gm}\left[ \frac{3 M}{4 \pi \rho} \right]^{\frac{1}{3}}
$$
$$
M^{\frac{2}{3}} > \frac{5kT}{Gm}\left[ \frac{3}{4 \pi \rho} \right]^{\frac{1}{3}}
$$

$$
M > \left[ \frac{5kT}{Gm}\right]^{\frac{3}{2}}\left[ \frac{3}{4 \pi \rho} \right]^{\frac{1}{2}}
$$

This value of $M$ is the threshold at which a globule of gas will start to contract and is called the Jean's mass $M\_J$:

$$
M\_J = \left[ \frac{5kT}{Gm}\right]^{\frac{3}{2}}\left[ \frac{3}{4 \pi \rho} \right]^{\frac{1}{2}}
$$

* If $M > M\_J$, gravity is stronger --> Collapse
* If $M < M\_J$, kinetic energy is stronger --> Expansion

References:

1. [Christhoper Lovell, 2016, Deriving the Jean's Mass](https://www.christopherlovell.co.uk/blog/2016/04/26/jeans-mass.html)
2. [Chris Mios, 2005, Gravitational Collapse of Gas Clouds](http://burro.astr.cwru.edu/Academics/Astr221/LifeCycle/jeans.html)

> **Comments:**
> **username_2**: I can't resist wondering how this applies to the famous  https://what-if.xkcd.com/4/  
> **username_3**: Pressure and gravity have different dimensions/units. They can never be equal.  
> **username_1**: @username_3 You are correct of course, pressure is the force (of gas molecules) per square metre and gravity is the force on a molecule or on a volume of space containing some mass. I will edit the answer accordingly.  
> **username_1**: I have removed the section as it was not really conducive to the argument.  

---

### Answer by username_4 on 2022-02-12. Score: 1

Whilst the answer given by @username_1 addresses one of the fundamental criteria for gravitational collapse here (the 'Jeans criterion') it contains some inaccuracies and omissions (which have been adopted from the cited references).

First of all, the equation assumed for the gravitational potential energy in the derivation implies a sphere of constant density. In hydrostatic equilibrium however (the assumption made in the reference this answer is based on) the density in an isothermal cloud must decrease like $1/R^2$ (as is not too difficult to show) which results in a gravitational potential energy

$$U=-\frac{GM^2}{R}$$

i.e. a factor $5/3$ larger than assumed here, with the Jeans Mass changing correspondingly.

For arbitrary density distributions, the potential energy is

$$U=-\frac{kGM^2}{R}$$

where $\frac{1}{2}\le k\lt \infty$ (see also [this SE post](https://physics.stackexchange.com/questions/341065/gravitational-potential-energy-of-any-spherical-distribution))

More importantly, the impression is given here that star formation would proceed as long as the Jeans criterion (in whatever form) is fulfilled. This crucially ignores the fact that for a closed system the total energy must be constant, so the potential energy lost during collapse must be turned into kinetic energy i.e. the temperature must increase (given that the thermal collision times/distances for atoms are usually orders of magnitude shorter than the free fall times/ dimensions of the cloud). This would stop any collapse rather soon.
From the equations

$$U=-\frac{GM^2}{R}$$
$$E=\frac{U}{2}$$

(where $E$ is the total energy as per the virial theorem) we can see that

$$R=-\frac{GM^2}{2E}$$

Since the total energy $E$ is constant, this means the radius $R$ must be (on average) constant as well.
Even if we assume that initially the kinetic energy i.e. the temperature is zero (in which case we would have $E=U$ instead) the cloud could collapse only to half its original size.

This means that for star formation to occur, they gas cloud must continuously lose energy. This can only happen through inelastic collisions between the atoms and molecules, with the lost energy radiated away in the resulting atomic decay processes. A theoretical problem is here that the initial temperature of the cloud is much too low for any electronic states of atoms to be excited from the ground state. Hence it is generally assumed that the excitation of vibrational and rotational states of certain molecules are responsible for the energy losses. It is however also possible that highly excited states ('Rydberg states') of hydrogen atoms contribute to this. Although their density will be very small, the excitation cross section of these Rydberg states is extremely large, which could result in a significant effect (I have theoretically examined the significance of highly excited Rydberg states for the [scattering of radio waves](https://www.researchgate.net/publication/253543274_Scattering_of_radio_waves_by_high_atomic_Rydberg_states), which is a different context and applied to different physical conditions, but the general principles should be transferable).


---

### Answer by username_5 on 2022-02-09. Score: -10

Regions of higher density mean that there are indeed some extra atoms of gas in there relative to their surroundings, and the gravitational pull (if you will ignore GR here and go with the Newtonian models) would not balance for a particular atom situated in the center of two spheres of mismatching number of gas that exert unequal forces. By that force, all the regions with the slightest of a "fluctuation" relative to the other surrounding regions, will slowly but steadily accumulate more matter and will start rotating due to unbalanced forces on it too by its neighbors, and then it will spin and then collapse into itself.

(the heat and temperature will be enough to trigger nuclear fusion reactions and the gravitational pressure from outside is then balanced from the exothermic reaction which produces heat and causes an outward tendency to expand: forming a star when after many oscillations, the protostar settles to a definite volume)

To answer *why indeed do density fluctuations occur in the first place*,

1. God made the universe with a specific set of *chosen (for unknown reasons)* like this and let it evolve thereafter: we as a species developed in those conditions as a *result-there might be other life developed in non-Earth conditions as well.*

2)Our observable universe maybe just one of the many with innumerable conditions-it is the way it is because it is. +chaotic boundary conditions (either the universe is spatially infinite or there are infinite universes-ours is just one of them)

3. The weak anthropic principle

So, the conditions may arise due to divine inspiration or as predicted by the weak anthropic principle.

The fluctuations themselves arise because the universe functions according to quantum mechanics *and* relativity, and with the uncertainty principle we know that the early universe must have been uncertain with just the minimum fluctuations in the *positions-(possibly causing clumps)* and velocity. The small irregularities would then have been blown up by the rapid expansion of the universe as the inflationary models suggest.

All changes and people who point out errors are welcome.

> **Comments:**
> **username_6**: Hi username_5, mentioning divine intervention before Jeans criterion on this (mainstream physics) site is probably going to frowned upon.  

---

