<p>It's more do do with having a higher pressure gradient than a higher pressure, though for a cloud of a set size, the two are equivalent.</p>

<p>For a cloud to be in equilibrium requires
<span class="math-container">$$ \frac{dP}{dr} = - G\frac{M\rho}{R^2},$$</span>
where <span class="math-container">$dP/dr$</span> is the pressure gradient and <span class="math-container">$M$</span> and <span class="math-container">$R$</span> are the mass and radius of the cloud and <span class="math-container">$\rho$</span> is its density.</p>

<p>It we just make a rough approximation of a linear pressure gradient and that the pressure on the outside is zero, then
<span class="math-container">$$ P_c \simeq G\frac{M\rho}{R},$$</span>
where <span class="math-container">$P_c$</span> is the central pressure, which is in turn, for a perfect gas <span class="math-container">$P_c = \rho k_B T_c/\mu$</span> where <span class="math-container">$T_c$</span> is the temperature and <span class="math-container">$\mu$</span> is the average mass of a particle in the cloud. </p>

<p>In order to collapse, the RHS has to get bigger than the LHS.  For a given mass, radius and density, then this means the temperature of the cloud must be less than some critical value.  i.e.
<span class="math-container">$$ G \frac{M\rho}{R} &gt; \frac{\rho k_B Tc}{\mu}$$</span>
<span class="math-container">$$ T_c &lt; \left(\frac{G\mu}{k_B}\right)\left( \frac{M}{R}\right)$$</span></p>

<p>Thus a hot cloud won't collapse and a cold one will. But one way to get the hotter cloud to collapse would be to increase <span class="math-container">$M$</span> (while keeping <span class="math-container">$R$</span> constant).</p>

CC BY-SA 4.0
2020-03-10T22:35:42.560Z
Somehow math always makes it better. :-)
StephenG - Help Ukraine
CC BY-SA 4.0
2020-03-09T23:00:52.543Z
2020-03-09T23:56:49.670Z
2020-03-09T23:56:49.670Z
ProfRob
<p>I was wondering why does a hot cloud need more mass to collapse than a cold cloud to form a protostar? Is it because there's a higher thermal pressure inside the hotter cloud than it is in a colder one?</p>

CC BY-SA 4.0
2020-03-10T14:42:22.147Z
In layman's terms higher temperature = faster molecules .. if you want those fast molecules to clump together .. you need higher gravitational pull = more mass
eagle275
<star-formation><protostar>
Why does a hot cloud need more mass to collapse?
space nerd
# Why does a hot cloud need more mass to collapse?

(Asked by: username_0 on 2020-03-09)

I was wondering why does a hot cloud need more mass to collapse than a cold cloud to form a protostar? Is it because there's a higher thermal pressure inside the hotter cloud than it is in a colder one?

**Tags:** star-formation, protostar
> **Comments:**
> **username_1**: In layman's terms higher temperature = faster molecules .. if you want those fast molecules to clump together .. you need higher gravitational pull = more mass  

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## 1 answers

### Answer by username_2 on 2020-03-09. Score: 13 | Accepted answer

It's more do do with having a higher pressure gradient than a higher pressure, though for a cloud of a set size, the two are equivalent.

For a cloud to be in equilibrium requires
$$ \frac{dP}{dr} = - G\frac{M\rho}{R^2},$$
where $dP/dr$ is the pressure gradient and $M$ and $R$ are the mass and radius of the cloud and $\rho$ is its density.

It we just make a rough approximation of a linear pressure gradient and that the pressure on the outside is zero, then
$$ P\_c \simeq G\frac{M\rho}{R},$$
where $P\_c$ is the central pressure, which is in turn, for a perfect gas $P\_c = \rho k\_B T\_c/\mu$ where $T\_c$ is the temperature and $\mu$ is the average mass of a particle in the cloud.

In order to collapse, the RHS has to get bigger than the LHS. For a given mass, radius and density, then this means the temperature of the cloud must be less than some critical value. i.e.
$$ G \frac{M\rho}{R} > \frac{\rho k\_B Tc}{\mu}$$
$$ T\_c < \left(\frac{G\mu}{k\_B}\right)\left( \frac{M}{R}\right)$$

Thus a hot cloud won't collapse and a cold one will. But one way to get the hotter cloud to collapse would be to increase $M$ (while keeping $R$ constant).

> **Comments:**
> **username_3**: Somehow math always makes it better. :-)  

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