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If stars are formed by the collapse of dust clouds under gravity, how is the pressure of the dust cloud overcome?

As more material gathers together, gravity will increase, but pressure will also increase. If I am not mistaken, both will increase as the volume shrinks, but gravity as a function of the square of the radius of the gas cloud, and pressure as a function of the cube of its radius. By this reasoning, we would not expect gravity to be able to overcome the hydrostatic pressure, and compress material together sufficiently to form a star.

What then, is the accepted explanation that allows stars to form under gravity, from dust clouds?

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    $\begingroup$ Squares and cubes are not "exponential"! $\endgroup$ – Henning Makholm Nov 4 '15 at 19:26
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    $\begingroup$ To expand on Henning's comment, "exponentially" means according to a function like function like $e^x$, while you have functions like $x^2$ and $x^3$ which rise "quadratically" and "cubically" respectively with the general category being described as "polynomially". $\endgroup$ – dmckee Nov 5 '15 at 3:24
  • $\begingroup$ Yes, I will correct to remove the word exponential. $\endgroup$ – user2800708 Nov 5 '15 at 12:06
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The answer lies in something called the virial theorem.

You are correct, a cloud that is in equilibrium will have a relationship between the temperature and pressure in its interior and the gravitational "weight" pressing inwards. This relationship is encapsulated in the virial theorem, which says (ignoring complications like rotation and magnetic fields) that twice the summed kinetic energy of particles ($K$) in the gas plus the (negative) gravitational potential energy ($\Omega$) equals zero. $$ 2K + \Omega = 0$$

Now you can write down the total energy of the cloud as $$ E_{tot} = K + \Omega$$ and hence from the virial theorem that $$E_{tot} = \frac{\Omega}{2},$$ which is negative.

If we now remove energy from the system, by allowing the gas to radiate away energy, such that $\Delta E_{tot}$ is negative, then we see that $$\Delta E_{tot} = \frac{1}{2} \Delta \Omega$$

So $\Omega$ becomes more negative - which is another way of saying that the star is attaining a more collapsed configuration.

Oddly, at the same time, we can use the virial theorem to see that $$ \Delta K = -\frac{1}{2} \Delta \Omega = -\Delta E_{tot}$$ is positive. i.e. the kinetic energies of particles in the gas (and hence their temperatures) actually become hotter. In other words, the gas has a negative heat capacity. Because the temperatures and densities are becoming higher, the interior pressure increases and may be able to support a more condensed configuration. However, if the radiative losses continue, then so does the collapse.

This process is ultimately arrested in a star by the onset of nuclear fusion.

So the key point is that collapse inevitably proceeds if energy escapes from the protostar. But warm gas radiates. The efficiency with which it does so varies with temperature and composition and is done predominantly in the infrared and sub-mm parts of the spectrum - through molecular vibrational and rotational transitions. The infrared luminosities of protostars suggest this collapse takes place on an initial timescale shorter than a million years.

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  • $\begingroup$ This answer needs to be pushed up. +1'd $\endgroup$ – Mindwin Nov 4 '15 at 19:58
  • $\begingroup$ Why does it radiate away some energy though? And, is it always possible to radiate away energy, I mean as the gas collapses might it get to dense to allow that to happen? $\endgroup$ – user2800708 Nov 4 '15 at 22:38
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    $\begingroup$ @user2800708 It radiates because it's a hot ball of gas. However, if the outer layers became more opaque to radiation then yes, that could slow the process of contraction down. Note that energy transport in the interior is usually by convection. $\endgroup$ – Rob Jeffries Nov 4 '15 at 23:27
  • $\begingroup$ Very nice, do we have some estimates for how long it would take to radiate away enough energy? $\endgroup$ – user2800708 Nov 5 '15 at 12:03
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    $\begingroup$ @SamuelWeir Now the question is on the Hot Network Questions list... I think it went virial. $\endgroup$ – Michael Nov 6 '15 at 1:54
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As gas clouds collapse, they increase in internal energy (measured by temperature). This is part of what causes their pressure to increase. As they increase in temperature, though, they also increase the amount of radiation they emit. As they emit radiation, their internal energy decreases and thus their pressure also decreases, allowing for further collapse.

This makes it seem like the temperature would decrease, but it turns out that the physics involved makes it so that when internal energy is removed from a collapsing gas cloud (by radiating the energy away), the amount of gravitational energy that is converted to internal energy is greater than the amount of internal energy that is radiated away. Thus, the temperature increases as the gas cloud collapses, even though the collapse is happening because energy is radiating away. In this way, collapsing gas clouds can be considered to have a negative specific heat: removing energy from the system actually increases the temperature, because of the gravitational energy that is released.

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  • $\begingroup$ Good point about the thermal pressure being limited by radiating energy away. I was thinking that another way of losing thermal energy would be by "evaporation" in which the most energetic atoms of the dust cloud would have enough kinetic energy to escape the gravitational pull of the cloud, thus carrying away some of the thermal energy of the cloud. Not sure how significant this effect is compared to radiation loss, however. $\endgroup$ – Samuel Weir Nov 4 '15 at 16:28
  • $\begingroup$ @RobJeffries, yes, thanks for pointing this out. I've edited my answer accordingly. $\endgroup$ – NeutronStar Nov 4 '15 at 17:56
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For a uniform, spherical distribution of mass (cloud of gas and dust) of radius $R$ and mass $M$ in absence of magnetic, radiation fields etc, we have $dm = 4\pi \rho r^2 dr$ and the potential energy of a spherical shell of inner radius $r$ and outer $r + d r$ is $dU = -G\frac{m(r)dm}{r}$, $m(r) = \frac{4}{3}\rho r^3$, and a simple integration yields, $$U(r) = -\frac{3}{5} \frac{GM^2}{R}.$$ For a mass distribution to collapse, according to the virial theorem, $2K + U <0$, with $K$ the kinetic energy of the mass, mainly due to thermal motion, i.e. $K = \frac{3}{2} NkT$, where $N$ is the number of molecules. If we express the mass of the cloud as $M = N \mu$ with $\mu$ the mean mass of the molecules, then the previous inequality for the cloud to collapse becomes $$M> M_{\text{Jeans}}, \quad M_{\text{Jeans}} \equiv \left( \frac{5kT}{G\mu} \right)^{3/2} \left( \frac{3}{4\pi \rho} \right)^{1/2},$$

the condition involving the Jeans limit for the mass of a cloud. It is obvious from this inequality that dense and cold clouds are easier to collapse and form a star.

The equation that relates the pressure and the density is known as the equation of state; it is in general unknown.

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  • $\begingroup$ But if the gas were adiabatic then $T \propto \rho^{2/3}$, the temperature rises and the Jeans mass increases. Isn't this what the question is about? $\endgroup$ – Rob Jeffries Nov 5 '15 at 7:57
  • $\begingroup$ Yes, I noticed that, and that's why I added the last sentence. $\endgroup$ – auxsvr Nov 5 '15 at 13:01
  • $\begingroup$ But you cannot use the Jean's mass argument unless you say something about the equation of state. An adiabatic cloud cannot "collapse and form a star" as you describe, because it will achieve a new equilibrium as the Jeans mass grows. The Jeans mass argument is an adjoint to my answer because it gives a necessary, but not sufficient, condition for the collapse to begin. $\endgroup$ – Rob Jeffries Nov 5 '15 at 13:26
  • $\begingroup$ The OP assumed certain equation of state in his question and wonders whether the cloud will collapse or not, therefore the virial theorem must be expressed in terms of the thermodynamic variables of the cloud. I'm not going further than this, because I'm not certain what the OP is claiming. $\endgroup$ – auxsvr Nov 5 '15 at 14:43
  • $\begingroup$ I think my statements about pressure as ^3 vs gravity as ^2 only hold at the surface of the gas cloud. I don't know how the density varies as you get deeper into a collapsing gas cloud, so... perhaps that could be part of the answer? that the cloud gets much more dense in the middle. $\endgroup$ – user2800708 Nov 6 '15 at 10:32
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The secret is to evacuate the heat, mainly by radiation. But for this you need dust or "metals", since H and He alone radiates very unefficiently. Paradoxically it is not so easy to collapse completely enough.

( BTW for dark mater there is no possible radiation to dissipate energy, which keeps it fuzzy and a lot less concentrated than ordinary mater.)

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  • $\begingroup$ By "black" do you mean dark matter? $\endgroup$ – David Moles Nov 4 '15 at 19:41
  • $\begingroup$ yep. fixed. :-D $\endgroup$ – Fabrice NEYRET Nov 4 '15 at 19:48
  • $\begingroup$ Would a cloud of H and He alone never collapse, or take longer than the timescales we think this happens on? Its sounds a bit chicken and egg, since the first stars would have formed when no metals existed, since metals are made in stars. $\endgroup$ – user2800708 Nov 6 '15 at 10:33
  • $\begingroup$ formally, it does radiate a bit, the point is to resist while many things around could tear appart. Also a bigger cloud will reach an higher temperature. anyway by many point (formation, stages in life, duration, end) the first generation of stars were very different. $\endgroup$ – Fabrice NEYRET Nov 6 '15 at 14:17

protected by Qmechanic Nov 5 '15 at 11:21

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