2
$\begingroup$

I know that stars are formed from dense regions in large gas clouds.

I know that when gravity causes the mass of the clump to get so big that its internal pressure can't sustain it, it collapses and becomes a star.

I know that this threshold is known as the Jeans mass.

What I don't know is exactly what happens during the gravitational collapse of a star and how the hydrogen/helium gas becomes a flaming ball of fire.

Edit:

How long does the process of gravitational collapse take?

$\endgroup$
  • $\begingroup$ My understanding of the fusion reaction is that the caloric density is actually quite low, like a compost pile. It's just that there's so much of it. $\endgroup$ – Mike Dunlavey Feb 27 '15 at 20:28
  • $\begingroup$ Regarding the "how long?" - See free-fall time. $\endgroup$ – HDE 226868 Feb 27 '15 at 21:16
  • 1
    $\begingroup$ @MikeDunlavey is correct. The idea of the star suddenly igniting in some conflagration is way off. It is more of a simmer or slow stewing. Only 250 W/m$^3$ is produced in the Sun's core. $\endgroup$ – Rob Jeffries Feb 28 '15 at 9:09
  • $\begingroup$ @HDE226868 Depending on what the OP means, the freefall time is unlikely to be relevant. The time between collapse initiation and nuclear ignition is mainly the KH timescale after the main phase of mass accretion. $\endgroup$ – Rob Jeffries Feb 28 '15 at 9:13
1
$\begingroup$

A star is neither "flaming" nor "fire" in the sense that we use those words about things on Earth. It's just a big, hot ball of ionized gas.

The only thing that happens "to" it is that it gets hotter and denser. At some point the temperature rises high enough to ionize the gas. Later still fusion becomes possible at non-vanishing rates.

The energy for the warming comes from gravity, and how much is well described by the virial theorem.

The proto-star continues to shrink and warm until the power produced by fusion matches that lost to radiation from the surface at which point the system is in equilibrium and stops collapsing.

$\endgroup$
6
$\begingroup$

Short answer: gravitational potential energy is converted into heat.


Let's look at the Sun as an example. Its mass is $M_\odot = 2.0\times10^{30}\ \mathrm{kg}$ and its radius is $R_\odot = 7.0\times10^8\ \mathrm{m}$. If its density were uniform, its gravitational binding energy would be $$ U_{\odot,\,\text{uniform}} = -\frac{3GM_\odot^2}{5R_\odot} = -2.3\times10^{41}\ \mathrm{J}. $$ In fact the Sun's mass is centrally concentrated, so $U_{\odot,\,\text{actual}} < U_{\odot,\,\text{uniform}}$.

Where did the Sun come from? Something like a giant molecular cloud with a density of $2\times10^{-15}\ \mathrm{kg/m^3}$. The mass of the Sun would thus have been extended over something like a sphere of radius $6\times10^{14}\ \mathrm{m}$, for a gravitational binding energy of $$ U_\text{cloud} = -3\times10^{35}\ \mathrm{J}, $$ which is negligible in comparison with $U_\odot$.

All of the $2.3\times10^{41}\ \mathrm{J}$ had to go somewhere, and the only place to dump energy is into heat. The gas particles gain velocity as they fall into the potential well, but they don't lose that velocity because they never climb back out of the well.

Not worrying about whether the heating is isobaric or isochoric or somewhere in between, the heat capacity of monatomic gas is about twice the ideal gas constant, or $8.3\times10^3\ \mathrm{J\,K^{-1}\,kg^{-1}}$. At this amount, in order to heat all of $M_\odot$ by the average temperature of the Sun (say $10^7\ \mathrm{K}$, somewhere between the core and surface temperatures), you would need about $1.7\times10^{41}\ \mathrm{J}$ of energy. There is enough energy released by gravitational collapse to heat the Sun to its current temperature. You can do a more detailed analysis taking into account how much cooling occurs during the collapse, but the steep temperature dependence of the Stephan-Boltzmann law makes it difficult to lose heat to space until the object is already hot. I'm also neglecting a factor of $2$ that comes from splitting the energy between heating the gas and compressing it.

Once the material is this hot, it simply glows like any blackbody emitter. The energy lost to space is replenished by nuclear fusion in the core. In fact, fusion acts as a regulator: too much of it and the star expands and cools, slowing down fusion; too little and the star collapses further, heating up more and increasing the fusion rate.

In summary, gravitational collapse provides the initial energy to heat a star. As it uses up this energy source, it begins to tap into fusion. Ultimately it reaches an equilibrium where the energy produced by fusion is balanced by the energy radiated into space.

$\endgroup$
5
$\begingroup$

I'm not going to attempt to usurp Chris White's perfectly good answer - but just fill in some detail and answer the edit.

For a star like the Sun, the collapse proceeds in 4 basic stages, each takes about 10 times as long as the previous one.

  1. Pseudo-spherical collapse of the cloud - not far from a free fall timescale, often quoted as a few $10^4$ years.

  2. Class I phase. Central protostar accretes from a disk that is fed from the surrounding envelope - a few $10^5$ years (for a star like the Sun). Most of the mass of the star is assembled in phases 1 and 2.

  3. Class II phase. Envelope is dispersed, accretion via a disk, proceeds onto the central protostar for a few million years (again, for a star like the Sun). The protostar would typically be 2-5 times its main sequence radius at this point.

  4. Class III diskless star. Mass accretion has almost ceased and the star contracts on the "Kelvin Helmholtz timescale" - basically its gravitational potential energy divided by its luminosity. This is the longest time step and is mass dependent. It is around 10 million years for a star like the Sun, but 100 million years for an M dwarf. Conversely, everything above happens much faster for higher mass stars, where the start of the collapse to ignition could take a few $10^5$ years in total.

$\endgroup$
0
$\begingroup$

the answer is actually very simple. Electrons, protons and neutrons and other subatomic particles don't exist in the degree of proximity that a collapsing start forced them into. The massive gravitational force overcomes the equilibrium forces that exist in matter in its "normal" state ("normal" here refers to the state of the gas/dust cloud before gravitational force reaches star-compression levels).

Gravitation-caused compression has to go somewhere, so the temperature of the collapsing body increases. Anyone who's ever pumped a bicycle tire with a hand-pump has experiences the gas-compression analog of molecular-compression.

Obviously, as the temperature rises, so does the radiation of electromagnetic energy also increases, but not fast enough to make the star cool.

You don't need equations and diameter/mass statistics for this answer - those speak to the magnitude of the effect which isn't really the OP's question

$\endgroup$
  • $\begingroup$ The other answers do a far better job to describe this process quantitatively and qualitatively, but this seems to attempt to answer (part of) the question, so the flag cast on this seems invalid. $\endgroup$ – ACuriousMind Feb 28 '15 at 13:49
  • $\begingroup$ Indeed. RPFeynman was fond of answering questions directly and using minimal overlay of formulae and abstract theory. He was able to communicate the essence of ideas to people without assuming they had the same level of familiarity as he. One of his books "Surely you're joking, Mr Feynman" displays his overall demeanor as favoring simple, direct answers until more complex, abstract mathematical descriptions are necessary. Kudos to those with the deep physics at hand. I have a PhD myself, but style my answers after Feynman's direct simplicity $\endgroup$ – user192127 Mar 1 '15 at 16:08
  • $\begingroup$ Except your direct simplicity is too simplistic. You imply that the interior gets hot because the star can't lose energy fast enough. The opposite is true. The interior gets hot because the star is able to cool efficiently. The star behaves like it has a negative heat capacity and the faster it cools, the hotter the interior gets and the smaller it becomes. $\endgroup$ – Rob Jeffries Apr 5 '17 at 22:37
  • $\begingroup$ disagree - it's exactly simple enough to capture the essence. "can't lose energy fast enough" is relative to some arbitrary rate of energy radiation. In some cases the star heats up (in those cases, the energy radiation rate - whatever it is - is not enough for the star's temperature to fall, so it heat up). In other cases, the rate of radiation is high enough that the star cools. $\endgroup$ – user192127 Apr 7 '17 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.