# How is hydrostatic pressure overcome when a star is formed?

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?

• Squares and cubes are not "exponential"! Nov 4, 2015 at 19:26
• 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". Nov 5, 2015 at 3:24
• Yes, I will correct to remove the word exponential. Nov 5, 2015 at 12:06

The answer lies in something called the virial theorem and the fact that the contraction of a gas cloud/protostar is not adiabatic - heat is radiated away as it shrinks.

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 which supplies the energy that is lost radiatively at the surface (or through neutrinos from the interior).

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 and this timescale is set by how efficiently energy can be removed from the system.

• This answer needs to be pushed up. +1'd Nov 4, 2015 at 19:58
• 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? Nov 4, 2015 at 22:38
• @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. Nov 4, 2015 at 23:27
• Very nice, do we have some estimates for how long it would take to radiate away enough energy? Nov 5, 2015 at 12:03
• @SamuelWeir Now the question is on the Hot Network Questions list... I think it went virial. Nov 6, 2015 at 1:54

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.

• 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.
– user93237
Nov 4, 2015 at 16:28
• @RobJeffries, yes, thanks for pointing this out. I've edited my answer accordingly. Nov 4, 2015 at 17:56

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.

• 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? Nov 5, 2015 at 7:57
• Yes, I noticed that, and that's why I added the last sentence. Nov 5, 2015 at 13:01
• 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. Nov 5, 2015 at 13:26
• 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. Nov 5, 2015 at 14:43
• 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. Nov 6, 2015 at 10:32

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.)

• By "black" do you mean dark matter? Nov 4, 2015 at 19:41
• yep. fixed. :-D Nov 4, 2015 at 19:48
• 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. Nov 6, 2015 at 10:33
• 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. Nov 6, 2015 at 14:17

Nebulae- An emission nebula is a nebula formed of ionized gases that emit light of various wavelengths. The most common source of ionization is high-energy ultraviolet photons emitted from a nearby hot star.A planetary nebula, abbreviated as PN or plural PNe, is a type of emission nebula consisting of an expanding, glowing shell of ionized gas ejected from red giant stars late in their lives. The term "planetary nebula" is a misnomer because they are unrelated to planets or exoplanets. An emission nebula is a nebula formed of ionized gases that emit light of various wavelengths. The most common source of ionization is high-energy ultraviolet photons emitted from a nearby hot star. The ions move inside the gas and force the atoms in the gas to gain or lose electrons which results in ionization of these gases.A reflection nebula is a cloud of interstellar gas and dust that reflects the light from other stars. This happens in the surroundings of stars that are not hot enough to excite the hydrogen atoms of the cloud (black dwarfs)(as is the case for emission nebulae that are emitting light, not just reflecting the light from stars).A dark nebula or absorption nebula is a type of interstellar cloud that is so dense that it obscures the visible wavelengths of light from objects behind it, such as background stars and emission or reflection nebulae.The nebulae pull each other due to their gravitational forces and form clumps of gases known as protostars. This force(gravitational potential)changes to heat energy. Over millions of years , the gases gain enough heat to start the process of fusion of hydrogen which is distributed throughout the protostar. The layers of gases exert pressure at the centre which is the core and hence the core is hot.Simce the core is pressurized, it contracts and the atoms come closer. These atoms collide with each other and gain energy(collisions of macroscopic bodies, some kinetic energy is turned into vibrational energy of the atoms, causing a heating effect, and the bodies are deformed.) , they become hot. At last , the temperature reaches where hydrogen fuses into helium(Nuclear fusion requires so much heat because it is needed to overcome the electrostatic repulsion between the nuclei in the reaction. A tokamak reactor uses strong magnetic fields to contain the fusion reaction.) The hydrogen is not only at the core but also at the parts forming an envelope of non- burning hydrogen.Small stars with mass of 1 and a half mass of sun fuse hydrogen to helium and this releases energy which is radiation pressure and counteracts the force of gravity and keeps the star from absorbing the points above the point of gravity. Slowly the core becomes hot enough to warm the outer layers and so they expand becoming a red giant / when the core becomes hot enough the outer envelope of hydrogen is fused which is present just outside the core. As gravity crushes the star, I understand that the star heats up as gravity crushes it. As a result, although the stellar core remains “dead” (no fusion occurs), a “shell” of gas around the stellar core becomes hot enough to begin fusing helium. Since the fusion occurs as a “shell” around the stellar core, the outward-push from the fusion is what pushes the star’s outer layers further. The result is that the star grows into a Red Giant. The radiation pressure by the core reduces in energy as it travels till the outer layers. The core becomes a helium inert core but the gas is still hot and at high pressure and hence the core immediately doesn't contract under the effectof its own gravity.The core contracts and atoms are brought near to each other and collide with each other more due to brownian motion , this increases the energy possessed by the atoms and this energy converts to energy and heats the star.The helium requires high amount of heat and this provides ideal conditions for helium to fuse into carbon followed by many elements till they reach iron. Once iron is reached, fusion is halted since iron is so tightly bound that no energy can be extracted by fusion. Iron can fuse, but it absorbs energy in the process and the core temperature drops. The star's gravity prevails and compresses the stars. Now , the fate of the star depends on its mass. A low or medium mass star (with mass less than about 8 times the mass of our Sun) will become a white dwarf. A typical white dwarf is about as massive as the Sun, yet only slightly bigger than the Earth. This makes white dwarfs one of the densest forms of matter, surpassed only by neutron stars and black holes.The white dwarf is supported by the paulie exclusion principle in which - In a degenerate gas, all quantum states are filled up to the Fermi energy. Most stars are supported against their own gravitation by normal thermal gas pressure, while in white dwarf stars the supporting force comes from the degeneracy pressure of the electron gas in their interior. In neutron stars, the degenerate particles are neutrons.A fermion gas in which all quantum states below a given energy level are filled is called a fully degenerate fermion gas. The difference between this energy level and the lowest energy level is known as the Fermi energy. In an ordinary fermion gas in which thermal effects dominate, most of the available electron energy levels are unfilled and the electrons are free to move to these states. As particle density is increased, electrons progressively fill the lower energy states and additional electrons are forced to occupy states of higher energy even at low temperatures. Degenerate gases strongly resist further compression because the electrons cannot move to already filled lower energy levels due to the Pauli exclusion principle. Since electrons cannot give up energy by moving to lower energy states, no thermal energy can be extracted. The momentum of the fermions in the fermion gas nevertheless generates pressure, termed "degeneracy pressure". The electrons occupy the lower energy levels first according to aufbau principle and then fill the higher orbitals , when they are pressurized into smaller states , the gravity cannot further compress the electrons due to same quantum numbers:-

1 Principal quantum number (n)

2 Azimuthal quantum number (ℓ)

3 Magnetic quantum number (mℓ)

4 Spin quantum number (s)

The white dwarf stars are white as they are still hot from their activities, eventually becoming cold by the radiation loss in the form of infrared(the atoms become hot and want to lose energy in the fastest way hence lose energy by infrared which is the feeling of warmth when we come closer to a hot object ) ‎ A red giant star with more than 7 times the mass of the Sun is fated for a more spectacular ending.

These high-mass stars go through some of the same steps as the medium-mass stars. First, the outer layers swell out into a giant star, but even bigger, forming a red supergiant. Next, the core starts to shrink, becoming very hot and dense. Then, fusion of helium into carbon begins in the core. When the supply of helium runs out, the core will contract again, but since the core has more mass, it will become hot and dense enough to fuse carbon into neon. In fact, when the supply of carbon is used up, other fusion reactions occur, until the core is filled with iron atoms.

Up to this point, the fusion reactions put out energy, allowing the star to fight gravity. However, fusing iron requires an input of energy, rather than producing excess energy. With a core full of iron, the star will lose the fight against gravity.

The core temperature rises to over 100 billion degrees as the iron atoms are crushed together.The electrons are compressed till the nucleus where they are attracted by the protons and the electrons collide with the protons forming neutrons and releasing a neutrino and the newly created neutrinos go flying outward, expelling the outer layers of the star in a gigantic explosion called a supernova (to be precise, a type II or core collapse supernova).See , the neutrons are in very large numbers as the star is very big. Now , the remnants of the supernova also depends on mass and then they are either neutron stars or black holes. One more condition of supernovae - Consider the typical momentum transfer exhibit found in many science museums, as depicted in the animation on the right. Rubber balls of different sizes are held at different heights. The balls are then let go at the same moment. Gravity pulls them all down and they all fall towards the ground. In the next few moments, the bottom ball hits the ground and bounces back, and then the balls start colliding. Momentum equals mass times velocity. This means that a heavy object going slow has as much momentum as a light object going fast. When two objects collide, they transfer some momentum. When a heavy slow object collides with a light object, it can give it a very high velocity because of the conservation of momentum. As this animation shows, by arranging the rubber balls from heaviest on the bottom to lightest on the top, momentum is transferred to ever lighter objects, meaning ever higher speeds. As a result, even though gravity is pulling all the balls downwards, the upper balls rebound at incredible speeds. This is all in keeping with the law of conservation of momentum. The lower balls are too heavy and too slow to fly off. They remain behind as the surviving core of the original system. On the other hand, the upper balls are blown away (in a science museum exhibit, they are captured at the top of the apparatus so that the demonstration can be rerun). This explosion of rubber balls occurs without any significant chemical or nuclear reactions taking place. This explosion is simply due to gravity and momentum transfer, i.e. a gravitational rebound. If you look closely at the animation, you see that the rebound takes the form of an outward shock wave that gains in intensity as it spreads.Heavier objects, however, also have more inertia, which means they resist moving more than lighter objects do therefore inertia causes these heavier objects to tranfer momentum to lighter objects.A supernova is the same kind of explosion as this rubber-balls demonstration. An aging star is composed of denser layers down towards the center, and thinner layers near the surface. The star's nuclear reactions typically balance out the force of gravity. But when the star runs out of fuel, the nuclear reactions slow down. This means that gravity is no longer balanced. Gravity begins collapsing the star. After the core of a collapsing star reaches a critical density, its pressure becomes strong enough to hold back the collapse. But, like the rubber balls, the star has been falling inwards and now bounces back. The outer layers are blown off into space in a giant explosion, spreading fertile dust clouds through-out the universe . But because of the momentum transfer, the star's core survives. The collapsing event has so intensely squeezed the star's core, that it transforms into something exotic. If the star started out with between 5 and 12 times the mass of our sun, the core becomes a big ball of neutrons called a neutron star. If the star started out with more than 12 times the mass of our sun, the core becomes a black hole. You may be tempted to argue that when a star explodes so that all that remains is a black hole, there is nothing left and the star has therefore been completely destroyed. But a black hole is not nothing. Black holes have mass, charge, angular momentum, and exert gravity. A black hole is just a star that is dense enough, and therefore has strong enough gravity, to keep light from escaping. The black hole created by a supernova is the leftover core of the star that exploded.

Black holes are very dense and bend spacetime , even light can't escape.

Do you know? - hot stars are blue ! The hotter the star, the shorter the wavelength of light it will emit. Since it is in middle of the radiation chain , it emits Blue light and uv rays. SInce less hot stars donot have much energy to emit blue light hence it emits light of red light.

From - https://stellarxplore.wixsite.com/universe (My website)