7
$\begingroup$

From my research, I found out that quasistars theoretically existed because of a black hole core whose radiation pressure counteracted gravity within the star. However, a few websites stated that quasistars do not exist now because there are metals contaminating hydrogen and helium.

Could somebody explain to me why metals (or a small displacement of hydrogen and helium) would influence a black hole's radiation pressure, or is it just that there's no where that that much mass (1000+ solar masses) can exist anywhere in one place now?

I looked on Wikipedia (I usually do this and then research on different websites if I find something interesting there) and here1.

$\endgroup$
  • $\begingroup$ It will be a lot easier for people to respond if you provide links to the statements you've made, e.g. , who claims a quasistar could exist, who did an analysis to show metallic hydrogen is always in a black hole, etc. $\endgroup$ – Carl Witthoft Sep 4 '18 at 18:48
  • $\begingroup$ Quasi star ukads.nottingham.ac.uk/abs/2006MNRAS.370..289B $\endgroup$ – Rob Jeffries Sep 4 '18 at 19:15
6
$\begingroup$

Gas clouds with masses much higher than $10^3\,M_\odot$ are plentiful in galaxies; the typical star-forming cloud (the so-called molecular clouds) have masses of $10^3\,M_\odot$ to $10^7\,M_\odot$. When quasistars (hypothetical stars powered not by nuclear fusion, but by accretion onto a central black hole) cannot exist today, it is because all gas in the Universe has become polluted with metals.

Stars form from collapsing gas clouds. In order for a region of a cloud to collapse, it must be sufficiently dense, and sufficiently cool; if it's too dilute, there's not enough gravity, and if it's too hot, the energy of the individual atoms counteracts the collapse, making the atoms escape.

Jeans' mass

This criterion is captured in the Jeans instability equation. The relation can be expressed in several ways; one way is to say that the mass of the cloud — or a small region of it — must exceed the "Jeans mass": $$ M_\mathrm{cloud} \gtrsim M_J \simeq 3\times10^4 \frac{T^{3/2}}{n^{1/2}}\,M_\odot, $$ where $T$ (in $K$) and $n$ (in $\mathrm{cm}^{-3}$) are the temperature and the number density of the gas.

From this equation you see that the cooler the gas is, the smaller the threshold. In other words, the smaller stars you can form. If the gas is not able to cool, only the largest clumps will collapse, and hence such stars will be very massive.

Gas cooling

So, how does the gas cool? Hot gas means that the particles have large velocities. If the particles collide, they may excite each other, bringing an electron to a higher state at the expense of slowing down — i.e. cooling. When the electron de-excites, a photon is emitted, which may leave the system. Thus, the kinetic energy of the atoms is converted into electromagnetic energy which escapes.

However, an electron is only excited if the energy of the collision matches closely the energy needed for the excitation. If the collisional energy is too high, or too low, the atoms simply bounce off of each other, maintaining their total energy (although one may transfer some energy to the other).

The effect of metals

If the gas consists only of hydrogen and helium, there are only a few available energies for excitation. Hydrogen happens to be able to cool efficiently around $T\sim10^4\,\mathrm{K}$, while helium cools efficiently around $T\sim10^5\,\mathrm{K}$, but at other temperatures, the gas tends to stay at its given temperature.

However, as soon as there a some metals, the many electrons of these metals, with their many possible transitions, allow for atoms with many possible energies to be excited. Thus, before a gas cloud of $M\sim10^3\,M_\odot$ collapses to form a $10^3\,M_\odot$ star, it will fragment into smaller pieces, forming smaller stars.

$\endgroup$
  • $\begingroup$ See also this answer for a discussion of the cooling function. $\endgroup$ – pela Sep 4 '18 at 19:03
  • 2
    $\begingroup$ I agree. Fuller answer than mine. $\endgroup$ – Rob Jeffries Sep 4 '18 at 19:13
4
$\begingroup$

If you build a very massive protostar, more than a thousand solar masses, then it is possible for the core of the protostar to collapse directly to a black hole whilst it is still surrounded by a massive envelope. The collapse will happen "inside out", so that the envelope collapses at a slower rate. However, there is a maximum rate at which black holes can grow, because the compressing material gets very hot and emits lots of radiation and the radiation pressure can stall the collapse (temporarily). This is a quasi star.

The key to a quasi star is it's large initial mass, which prevents the envelope being "blown away" by the initial release of energy during the black hole formation. Such massive protostars can only be built in the early universe from pristine material. If the material is polluted with heavier elements then it can cool more readily - the heavier atoms can form molecules and radiate away energy. This cooling allows a large cloud to fragment into much smaller pieces, so that in the present day universe, the collapse of such a large cloud would not lead to one massive protostar, but a cluster of smaller protostars.

$\endgroup$
  • $\begingroup$ Good point about the formation of molecules. I suppose that actually dominates over collisional excitation at very low temperatures. $\endgroup$ – pela Sep 5 '18 at 15:58
0
$\begingroup$

The addition of metals (i.e., elements heavier than helium) to a stellar mixture makes it less transparent to radiation. Basically, hydrogen and helium have relatively simple and uncrowded spectra, but the "metals" add many new spectral lines and the mixture absorbs much more light and is more efficiently heated by it and picks up more momentum from it, also..

The gas in the early universe then had less metals, and was thus less affected by the radiation of a condensing new star. Consequently, the star could grow to higher mass before its radiation cut off the inflow of gasses allowing it to grow. (Today, the upper limit for star formation is around 100 Ssolar masses; with a mixture of H and He only, it appears to be as high as 250 solar masses.) See the Wikipedia article for a good explanation. These super-large stars are needed to form a quasistar, and can only form from unsullied H/He. So quasistars (if they exist at all) could only form very early in the evolution of the universe.

$\endgroup$
  • $\begingroup$ This isn't right. The reason is fragmentation. $\endgroup$ – Rob Jeffries Sep 4 '18 at 19:12

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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