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I Questions about the lifecycle of stars

  1. Mar 21, 2017 #1

    I have been following lectures found here on the end of the lifecycle of stars (while trying my best not to get distracted by his bow tie) and I have some questions on this. Some of them being general questions and some a bit more detailed.

    1. First a more general quesion: I’ve read that adding mass to a small planet would make its volume increase but after it reaches a certain mass, adding more mass to it would actually make the volume shrink. Is this because eventually the mass is not proportional to the pressure force inside?

    2. Why does the fusion process of a larger star end faster than a smaller one? I understand it’s because of the larger mass but isn’t there also more fusion particles to go through? Or is the fusion rate not proportional to the mass?

    3. According to this video, when the core gets depleted from H+, it collapses while at the same time the star as a whole expands and gets larger. I understand that this is caused by the H-burning shell around the core. However, I would expect more that if the core gets depleted from H+, the star as a whole should shrink in size until a new equilibrium has been reached, that equilibrium being caused by that arisen H-burning shell. Why isn’t it like that?

    4. The lecturer explains here that fusion of helium to carbon releases energy. How can this be while I can clearly see that there is no net loss of mass before and after fusion? Is it because of the increased release of binding energy?

    5. At 5:00 here he explains that the star shrinks in size while the core expands because of the Heliumflash. I don’t quite understand why the star shrinks at that moment.

    6. I understand that for a low mass star, the star eventually sheds its outer layers leaving a core behind that is being prevented from collapsing because of the electron degeneracy pressure (a white dwarf). However, in the case of a very large star that turns into a white dwarf that exceeds the Chandrasekhar limit, the electron degeneracy force is not strong enough to prevent collapse. My question is, how was the electron degeneracy force able to prevent the star’s core from collapsing until it became a white dwarf in the first place? Shouldn’t the core collapse happen even before a large star sheds its outer layers and becomes a white dwarf?

    7. If a white dwarf exceeds the Chandrasekhar limit, wat determines it to undergo carbon fusion and explode or to implode and become a neutron star?

    Apologies for the long post. I'd appreciate a lot if someone could shed some light on these.
  2. jcsd
  3. Mar 21, 2017 #2
    Yes, the gravitational force overwhelms the outward pressure of heat and causes increased density.

    Big stars have a lot more gravity pushing down on the core, so the amount of the core which can actually fuse is much larger. Fusion ONLY happens in the core, so a bigger core means that it'll burn fuel faster.[/QUOTE]

    After hydrogen is finished burning, the star shrinks, but that process squeezes the core even tighter and causes helium to start fusing. The produces more energy than the hydrogen reaction, which causes the entire star to heat back up again and expand.

    Yes, the majority of energy released in fusion is binding energy. Carbon nuclei are packed very tightly, so they're difficult to break apart.

    The hydrogen fusion has slowed. There is a delay between when the hydrogen is being used up and the core getting hot enough to fuse helium. The star has to shrink slightly in order to compress the core enough to heat it up.

    Low mass stars do not turn into white dwarfs, they turn into black ones. Medium sized stars like ours collapse once the core runs out of fuel. The heat of the fusion is what props the star up, not degeneracy pressure. Degeneracy pressure only comes into play after fusion has stopped, when gravity is pulling on the star, but there isn't enough heat to hold it up. That hot core, is not super dense, and comes no where near the pressures needed to reach degenerate matter, even in the biggest stars.

    It's mass. Carbon fusion causes stars to rapidly cool, which compresses them. The core shrinks first and then all of that mantle comes crashing down on it. If there isn't enough mass above to overcome degeneracy pressure, it'll hit the core like a brick wall and bounce off, which causes an explosion. If it's got enough mass, it'll crunch it into a black hole, which accelerates the in falling material to nearly the speed of light, which causes an even bigger explosion.
  4. Mar 21, 2017 #3
    @newjerseyrunner : Thank you so much for your answers. A few things I wanted to say:

    But if bigger stars also have a bigger core, doesn't that also mean that there are more atom nuclei to fuse so that it compensates for the increased fuel rate? Or is the relation between fuel rate and mass not proportional?

    According to the video (not so sure about the accuracy of the lecturer), before the core starts fusing helium, a burning H-shell around the core arises first and that seems to be the first cause of expanding the volume of the star. So the video says: core gets depleted from H+ -> whole star shrinks in size -> shell around core gets heated enough to fuse H+ -> star expands on the outside while the core shrinks even more until degeneracy pressure prevents that core from shrinking further. My thought however is that the arisen H-burning shell should merely stop the shrinking of the star and not expand it, unless the fusion in the H-burning shell releases a sudden burst of huge energy. Perhaps that's the cause of the initial expansion of the star.

    Note that the lecturer says that the star is shrinking in size at the same time as the core fuses helium which I find confusing. The energy that the fusion of helium releases is merely spent on the expansion of the core and not on the star as a whole which I don't understand.

    I'm not so sure about the reliability of this video, but it says that in the red giant phase, the core already gets halted from collapsing by degeneracy pressure.

    Shouldn't it turn into a neutron star in that case first? From what I read, a black hole arises when gravity overcomes the neutron degeneracy pressure, along with other types of degeneracy pressures in a neutron star. See here
  5. Mar 21, 2017 #4
    It's not proportional. If you double the size of the star, you triple the size of the core. (Not really those numbers, just an example.)

    Stars are huge, so it takes time for these things to happen. When its running low on hydrogen, the power output slows down. But there is still a huge amount of hot material holding onto heat, so it takes a while for the entire star to cool enough to start to collapse, and it will not do so uniformly. As the core crunches down, the inner most layers of the mantel fall in on it first, this creates a super hot boundary layer which can start up fusion again.

    It has to. In order for the core to increase in temperature high enough to burn helium, it has to become denser.

    Once shrunk, it's actually much harder to push it back out because all of the matter is closer and therefore has higher gravity. This allows the temperature to skyrocket. The power of the star increases by a factor of more than a thousand. This causes the outer layers to heat up and expand so much, that gravity is so reduced by the distance that it can sort of just blow away. This red giant phase is held up by heat, which requires fusion to still be happening.

    Yes, I misspoke. Neutron degeneracy pressure is far stronger than electron degeneracy pressure.
  6. Mar 22, 2017 #5
    No, it does not.

    The rapid collapse happens only when during slow core shrinkage (which slowly progresses during most of the star's life - our Sun does it right now too) new processes start happening which either consume energy (iron fusion into heavier elements) or which do generate energy, but not enough to compensate for increasing pressure (say, when rise in temperature is enough for gamma->e+e- pair production). Carbon fusion is not one of those processes.
  7. Mar 22, 2017 #6


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    The shell of hydrogen undergoing fusion has a higher temperature, and I believe also more volume, than the hydrogen burning core did and releases more energy. This increase in energy causes the star to expand.

    I'm guessing that after the helium flash occurs and the core expands, the hydrogen burning shell somehow burns less fuel, lowering the energy output of the star even though the core is now burning helium.

    That's not something I've ever heard of. Low mass stars should turn into white dwarfs over very, very long periods of time and then cool to become black dwarfs.

    My understanding is that in a white dwarf that exceeds the Chandrasekhar limit, the entire star is degenerate and the addition of mass from a nearby companion slowly raises the temperature of the star. Once the temperature reaches a certain limit, the entire star undergoes runaway carbon fusion and blows itself apart in a supernova. There are no outer layers that come crashing down and no remnant is left behind.
  8. Mar 22, 2017 #7
    Proportional to what?

    And it does not have to be heat resisting pressure.

    Look at it this way: Imagine a self-gravitating drop of liquid (which planets are like). For simplification assume that it also has a constant density.
    The pressure at the centre is the weight of the liquid column from surface to centre. Sure, gravity goes to zero at the centre (because of symmetry) - but the pressure integrated over the column is finite and nonzero.

    Now, make the liquid 8 times denser by compressing the drop at unchanged total mass to 2 times smaller linear size.
    In that case, the surface gravity of the drop increases 4 times (because of inverse square law of gravity). Since the density of the liquid was increased 8 times, the weight of an 1 m column was increased 32 times. But since the depth of column from surface to centre was halved, the pressure at the centre was increased 16 times.

    Now, try making the liquid 8 times denser but this time by adding mass to the drop at constant size.
    In this case, the surface gravity of the drop increases 8 times (gravity is proportional to mass). Since the density of liquid was still increased 8 times, the weight of the 1 m column is increased 64 times, and since now the depth of column is unchanged, the pressure at the centre was increased also 64 times.

    The implications?
    If the compressibility of a substance is such that increasing density 8 times requires increasing pressure 16 times or less then an amount of that substance can never be at equilibrium. If its pressure is less than its gravity, it will collapse without limit; if its pressure exceeds its gravity, it will expand without limit.

    Note that isothermal perfect gas will increase its pressure only 8 times when its density is increased 8 times. Therefore an isothermal perfect gas can never be in equilibrium with its gravity.

    Contrast water. Water, like all substances, are compressible. But increasing the pressure of water from 1 atmosphere to 16 atmospheres will not increase its density 8 times.
    Nor will increasing the density of water from 1 atmosphere to 64 atmospheres increase its density 8 times.
    Low pressure solids and liquids are stiff - they are compressed only slightly by increased pressure.

    But as pressure is increased with increasing mass of the planet...

    For actual numbers, water is compressed by approximately 4 % when its pressure is increased from 1 atm to 1000 atm. In a certain sense "compressibility" decreases with pressure. As in, if you take water at pressure 1 bar and increase its pressure by 100 atmospheres, its density will increase by about 0,45 % Whereas increasing the pressure from 4900 bar to 5000 bar only increases the density by 0,16 %.

    But in another relevant sense, the compressibility of water will increase with pressure. If you increase the pressure of water 100 times, from 1 atm to 100 atm, the density is increased just about 0,45 % - yet when you increase water pressure also 100 times, from 50 atm to 5000 atm, the density increase is over 10 % or so.

    Now, if you reach a point where increase of pressure by 100 times causes the density to increase not 10 % but 900 % - 10 times - then the drop of water will shrink with each added drop.
    And when you reach the point where increase of pressure by 100 times causes the density to increase 32 times then the drop of water will collapse in a neutron star.
  9. Mar 22, 2017 #8
    It's density. As you add more matter and close in on the Chandrasekhar limit, the density is increasing, fast. The star is shrinking quite a bit.

    Carbon fusion rate depends on temperate and density, so density increase, even at constant temperature, eventually reaches a point where heat from carbon fusion creates feedback loop: fusion rate increases, temp increases, but pressure increase due to temperature is small compared to degeneracy pressure, thus the star does not significantly expand and does not reduce density. Feedback loop -> KABOOM.
  10. Mar 23, 2017 #9
    Oh, okay, I understand where I went astray from what was asked. It should be made clear that this is not a normal evolution of a white dwarf, but that if a white dwarf with a partner which it cannabilizes. Matter streams onto it slowly, and gravity is so rediculously high at the surface that it fuses until it reaches carbon. That builds up a thin layer of carbon that's all just below its critical density. Once enough carbon piles on it, the inner most carbon has enough energy to fuse. This is like throwing a sugar cube into superheated water. The carbon fuses. All of it, all at once.
  11. Mar 23, 2017 #10

    Ken G

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    Yes, the fusion rate goes as a fairly high power of the mass, roughly like mass to the 3 or 4 power, depending on what the mass is. The main reason for this is that stars are big leaky buckets of light, and higher mass stars are bigger leakier buckets. The fusion rate simply self-regulates to replace what is leaking out, so it is the leaking out rate that determines the fusion rate, not the other way around (a lot of places make this mistake, they think the high luminosity is due to something about fusion, but that's incorrect because high-mass stars already have high luminosity even before the star begins to fuse anything).
    It's because the temperature in the shell is determined by the dense core, which is a very different situation from when there is core burning. As I mentioned, core burning self-regulates to whatever is the luminosity of the star, but since fusion is very T sensitive, it doesn't need to change T much to accomodate almost any luminosity. But shell burning does not self-regulate its T, it is stuck with the gravitational environment set by the core, so tends to get very hot as the core contracts. When the shell gets very hot, its fusion rate goes through the roof, a situation the star could not tolerate except that all this heat goes into the envelope and puffs it out. When the envelope puffs out, its weight drops drastically, which drastically reduces the pressure in the H-burning shell. So what ends up happening is, shell burning self-regulates the amount of mass that is undergoing fusion, not the temperature of the fusion the way core burning does. That requires that the envelope must puff out more and more as the core contracts and the shell temperature rises. That's what causes a red giant.
    There is a loss of mass, but since E = mc^2, and c^2 is huge, it doesn't take much loss of m. But yes, the loss of m can be viewed in terms of the nuclear binding energy. Remember, the mass of a nucleon comes mostly from the energy of its quarks and gluons, not from the rest mass of its quarks which is quite small.
    It's because the helium flash expands the core, which reduces the temperature of shell burning. The shell quiets down, and no longer needs to reduce the weight on it. The envelope is allowed to sink back down, sort of like how oatmeal sinks back down when you cease microwaving it.
    Massive stars can have cores that exceed the Chandra mass by having their cores not be highly contracted. The Chandra mass only applies to a core that has lost so much heat and contracted so much that it is highly degenerate. For an ideal gas, whether it can support itself against gravity merely depends on where it is in its process of contraction. But when the core is made of iron and cannot fuse, it eventually goes relativistic as it contracts, and whether it is degenerate or still an ideal gas, it then collapses, and you get a core collapse supernova.
    It's whether it is made of mostly carbon, as for a normal white dwarf, or mostly iron, as for the core of a massive star. Only carbon undergoes thermonuclear runaway, iron implodes into a neutron star.
  12. Mar 24, 2017 #11
    Wow, thank you so much on those clear answers!

    Just to make sure I understand everything...

    Is this because the expanding of the core makes the core decrease in temperature and thus the H burning shell around it also decreases in temperature?

    So the masses written down on the left side next to each fusing atom symbol:


    Are rest masses of the total quarks of each atom?

    Thanks for you answers @Drakkith . The burning shell somehow burning less fuel is apparently caused by the shell decreasing in temperature which is caused by the core expanding, as @Ken G explained, I think.
  13. Mar 24, 2017 #12
    Not all answers here are correct... in general, late stage nuclear burning process has complex dependencies on temperature, pressure and density, and the species being burnt. For example, triple-alpha process is _extremely_ temperature-dependent (proportional to T^40 !) - that's why helium flashes commonly happen, and "hydrogen flashes" do not.
    If you'd try to just guess qualitatively how the old, layered star would behave, you can easily guess it wrong.
    Definitive scientific work in this area requires precise numerical simulation.
  14. Mar 24, 2017 #13


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    No. Forget quarks for the moment, they're not needed to understand this.
    The number on the left denote the number of protons (bottom left), which determine element type, and number of nucleons (protons+neutrons, top left), which determine isotopes. However, the latter is not the actual mass (atomic mass, ma) of the particular type of an atom for any but the carbon C-12 isotope, which was historically used as the baseline. For example, a hydrogen atom, with one proton only, has ma=1.008u (units are 1/12th of the mass of a C-12 atom).
    The difference comes from binding energy of the nucleons. A couple of deuterium atoms will have higher total atomic mass than one helium atom, despite having the same number of protons and neutrons. That's the mass defect that is the source of energy release in fusion reactions. It becomes reversed once you reach iron-nickel neighbourhood on the periodic table, as atoms heavier than those are more weakly bound, so fusing them requires energy (whereas splitting them releases energy, as in atomic plants and bombs).

    In the equation posted above, the sum of the atomic masses of berylium-8 and helium-4 is higher than the ma of carbon-12. That excess is released in the form of electrons and kinetic energy, as shown on the right-hand side of the equation.

    You can find atomic mass listed in more detailed periodic tables of elements. Just be careful not to confuse it with the often listed relative atomic mass, which a somewhat different, if related, concept.
    More here:
  15. Mar 24, 2017 #14
    Ah of course! I knew this but totally forgot it when I asked the question. What I really meant asking is what you now have explained; that having the same number of neutrons and protons doesn't necessarily mean having the same atomic mass.

    Can I say that, since nucleon mass mostly comes from the energy of the quarks and gluons and since fusing Helium into Carbon releases energy, that the total mass of of a Carbon atom is less than 3 Helium atoms? And does this mean that the total mass of 1 quark in a Carbon atom is less than 1 quark in a Helium atom?
  16. Mar 24, 2017 #15

    Ken G

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    Essentially, yes. There is a detail-- any degeneracy in the core will suppress its temperature relative to the kinetic energy of its particles (that's what degeneracy does, it is a thermodynamic effect that suppresses temperature), whereas the shell is normally not degenerate at all. So to be more precise, we should say that the shell comes to a temperature that is equivalent to the temperature the core would have if it were not at all degenerate, which can be different from the actual temperature of the core if the core has some degeneracy. Since the degeneracy is lifted by the helium flash, this distinction becomes less important post-flash than it was pre-flash.
    Yes. Indeed, you can verify this, some periodic tables will include the total mass in fine print.
    It's not really clear what you mean by the total mass of a quark, because to talk about the mass of a particle we would normally go into the particle's frame and give its rest mess. But you can have a system at rest that is comprised of particles that are not at rest, so the system can have a rest mass that includes the kinetic energy of those particles. That's what is happening in a nucleus.
  17. Mar 24, 2017 #16
    That´s not the full form of the dependence.
    A simpler explanation might be this:
    Protium fusion is a weak process whose speed is inherently limited by weak interaction. Which is why it cannot run away, even at very high temperatures - its timescale remains long compared to free fall timescale.
    Helium fusion is a strong or at least electromagnetic process. At sufficiently high temperatures, it can get faster than free fall timescale, and run away.
  18. Mar 24, 2017 #17

    Ken G

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    But note that helium flash runaway begins at temperatures where helium fusion is still relatively slow. It has to be in runaway mode even when the timescale is longer than the dynamical time, since otherwise it would never make it to the phase where the fusion time is shorter than the dynamical time. So the reason helium fusion runs away is not that the temperature gets high, it's that the electrons are degenerate, so expansion work is not done by them and not by the helium ions. When helium ions don't need to do the expansion work, they don't lose energy as the gas expands, which allows the temperature to continue to rise even as the gas expands on dynamical timescales. That's what allows the runaway to proceed through the slower phases. Indeed, degeneracy is lifted by the time the fusion time approaches the dynamical time, so in that sense the dynamical timescale does succeed in preventing complete runaway of the helium flash.
  19. Mar 24, 2017 #18
    Well, if you compress and/or heat it sufficiently, collision rate goes up and even "slow" process can in fact occur rather quickly. When hydrogen is falling on a neutron star in a binary system, it fuses VERY fast (the temperature of hydrogen impacting the NS is, very roughly, 10^12 K).
  20. Mar 24, 2017 #19
    Collision rate goes up, beta decay rate does not. You can speed up the reaction
    but the half life of reaction
    is still 10 minutes.
    Whereas the reaction
    is a strong process with no weak rate limiting step, and can be sped up as the carbon white dwarf undergoes a thermal runaway.
  21. Mar 24, 2017 #20
    Then how hydrogen fuses basically instantaneously (in less than a second) on the neutron star surface?
  22. Mar 28, 2017 #21
    I'm trying to understand this but how does the core "know" to regulate the fusion rate based on the leakage? Also, doesn't this description imply that a star that expands and decreases in density would increase the fusion rate in the core because of the increased surface and thus increased leakage? How is that possible while fusion rate actually increases with an increase in density?

    Ah ok, so let's talk about the rest mass of 1 quark then if it's comprised of particles that aren't necessarily at rest. Does this mean that the rest mass of 1 quark in a carbon atom is lighter than a quark in a Helium atom?
  23. Mar 28, 2017 #22

    Ken G

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    The self-regulation works because if the fusion rate is too fast, heat builds up, causing the pressure to rise and the gas expands, which dials the temperature back down and reduces the fusion rate. If it falls too much, then net heat loss occurs, causing the pressure to drop and the gas contracts, raising the temperature and fusion rate.
    You might think the rate the star loses light depends on the surface area, but that assumes the surface T stays fixed, and it doesn't-- it self-regulates to match the rate that light is diffusing up from the interior. The star can have any radius and still be in force balance, that's clear from its history of gradual contraction, but the radius does not determine the luminosity-- or at least only weakly affects it. This is because there are two competing factors that determine the leakage rate-- the timescale for the light to escape, and the amount of light in the bucket in the first place. When the radius is larger, the light leaks out in less time, but there is less total light (the light energy content scales like the 4th power of internal temperature times the 3rd power of radius, but the internal temperature itself is inversely proportional to radius when in force balance, so the net result is that the energy content is inversely proportional to radius). These two effects tend to cancel each other, because the leakage time depends not just on the area, but also on how far the light has to go, so the leakage time is also inversely proportional to radius if the opacity per gram stays the same (which it approximately does).

    That's why as pre-main-sequence stars evolve, they approach the main-sequence along tracks of nearly constant luminosity (called "Henyey tracks"). Thus, the luminosity is not only not determined by nuclear fusion physics, it's hardly even affected by it (you'll see that totally wrong in a lot of places, by the way). What is affected by nuclear fusion is the stellar radius-- the gradual contraction is paused throughout the main-sequence phase, because that's the phase where fusion is self-regulated to replace the leaking energy, and so the star suffers no net loss of heat. Prior to fusion, net loss of heat implies contraction.
    That's how the self-regulation works.
    The rest mass of a quark is the same in all nuclei, what changes is the energy associated with each quark, about 98% of which is the energy of the gluons that confine the quarks. This means that your mass is largely E/c^2, where E is the total energy of all the gluons in your nuclei, but the gluon energy per nucleon is higher in H and He than it is in C and O.
  24. Mar 28, 2017 #23
    The "rest mass of a quark", unlike e.g. electron's rest mass, is not an easily definable concept.
  25. Mar 28, 2017 #24
    Great, thanks for the detailed explanation. Here's my summary to see if i understand this correctly (apologies if I still don't).

    If fusion rate goes up -> star expands in radius. This expansion has 2 consequences:
    1. Leakage of light per unit time increases
    2. Internal energy content declines because its net proportionality is inversely to the radius

    These 2 effects cancel out. That in combination with the fact that leakage of light per unit time is also inversely proportional to the radius (light has to travel more from the inside) gives a net result of less light leakage after expansion -> fusion rate declines after expansion.

    But that's why I asked for the total mass of a quark to include the energy of the gluons that confine the quarks. Since I understood that the mass of an atom as a whole is determined by the amount of binding energy of its constituents, I thought I can extrapolate that to 1 quark and say that its mass is also determined by the energycontent of its gluons. Hence, me concluding that a quark in a Helium atom is heavier than in a Carbon atom.
  26. Mar 28, 2017 #25

    Ken G

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    Good point, since quarks cannot be isolated. Still, one will find statements like this in the Wiki on quarks: " For example, a proton has a mass of approximately 938 MeV/c2, of which the rest mass of its three valence quarks only contributes about 9 MeV/c2."
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