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I What % of the heavy elements are produced by kilonovas vs. supernovas?

  1. Nov 5, 2017 #21
    Without a doubt neutron star mergers were much more common in the early universe than they are today. Population III stars were much larger than Pop. II and I stars, so the percentage of neutron stars would have been proportionally increased - as would the number of supernovae in the early universe. Both combined could explain the elements heavier than iron, but neutron mergers on their own wouldn't cut it. I also don't understand how "low-mass stars" can contribute any element heavier than iron since they do not eject their outer envelope at relativistic speeds and it is not degenerate material being ejected.
     
  2. Nov 5, 2017 #22

    phyzguy

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    So you have done the calculation of the amount of r-process elements which are created in these events and compared it to the observed abundances? If so, please share that calculation with the rest of us. If not, maybe you shouldn't state the conclusion until you have done the calculation. The people at LIGO in this paper have done the calculation, and they conclude (see a quote from that paper below) that these events can produce all of the observed r-process elements. Since you have a different conclusion, can you please explain what they have done wrong?


    "Our results suggest that dynamical ejecta from rare NS mergers could be an important and inhomogeneous source of r-process elements in the galaxy (Ji et al. 2016 ;
    Beniamini et al. 2016 ). If more than 10% of themass ejected from mergers is converted to r-process elements, our prediction for average r-process density in the local universe is consistent with the Galactic abundance."
     
  3. Nov 5, 2017 #23

    phyzguy

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    Perhaps another error in your thinking is that neutron stars are not as rare as you seem to think. NASA estimates that there are on the order of 1 billion neutron stars in our galaxy.
     
  4. Nov 5, 2017 #24
    Well, then NASA has a problem. Because they also estimate that there are on average of three supernovae every century in the Milky Way galaxy. Even if every supernovae in the Milky Way from the beginning of time until today became a neutron star it would not be nearly enough to be one billion neutron stars. In order for there to be a billion neutron stars in the Milky Way galaxy there would have to be at least one supernova in the Milky Way galaxy on average every 13.8 years, and every supernova would have to result in a neutron star. Clearly one, or both, of those estimates is flat out wrong.
     
    Last edited: Nov 5, 2017
  5. Nov 5, 2017 #25

    phyzguy

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    If the current supernova rate of about 1 every 33 years had existed for all of cosmic history, and the Milky Way is about 10 billion years old, then we would expect there to be about 10 billion/33 = 333 million neutron stars. This is not so different from the 1 billion order of magnitude estimate I gave. However, it is well known that this is not the case. Looking at the image below, from this site, we can see that in the past stars were forming (and dying!) at a rate more than 10X what we see today. Thus there are many more "dead" stars, like neutron stars than you would predict by your simple calculation. I urge you to learn some astronomy before you start attacking what is known.


    vs_sfr.png
     
  6. Nov 6, 2017 #26
    I'm starting to think that there is a typo in that graphic. That is to say, what they should've labled "low-mass stars" should have been labled "high-mass stars" and vice-versa. Then that graphic makes more sense.
     
  7. Nov 6, 2017 #27
    When you say a neutron is unbounded, do you mean that it is loosely orbiting the nucleus? Then is keeping the neutron bound, how they stabilize the neutron within the nucleus? If an H-3 Tritium is a bound state of neutrons, then why does Tritium still decay at some point?
     
  8. Nov 6, 2017 #28
    And also not all of those supernovas would result in neutrons stars, some percentage of them would produce a black hole instead.
     
  9. Nov 6, 2017 #29

    phyzguy

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    Of course. But given the star formation rate in the past more than 10X what we see today, I don't think that NASA's estimate of 1 billion neutron stars in the Milky Way is unreasonable. Also, in the paper I quoted earlier in Post #22, the LIGO team did the analysis in detail and came to the conclusion that neutron star mergers can explain the r-process elements. If you're going to argue with their conclusions, as |Glitch| is, I think you need to examine their analysis in detail and explain where you think they are wrong.
     
  10. Nov 6, 2017 #30
    Wouldn't "high-mass stars" be the same category as their "exploding massive stars?" There is more than just a typo problem with that graphic.
     
  11. Nov 7, 2017 #31

    stefan r

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    S-process, "slow neutron capture process".
    Third dredge up, Material from the core shows on the surface of AGB stars. Visible on the surface also means present throughout the convective zone.
    Planetary nebula, Stars eject most of the material that was in a convective zone out into space.

    The core is degenerate in AGB stars before helium flashes. Core gets a lot hotter and expands.

    The s-process is reproducible in laboratories on earth. Get a pure isotope, bombard it with neutrons, and measure what you got. Isotopes that are stable and are in the s-process sequence are much more abundant than isotopes that are not in the sequence. Even if the non-s-process isotope is more stable than the s-process isotope the s-process isotope is more common.

    Gold has abundance in universe of 6 x 10-10. Excluding dark matter the milky way has less than 3 x 1011 solar mass. So gold mass in the milky way should be about 180 solar masses. The black hole in the center of the milky way has mass 4.1 x 106 solar mass. If the black hole formed from only neutron stars merging (unlikely) it would have ejected 41,000 solar mass of heavy elements. That is about the right order of magnitude. A lot of that material should have fallen back in but there are also other black holes.

    Rapidly spinning neutron stars would have different collision dynamics. In some cases that should mean a lot more ejected mass.

    When a neutron star drops into a small black hole does it get disrupted? How much of that would eject?
     
  12. Nov 7, 2017 #32
  13. Nov 9, 2017 #33
    Easier to read, but still just as perplexing. In fact, a bit more perplexing, this chart shows a 7th colour (goldish) which is not even explained where the origin of this one is!
     
  14. Nov 10, 2017 #34
  15. Nov 11, 2017 #35
    No, what I mean is that it is not orbiting the nucleus at all, even loosely - it bounces off in a single collision and never returns.
    A neutron is stabilized if a neutron is more strongly bound than a proton would be bound in its place - and stable if it is more strongly bound at least by the margin of neutron decay energy, which is 782 keV.
    On the other hand, a loosely orbiting neutron can be actually destabilized. Because a loosely orbiting neutron may decay into a tightly orbiting proton. A process which can release much more energy and happen much faster than decay of free neutron to a free proton.
    Because He-3 is also a bound state. And, as it happens, although the neutron in T is stabilized - the neutron in T is actually orbiting less loosely than the proton in He-3 - it is not stabilized quite enough. Free neutron has decay energy of 782 keV and half-life of 10 minutes. Triton has decay energy of mere 18 keV, and half-life of 12 years.
     
  16. Nov 19, 2017 #36
    You're talking about a neutron bullet hitting a nucleus, I assume?

    How is a loosely orbiting neutron determined vs. a tightly orbiting one?

    When you say the neutron in T is "orbiting less loosely" than the proton in He-3, shouldn't that be "more loosely"?
     
  17. Nov 19, 2017 #37

    stefan r

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    "more loosely" means "has more energy"? A free neutron decays with 782 kev then a "more loosely" one should decay at more than 782 kev.
     
  18. Nov 20, 2017 #38
    I'm still having some problems with this terminology. If an H3 nucleus is "less loose" than an He3 nucleus, then that would indicate that H3 < He3 in terms of their energy levels. So having less energy would indicate that it's closer to the ground state, thus closer to stability, right?
     
  19. Nov 20, 2017 #39

    stefan r

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    A container is more tightly bound to earth in a valley, it is more loosely bound on top of a tower or "cliff". The ground state is at the bottom and falling off the cliff releases energy. Climbing up adsorbs energy.
    If the container is also a rocket then you can have a chemical transition. There is a higher energy state of nonreacted chemical in the container and lower energy state of empty container with released chemical.
    When we fire the rocket out of the valley to the top of the cliff it goes from a tightly bound condition with respect to earth and lands on top of the cliff in a position that is loosely bound.
    The payload clears the top of the cliff and lands with impact energy. The difference in impact energy of a rocket fired on level ground verses launch from the valley can be measured. So if we have a standardized rocket we could estimate the height of cliffs and also determine whether rocket fired up the cliff or down.

    Using the analogy to clarify terms. The neutron is our loaded canister. The electron and antinuetrino is our rocket exhaust. The proton is the payload. The 3H and 3He are positions. Snorkack says that a free(in the plains) neutron (rocket) gains 782 keV(altitude). A 3H decay gains 18 keV. So the "launch pad" was in a valley below a 764 keV "cliff". The proton is sitting in position at the "top" of the "cliff" which is more loosely bound and higher energy. Overall the launch releases 18 keV so the overall final condition is lower energy than the initial conditions despite the components being more loosely bound.

    The analogy breaks horribly if you start talking about reversibility or uncertainty of position and momentum. It is possible that the payload randomly finds itself up the cliff and that allows the fuel to combust and the payload cannot fall back off the cliff without the fuel because it is repelled by the rocks on the valley floor. It is not a rocket it is a neutron. But the term "loosely bound" with respect to a force can be used with multiple types of forces. Gravity, electro-magnetic, and nuclear forces can bind things. We only measure the heat and the components.
     
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