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

  1. 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.
     
  2. 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?
     
  3. 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.
     
  4. 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.
     
  5. 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.
     
  6. 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?
     
  7. Nov 7, 2017 #32

    Buzz Bloom

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  8. 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!
     
  9. Nov 10, 2017 #34
  10. 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.
     
  11. 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"?
     
  12. 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.
     
  13. 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?
     
  14. 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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