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

[stefan r]

... 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.

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.

It is extremely difficult to believe that all the gold, platinum, uranium, etc. in the universe was created only by neutron star mergers. Considering that neutron stars themselves are already rare, and mergers of neutron stars are exceedingly rare, and you are talking about only between 0.1% and 1% of that ejected degenerate material creating everything heavier than iron in the universe. I realize that elements heavier than iron are not common, but even a billion times rarer than hydrogen is still more common than the process I just described.

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?

 

[Buzz Bloom]

Here is a link to a version of the chart in #16 that is a bit easier to read.

https://apod.nasa.gov/apod/ap171024.html​

 

[bbbl67]

Here is a link to a version of the chart in #16 that is a bit easier to read.

https://apod.nasa.gov/apod/ap171024.html​

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!

 

[glappkaeft]

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!

Go to the original source, it's better updated and has tool tip annotations:
https://upload.wikimedia.org/wikipedia/commons/3/31/Nucleosynthesis_periodic_table.svg

There is also the blog posted that the chart is based on:
http://blog.sdss.org/2017/01/09/origin-of-the-elements-in-the-solar-system/

 

[snorkack]

When you say a neutron is unbounded, do you mean that it is loosely orbiting the nucleus?

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.

Then is keeping the neutron bound, how they stabilize the neutron within the nucleus?

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.

If an H-3 Tritium is a bound state of neutrons, then why does Tritium still decay at some point?

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.

 

[bbbl67]

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.

You're talking about a neutron bullet hitting a nucleus, I assume?

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.

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

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.

When you say the neutron in T is "orbiting less loosely" than the proton in He-3, shouldn't that be "more loosely"?

 

[stefan r]

When you say the neutron in T is "orbiting less loosely" than the proton in He-3, shouldn't that be "more loosely"?

"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.

 

[bbbl67]

"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.

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?

 

[stefan r]

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?

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.