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Questionable change?

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The paragraph "For lower mass stars (about 100 solar masses and below), other conditions keep the star's core stable. Pair production does not cause an instability. These stars collapse in ordinary Supernova." was changed to "For lower mass stars (about 100 solar masses and below), the gamma rays are not energetic enough to produce electron-positron pairs, and if these stars become supernova they do so via other means.". I'm wondering if the latter is true - are electron-positron pairs not produced at all in regular supernovae? That seems unlikely to me. I would guess they are produced in inadequate numbers to cause the runaway reaction that leads to total explosion. --Keflavich 05:13, 13 May 2007 (UTC)[reply]

Energy source

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So I'll ask a dumb question:

In a typical core-collapse supernova, the core is made of iron and any thermonuclear reactions would not produce a surplus of energy that could drive an explosion. So the supernova energy comes from the gravitational energy of the collapse, leaving a remnant behind due to the requirement to balance the energy. Is this right? If so, what generates the extra energy in the core needed to gravitationally decouple a pair-instability supernova? Does an explosive reaction occur outside the collapsing iron core?

This is unclear to me from the text. Thanks. — RJH (talk) 15:37, 15 May 2007 (UTC)[reply]

Based on articles 1 and 3, a pair-instability supernova could not occur with an iron core. The explosion results from explosive burning of oxygen and silicon, which probably wouldn't occur if enough of the core had already burned. Anyway, the supernova described on this page really only applies to zero-metallicity population III stars. --Keflavich 18:10, 15 May 2007 (UTC)[reply]
Low metallicity, not zero; follow the refs out [1] and you find tables of how far up the metallicity scale the pair-instability effect can still happen. There's a plateau at which the pair instability effect happens for arbitrarily large mass stars (which at zero metallicity, will direct collapse to black holes via endothermic photodisintegrative processes). Beyond that a bit, the pair-instability effect stops entirely. Most of these are Pop III stars, but Eta Carinae and the SN 2006gy supernova are newer... Georgewilliamherbert 18:46, 15 May 2007 (UTC)[reply]
Thanks for pointing that out... I never even read the y axis before. --Keflavich 19:21, 15 May 2007 (UTC)[reply]
Thank you, I believe I understand it now. So it looks like it would be correct for the lead to say "low metallicity" rather than moderate. — RJH (talk) 15:13, 17 May 2007 (UTC)[reply]
Could population II stars have low enough metallicities to form pair instability supernovae? I'm thinking of the blue stragglers in globular clusters, which are considered Population II--Robert Treat (talk) 06:54, 25 January 2009 (UTC)[reply]
It seems like everyone's discussing population III stars as the only possible medium for this form of supernova. Correct me if I'm wrong, but are pop III stars only theoretical in existence, and wouldn't they have to be incredibly far away to still be 'in existence'? Surely due to their high mass and short lifetimes, they'd need to be at least as far away as we can possibly see in order to still exist. Therefore, how can SN 2006gy possibly have formed from a pop III. star when it is fewer than 100Mpc away? —Preceding unsigned comment added by The Smurfmeister (talkcontribs) 14:52, 16 March 2009 (UTC)[reply]
From what I've read at Wikipedia and other sites, many of the Population III stars would have been too massive to explode as pair-instability supernovae and would have collapsed into black holes through photodisintegration. They could still have deposited the first metals, though.
Not all of the star’s mass is swallowed up in a black hole’s formation. A portion of mass escapes from the poles in the form of immensely powerful jets of gas and other substances, called relativistic jets because of their high velocities, and these jets could have seeded the universe with the first metal deposits. Radiation in the form of gamma-ray bursts also escapes during the black hole’s formation--Robert Treat (talk) 20:45, 26 July 2010 (UTC).[reply]

Correct a reference

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I'm new to Wikipedia, and so I'm not sure how to correct the Fryer link in the references section. arXiv.org states that documents should not be linked out of the cache, but rather against the main document entry page here: http://arxiv.org/abs/astro-ph/0007176v1

Wjhudson 17:10, 18 May 2007 (UTC) wjhudson[reply]

You could have just edited the URL that shows up in the reference (it's located at the top, inline with where it first appears, in a macro that puts the [3] tag in and then puts the detailed ref at the bottom). I've made the change you list now. That was a good spot. Thanks for noticing that; I think the ref being the cached one was my fault. Georgewilliamherbert 20:02, 18 May 2007 (UTC)[reply]
Thanks George. It looks like you're also the person I need to thank for linking to my blog entry on SN2006gy in the external references section. I am happy you found it useful. Wjhudson 18:05, 19 May 2007 (UTC)wjhudson[reply]

Thermonuclear?

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"As described in the introduction, the results of pair creation interactions are pairs of electrons and positrons. These particles are released into the star's core and usually recombine (releasing another gamma ray) in very short time periods."

Surely they annihilate themselves rather than recombine? Recombine suggests to me that a particle is left when the process is over. This of course leads to the question that is the runaway reaction that blows the star apart indeed a thermonuclear event as the introduction describes or matter/antmatter annihilation? Or is the reaction primarily thermonucear driven by the increasing temperature and pressure caused by the collapse associated with increasing pair production? --LiamE (talk) 16:46, 22 January 2008 (UTC)[reply]

  • Recombination generally happens first for positronium (positron + electron "atom"), then after a series of level transitions the positron and electron annihilate in the center; normal positronium has a decay lifetime of something like 1 microsecond. However, I'd guess conditions inside a supernova are different (energetic) enough that the positron and electron probably do not form an atom before annihilating most of the time. --Keflavich (talk) 00:29, 17 February 2008 (UTC)[reply]


Explanations

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This article doesn't explain why a black hole is not formed, and what triggers the sudden occurrence of the pair-instability state.Andrewjlockley (talk) 15:48, 27 February 2010 (UTC)[reply]

A black hole is not formed because the thermal energy of the explosion exceeds the gravitational binding energy of the star. e+ / e- pair creation happens when very high energy gamma rays interact with nuclei, and such high energy gamma rays happen when the star is massive enough that the core gets hot enough to produce them. Less massive stars don't reach such core temperatures, because their gravitational well is less deep, so more energy is able to escape, keeping things cooler. There is some discussion of this in some of the related articles. 66.127.53.162 (talk) 20:24, 22 April 2010 (UTC)[reply]

sn2007bi

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This article about sn2007bi is interesting but maybe not directly usable in WP. It might point to something more suitable. 66.127.53.162 (talk) 08:43, 22 April 2010 (UTC)[reply]

Mean free path ?

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I'm unclear why this is relevant. I do not believe the reduction in gamma mean free path with increasing temperature is responsible for the instability. Rather, it is the fact that increasing temperature shifts the equilibrium between gammas and pairs in favor of pairs. This reduces gas pressure and contracts and heats the star, shifting the equilibrium yet further in favor of pairs, and so. So I think the discussion of mean free paths is irrelevant and unhelpful. Yaush (talk) 18:46, 1 March 2011 (UTC)[reply]

It's been a little while, but I've read all the sources (and understood them generally, though I can't directly do all the math involved). They mentioned MFP. Just producing more pairs doesn't matter - pairs recombine to form gammas very quickly. But pair production significantly affects MFP, since the recombination emits in a random direction, i.e. MFP is strongly affected by the pair production. Georgewilliamherbert (talk) 22:20, 1 March 2011 (UTC)[reply]
It's not the rate of pair production alone that matters. It's the balance between pair production and pair annihilation. The rate of pair production increases with temperature faster than the rate of pair annihilation, so at any given moment, more of the energy of the star is in the form of pairs. Since the pairs have mass (about an MeV per pair) and the gammas do not, more of the energy of the star as pairs means less available as kinetic energy to provide pressure.Yaush (talk) 23:09, 1 March 2011 (UTC)[reply]
You are entirely correct, and I’ve revised the article to reflect this. — Preceding unsigned comment added by 81.151.194.209 (talk) 21:08, 2 December 2012 (UTC)[reply]

Readability Edits

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I've made a few edits to this article with the intent to construct a more readable narrative from (what seem to me to be) somewhat disconnected scientific observations.

Not pretending to have any scientific expertise, in this area or otherwise, I have endeavored at all times to maintain the scientific content, and avoided making edits to material appearing to admit of more than a single correct explanation.

If there are times when I have erred, I apologize, but I do think that the article could be made rather more comprehensible to a layperson without compromising its scientific rigor.Drolz09 03:37, 17 October 2013 (UTC)[reply]

I don't think the edits are quite right yet. However, it was probably me that wrote much (certainly some) of the previous wording, so I'll wait and see if someone else can tweak it up before I have a go. I'd probably just make it unreadable for the layman again :lol: Lithopsian (talk) 13:12, 17 October 2013 (UTC)[reply]
Also, this article doesn't explain why pair-instability causes a net reduction in counter-gravitational pressure. It is not clear to me why the positrons created in this reaction wouldn't rapidly compensate for the energy converted into mass by annihilating upon contact with electrons. I've no doubt there is a good and perhaps obvious reason for this, but can't find such here.'''''Drolz09 03:46, 17 October 2013 (UTC)[reply]
Pair creation from gamma rays is not an all-or-nothing event where suddenly positrons stop annihilating and start being created. Like all events inside stars, it is a balance of annihilations and creations, and the balance shifts depending on conditions. In this case, the creation side of things is extremely dependant on the temperature, in fact on the number of high energy gamma rays that exist, so strongly so that people tend to treat it as something that "switches on" at a particular point. The annihilation side of things is relatively constant, with the average life of a positron not changing much over that range of conditions. Net result, as the temperature increases a whole lot more photons turn into positrons and electrons on a time scale far shorter than the annihilation half-life. Fewer high energy photons, less internal pressure, which leads to collapse for a number of reasons. Any drop in internal pressure and temperature prevents fusion of the little remaining fuel, feeding back into the collapse, while degeneracy of such a large core is insufficient to halt the collapse. As with other types of supernova, the collapse then causes a massive temperature spike and runaway burning (much of it endothermic) of the core. Bang! Detailed models are still being argued over. Lithopsian (talk) 13:12, 17 October 2013 (UTC)[reply]
So, here's my effort at fashioning a layman-readable narrative for the article. Is it at least close to accurate? Since I am not myself a physicist, any corrections/suggestions would be much appreciated:

Nuclear fusion in a star's core produces gamma ray photons. When such a photon interacts with an atomic nucleus, it may convert into a positron and electron; the positron-electron pair then annihilate, changing back into a photon. At very high temperatures, pair production becomes much more frequent, to the point that positron-electron pairs are created faster than they can annihilate. When the stellar core reaches these temperatures, it begins "losing photons" to pair production. This causes the photon pressure to drop. Since photon pressure is what supports the core against gravity, the drop allows gravity to compress the core more tightly, which raises the temperature further.

As temperature rises in the core, two things happen. First, pair production becomes even more prevalent. Second, nuclear fusion accelerates. Normally, the increase in fusion would result in increased photon pressure, causing the core to expand and cool. The loss of photon pressure to pair production disables this safety valve, so the core continues to shrink. The result is a runaway fusion reaction, which rapidly heats the core to the point that its thermal energy overcomes gravity. The core then explodes, tearing apart the entire star. 129.74.116.38 (talk) 16:02, 27 January 2014 (UTC)[reply]

Remnants Section Contradicts Other Information In the Article and is Redundant

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The remnants section contradicts the information given previously in the article, which says the existence of a remnant is conditioned on the mass of the star. Given that the evolution by mass paragraphs above describe the remnant for each mass range, I think this section should probably be deleted as redundant.


Bgovern (talk) 04:29, 15 August 2022 (UTC)[reply]