Talk:Nuclear reaction

message: Hello Patrick, I think that should be that one particle is impossible for an exothermic reaction, and has very low probability for an endothermic reaction.
Consider the reference frame in which, before the collision, all the incoming particles' momenta sum to zero (i.e., the "zero momentum frame"). By conservation of momentum, they still must do so after the collision. But if we have zero momentum and only one particle, that therefore means it must be stationary in this frame. But then what about conservation of energy? Since the system initially had kinetic energy but now has zero kinetic energy, we can only balance energy if the reaction is endothermic, with the kinetic energy of the incoming particles in this frame exactly equal to the energy absorbance - which in turn means that given the position and velocity of one particle, for such a "sticking" reaction to occur the position and velocity of the other particle must exactly equal a calculated value. If the energy absorbance of the reaction was precisely defined, this would have zero probability, but if it has some width there will be a small probability of such a reaction. (Of course, on the macroscopic scale our absorbance line can be extremely broad, so such a collision is much more likely to occur with macroscopic particles.) However as soon as we have at least two particles, we can balance momentum, kinetic energy and reaction energy, of any amounts, in any frame. Hence, an exothermic reaction must have at least two products, an endothermic reaction might in principle have only one but in practice rarely if ever does. Or something like that; it's late here. Securiger 16:30, 22 Nov 2004 (UTC)
thread_title:  [[Two particles]] 

message: Why is that Nuclear reaction is one of the high-importance articles while Nuclear fusion, which is just one reaction type, is a top one? --Crwx (talk) 16:16, 21 April 2012 (UTC)
thread_title:  importance scale 

message: Nuclear reactions should transmute the nuclei into different species.  The article in general is rather clear on this point, but under Notable Types it has inelastic scattering.  This is decidedly not a nuclear reaction.  Unfortunately, the problem is that the main article does not even mention nuclear physics at all!  So, I'm not sure what to do about this.  Most likely the best case is to move all the scattering material to the scattering page.  I do think, however, that some kind of note somewhere in the article that elastic and inelastic scattering are not nuclear reactions, with links to the respective articles, is a good idea.  The elastic scattering page also does not mention nuclear reactions.  At least I'll go fix that now.  DAID (talk) 10:45, 27 September 2012 (UTC)
replies: text: Actually in the main article on inelastic scattering, the third and fourth words are nuclear physics, which can be taken as a good reason to add a section on Nuclear physics. I suggest moving the section Inelastic scattering of this article to the end of the Inelastic scattering article, with a name change to Nuclear physics and a slight rewrite to put it in the context of that article. As for this article, Inelastic scattering should be in the See also list at the end, which is for related subjects that don't quite fit the topic of the article. Dirac66 (talk) 14:17, 27 September 2012 (UTC)
replies: text: In regard to 'nuclear physics' being on that page, please check the timestamp of this post and those changes.  I did that, right after this post.  As I said I would.  DAID (talk) 17:05, 27 September 2012 (UTC)
replies: text: OK, sorry. I hadn't noticed you had just added the phrase nuclear physics today. I still think that since inelastic scattering is not actually a reaction, the section in this article should be moved to the other article. Dirac66 (talk) 18:06, 27 September 2012 (UTC)
replies: text:  Most people would say a reaction has taken place if the products are different from the reactants. Sometimes in inelastic scattering that is the case, as when a particle bangs off a nucleus and leaves some energy behind, leaving that nucleus in an excited state (from which it can decay to something else). Exciting a nucleus by hitting it with something is a reaction. SBHarris 20:40, 27 September 2012 (UTC)
replies: text: For inelastic scattering there is always a transfer of energy by definition. Scattering without energy transfer is called elastic. Dirac66 (talk) 21:02, 27 September 2012 (UTC)
replies: text:  Well, that's complicated, for "elastic" as applied to scattering is not the same as "elastic" as applied to collisions. And the only easy generalization is that all nuclear reactions are inelastic for the incoming particle, but very many inelastic scattering events produce no nuclear reactions, and some of these are even kinetically elastic as well.  See inelastic scattering. A transfer of simple kinetic energy from one particle to another is kinetically elastic (K-E) since the sum of KE is conserved in COM frame and not transformed to other types of energy-- but in nuclear terms it is IN-elastic scattering (I-S), because the energy of the incoming particle is not conserved. Momentum-transfer results in I-S, but does not imply anything else, as some I-S is K-E, and some is not. The article referenced uses the example of Compton scattering, but that's for electron targets, and we can easily generalize to nuclear interactions such as Bremsstrahlung and pair production. Neither of these are "elastic scattering" events (and they aren't kinetically elastic either, since some KE gets turned into radiation or new particles). Both involve interactions between an electron (classically) and a nucleus. But because the nucleus and the electron emerge unchanged from the event, neither of these are nuclear reactions, either. So nuclear reactions are a subset of inelastic scattering off nuclei, and inelastic scattering interactions in turn may be kinetically elastic or not. The Venn diagram is three overlapping circles.  You might consider the simple straight-on collision of a neutron hitting a deuteron, after which the neutron is greatly slowed and the deuterium gets (let me see) 8/9ths of the neutron's initial kinetic energy. This is kinetically elastic, but certainly an inelastic scattering event for the neutron, since it loses all but 1/8th of its energy. But it's not a nuclear reaction. If it produces a lepton-pair now it becomes not only inelastic scattering but it's kinetically inelastic now as well, since new particles are created and even the total kinetic energy of the reactants is not conserved. Still not a nuclear reaction, though! But if the deuteron is broken up, now you have a kinetically inelastic event, an inelastic scattering, AND now (finally) also a nuclear reaction. SBHarris 01:22, 28 September 2012 (UTC)

text: OK, I hadn't realized there are several different meanings for the word inelastic. If I understand you correctly, the point is that some but not all inelastic scattering events lead to nuclear reactions. So perhaps what this article needs is a section named something like Reactions induced by inelastic scattering, with appropriate explanations and examples, rather than just Inelastic scattering. The current text of the Inelastic scattering section is about inelastic scattering in general rather than reactions, so I still think that text belongs in the other article. But I think here I'll limit myself to suggestions and leave the revision to you and DAID, since I am rather out of my depth on this subject. Dirac66 (talk) 00:47, 29 September 2012 (UTC)
replies: text: Agree with your suggested change. We should move the inelastic scattering section to that article, as only a small subset of inellastic scattering produces nuclear reactions. If you look at the discussion down below, there is some question of whether any scattering but nuclei-nuclei scattering should be called a nuclear reaction, since the idea is that this must be a reaction between TWO NUCLEI. I dunno. I think induced nuclear fission (as in a reactor) is the most famous nuclear reaction and it doesn't feature reaction betweeen two nuclei, since a neutron is not a nucleus. At least one of the reactants must be a nucleus, anyway (protons count as H-1 nuclei). I don't think we must jigger things to include the triple alpha process, inasmuch as this can be seen as rapid sequential reaction between two nuclei followed almost immediately (less than the life time of Be-8) by another. I'll take the radioactive part out of the lede now, since we're all agreed on that. SBHarris 01:08, 29 September 2012 (UTC)
thread_title:  Scattering is not a reaction