SciForums.com > Science > Physics & Math > Particle Interaction problem PDA View Full Version : Particle Interaction problem Post ReplyCreate New Thread neelakash02-06-08, 11:11 PMI have already solved almost all of the following questions.Just want to check them and improve reasoning.So, please do not blame me for asking you to do it for me. Check if the following interactions are possible or forbidden and indicate the interactions when applicable. \ Q ---charge conservation test \ L ---Lecton number test \ B ---Baryon number test \ I_3 ---Isospin test --------------------------------- Isospin \ I_3 is violated in EM and Weak interaction ---------------------------------------------------------- Strangeness \ S is violated \Delta\ S=1 in weak interaction ----------------------------------------------------------- (i) \pi^-\ + \ p \rightarrow\ k^0 \ + \Lambda^0---Allowed [Q,B,S,I_3---all satisfied---Strong interaction] (ii) \pi^-\ + \ p \rightarrow\ k^0 \ + \Sigma^0---Allowed [Q,B,S,I_3---all satisfied---Strong interaction] [(I am doubtful if I have correctly said that they are strong interaction.Are they?] (iii) \Sigma^0\rightarrow\Lambda^0\ + \gamma ---Allowed [Q,B,S,I_3---all satisfied---strong interaction] (iv) \Sigma^+\rightarrow\ p \ + \gamma ---Not Allowed [I_3 non-conservation and ∆S=1---it looks that this might be allowed via weak interaction.But I found in book that it is not allowed. Is it bacause there is no Leptons?] (v) \pi^0\rightarrow\gamma\ + \gamma ---Allowed.[I suspect it is EM interaction.But any strong evidence?] (vi) \ K^+ \rightarrow\pi^0\ + \mu^+\ + \nu_\mu ---Allowed[Q,L,B satisfied---∆S=1--weak interaction because nutrino is involved only in weak interaction]. (vii) \Xi^-\rightarrow\Lambda^0\ + \pi^- ---Allowed[Q,B satisfied.∆S=1 as in weak interaction] (viii) \Xi^-\rightarrow\ n \ + \pi^- ---Not allowed [∆S=2---never allowed in any interaction] (ix) \Lambda^0\rightarrow\pi^+\ + \pi^- ---Not Allowed.[B,S non-conservation] (x) \ n \rightarrow\ e^+ \ + \ e^- ---Not allowed.B non-conservation and mass non-conservation (n,e are stable particles). BenTheMan02-07-08, 12:43 AMCheck (x). Just baryon number is violated. There is no conservation of mass!!! BenTheMan02-07-08, 12:57 AMAlso, you can check your answers here. http://pdglive.lbl.gov/listings1.brl?exp=Y The process (iv) IS allowed to happen, but it is highly suppressed (only happens about 1% of the time). The reason the book says that it is not allowed to happen is because it doesn't happen at tree level. The \Sigma^+ is made of uus quarks. The proton is uud. The process is actually (in terms of quarks) s\rightarrow d\gamma. This is called a flavor changing neutral current---that is, the s and d quarks have the same charge, so the thing that changes s to d must be neutral, i.e. uncharged. And to first approximation, there ARE no flavor changing neutral currents in the Standard Model. What CAN happen though, is that s can emit a W boson, which can change to a c quark, which can then reabsorb the W boson and change to a d quark, with a small probability. This is probably more physics than you are learning, so it may be new to you. Think of it this way---you presumably know what a Taylor series is: e^x = 1 + x + \frac{1}{2}x^2 + \ldots And you also know that to have a GOOD Taylor series converges, which means that each higher order term is smaller than the preceding one. Particle physics is EXACTLY like that---we do an expansion in loops in our Feynman diagrams: http://content.answers.com/main/content/img/McGrawHill/Encyclopedia/images/CE255550FG0010.gif The big picture at the top is called the tree level result, or the classical approximation. The second term in the Taylor series consists of one loop, the third term is two loops, etc. All of this to say, to leading order (tree level, classical approximation), the process (iv) is forbidden! But it DOES happen at one loop level. God I love physics. BenTheMan02-07-08, 01:04 AMHere's what a flavor changing neutral current (FCNC) looks like: http://upload.wikimedia.org/wikipedia/en/thumb/5/56/Tau-decay-fcnc.png/180px-Tau-decay-fcnc.png Ignore the bottom graph for now. Now replace \tau with s, \nu with c and e with d. neelakash02-07-08, 09:01 AMCheck (x). Just baryon number is violated. There is no conservation of mass!!! Yes,I know there is no conservation law for mass.But there are some interactions which exhibit non-conservation of mass within a time interval compatible with uncertainty principle.That is Uncertainty principle allows for that non-conservation only for that limited time interval. But here,n and e are stable particles, their existence is not limited by finite time---so that mass non-conservation will not be applicable. I do not know if this is correct.But I thought this way. neelakash02-07-08, 09:16 AMHi Ben, thank you for that link.Looks like a nice webpage. Specifically,I want to know how to understand-given an interaction,what type of interaction it is? That is how to understand if it is weak/strong or EM? Generally I follow the rules:(i)whenever there is \nu, it must be an weak interaction.(ii)For isospin conservation, the interaction must be strong interaction;(iii)strangness always change by one unit in weak interaction etc... Also there are some rule of thumbs: (i) weak interaction necessarily involves leptons (ii)strong interaction always involve mesons...(iii)EM interaction always affects charged particles...----------But I do not know if these rules are correct at all.Please discuss this point clearly. BenTheMan02-07-08, 09:20 AMYes,I know there is no conservation law for mass.But there are some interactions which exhibit non-conservation of mass within a time interval compatible with uncertainty principle.That is Uncertainty principle allows for that non-conservation only for that limited time interval. But here,n and e are stable particles, their existence is not limited by finite time---so that mass non-conservation will not be applicable. I do not know if this is correct.But I thought this way. Well, I don't think it's correct. Remember, E^2 = m^2c^4 + p^2c^2....you can ALWAYS exchange mass for kinetic energy. For example, one of the biggest signatures for grand unified theories is proton decay, via the cannel: p \rightarrow \pi^0 + e^+ baryon and lepton number are violated (this is a general feature of GUTs), and mass is not conserved. What you should do is apply this logic to some of the other problems you have, and see if THEY conserve mass. And the neutron is not a stable particle. It's half-life outside of a neucleus is something like 14 minutes. neelakash02-07-08, 09:22 AMWow!!! That's really a difficult question...atleast in my level. I heard about the flavour change neutral current interactions...but this is the first encounter. Thanks. neelakash02-07-08, 09:26 AMWhat you should do is apply this logic to some of the other problems you have, and see if THEY conserve mass. And the neutron is not a stable particle. It's half-life outside of a neucleus is something like 14 minutes. I see...It was a gross mistake... Thank you very much. BenTheMan02-07-08, 09:34 AMI see...It was a gross mistake... Thank you very much. Now worries---you were probably taught about conservation of mass in chemistry class or something, where it is an approximate conservation law. There are no such things in particle physics. BenTheMan02-07-08, 09:36 AMSpecifically,I want to know how to understand-given an interaction,what type of interaction it is? That is how to understand if it is weak/strong or EM? In general, EM interactions change charge, weak interactions change flavor (i.e. u to d) and strong interactions change color (red quark to blue quark). AlphaNumeric02-07-08, 09:52 AMGenerally I follow the rules:(i)whenever there is \nu, it must be an weak interaction.(ii)For isospin conservation, the interaction must be strong interaction;(iii)strangness always change by one unit in weak interaction etc...Remember that isospin is a very good but approximate symmetry. The existence of pions pays testiment to that, since they are pseudo-Goldstone bosons (though you might not know what that means yet). Basically, when you do things like work out the multiplets for baryons or mesons and plot them on a mass vs Y+L (or whatever it is) diagram, you find that the symmetry isn't perfect. You don't find that you get say 3 particles of the same mass but different Y+L (as a perfect symmetry would imply in a octo-plet), their masses have minor deviations. This leads to isospin breaking. If you have the textbook by Georgi on Lie Algebras in Particle Physics he specifically goes through rejigging the QCD Lagrangian (at least the light quark section) to show how there's an isospin conserving section and a non-isospin conserving section. The non-conserving section is due to the difference between the various quark masses. If they had m_{d}=m_{u} then there'd be a perfect symmetry. Since there isn't you get processes within the SM which break isospin but typically they are supressed due to the small difference between the quark masses. It slips my mind wether they are loop processes or tree level. Loop processes are generally further suppressed from tree level (at least until you get to string theory where an interesting anomaly cancellation occurs). BenTheMan02-07-08, 12:32 PMAlso there are some rule of thumbs: (i) weak interaction necessarily involves leptons (ii)strong interaction always involve mesons...(iii)EM interaction always affects charged particles...----------But I do not know if these rules are correct at all.Please discuss this point clearly. Well, I'd say that (i) and (ii) are wrong. For (i), Weak interactions can change flavor, for example and up quark to a down quark, as in beta decay. Weak interactions always involve a change in flavor. For (ii), it may be true at this level, but is not true in general. The strong interactions affect color, which is another charge carried by quarks. neelakash02-07-08, 02:31 PMThen how to see if a given interaction is strong or EM? neelakash02-07-08, 02:38 PMLet us return to the interactions: (i) \bar{u}\ d \ + \ u \ u \ d \rightarrow\ d \bar{s}\ + \ u \ d \ s Similar with the (ii) So, (i) and (ii) are weak interactions??? neelakash02-07-08, 03:57 PMInteraction (iii) preserves the flavour---Is it strong interaction?You told charge is not conserved in EM interaction(---does it mean the charge of the quarks)??? (iv) Already explained by you (v) \ u \bar{u}\rightarrow\gamma\ + \gamma Here, quark charge is not conserved---so, it is EM interaction,right? (vi) \ u \bar{s}\rightarrow\ u \bar{u}\ + \nu\ + \nu_\mu It is a weak interaction? (vii) \ d \ s \ s \rightarrow\ u \ d \ s \ + \ d \bar{u} Same; weak interaction neelakash02-07-08, 04:04 PMMay be I am wrong in predicting with my sleepy eyes...just give me some time.It is time to bed... BenTheMan02-07-08, 05:51 PMHere, quark charge is not conserved---so, it is EM interaction,right? I'm not sure what quark charge'' you're talking about---do you mean color''? Either way, this interaction IS an EM interaction. the rest of your assesments are correct. AlphaNumeric02-07-08, 06:11 PMLet us return to the interactions: (i) \bar{u}\ d \ + \ u \ u \ d \rightarrow\ d \bar{s}\ + \ u \ d \ s Similar with the (ii) So, (i) and (ii) are weak interactions???You can see in (i) that, when you strip away the unchanged particles, you have \bar{u} +\ u \rightarrow \bar{s}\ + \ s This can be facilitated via the electromagnetic interaction, since \bar{u} +\ u \rightarrow \gamma \rightarrow \bar{s}\ + \ s provided the colour factors are equal and opposite. BenTheMan02-07-08, 06:22 PMoops. yeah. what he said. neelakash02-07-08, 11:18 PM(i) and (ii) are the same: \bar{u}\ d \ + \ u \ u \ d \rightarrow\ d \bar{s}\ + \ u \ d \ s This results in \bar{u}\ + \ u \rightarrow\bar{s}\ + \ s Here, both sides have neutral electric charge,neutral colour charge [quark-antiquark pair must have opposite colour charges],but flavour changed---so how can it be EM interaction??? neelakash02-07-08, 11:27 PM(iii) \ u \ d \ s \rightarrow\ u \ d \ s \ + \gamma This is EM or strong interaction???But I do not see the violation of electric or colour charge!!!Since strangeness is conserved it should be strong interaction, right?But how to reconcile with Ben's argument? neelakash02-07-08, 11:30 PM(iv) If allowed: \ u \ u \ s \rightarrow\ u \ u \ d \ + \gamma s changes to d---so undoubtedly weak interaction.Also \Delta\ S \ = \ 1 neelakash02-07-08, 11:34 PM(v) \ u \bar{u}\rightarrow\gamma\ + \gamma Same problem: electric charge and colour charge are conserved...(am I making a mistake?)---but no quarks on the right side!!!what should it be? neelakash02-07-08, 11:36 PM(vi) No problem: weak interaction neelakash02-07-08, 11:40 PM(vii) This is effectively: \ s \rightarrow\ u \ + \ d \bar{u} Clearly strangeness changes by one unit---and it changes flavour and makes u and d quarks and antiquarks...So, it should be weak interaction. AlphaNumeric02-08-08, 05:43 AMHere, both sides have neutral electric charge,neutral colour charge [quark-antiquark pair must have opposite colour charges],but flavour changed---so how can it be EM interaction???It's only electomagnetic if u and u' (and thus s and s') have opposite colour charges. If they don't, then the process is gluonic. ie let c be some colour charge r, g or b and \bar{c'} be \bar{b}, \bar{r} or \bar{g} where c \not= c' u_{c} + \bar{u}_{\bar{c}} \to \gamma \to s_{c} + \bar{s}_{\bar{c}} or u_{c} + \bar{u}_{\bar{c'}} \to g_{c\bar{c'} }\to s_{c} + \bar{s}_{\bar{c'}} Since there's an overall colour charge in the second one, the intermediate bosonic state must be able to carry colour. As with any purely electromagnetic process, if the energy is high enough then Z bosons are likely to be produced, so it could be weak interactions too : u_{c} + \bar{u}_{\bar{c}} \to Z \to s_{c} + \bar{s}_{\bar{c}} So depending on the colour of the quarks, it could be any interaction!! Here, both sides have neutral electric charge,neutral colour charge [quark-antiquark pair must have opposite colour charges],but flavour changed---so how can it be EM interaction???There's no flavour changing, the overall flavour on both sides is zero. If colours are opposite then since there's no baryon, lepton or anything 'number' on the left hand side, the process could turn into any matter/antimatter pair. (v)Similar logic to above, except here you know there's no colour charge since the RHS has no colour. u + \bar{u} \to u + \bar{u} + \gamma \to 2\gamma This is EM or strong interaction???But I do not see the violation of electric or colour charge!!!Since strangeness is conserved it should be strong interaction, right?But how to reconcile with Ben's argument?There's never a violation of electromagnetic charge. Or colour. They are difference from things like lepton or baryon number. EM and strong charges are the result of gauge symmetries. Such symmetries essentially make the photon and gluon massless (though you wouldn't know how to prove that yet). If they were violated the photon and gluon wouldn't be massless (they'd be masses due to Goldstone's theorem). u, d and s are charged. They interact with the electromagnetic field. Just as an electron can emit a photon, so can u, d and s (and any other charged particle, tautologically). The process is nothing more than one of the particles emitting a photon and carrying on as normal. Since the photon (and the Z and the graviton) has no quantum numbers of any kind, when a particle emits it, it doesn't change (other than 4-momentum). Provided a particle is EM charged, it can emit as many photons as it likes without having to consider various conservation numbers (other than 4-momentum). neelakash02-08-08, 10:19 AMI need to review more.Possibly,I could not follow your reply.Just give me some time. neelakash02-10-08, 01:21 AM(i) Till now,I am not sure.Looks like EM interaction. (ii)Strong interaction (iii)EM interaction (iv)weak interaction allowed. (v) EM interaction (vi) Weak interaction (vii)Weak interaction ---I am almost sure about (ii)-(vii) apart from (i).I know strange particles are produced via strong interaction---that conserves strangeness and decay via weak interaction.In fact I have seen the feynmann diagram in a book and it involved a gluon.But from conservation principles,it looks like EM interaction. [Strong interaction conserves all quantum numbers, EM interaction violates Isospin I and G quantum number, and weak interaction violates I,(I_3), S,C, P,CP in a special case,T and G.] neelakash02-10-08, 02:06 AMHmmm... Now,I found a way to say (i) is a strong interaction. Grifiiths says in his book-the characteristic time of an interaction depends on the mass difference of mother and daughters.Now, mass difference is ~530 MeV.Now, just put in energy-time uncertainty principle---the characteristic time of interaction is ~(10^-23) s Hence,nothing else, this is a strong interaction. AlphaNumeric02-11-08, 06:12 PMDoes that automatically exclude electromagnetic interactions? In LEP collisions between electrons and positrons, the first particles created by the annihilations were Zs and photons which then turns into all sorts. Those Zs and photons were never seen, they turned into other particles practically instantly. Besides, as I mentioned, if the colour charges on the u and u' match then the intermediate state will have no colour charge, it would be a colour singlet. Gluons are never colour singlets, otherwise they wouldn't experience total confinement. If the colour charges are not matched then it must be strong, because the intermediate boson must carry colour charge, only the gluon does that. So it will be dependent on the colour charge of the u and u'. neelakash02-12-08, 02:00 AMI am too small a kid to deny your opinion;but I did not understand many of the things you told (may be I will understand them in future).However, as far as I know 10^(-23) second is a characteristic time for strong interaction. Actually, the conservation laws told me it is an EM interaction---[EM interaction violates Isospin I quantum number---as it is here].So, in my opinion, it might be strong interaction as well as an EM interaction. (Examiners are not always happy with multiple answers.) AlphaNumeric02-12-08, 08:52 AMColour is conserved. If the total colour initially is zero, the intermediate boson must have no colour. Gluons have colour. Hence they only mediate the process if there isn't a cancellation of colour between u and u'. If you explained that, the examiner should accept it. Personally I've never used the mass change and time of interaction to consider which one it is. That might make a certain process more likely than others, but due to the nature of quantum field theory, you have to factor in ALL possibilities. The mathematics of QFT will weight the processes accordingly (ie as your total collision energy in an electron+positron process goes up electroweak theory says the contributions from Z boson processes becomes more and more important rather than just considering the photon only, hence why QED is not the end of the story). neelakash02-13-08, 08:50 AMHmmm... I asked Dr. Griffiths about this interaction.He suggests that a single pair of particles can undergo various kinds of interactions.Thus, this nteraction can proceed either by EM or by strong interaction. Post ReplyCreate New Thread