## Re: How real are the "Virtual" partticles?

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Igor Khavkine wrote...\n&gt;Masslessness, in the standard meaning of the word, is equivalent to\nmotion\n&gt;with the speed of light (and only the speed of light). Photons travel\nonly\n&gt;at the speed of light, thus they are massless. All fermions that we\nknow\n&gt;(electrons, protons, neutrinos) seem to have mass (i.e. cannot travel\nat\n&gt;the speed of light).\n\n&gt;Does this answer your question?\n\nYes. It also gives me some insight into how ridiculously easily\nanswered all my questions are... they cause me so much confusion and\nyet the answers are so obvious.\n\nI have another one though, forgive me if it\'s also that elementary...\nI read that the leptons correspond one for one with the quarks. Does\nthis mean each has the same mass and charge, and spin as a quark, and\nif so, why is it that quark masses were revealed much more easily than\nlepton masses?\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote...
>Masslessness, in the standard meaning of the word, is equivalent to

motion
>with the speed of light (and only the speed of light). Photons travel

only
>at the speed of light, thus they are massless. All fermions that we

know
>(electrons, protons, neutrinos) seem to have mass (i.e. cannot travel

at
>the speed of light).

Yes. It also gives me some insight into how ridiculously easily
answered all my questions are... they cause me so much confusion and
yet the answers are so obvious.

I have another one though, forgive me if it's also that elementary...
I read that the leptons correspond one for one with the quarks. Does
this mean each has the same mass and charge, and spin as a quark, and
if so, why is it that quark masses were revealed much more easily than
lepton masses?



"mathman" wrote in message news:mathman.1labsu@physicsforums.com... | There are at least 2 classes of virtual particles. First there are | particles associated with the vacuum of space - popping in and out of | existence all of the time. A major observation of their presence is by | the Casimir effect. Look it up in Google and get a description of what $| it$ is all about. Second there are particles associated with particle | interactions. A simple example is the repulsive force between | electrons, described by photons going between them. Aren't your "2 classes" just virtual fermions and virtual bosons? FrediFizzx



Franz Heymann wrote: > "Eugene Stefanovich" wrote in message > news:42164384.3090301@synopsys.com... > > [snip] > > >>There is a more consistent way to look at interactions: QFT can be >>reformulated in terms of real particles and instantaneous potentials >>acting between them. >>All experimental predictions (e.g., the S-matrix) remain the same, >>and quite a lot of "invisible" stuff gets removed from the theory: >>No retardation, no virtual particles, no fields. > > > I'll go for the virtual particles rather than some mysterious action > at a distance. > There is a way to distinguish these two approaches in experiment. Take two charged macroscopic particles (e.g., two specks of dust) in vacuum. Arrange a slow collision of these particles and measure their trajectories with a good time resolution. We are interested, in particular, in the dependence of particle momenta on time. At the same time measure momenta of all real photons emitted by this couple of charges (there could be "soft" photons whose detection is tricky). If there are virtual particles, we should see some imbalance of the total momentum (specks of dust + photons): a part of the total momentum belongs to invisible virtual particles which carry the interaction between the two charges. If there are no virtual particles (interaction propagates instantaneously) we should see that the total momentum of real particles is conserved at all times, because there are no "virtual" degrees of freedom which can carry the extra momentum. I am not sure if current experimental resolution is good enough to figure out which of the two approaches is better. Eugene Stefanovich.



On Thu, 03 Mar 2005 22:25:43 $+0000,$ Sci~Girl wrote: > Igor Khavkine wrote... >>Does this answer your question? > > Yes. It also gives me some insight into how ridiculously easily answered > all my questions are... they cause me so much confusion and yet the > answers are so obvious. All questions are non-trivial until answered and obvious afterward. > I have another one though, forgive me if it's also that elementary... I > read that the leptons correspond one for one with the quarks. Does this > mean each has the same mass and charge, and spin as a quark, and if so, > why is it that quark masses were revealed much more easily than lepton > masses? Eh, that is not true. Elementary particles are grouped into generations. Each generation has two leptons and two quarks. Perhaps you can call this grouping "correspondence". For example, the first generation has the electron, the electron neutrino, the up quark and the down quark. Each of these particles has different charges for the electric, weak, and strong forces. Each fundamental fermion that we know of (lepton or quark) carries spin 1/2. Lepton masses are quite well known (for instance, the electron mass has been known for a long time) while quark masses are harder to determine because they can only be found bound inside baryons (composite particles such as protons, neutrons, and others). But quark masses have also been determined. Hope you can extract answers to your question from the above. Igor



Sci~Girl wrote: > I have another one though, forgive me if it's also that elementary... > I read that the leptons correspond one for one with the quarks. Does > this mean each has the same mass and charge, and spin as a quark, and > if so, why is it that quark masses were revealed much more easily than > lepton masses? Igor Khavkine gave a nice, but I'd also like to say a few words about why physicists say there is a correspondence - the theory is only renormalizable if the number of lepton pairs equals the number quark pairs. More speccifically, we needto have the sum of the charges in a lepton pair plus 3 times the sum of the charges in the corresponding quark pair equals zero. For example, onsider the lepton pair consisting of the the electon (charge -1) and the electron neutrino (charge 0), and quark pair consisting of the up quark (charge $+2/3)$ and the down quark (charge $-1/3)$. $$-1 ++ 3(2/3 - 1/3) =$$ Regards, George



To Igor Khavkine in response to his response to my question: No wait. Back to the question about photons and masslessness (I can't believe I didn't catch this earlier, too, I've been really out of it lately, screwing up in virtually $-get$ the joke?- everything). Mass INCREASES as something approaches the speed of light, when viewed from another frame of reference, until it eventually becomes infinite. Infinity does not equal zero (I know this will start some sort of philosophical argument, but for now and for these purposes, I won't dwell on it). Also, how can something without mass move, and accelerate or slow down depending on what it passes through? Only things with mass can react when force is applied, and some force is needed to get the photons moving in the first place. I still think the photon must have a mass, even if it is small to an incredibly extreme extent. I am willing to change my opinion, though, if necessary, when there's a really good, really definite reason why it cannot have a mass.



And yes, that does answer my question regarding leptons and quarks... baryons are made up of three quarks, correct? I'm memorizing the "Elementary Particle Physics Glossary" I found on a website somewhere, through a link from www.howstuffworks.com, which is my absolute all-time favorite website.



"mathman" wrote in message news:mathman.1labsu@physicsforums.com... > There are at least 2 classes of virtual particles. First there are > particles associated with the vacuum of space - popping in and out of > existence all of the time. A major observation of their presence is by > the Casimir effect. Look it up in Google and get a description of what > it is all about. Second there are particles associated with particle > interactions. A simple example is the repulsive force between > electrons, described by photons going between them. I know of no difference betwen these two categories of virtual particles. To the best of my knowledge they are described by the same theoretical approach. Those which correspond to vacuum polarisation are also illustrated by Feynman diagrams. -- Franz "A first-rate laboratory is one in which mediocre scientists can produce outstanding work" P.M.S. Blackett



"Eugene Stefanovich" wrote in message news:4227CD38.80109@synopsys.com... > Franz Heymann wrote: > > "Eugene Stefanovich" wrote in message > > news:42164384.3090301@synopsys.com... > > > > [snip] > > > > > >>There is a more consistent way to look at interactions: QFT can be > >>reformulated in terms of real particles and instantaneous potentials > >>acting between them. > >>All experimental predictions (e.g., the S-matrix) remain the same, > >>and quite a lot of "invisible" stuff gets removed from the theory: > >>No retardation, no virtual particles, no fields. > > > > > > I'll go for the virtual particles rather than some mysterious action > > at a distance. > > > > There is a way to distinguish these two approaches in experiment. > Take two charged macroscopic particles (e.g., two specks of dust) > in vacuum. Arrange a slow collision of these particles and measure > their trajectories with a good time resolution. We are interested, > in particular, in the dependence of particle momenta on time. > At the same time measure momenta of all real photons emitted by this > couple of charges (there could be "soft" photons whose detection > is tricky). If there are virtual particles, we should see some imbalance > of the total momentum (specks of dust + photons): a part of the total > momentum belongs to invisible virtual particles which carry the > interaction between the two charges. If there are no virtual particles > (interaction propagates instantaneously) we should see that the > total momentum of real particles is conserved at all times, because > there are no "virtual" degrees of freedom which can carry the extra > momentum. > > I am not sure if current experimental resolution is good enough > to figure out which of the two approaches is better. The scattering would, of course, be experimentally totally indistinguishable from a classical event. -- Franz "A first-rate laboratory is one in which mediocre scientists can produce outstanding work" P.M.S. Blackett



"Arnold Neumaier" wrote in message news:4225B962.6070905@univie.ac.at... > Franz Heymann wrote: > > "Arnold Neumaier" wrote in message > > news:421F2395.9070900@univie.ac.at... > > > >>PD wrote: > >> > >> > >>>If I see a peak in the invariant mass of a combination of two electrons > >>>seen in a detector, then I take it that I've seen evidence of a > >>>particle decaying into those two electrons, especially if other > >>>parameters of the final state points to a fixed set of quantum numbers > >>>(such as $spin=0 or spin=1)$. If that peak has some width to it, that's > >>>evidence of the finite lifetime of the decaying particle, and indeed > >>>any decay that is "off-peak" in this invariant mass can be said to > >>>point to the decay of a virtual particle. Neutral pions, $J/\psi's,$ Z > >>>bosons, all fit this description. > >>> > >>>This doesn't seem to me to be relegated to a "mathematical trick". > >>>Virtual particles are as "real" as real particles -- the boundary is a > >>>soft, fuzzy one. > >> > >>You mix up unstable particles and virtual particles. These are distinct > >>concepts. > >> > >>What you see as a peak in a spectrum is a resonance, the signature of an > >>unstable particle. > > > > And if you turn the Feynman diagram which describes the production of > > this unstable particle through 90 deg, why does this real particle > > magically becomes a virtual particle > > It doesn't! > > Unstable particles are modelled exactly like stable particles, > namely as external lines in a Feynman diagram. > Virtual particles in Feynman diagrams are exactly those > which are not given by external lines. > > Hence what is real and what is virtual is not affected by a > diagram rotation - this only affects what is input oand what is output. I am not happy with that. Let me illustrate my unhappiness with a specific example. Take the resonant elastic scattering case: $$\pi + p[/itex] --> Z --> $\pi + p$$ I hope we both agree that the Z is produced as a real but unstable particle in the s-channel. If we turn this diagram on its side, we get p + pbar --> [itex]\pi + \pi$ There is now a Z exchanged in the t-channel. That object will not be on the mass shell of a real Z, since its 4-momentum will depend on the four-momentum exchange in the interaction. -- Franz "A first-rate laboratory is one in which mediocre scientists can produce outstanding work" P.M.S. Blackett



Franz Heymann wrote: > "Arnold Neumaier" wrote in message > news:4225B962.6070905@univie.ac.at... > >>Franz Heymann wrote: >> >>>"Arnold Neumaier" wrote in message >>>news:421F2395.9070900@univie.ac.at... >>> >>> >>>>PD wrote: >>>> >>>> >>>> >>>>>If I see a peak in the invariant mass of a combination of two > > electrons > >>>>>seen in a detector, then I take it that I've seen evidence of a >>>>>particle decaying into those two electrons, especially if other >>>>>parameters of the final state points to a fixed set of quantum > > numbers > >>>>>(such as $spin=0 or spin=1)$. If that peak has some width to it, > > that's > >>>>>evidence of the finite lifetime of the decaying particle, and > > indeed > >>>>>any decay that is "off-peak" in this invariant mass can be said > > to > >>>>>point to the decay of a virtual particle. Neutral pions, $J/\psi's,$ > > Z > >>>>>bosons, all fit this description. >>>>> >>>>>This doesn't seem to me to be relegated to a "mathematical > > trick". > >>>>>Virtual particles are as "real" as real particles -- the boundary > > is a > >>>>>soft, fuzzy one. >>>> >>>>You mix up unstable particles and virtual particles. These are > > distinct > >>>>concepts. >>>> >>>>What you see as a peak in a spectrum is a resonance, the signature > > of an > >>>>unstable particle. >>> >>>And if you turn the Feynman diagram which describes the production > > of > >>>this unstable particle through 90 deg, why does this real particle >>>magically becomes a virtual particle >> >>It doesn't! >> >>Unstable particles are modelled exactly like stable particles, >>namely as external lines in a Feynman diagram. >>Virtual particles in Feynman diagrams are exactly those >>which are not given by external lines. >> >>Hence what is real and what is virtual is not affected by a >>diagram rotation - this only affects what is input oand what is > > output. > > I am not happy with that. > Let me illustrate my unhappiness with a specific example. > Take the resonant elastic scattering case: > > $\pi + p$ --> Z --> $\pi + p$ > > I hope we both agree that the Z is produced as a real but unstable > particle in the s-channel. No. Z is a virtual particle since it is an intermediate line. The line appears only in the standard perturbative formalism for the computation of the $in/out$ cross section not involving Z, and would not be there if that cross section was computed by another method. That's why it is virtual only. One can also see it by computing the momentum balance and check that the momentum does not lie on the Z's (complex) mass shell, as would be needed for a real Z. Arnold Neumaier



On Sun, 06 Mar 2005 07:50:55 $+0000,$ Sci~Girl wrote: > To Igor Khavkine in response to his response to my question: > > No wait. Back to the question about photons and masslessness (I can't > believe I didn't catch this earlier, too, I've been really out of it > lately, screwing up in virtually $-get$ the joke?- everything). Mass > INCREASES as something approaches the speed of light, when viewed from > another frame of reference, until it eventually becomes infinite. Infinity > does not equal zero (I know this will start some sort of philosophical > argument, but for now and for these purposes, I won't dwell on it). Your confusion stems from the common misuse of the term mass in elementary treatments of relativity. A particle's motion can be described by two quantities, it's energy E and its momentum p. Neither of them is invariant when viewed from different reference frames. However, the combination $$m^2 = E^2 - p^2$$ is invariant. The quantity m obtained from the above equation is called the mass and is the same for all observers. If a particle is moving very fast in your reference frame, you perceive a large momentum p. From the above formula, you can see that the energy will also have to be large $E = \sqrt(m^2 + p^2)$. (Eh, I'm using units c=1 for simplicity.) On the other hand, if a particle is stationary in your reference frame, it has $p =$ and $E = m.$ As you can see, in this case the particle's energy can be identified with its mass. What some elementary treatments of relativity do next is to call E (divided by $c^2$ in usual units) the relativistic mass (in all reference frames) and denote it by something like m'. However, most physicists consider the introduction of m' redundant and just call it energy. The term mass is reserved for the invariant quantity I defined earlier. > Also, how can something without mass move, and accelerate or slow down > depending on what it passes through? Only things with mass can react when > force is applied, and some force is needed to get the photons moving in > the first place. I still think the photon must have a mass, even if it is > small to an incredibly extreme extent. I am willing to change my opinion, > though, if necessary, when there's a really good, really definite reason > why it cannot have a mass. Short answer, it can't. That's why photons only have one speed, the speed of light. No force is needed to get photons moving just as no force is needed to get massive particles moving in the first place. All matter "moves". However, the "motion" can be purely directed along the time axis into the future. Force, or impulse, is necessary to *change* the direction of motion. This applies to photons as well as massive particles. However, it is hard to apply a force to a photon because of the way it interacts with other matter. If you think about light slowing down when it goes through a medium, you must consider what happens at the microscopic scale. There, you'll see that photons get absorbed and re-emitted by atoms with a certain delay between the two events. In between emission and absorption, photons still move with the speed of light, but the delay and all the bouncing around create an apparent retardation in the speed of light. What I gave you above are reasons to think that a massless photon is not an inconsistent thing to have. There are various high brow theoretical reasons why a photon should have exactly zero mass, but they are no substitute for experimental checks. However, experiment provides no evidence that would lead you to doubt the theory. Upper bounds on the photon mass are extremely tiny. The Particle Data Group cites something like $10^(-16)$ eV, which is about $10^20$ times smaller than the mass of the electron. Hope this helps. Igor



"Arnold Neumaier" wrote in message news:422AC49F.3030502@univie.ac.at... > Franz Heymann wrote: > > "Arnold Neumaier" wrote in message > > news:4225B962.6070905@univie.ac.at... > > > >>Franz Heymann wrote: > >> > >>>"Arnold Neumaier" wrote in message > >>>news:421F2395.9070900@univie.ac.at... > >>> > >>> > >>>>PD wrote: > >>>> > >>>> > >>>> > >>>>>If I see a peak in the invariant mass of a combination of two > > > > electrons > > > >>>>>seen in a detector, then I take it that I've seen evidence of a > >>>>>particle decaying into those two electrons, especially if other > >>>>>parameters of the final state points to a fixed set of quantum > > > > numbers > > > >>>>>(such as $spin=0 or spin=1)$. If that peak has some width to it, > > > > that's > > > >>>>>evidence of the finite lifetime of the decaying particle, and > > > > indeed > > > >>>>>any decay that is "off-peak" in this invariant mass can be said > > > > to > > > >>>>>point to the decay of a virtual particle. Neutral pions, $J/\psi's,$ > > > > Z > > > >>>>>bosons, all fit this description. > >>>>> > >>>>>This doesn't seem to me to be relegated to a "mathematical > > > > trick". > > > >>>>>Virtual particles are as "real" as real particles -- the boundary > > > > is a > > > >>>>>soft, fuzzy one. > >>>> > >>>>You mix up unstable particles and virtual particles. These are > > > > distinct > > > >>>>concepts. > >>>> > >>>>What you see as a peak in a spectrum is a resonance, the signature > > > > of an > > > >>>>unstable particle. > >>> > >>>And if you turn the Feynman diagram which describes the production > > > > of > > > >>>this unstable particle through 90 deg, why does this real particle > >>>magically becomes a virtual particle > >> > >>It doesn't! > >> > >>Unstable particles are modelled exactly like stable particles, > >>namely as external lines in a Feynman diagram. > >>Virtual particles in Feynman diagrams are exactly those > >>which are not given by external lines. > >> > >>Hence what is real and what is virtual is not affected by a > >>diagram rotation - this only affects what is input oand what is > > > > output. > > > > I am not happy with that. > > Let me illustrate my unhappiness with a specific example. > > Take the resonant elastic scattering case: > > > > $\pi + p$ --> Z --> $\pi + p$ > > > > I hope we both agree that the Z is produced as a real but unstable > > particle in the s-channel. > > No. Z is a virtual particle since it is an intermediate line. You are doing nothing to alleviate my unhappiness. If the Z were a virtual prticle, it would not be on its mass shell and a mass measurement would not be possible by studying the reaction. My undertanding is that the line corresponding to the Z is the trajectory of a real Z travelling from its birth to its decay. It just happens that its life time is exceedingly short. My understanding is that the s-channel Feynman diagram which produces the excited Z is simply a conflation of two separate Feynman diagrams, one in which $p+\pi$ interact to produce the Z, followed by one in which the Z decays. (It occurs to me to warn readers that the Z of which we speak is not the Z0 boson, but a nucleon excited state.) > The line appears only in the standard perturbative formalism for the > computation of the $in/out$ cross section not involving Z, and would not > be there if that cross section was computed by another method. > That's why it is virtual only. But it is observed, and its (complex) mass is actually measured. > One can also see it by computing > the momentum balance and check that the momentum does not lie > on the Z's (complex) mass shell, as would be needed for a real Z. Oh? Sorry, but the Z is no less real than the pi0. If the pi0 can decay into two photons, then it can also be created bytwo photons. Feel free , therefore to draw a Feynman diagram showing a pi0 produced in the s-channel by an interaction of two photons, and decaying. They both happen to have imaginary parts to their masses. -- Franz "A first-rate laboratory is one in which mediocre scientists can produce outstanding work" P.M.S. Blackett



Franz Heymann wrote: > "Arnold Neumaier" wrote in message > news:422AC49F.3030502@univie.ac.at... > >>Franz Heymann wrote: >> >>>>Unstable particles are modelled exactly like stable particles, >>>>namely as external lines in a Feynman diagram. >>>>Virtual particles in Feynman diagrams are exactly those >>>>which are not given by external lines. >>>> >>>>Hence what is real and what is virtual is not affected by a >>>>diagram rotation - this only affects what is input oand what is >>> >>>output. >>> >>>I am not happy with that. >>>Let me illustrate my unhappiness with a specific example. >>>Take the resonant elastic scattering case: >>> $>>>\pi + p$ --> Z --> $\pi + p$ >>> >>>I hope we both agree that the Z is produced as a real but unstable >>>particle in the s-channel. >> >>No. Z is a virtual particle since it is an intermediate line. > > You are doing nothing to alleviate my unhappiness. > If the Z were a virtual prticle, it would not be on its mass shell and > a mass measurement would not be possible by studying the reaction. It is not on its (complex) mass shell when it is in an internal line as in your diagram. When measured, one has instead a diagram where the Z is detected, and hence on an external line of Feynman diagrams associated to the coresponding cross section! In ordinary scattering experiemnts the existence of the Z shows in resonances of the cross section. This is evidence for the presence of the Z in the action, and can be used to calculate properties of Z, but it is Not a measurement of Z as a particle. > My undertanding is that the line corresponding to the Z is the > trajectory of a real Z travelling from its birth to its decay. This interpretation of Feynman diagrams is popular but completely unfounded, and gives misleading conclusions if carried beyond the most superficial view. > But it is observed, and its (complex) mass is actually measured. It is _not_ observed in an experiment where the Feynman diagram you drew is part of the calculation. $It _can_$ be observed, but only in a process analyzed with Feynman diagrams where the Z is an external line. >>One can also see it by computing >>the momentum balance and check that the momentum does not lie >>on the Z's (complex) mass shell, as would be needed for a real Z. > > Oh? Sorry, but the Z is no less real than the pi0. > If the pi0 can decay into two photons, then it can also be created > bytwo photons. For the pi0 the situation is the same. If measured, the cross section for pi0 production or decay comes from an analysis with external pi0 lines, while in other processes it is a virtual particle without objective meaning. Arnold Neumaier



"Arnold Neumaier" wrote in message news:422EB74C.5020806@univie.ac.at... > Franz Heymann wrote: > > "Arnold Neumaier" wrote in message > > news:422AC49F.3030502@univie.ac.at... > > > >>Franz Heymann wrote: > >> > >>>>Unstable particles are modelled exactly like stable particles, > >>>>namely as external lines in a Feynman diagram. > >>>>Virtual particles in Feynman diagrams are exactly those > >>>>which are not given by external lines. > >>>> > >>>>Hence what is real and what is virtual is not affected by a > >>>>diagram rotation - this only affects what is input oand what is > >>> > >>>output. > >>> > >>>I am not happy with that. > >>>Let me illustrate my unhappiness with a specific example. > >>>Take the resonant elastic scattering case: > >>> > $>>>\pi + p$ --> Z --> $\pi + p$ > >>> > >>>I hope we both agree that the Z is produced as a real but unstable > >>>particle in the s-channel. > >> > >>No. Z is a virtual particle since it is an intermediate line. > > > > You are doing nothing to alleviate my unhappiness. > > If the Z were a virtual prticle, it would not be on its mass shell and > > a mass measurement would not be possible by studying the reaction. > > It is not on its (complex) mass shell when it is in an internal line as > in your diagram. > > When measured, one has instead a diagram where the Z is detected, > and hence on an external line of Feynman diagrams associated to > the coresponding cross section! > > In ordinary scattering experiemnts the existence of the Z shows in > resonances of the cross section. This is evidence for the presence > of the Z in the action, and can be used to calculate properties of Z, > but it is Not a measurement of Z as a particle. > > > > My undertanding is that the line corresponding to the Z is the > > trajectory of a real Z travelling from its birth to its decay. > > This interpretation of Feynman diagrams is popular but completely > unfounded, and gives misleading conclusions if carried beyond the > most superficial view. > > > > But it is observed, and its (complex) mass is actually measured. > > It is _not_ observed in an experiment where the Feynman diagram > you drew is part of the calculation. > > $It _can_$ be observed, but only in a process analyzed with > Feynman diagrams where the Z is an external line. I maintain that it is an external line which has its far end terminated in the decay of the resonance. In atomic physics, a photon may be absorbed by an atom in its ground state. The excited state very quickly decays back to the ground state by the emission of a photon. Are you now saying that the excited state is not a real, but a virtual state? > >>One can also see it by computing > >>the momentum balance and check that the momentum does not lie > >>on the Z's (complex) mass shell, as would be needed for a real Z. > > > > Oh? Sorry, but the Z is no less real than the pi0. > > If the pi0 can decay into two photons, then it can also be created > > bytwo photons. > > For the pi0 the situation is the same. If measured, the cross > section for pi0 production or decay comes from an analysis with > external pi0 lines, while in other processes it is a virtual > particle without objective meaning. I fear me I will never understand you, and you will never understand me. I suggest we discontinue this conversation after your next reply to me, if indeed you wish to make one. -- Franz "A first-rate laboratory is one in which mediocre scientists can produce outstanding work" P.M.S. Blackett



Franz Heymann wrote: > "Arnold Neumaier" wrote in message > news:422EB74C.5020806@univie.ac.at... > >>Franz Heymann wrote: >> >>>"Arnold Neumaier" wrote in message >>>news:422AC49F.3030502@univie.ac.at... >>> >>>>Franz Heymann wrote >>>> >>>>>>Unstable particles are modelled exactly like stable particles, >>>>>>namely as external lines in a Feynman diagram. >>>>>>Virtual particles in Feynman diagrams are exactly those >>>>>>which are not given by external lines. >>>>>> >>>>>>Hence what is real and what is virtual is not affected by a >>>>>>diagram rotation - this only affects what is input oand what is >>>>> >>>>>output. >>>>> >>>>>I am not happy with that. >>>>>Let me illustrate my unhappiness with a specific example. >>>>>Take the resonant elastic scattering case: >>>>> $>>>>>\pi + p$ --> Z --> $\pi + p$ >>>>> >>>>>I hope we both agree that the Z is produced as a real but unstable >>>>>particle in the s-channel. >>>> >>>>No. Z is a virtual particle since it is an intermediate line. >>> >>>You are doing nothing to alleviate my unhappiness. >>>If the Z were a virtual prticle, it would not be on its mass shell and >>>a mass measurement would not be possible by studying the reaction. >>It is not on its (complex) mass shell when it is in an internal line as >>in your diagram. >> >>When measured, one has instead a diagram where the Z is detected, >>and hence on an external line of Feynman diagrams associated to >>the coresponding cross section! >> >>In ordinary scattering experiments the existence of the Z shows in >>resonances of the cross section. This is evidence for the presence >>of the Z in the action, and can be used to calculate properties of Z, >> >>>But it is observed, and its (complex) mass is actually measured. >> >>It is _not_ observed in an experiment where the Feynman diagram >>you drew is part of the calculation. >> >>It _can_ be observed, but only in a process analyzed with >>Feynman diagrams where the Z is an external line. > > I maintain that it is an external line which has its far end > terminated in the decay of the resonance. Fine; then we agree. For an external line remains external under rotating the diagram. Initially you argued that rotation changes virtual particles to real ones; it was only that which I objected to. > In atomic physics, a photon may be absorbed by an atom in its ground > state. The excited state very quickly decays back to the ground state > by the emission of a photon. > > Are you now saying that the excited state is not a real, but a virtual > state? No. There is a big difference between the two. real = external line = on-shell = observable virtual = internal line = off-shell = ficiticous excited states are bound states and usually appear _only_ as external lines. Arnold Neumaier