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Re: How real are the "Virtual" partticles? |
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| Mar3-05, 04:25 PM | #18 |
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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>Masslessness, in the standard meaning of the word, is equivalent to\nmotion\n>with the speed of light (and only the speed of light). Photons travel\nonly\n>at the speed of light, thus they are massless. All fermions that we\nknow\n>(electrons, protons, neutrinos) seem to have mass (i.e. cannot travel\nat\n>the speed of light).\n\n>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"> 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). >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. 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? |
| Mar4-05, 09:13 AM | #19 |
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<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>"mathman" <mathnucl@optonline.net> wrote in message\nnews:mathman.1labsu@physicsforums.com...\n| There are at least 2 classes of virtual particles. First there are\n| particles associated with the vacuum of space - popping in and out of\n| existence all of the time. A major observation of their presence is\nby\n| the Casimir effect. Look it up in Google and get a description of\nwhat\n| it is all about. Second there are particles associated with particle\n| interactions. A simple example is the repulsive force between\n| electrons, described by photons going between them.\n\nAren\'t your "2 classes" just virtual fermions and virtual bosons?\n\nFrediFizzx\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"mathman" <mathnucl@optonline.net> 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 [itex]| it[/itex] 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 |
| Mar4-05, 09:13 AM | #20 |
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<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>Franz Heymann wrote:\n> "Eugene Stefanovich" <eugenev@synopsys.com> wrote in message\n> news:42164384.3090301@synopsys.com...\n>\n> [snip]\n>\n>\n>>There is a more consistent way to look at interactions: QFT can be\n>>reformulated in terms of real particles and instantaneous potentials\n>>acting between them.\n>>All experimental predictions (e.g., the S-matrix) remain the same,\n>>and quite a lot of "invisible" stuff gets removed from the theory:\n>>No retardation, no virtual particles, no fields.\n>\n>\n> I\'ll go for the virtual particles rather than some mysterious action\n> at a distance.\n>\n\nThere is a way to distinguish these two approaches in experiment.\nTake two charged macroscopic particles (e.g., two specks of dust)\nin vacuum. Arrange a slow collision of these particles and measure\ntheir trajectories with a good time resolution. We are interested,\nin particular, in the dependence of particle momenta on time.\nAt the same time measure momenta of all real photons emitted by this\ncouple of charges (there could be "soft" photons whose detection\nis tricky). If there are virtual particles, we should see some imbalance\nof the total momentum (specks of dust + photons): a part of the total\nmomentum belongs to invisible virtual particles which carry the\ninteraction between the two charges. If there are no virtual particles\n(interaction propagates instantaneously) we should see that the\ntotal momentum of real particles is conserved at all times, because\nthere are no "virtual" degrees of freedom which can carry the extra\nmomentum.\n\nI am not sure if current experimental resolution is good enough\nto figure out which of the two approaches is better.\n\nEugene Stefanovich.\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Franz Heymann wrote:
> "Eugene Stefanovich" <eugenev@synopsys.com> 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. |
| Mar4-05, 09:14 AM | #21 |
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<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>On Thu, 03 Mar 2005 22:25:43 +0000, Sci~Girl wrote:\n> Igor Khavkine wrote...\n\n>>Does this answer your question?\n>\n> Yes. It also gives me some insight into how ridiculously easily answered\n> all my questions are... they cause me so much confusion and yet the\n> answers are so obvious.\n\nAll questions are non-trivial until answered and obvious afterward.\n\n> I have another one though, forgive me if it\'s also that elementary... I\n> read that the leptons correspond one for one with the quarks. Does this\n> mean each has the same mass and charge, and spin as a quark, and if so,\n> why is it that quark masses were revealed much more easily than lepton\n> masses?\n\nEh, that is not true. Elementary particles are grouped into generations.\nEach generation has two leptons and two quarks. Perhaps you can call this\ngrouping "correspondence". For example, the first generation has the\nelectron, the electron neutrino, the up quark and the down quark. Each of\nthese particles has different charges for the electric, weak, and strong\nforces. Each fundamental fermion that we know of (lepton or quark) carries\nspin 1/2.\n\nLepton masses are quite well known (for instance, the electron mass has\nbeen known for a long time) while quark masses are harder to determine\nbecause they can only be found bound inside baryons (composite particles\nsuch as protons, neutrons, and others). But quark masses have also been\ndetermined.\n\nHope you can extract answers to your question from the above.\n\nIgor\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Thu, 03 Mar 2005 22:25:43 [itex]+0000,[/itex] 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 |
| Mar6-05, 01:50 AM | #22 |
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<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>Sci~Girl wrote:\n> I have another one though, forgive me if it\'s also that elementary...\n> I read that the leptons correspond one for one with the quarks. Does\n> this mean each has the same mass and charge, and spin as a quark, and\n> if so, why is it that quark masses were revealed much more easily than\n> lepton masses?\n\nIgor Khavkine gave a nice, but I\'d also like to say a few words about\nwhy physicists say there is a correspondence - the theory is only\nrenormalizable if the number of lepton pairs equals the number quark\npairs. More speccifically, we needto have the sum of the charges in a\nlepton pair plus 3 times the sum of the charges in the corresponding\nquark pair equals zero.\n\nFor example, onsider the lepton pair consisting of the the electon\n(charge -1) and the electron neutrino (charge 0), and quark pair\nconsisting of the up quark (charge +2/3) and the down quark\n(charge -1/3).\n\n-1 + 0 + 3(2/3 - 1/3) = 0\n\nRegards,\nGeorge\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>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 [itex]+2/3)[/itex] and the down quark (charge [itex]-1/3)[/itex]. [tex]-1 ++ 3(2/3 - 1/3) =[/tex] Regards, George |
| Mar6-05, 01:50 AM | #23 |
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<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>To Igor Khavkine in response to his response to my question:\n\nNo wait. Back to the question about photons and masslessness (I can\'t\nbelieve I didn\'t catch this earlier, too, I\'ve been really out of it\nlately, screwing up in virtually -get the joke?- everything). Mass\nINCREASES as something approaches the speed of light, when viewed from\nanother frame of reference, until it eventually becomes infinite.\nInfinity does not equal zero (I know this will start some sort of\nphilosophical argument, but for now and for these purposes, I won\'t\ndwell on it).\n\nAlso, how can something without mass move, and accelerate or slow down\ndepending on what it passes through? Only things with mass can react\nwhen force is applied, and some force is needed to get the photons\nmoving in the first place. I still think the photon must have a mass,\neven if it is small to an incredibly extreme extent. I am willing to\nchange my opinion, though, if necessary, when there\'s a really good,\nreally definite reason why it cannot have a mass.\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>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 [itex]-get[/itex] 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. |
| Mar6-05, 01:51 AM | #24 |
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<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>And yes, that does answer my question regarding leptons and quarks...\nbaryons are made up of three quarks, correct? I\'m memorizing the\n"Elementary Particle Physics Glossary" I found on a website somewhere,\nthrough a link from www.howstuffworks.com, which is my absolute\nall-time favorite website.\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>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. |
| Mar6-05, 01:53 AM | #25 |
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<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>"mathman" <mathnucl@optonline.net> wrote in message\nnews:mathman.1labsu@physicsforums.com...\n> There are at least 2 classes of virtual particles. First there are\n> particles associated with the vacuum of space - popping in and out\nof\n> existence all of the time.\n\nA major observation of their presence is by\n> the Casimir effect. Look it up in Google and get a description of\nwhat\n> it is all about. Second there are particles associated with\nparticle\n> interactions. A simple example is the repulsive force between\n> electrons, described by photons going between them.\n\nI know of no difference betwen these two categories of virtual\nparticles.\nTo the best of my knowledge they are described by the same theoretical\napproach.\nThose which correspond to vacuum polarisation are also illustrated by\nFeynman diagrams.\n\n--\nFranz\n"A first-rate laboratory is one in which mediocre scientists can\nproduce outstanding work"\nP.M.S. Blackett\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"mathman" <mathnucl@optonline.net> 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 |
| Mar6-05, 01:54 AM | #26 |
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<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>"Eugene Stefanovich" <eugenev@synopsys.com> wrote in message\nnews:4227CD38.80109@synopsys.com...\n> Franz Heymann wrote:\n> > "Eugene Stefanovich" <eugenev@synopsys.com> wrote in message\n> > news:42164384.3090301@synopsys.com...\n> >\n> > [snip]\n> >\n> >\n> >>There is a more consistent way to look at interactions: QFT can be\n> >>reformulated in terms of real particles and instantaneous\npotentials\n> >>acting between them.\n> >>All experimental predictions (e.g., the S-matrix) remain the same,\n> >>and quite a lot of "invisible" stuff gets removed from the theory:\n> >>No retardation, no virtual particles, no fields.\n> >\n> >\n> > I\'ll go for the virtual particles rather than some mysterious\naction\n> > at a distance.\n> >\n>\n> There is a way to distinguish these two approaches in experiment.\n> Take two charged macroscopic particles (e.g., two specks of dust)\n> in vacuum. Arrange a slow collision of these particles and measure\n> their trajectories with a good time resolution. We are interested,\n> in particular, in the dependence of particle momenta on time.\n> At the same time measure momenta of all real photons emitted by this\n> couple of charges (there could be "soft" photons whose detection\n> is tricky). If there are virtual particles, we should see some\nimbalance\n> of the total momentum (specks of dust + photons): a part of the\ntotal\n> momentum belongs to invisible virtual particles which carry the\n> interaction between the two charges. If there are no virtual\nparticles\n> (interaction propagates instantaneously) we should see that the\n> total momentum of real particles is conserved at all times, because\n> there are no "virtual" degrees of freedom which can carry the extra\n> momentum.\n>\n> I am not sure if current experimental resolution is good enough\n> to figure out which of the two approaches is better.\n\nThe scattering would, of course, be experimentally totally\nindistinguishable from a classical event.\n\n--\nFranz\n"A first-rate laboratory is one in which mediocre scientists can\nproduce outstanding work"\nP.M.S. Blackett\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Eugene Stefanovich" <eugenev@synopsys.com> wrote in message
news:4227CD38.80109@synopsys.com... > Franz Heymann wrote: > > "Eugene Stefanovich" <eugenev@synopsys.com> 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 |
| Mar6-05, 01:54 AM | #27 |
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<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>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\nnews:4225B962.6070905@univie.ac.at...\n> Franz Heymann wrote:\n> > "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\n> > news:421F2395.9070900@univie.ac.at...\n> >\n> >>PD wrote:\n> >>\n> >>\n> >>>If I see a peak in the invariant mass of a combination of two\nelectrons\n> >>>seen in a detector, then I take it that I\'ve seen evidence of a\n> >>>particle decaying into those two electrons, especially if other\n> >>>parameters of the final state points to a fixed set of quantum\nnumbers\n> >>>(such as spin=0 or spin=1). If that peak has some width to it,\nthat\'s\n> >>>evidence of the finite lifetime of the decaying particle, and\nindeed\n> >>>any decay that is "off-peak" in this invariant mass can be said\nto\n> >>>point to the decay of a virtual particle. Neutral pions, J/psi\'s,\nZ\n> >>>bosons, all fit this description.\n> >>>\n> >>>This doesn\'t seem to me to be relegated to a "mathematical\ntrick".\n> >>>Virtual particles are as "real" as real particles -- the boundary\nis a\n> >>>soft, fuzzy one.\n> >>\n> >>You mix up unstable particles and virtual particles. These are\ndistinct\n> >>concepts.\n> >>\n> >>What you see as a peak in a spectrum is a resonance, the signature\nof an\n> >>unstable particle.\n> >\n> > And if you turn the Feynman diagram which describes the production\nof\n> > this unstable particle through 90 deg, why does this real particle\n> > magically becomes a virtual particle\n>\n> It doesn\'t!\n>\n> Unstable particles are modelled exactly like stable particles,\n> namely as external lines in a Feynman diagram.\n> Virtual particles in Feynman diagrams are exactly those\n> which are not given by external lines.\n>\n> Hence what is real and what is virtual is not affected by a\n> diagram rotation - this only affects what is input oand what is\noutput.\n\nI am not happy with that.\nLet me illustrate my unhappiness with a specific example.\nTake the resonant elastic scattering case:\n\npi + p --> Z --> pi + p\n\nI hope we both agree that the Z is produced as a real but unstable\nparticle in the s-channel.\nIf we turn this diagram on its side, we get\n\np + pbar --> pi + pi\n\nThere is now a Z exchanged in the t-channel. That object will not be\non the mass shell of a real Z, since its 4-momentum will depend on\nthe four-momentum exchange in the interaction.\n\n--\nFranz\n"A first-rate laboratory is one in which mediocre scientists can\nproduce outstanding work"\nP.M.S. Blackett\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message
news:4225B962.6070905@univie.ac.at... > Franz Heymann wrote: > > "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> 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 [itex]spin=0 or spin=1)[/itex]. 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, [itex]J/\psi's,[/itex] 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: [tex]\pi + p[/itex] --> Z --> [itex]\pi + p[/tex] 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[/itex] 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 |
| Mar7-05, 03:45 PM | #28 |
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<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>Franz Heymann wrote:\n> "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\n> news:4225B962.6070905@univie.ac.at...\n>\n>>Franz Heymann wrote:\n>>\n>>>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\n>>>news:421F2395.9070900@univie.ac.at...\n>>> ;\n>>>\n>>>>PD wrote:\n>>>>\n>>>>\n>>>>\n>>& gt;>>If I see a peak in the invariant mass of a combination of two\n>\n> electrons\n>\n>>>>>seen in a detector, then I take it that I\'ve seen evidence of a\n>>>>>particle decaying into those two electrons, especially if other\n>>>>>parameters of the final state points to a fixed set of quantum\n>\n> numbers\n>\n>>>>>(such as spin=0 or spin=1). If that peak has some width to it,\n>\n> that\'s\n>\n>>>>>evidence of the finite lifetime of the decaying particle, and\n>\n> indeed\n>\n>>>>>any decay that is "off-peak" in this invariant mass can be said\n>\n> to\n>\n>>>>>point to the decay of a virtual particle. Neutral pions, J/psi\'s,\n>\n> Z\n>\n>>>>>bosons, all fit this description.\n>>>>>\n>>>>>This doesn\'t seem to me to be relegated to a "mathematical\n>\n> trick".\n>\n>>>>>Virtual particles are as "real" as real particles -- the boundary\n>\n> is a\n>\n>>>>>soft, fuzzy one.\n>>>>\n>>>>You mix up unstable particles and virtual particles. These are\n>\n> distinct\n>\n>>>>concepts.\n>>>>\n>>&g t;>What you see as a peak in a spectrum is a resonance, the signature\n>\n> of an\n>\n>>>>unstable particle.\n>>>\n>>>And if you turn the Feynman diagram which describes the production\n>\n> of\n>\n>>>this unstable particle through 90 deg, why does this real particle\n>>>magically becomes a virtual particle\n>>\n>>It doesn\'t!\n>>\n>>Unstable particles are modelled exactly like stable particles,\n>>namely as external lines in a Feynman diagram.\n>>Virtual particles in Feynman diagrams are exactly those\n>>which are not given by external lines.\n>>\n>>Hence what is real and what is virtual is not affected by a\n>>diagram rotation - this only affects what is input oand what is\n>\n> output.\n>\n> I am not happy with that.\n> Let me illustrate my unhappiness with a specific example.\n> Take the resonant elastic scattering case:\n>\n> pi + p --> Z --> pi + p\n>\n> I hope we both agree that the Z is produced as a real but unstable\n> particle in the s-channel.\n\nNo. Z is a virtual particle since it is an intermediate line.\nThe line appears only in the standard perturbative formalism for the\ncomputation of the in/out cross section not involving Z, and would not\nbe there if that cross section was computed by another method.\nThat\'s why it is virtual only. One can also see it by computing\nthe momentum balance and check that the momentum does not lie\non the Z\'s (complex) mass shell, as would be needed for a real Z.\n\n\nArnold Neumaier\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Franz Heymann wrote:
> "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message > news:4225B962.6070905@univie.ac.at... > >>Franz Heymann wrote: >> >>>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> 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 [itex]spin=0 or spin=1)[/itex]. 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, [itex]J/\psi's,[/itex] > > 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: > > [itex]\pi + p[/itex] --> Z --> [itex]\pi + p[/itex] > > 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 [itex]in/out[/itex] 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 |
| Mar7-05, 03:47 PM | #29 |
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<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>On Sun, 06 Mar 2005 07:50:55 +0000, Sci~Girl wrote:\n\n> To Igor Khavkine in response to his response to my question:\n>\n> No wait. Back to the question about photons and masslessness (I can\'t\n> believe I didn\'t catch this earlier, too, I\'ve been really out of it\n> lately, screwing up in virtually -get the joke?- everything). Mass\n> INCREASES as something approaches the speed of light, when viewed from\n> another frame of reference, until it eventually becomes infinite. Infinity\n> does not equal zero (I know this will start some sort of philosophical\n> argument, but for now and for these purposes, I won\'t dwell on it).\n\nYour confusion stems from the common misuse of the term mass in elementary\ntreatments of relativity. A particle\'s motion can be described by two\nquantities, it\'s energy E and its momentum p. Neither of them is invariant\nwhen viewed from different reference frames. However, the combination\n\nm^2 = E^2 - p^2\n\nis invariant. The quantity m obtained from the above equation is called\nthe mass and is the same for all observers. If a particle is moving very\nfast in your reference frame, you perceive a large momentum p. From the\nabove formula, you can see that the energy will also have to be large\nE = sqrt(m^2 + p^2). (Eh, I\'m using units c=1 for simplicity.) On the\nother hand, if a particle is stationary in your reference frame, it has\np = 0 and E = m. As you can see, in this case the particle\'s energy can be\nidentified with its mass.\n\nWhat some elementary treatments of relativity do next is to call E\n(divided by c^2 in usual units) the relativistic mass (in all reference\nframes) and denote it by something like m\'. However, most physicists\nconsider the introduction of m\' redundant and just call it energy. The\nterm mass is reserved for the invariant quantity I defined earlier.\n\n> Also, how can something without mass move, and accelerate or slow down\n> depending on what it passes through? Only things with mass can react when\n> force is applied, and some force is needed to get the photons moving in\n> the first place. I still think the photon must have a mass, even if it is\n> small to an incredibly extreme extent. I am willing to change my opinion,\n> though, if necessary, when there\'s a really good, really definite reason\n> why it cannot have a mass.\n\nShort answer, it can\'t. That\'s why photons only have one speed, the speed\nof light. No force is needed to get photons moving just as no force is\nneeded to get massive particles moving in the first place. All matter\n"moves". However, the "motion" can be purely directed along the time axis\ninto the future. Force, or impulse, is necessary to *change* the direction\nof motion. This applies to photons as well as massive particles. However,\nit is hard to apply a force to a photon because of the way it interacts\nwith other matter.\n\nIf you think about light slowing down when it goes through a medium, you\nmust consider what happens at the microscopic scale. There, you\'ll see\nthat photons get absorbed and re-emitted by atoms with a certain delay\nbetween the two events. In between emission and absorption, photons still\nmove with the speed of light, but the delay and all the bouncing around\ncreate an apparent retardation in the speed of light.\n\nWhat I gave you above are reasons to think that a massless photon is not\nan inconsistent thing to have. There are various high brow theoretical\nreasons why a photon should have exactly zero mass, but they are no\nsubstitute for experimental checks. However, experiment provides no\nevidence that would lead you to doubt the theory. Upper bounds on the\nphoton mass are extremely tiny. The Particle Data Group cites something\nlike 10^(-16) eV, which is about 10^20 times smaller than the mass of the\nelectron.\n\nHope this helps.\n\nIgor\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Sun, 06 Mar 2005 07:50:55 [itex]+0000,[/itex] 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 [itex]-get[/itex] 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 [tex]m^2 = E^2 - p^2[/tex] 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 [itex]E = \sqrt(m^2 + p^2)[/itex]. (Eh, I'm using units c=1 for simplicity.) On the other hand, if a particle is stationary in your reference frame, it has [itex]p = [/itex] and [itex]E = m.[/itex] 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 [itex]c^2[/itex] 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 [itex]10^(-16)[/itex] eV, which is about [itex]10^20[/itex] times smaller than the mass of the electron. Hope this helps. Igor |
| Mar9-05, 02:03 AM | #30 |
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<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>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\nnews:422AC49F.3030502@univie.ac.at...\n> Franz Heymann wrote:\n> > "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\n> > news:4225B962.6070905@univie.ac.at...\n> >\n> >>Franz Heymann wrote:\n> >>\n> >>>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\n> >>>news:421F2395.9070900@univie.ac.at...\n> >>>\n> >>>\n> >>>>PD wrote:\n> >>>>\n> >>>>\n> >>>>\n> >>>>>If I see a peak in the invariant mass of a combination of two\n> >\n> > electrons\n> >\n> >>>>>seen in a detector, then I take it that I\'ve seen evidence of a\n> >>>>>particle decaying into those two electrons, especially if other\n> >>>>>parameters of the final state points to a fixed set of quantum\n> >\n> > numbers\n> >\n> >>>>>(such as spin=0 or spin=1). If that peak has some width to it,\n> >\n> > that\'s\n> >\n> >>>>>evidence of the finite lifetime of the decaying particle, and\n> >\n> > indeed\n> >\n> >>>>>any decay that is "off-peak" in this invariant mass can be said\n> >\n> > to\n> >\n> >>>>>point to the decay of a virtual particle. Neutral pions,\nJ/psi\'s,\n> >\n> > Z\n> >\n> >>>>>bosons, all fit this description.\n> >>>>>\n> >>>>>This doesn\'t seem to me to be relegated to a "mathematical\n> >\n> > trick".\n> >\n> >>>>>Virtual particles are as "real" as real particles -- the\nboundary\n> >\n> > is a\n> >\n> >>>>>soft, fuzzy one.\n> >>>>\n> >>>>You mix up unstable particles and virtual particles. These are\n> >\n> > distinct\n> >\n> >>>>concepts.\n> >>>>\n> >>>>What you see as a peak in a spectrum is a resonance, the\nsignature\n> >\n> > of an\n> >\n> >>>>unstable particle.\n> >>>\n> >>>And if you turn the Feynman diagram which describes the\nproduction\n> >\n> > of\n> >\n> >>>this unstable particle through 90 deg, why does this real\nparticle\n> >>>magically becomes a virtual particle\n> >>\n> >>It doesn\'t!\n> >>\n> >>Unstable particles are modelled exactly like stable particles,\n> >>namely as external lines in a Feynman diagram.\n> >>Virtual particles in Feynman diagrams are exactly those\n> >>which are not given by external lines.\n> >>\n> >>Hence what is real and what is virtual is not affected by a\n> >>diagram rotation - this only affects what is input oand what is\n> >\n> > output.\n> >\n> > I am not happy with that.\n> > Let me illustrate my unhappiness with a specific example.\n> > Take the resonant elastic scattering case:\n> >\n> > pi + p --> Z --> pi + p\n> >\n> > I hope we both agree that the Z is produced as a real but unstable\n> > particle in the s-channel.\n>\n> No. Z is a virtual particle since it is an intermediate line.\n\n\nYou are doing nothing to alleviate my unhappiness.\nIf the Z were a virtual prticle, it would not be on its mass shell and\na mass measurement would not be possible by studying the reaction.\n\nMy undertanding is that the line corresponding to the Z is the\ntrajectory of a real Z travelling from its birth to its decay.\nIt just happens that its life time is exceedingly short.\nMy understanding is that the s-channel Feynman diagram which produces\nthe excited Z is simply a conflation of two separate Feynman diagrams,\none in which p+pi interact to produce the Z, followed by one in which\nthe Z decays.\n\n(It occurs to me to warn readers that the Z of which we speak is not\nthe Z0 boson, but a nucleon excited state.)\n\n> The line appears only in the standard perturbative formalism for the\n> computation of the in/out cross section not involving Z, and would\nnot\n> be there if that cross section was computed by another method.\n> That\'s why it is virtual only.\n\nBut it is observed, and its (complex) mass is actually measured.\n\n> One can also see it by computing\n> the momentum balance and check that the momentum does not lie\n> on the Z\'s (complex) mass shell, as would be needed for a real Z.\n\nOh? Sorry, but the Z is no less real than the pi0.\nIf the pi0 can decay into two photons, then it can also be created\nbytwo photons.\nFeel free , therefore to draw a Feynman diagram showing a pi0 produced\nin the s-channel by an interaction of two photons, and decaying.\n\nThey both happen to have imaginary parts to their masses.\n\n--\nFranz\n"A first-rate laboratory is one in which mediocre scientists can\nproduce outstanding work"\nP.M.S. Blackett\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message
news:422AC49F.3030502@univie.ac.at... > Franz Heymann wrote: > > "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message > > news:4225B962.6070905@univie.ac.at... > > > >>Franz Heymann wrote: > >> > >>>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> 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 [itex]spin=0 or spin=1)[/itex]. 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, [itex]J/\psi's,[/itex] > > > > 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: > > > > [itex]\pi + p[/itex] --> Z --> [itex]\pi + p[/itex] > > > > 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 [itex]p+\pi[/itex] 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 [itex]in/out[/itex] 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 |
| Mar11-05, 02:28 AM | #31 |
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<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>Franz Heymann wrote:\n> "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\n> news:422AC49F.3030502@univie.ac.at...\n>\n>>Franz Heymann wrote:\n>>\n>>>>Unstable particles are modelled exactly like stable particles,\n>>>>namely as external lines in a Feynman diagram.\n>>>>Virtual particles in Feynman diagrams are exactly those\n>>>>which are not given by external lines.\n>>>>\n>>>>Hence what is real and what is virtual is not affected by a\n>>>>diagram rotation - this only affects what is input oand what is\n>>>\n>>>output.\n>>>\n>>>I am not happy with that.\n>>>Let me illustrate my unhappiness with a specific example.\n>>>Take the resonant elastic scattering case:\n>>>\n>>>pi + p --> Z --> pi + p\n>>>\n>>>I hope we both agree that the Z is produced as a real but unstable\n>>>particle in the s-channel.\n>>\n>>No. Z is a virtual particle since it is an intermediate line.\n>\n> You are doing nothing to alleviate my unhappiness.\n> If the Z were a virtual prticle, it would not be on its mass shell and\n> a mass measurement would not be possible by studying the reaction.\n\nIt is not on its (complex) mass shell when it is in an internal line as\nin your diagram.\n\nWhen measured, one has instead a diagram where the Z is detected,\nand hence on an external line of Feynman diagrams associated to\nthe coresponding cross section!\n\nIn ordinary scattering experiemnts the existence of the Z shows in\nresonances of the cross section. This is evidence for the presence\nof the Z in the action, and can be used to calculate properties of Z,\nbut it is Not a measurement of Z as a particle.\n\n\n> My undertanding is that the line corresponding to the Z is the\n> trajectory of a real Z travelling from its birth to its decay.\n\nThis interpretation of Feynman diagrams is popular but completely\nunfounded, and gives misleading conclusions if carried beyond the\nmost superficial view.\n\n\n> But it is observed, and its (complex) mass is actually measured.\n\nIt is _not_ observed in an experiment where the Feynman diagram\nyou drew is part of the calculation.\n\nIt _can_ be observed, but only in a process analyzed with\nFeynman diagrams where the Z is an external line.\n\n\n\n>>One can also see it by computing\n>>the momentum balance and check that the momentum does not lie\n>>on the Z\'s (complex) mass shell, as would be needed for a real Z.\n>\n> Oh? Sorry, but the Z is no less real than the pi0.\n> If the pi0 can decay into two photons, then it can also be created\n> bytwo photons.\n\nFor the pi0 the situation is the same. If measured, the cross\nsection for pi0 production or decay comes from an analysis with\nexternal pi0 lines, while in other processes it is a virtual\nparticle without objective meaning.\n\n\nArnold Neumaier\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Franz Heymann wrote:
> "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> 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: >>> [itex]>>>\pi + p[/itex] --> Z --> [itex]\pi + p[/itex] >>> >>>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. [itex]It _can_[/itex] 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 |
| Mar15-05, 12:17 PM | #32 |
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<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>\n\n"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\nnews:422EB74C.5020806@univie.ac.at...\n> Franz Heymann wrote:\n> > "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\n> > news:422AC49F.3030502@univie.ac.at...\n> >\n> >>Franz Heymann wrote:\n> >>\n> >>>>Unstable particles are modelled exactly like stable particles,\n> >>>>namely as external lines in a Feynman diagram.\n> >>>>Virtual particles in Feynman diagrams are exactly those\n> >>>>which are not given by external lines.\n> >>>>\n> >>>>Hence what is real and what is virtual is not affected by a\n> >>>>diagram rotation - this only affects what is input oand what is\n> >>>\n> >>>output.\n> >>>\n> >>>I am not happy with that.\n> >>>Let me illustrate my unhappiness with a specific example.\n> >>>Take the resonant elastic scattering case:\n> >>>\n> >>>pi + p --> Z --> pi + p\n> >>>\n> >>>I hope we both agree that the Z is produced as a real but\nunstable\n> >>>particle in the s-channel.\n> >>\n> >>No. Z is a virtual particle since it is an intermediate line.\n> >\n> > You are doing nothing to alleviate my unhappiness.\n> > If the Z were a virtual prticle, it would not be on its mass shell\nand\n> > a mass measurement would not be possible by studying the reaction.\n>\n> It is not on its (complex) mass shell when it is in an internal line\nas\n> in your diagram.\n>\n> When measured, one has instead a diagram where the Z is detected,\n> and hence on an external line of Feynman diagrams associated to\n> the coresponding cross section!\n>\n> In ordinary scattering experiemnts the existence of the Z shows in\n> resonances of the cross section. This is evidence for the presence\n> of the Z in the action, and can be used to calculate properties of\nZ,\n> but it is Not a measurement of Z as a particle.\n>\n>\n> > My undertanding is that the line corresponding to the Z is the\n> > trajectory of a real Z travelling from its birth to its decay.\n>\n> This interpretation of Feynman diagrams is popular but completely\n> unfounded, and gives misleading conclusions if carried beyond the\n> most superficial view.\n>\n>\n> > But it is observed, and its (complex) mass is actually measured.\n>\n> It is _not_ observed in an experiment where the Feynman diagram\n> you drew is part of the calculation.\n>\n> It _can_ be observed, but only in a process analyzed with\n> Feynman diagrams where the Z is an external line.\n\nI maintain that it is an external line which has its far end\nterminated in the decay of the resonance.\n\nIn atomic physics, a photon may be absorbed by an atom in its ground\nstate. The excited state very quickly decays back to the ground state\nby the emission of a photon.\n\nAre you now saying that the excited state is not a real, but a virtual\nstate?\n\n> >>One can also see it by computing\n> >>the momentum balance and check that the momentum does not lie\n> >>on the Z\'s (complex) mass shell, as would be needed for a real Z.\n> >\n> > Oh? Sorry, but the Z is no less real than the pi0.\n> > If the pi0 can decay into two photons, then it can also be created\n> > bytwo photons.\n>\n> For the pi0 the situation is the same. If measured, the cross\n> section for pi0 production or decay comes from an analysis with\n> external pi0 lines, while in other processes it is a virtual\n> particle without objective meaning.\n\nI fear me I will never understand you, and you will never understand\nme. I suggest we discontinue this conversation after your next reply\nto me, if indeed you wish to make one.\n\n--\nFranz\n"A first-rate laboratory is one in which mediocre scientists can\nproduce outstanding work"\nP.M.S. Blackett\n\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message
news:422EB74C.5020806@univie.ac.at... > Franz Heymann wrote: > > "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> 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: > >>> > [itex]>>>\pi + p[/itex] --> Z --> [itex]\pi + p[/itex] > >>> > >>>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. > > [itex]It _can_[/itex] 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 |
| Mar15-05, 12:22 PM | #33 |
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<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>\nEugene Stefanovich wrote:\n> Igor Khavkine wrote:\n>\n> >\n> > Guessing the form of the quantum Hamiltonian is called\nquantization.\n> > Afterward, you keep the Hamiltonian *operator* the same while its\n> > *matrix elements* may change because the basis in which they are\nwritten\n> > is may change. This is true regardless of how many times you claim\nthat\n> > your unitary transformation changes the Hamiltonian.\n>\n> There are two equivalent approaches to dressing. One approach\npromoted\n> by Shirokov and Shebeko is, as you said, to keep the Hamilton\noperator\n> of QED the same while change the definition of particles (from bare\n> particles\n> to dressed particles) and rewrite the Hamiltonian in terms of dressed\n> particle operators. Another approach is to keep the original particle\n> operators (just call them dressed) and to apply the unitary dressing\n> transformation directly to the Hamiltonian. Both approaches are\n> completely\n> equivalent (just as the Schroedinger and Heisenberg pictures of\nquantum\n> mechanics are equivalent). This is discussed in subsection 12.1.10 of\n> my book. If you prefer to use the language of the first approach,\nthis\n> is OK with me. The confusion arose because I used the language of the\n> 2nd approach in my book. I say that the unitary dressing\ntransformation\n> changes the Hamiltonian because the functional dependence of the\n> Hamiltonian on the dressed particle operators is different from\n> the functional dependence of the original QED Hamiltonian on the bare\n> particle operators.\n\nThere to "approaches" as you call them are exactly the same. What is\nconfusing that the meaning of your words does not coincide with what\nyou actually want to say. You give Shirokov and Shebeko as a reference\nfor the first approach. I can support that with any textbook on linear\nalgebra. Can you give a reference to your approach besides the PDF file\nfrom your website? Part of good communication is the use of language\nthat is appropriate and unambiguous for both parties.\n\n> >>You are right, I do need bare particles BEFORE the correct\nHamiltonian is\n> >>written. I need them to go through the 3 steps mentioned above. I\nhate to\n> >>do that, but I do not have other choice at the moment. My point was\nthat I\n> >>do not need bare or virtual particles AFTER the RQD Hamiltonian is\n> >>written.\n> >\n\n> > There is absolutely no relation between bare particles and what are\n> > called "virtual particles". The latter are simply not particles,\nthey\n> > are squiggles on paper.\n>\n> There is a relation. When you draw a real electron in QED you first\ndraw\n> a line corresponding to a "bare" electron. Then you add lines of\nvirtual\n> photons that begin and end on the bare line. Then you add small\n> virtual electron-positron loops to the virtual photon lines, etc.\n> So, loosely speeking, in QED real electron = bare electron + coat of\n> virtual particles.\n\nThat is one way to look at it, but is not the only one. I need not draw\nbare particle lines at all. If I wish, I can only draw dressed lines.\nIn this formulation scattering amplitudes can still be expanded in\nterms of Feynman diagrams, but only dressed propagators will be used.\nWherever Feynman diagrams will have internal lines and loops,\nintegration over the momenta of these internal lines is still required.\nThese internal lines whose momenta must be integrated over are commonly\nreferred to as "virtual particles". In this formulation, there are no\nbare particles, but still there are virtual particles. Conclusion: bare\nparticles have nothing to do with virtual particles and vice versa.\nAnd, I\'m afraid I\'m repeating myself, "virtual particles" are not\nparticles at all!\n\n> According to classical electromagnetic theory, electromagnetic waves\n> (= real photons in my language) can be created only by charges in\n> accelerated motion. Now, let us take a van-der-Graaf generator with\ntwo\n> highly charged metallic balls. There is no movement of charges\n> accelerated or otherwise. So, there are no electromagnetic waves in\nthis\n> case. If you put the generator in a dark room, place photographic\n> plates everywhere around it, and wait for 100 years, you\'ll not find\na\n> single blackened spot on any of the plates. So, there was no single\n> real photon in the room. However, there was a huge electrostatic\n> attraction between the two\n> balls. How are you going to apply your statement "Electromagnetic\nwaves\n> == electromagnetic interaction" in this case?\n\nYes, I will apply that statement. And if you will indeed perform the\nexperminet with a van der Graaf generator, you will see spots on the\nplates. Well, perhaps the photographic paper is not sensitive for low\nenough fequencies, in that case I suggest a simple radio reciever. You\nsay that electromagnetic radiation is associated with accelerating\ncharges. But tell me, how do you charge up a van der Graaf generator\nwithout accelerating any charges? Even if you assume that the electric\nfields have settled into a static configuration, the statement that\nthis interaction can still be explained with electromagnetic waves is\nnothing but the statement that the electric field can be written in a\nFourier decomposition in terms of different frequency components.\n\nIf you\'ve ever witnessed interference in radio or TV signals, have you\never wondered why it is called "static"?\n\nAnd if you think of the thought experiment that I propose to you (take\nto relatively stationary particles, perturb one of them at event (x,t)\nand measure the event (x\',t\') where the second particle will start\nfeeling the effects) you\'ll see that particle one cannot be perturbed\nwithout suffering acceleration and thus emitting radiation. There is\nnothing mysterious that the space-time separation between events (x,t)\nand (x\',t\') can only be light-like.\n\n> As I understand, you think that the Coulomb interaction was\n> transmitted by some material carrier (field) which cannot be detected\n> by itself (it doesn\'t blacken photographic plates, it doesn\'t leave\n> traces in bubble chambers, etc). In other words, the field is a form\n> of matter quite different from real particles (photons, electrons,\n> neutrinos, etc). I am not talking about the transverse\nelectromagnetic\n> field which freely propagates in space and, in my language, is\n> equivalent to the flow of real particles photons. I am talking about\n> the static Coulomb field. The only way to detect such a field is by\nits\n> action on charged particles.\n\nBlackening photographic plates is not the only way to detect a field.\nYour eyes, your radio, and charged particles in the ionosphere do a\ngood job as well. The electromagnetic field is detectable and changes\nin the electromagnetic field and the field itself *are* electromagnetic\nwaves, that is can they can be decomposed in oscillating Fourier modes.\n\n> My interpretation is that there is no such field. The interactions\n> between charged particles can be described more economically by\n> using direct instantaneous interparticle potentials\n> (Coulomb and magnetic). I understand that currently there is no\n> convincing experiment that allows us to distinguish between our two\n> interpretations. So, I admit that your point of view has a right to\n> exist. All I am asking from you is to admit that my point of view\n> does not\n> violate any sacred physical principle, does not contradict\nexperiment,\n> and also has a right for existence. Let future experiments decide who\n> was right.\n\nYour interpretation is at odds with not only with the conventional one,\nbut from the above I can also see that you do not recognize well\nestablished experimental evidence. So far I have not seen you propose\nan experiment that would be able to falsify your hypothesis. Care to\nspeculate?\n\nIgor\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Igor Khavkine wrote: > > > > > Guessing the form of the quantum Hamiltonian is called quantization. > > Afterward, you keep the Hamiltonian *operator* the same while its > > *matrix elements* may change because the basis in which they are written > > is may change. This is true regardless of how many times you claim that > > your unitary transformation changes the Hamiltonian. > > There are two equivalent approaches to dressing. One approach promoted > by Shirokov and Shebeko is, as you said, to keep the Hamilton operator > of QED the same while change the definition of particles (from bare > particles > to dressed particles) and rewrite the Hamiltonian in terms of dressed > particle operators. Another approach is to keep the original particle > operators (just call them dressed) and to apply the unitary dressing > transformation directly to the Hamiltonian. Both approaches are > completely > equivalent (just as the Schroedinger and Heisenberg pictures of quantum > mechanics are equivalent). This is discussed in subsection 12.1.10 of > my book. If you prefer to use the language of the first approach, this > is OK with me. The confusion arose because I used the language of the > 2nd approach in my book. I say that the unitary dressing transformation > changes the Hamiltonian because the functional dependence of the > Hamiltonian on the dressed particle operators is different from > the functional dependence of the original QED Hamiltonian on the bare > particle operators. There to "approaches" as you call them are exactly the same. What is confusing that the meaning of your words does not coincide with what you actually want to say. You give Shirokov and Shebeko as a reference for the first approach. I can support that with any textbook on linear algebra. Can you give a reference to your approach besides the PDF file from your website? Part of good communication is the use of language that is appropriate and unambiguous for both parties. > >>You are right, I do need bare particles BEFORE the correct Hamiltonian is > >>written. I need them to go through the 3 steps mentioned above. I hate to > >>do that, but I do not have other choice at the moment. My point was that I > >>do not need bare or virtual particles AFTER the RQD Hamiltonian is > >>written. > > > > There is absolutely no relation between bare particles and what are > > called "virtual particles". The latter are simply not particles, they > > are squiggles on paper. > > There is a relation. When you draw a real electron in QED you first draw > a line corresponding to a "bare" electron. Then you add lines of virtual > photons that begin and end on the bare line. Then you add small > virtual electron-positron loops to the virtual photon lines, etc. > So, loosely speeking, in QED real electron = bare electron + coat of > virtual particles. That is one way to look at it, but is not the only one. I need not draw bare particle lines at all. If I wish, I can only draw dressed lines. In this formulation scattering amplitudes can still be expanded in terms of Feynman diagrams, but only dressed propagators will be used. Wherever Feynman diagrams will have internal lines and loops, integration over the momenta of these internal lines is still required. These internal lines whose momenta must be integrated over are commonly referred to as "virtual particles". In this formulation, there are no bare particles, but still there are virtual particles. Conclusion: bare particles have nothing to do with virtual particles and vice versa. And, I'm afraid I'm repeating myself, "virtual particles" are not particles at all! > According to classical electromagnetic theory, electromagnetic waves > (= real photons in my language) can be created only by charges in > accelerated motion. Now, let us take a van-der-Graaf generator with two > highly charged metallic balls. There is no movement of charges > accelerated or otherwise. So, there are no electromagnetic waves in this > case. If you put the generator in a dark room, place photographic > plates everywhere around it, and wait for 100 years, you'll not find a > single blackened spot on any of the plates. So, there was no single > real photon in the room. However, there was a huge electrostatic > attraction between the two > balls. How are you going to apply your statement "Electromagnetic waves > == electromagnetic interaction" in this case? Yes, I will apply that statement. And if you will indeed perform the experminet with a van der Graaf generator, you will see spots on the plates. Well, perhaps the photographic paper is not sensitive for low enough fequencies, in that case I suggest a simple radio reciever. You say that electromagnetic radiation is associated with accelerating charges. But tell me, how do you charge up a van der Graaf generator without accelerating any charges? Even if you assume that the electric fields have settled into a static configuration, the statement that this interaction can still be explained with electromagnetic waves is nothing but the statement that the electric field can be written in a Fourier decomposition in terms of different frequency components. If you've ever witnessed interference in radio or TV signals, have you ever wondered why it is called "static"? And if you think of the thought experiment that I propose to you (take to relatively stationary particles, perturb one of them at event (x,t) and measure the event (x',t') where the second particle will start feeling the effects) you'll see that particle one cannot be perturbed without suffering acceleration and thus emitting radiation. There is nothing mysterious that the space-time separation between events (x,t) and (x',t') can only be light-like. > As I understand, you think that the Coulomb interaction was > transmitted by some material carrier (field) which cannot be detected > by itself (it doesn't blacken photographic plates, it doesn't leave > traces in bubble chambers, etc). In other words, the field is a form > of matter quite different from real particles (photons, electrons, > neutrinos, etc). I am not talking about the transverse electromagnetic > field which freely propagates in space and, in my language, is > equivalent to the flow of real particles photons. I am talking about > the static Coulomb field. The only way to detect such a field is by its > action on charged particles. Blackening photographic plates is not the only way to detect a field. Your eyes, your radio, and charged particles in the ionosphere do a good job as well. The electromagnetic field is detectable and changes in the electromagnetic field and the field itself *are* electromagnetic waves, that is can they can be decomposed in oscillating Fourier modes. > My interpretation is that there is no such field. The interactions > between charged particles can be described more economically by > using direct instantaneous interparticle potentials > (Coulomb and magnetic). I understand that currently there is no > convincing experiment that allows us to distinguish between our two > interpretations. So, I admit that your point of view has a right to > exist. All I am asking from you is to admit that my point of view > does not > violate any sacred physical principle, does not contradict experiment, > and also has a right for existence. Let future experiments decide who > was right. Your interpretation is at odds with not only with the conventional one, but from the above I can also see that you do not recognize well established experimental evidence. So far I have not seen you propose an experiment that would be able to falsify your hypothesis. Care to speculate? Igor |
| Mar16-05, 10:18 AM | #34 |
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<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>\nFranz Heymann wrote:\n> "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\n> news:422EB74C.5020806@univie.ac.at...\n>\n>>Franz Heymann wrote:\n>>\n>>>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message\n>>>news:422AC49F.3030502@univie.ac.at...\n>>> ;\n>>>>Franz Heymann wrote\n>>>>\n>>>>>>Unstable particles are modelled exactly like stable particles,\n>>>>>>namely as external lines in a Feynman diagram.\n>>>>>>Virtual particles in Feynman diagrams are exactly those\n>>>>>>which are not given by external lines.\n>>>>>>\n>>>>>>Hence what is real and what is virtual is not affected by a\n>>>>>>diagram rotation - this only affects what is input oand what is\n>>>>>\n>>>>>output.\n>>>> >\n>>>>>I am not happy with that.\n>>>>>Let me illustrate my unhappiness with a specific example.\n>>>>>Take the resonant elastic scattering case:\n>>>>>\n>>>>>pi + p --> Z --> pi + p\n>>>>>\n>>>>>I hope we both agree that the Z is produced as a real but unstable\n>>>>>particle in the s-channel.\n>>>>\n>>>>No. Z is a virtual particle since it is an intermediate line.\n>>>\n>>>You are doing nothing to alleviate my unhappiness.\n>>>If the Z were a virtual prticle, it would not be on its mass shell and\n>>>a mass measurement would not be possible by studying the reaction.\n>>It is not on its (complex) mass shell when it is in an internal line as\n>>in your diagram.\n>>\n>>When measured, one has instead a diagram where the Z is detected,\n>>and hence on an external line of Feynman diagrams associated to\n>>the coresponding cross section!\n>>\n>>In ordinary scattering experiments the existence of the Z shows in\n>>resonances of the cross section. This is evidence for the presence\n>>of the Z in the action, and can be used to calculate properties of Z,\n>>\n>>>But it is observed, and its (complex) mass is actually measured.\n>>\n>>It is _not_ observed in an experiment where the Feynman diagram\n>>you drew is part of the calculation.\n>>\n>>It _can_ be observed, but only in a process analyzed with\n>>Feynman diagrams where the Z is an external line.\n>\n> I maintain that it is an external line which has its far end\n> terminated in the decay of the resonance.\n\nFine; then we agree. For an external line remains external under\nrotating the diagram.\n\nInitially you argued that rotation changes virtual particles\nto real ones; it was only that which I objected to.\n\n\n> In atomic physics, a photon may be absorbed by an atom in its ground\n> state. The excited state very quickly decays back to the ground state\n> by the emission of a photon.\n>\n> Are you now saying that the excited state is not a real, but a virtual\n> state?\n\nNo. There is a big difference between the two.\n\nreal = external line = on-shell = observable\nvirtual = internal line = off-shell = ficiticous\n\nexcited states are bound states and usually appear _only_\nas external lines.\n\n\nArnold Neumaier\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"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Franz Heymann wrote:
> "Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> wrote in message > news:422EB74C.5020806@univie.ac.at... > >>Franz Heymann wrote: >> >>>"Arnold Neumaier" <Arnold.Neumaier@univie.ac.at> 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: >>>>> [itex]>>>>>\pi + p[/itex] --> Z --> [itex]\pi + p[/itex] >>>>> >>>>>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 |
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