pmb Posted May 28, 2012 Posted May 28, 2012 So light does not have mass.. But wen it travles at the speed of light ("c"), it gains mass and the time slows down.. (Relativity) Right?? That have zero mass in what sense? Precisely photons travel to the speed of light because have not mass (m=0). As you said above Einstein showed that a massive particle with nonzero mass cannot travel at the speed of light because its mass would become infinite (which means that you need an infinite energy to accelerate one electron up to c, and this is impossible). Your use of the term "mass" is contrary. When you choose to say that light has zero mass then you have chosen to use the term "mass" to mean "proper mass". The proper mass of a particle does not change its value with c. Therefore the assertion that Einstein showed that a massive particle can't travel at c because its mass would become infinite is then am invalid statement.
juanrga Posted May 28, 2012 Posted May 28, 2012 (edited) Your use of the term "mass" is contrary. When you choose to say that light has zero mass then you have chosen to use the term "mass" to mean "proper mass". The proper mass of a particle does not change its value with c. Therefore the assertion that Einstein showed that a massive particle can't travel at c because its mass would become infinite is then am invalid statement. My use of the term mass is standard. The term "proper mass" must be a term found in old classical relativistic textbooks, but is not found in modern literature dealing with QFT, particle physics, and the like. Also I do not find any advantage on introducing two or three definitions of mass, therefore, when I use the term mass it has a well-defined standard meaning. I think that you misunderstood me. I know that my m is invariant, this is a reason which I use this concept of mass. But I was replying to qft123. When I replied to him, I was replying to what he wrote about Einstein. See the post #6 where he writes about Einstein (relativistic) mass and about how a massive particle would obtain an infinite (relativistic) mass when moving at c. Edited May 28, 2012 by juanrga
pmb Posted May 28, 2012 Author Posted May 28, 2012 PMB Light do not have any proper mass... But wen it travels at the speed of light "c", it gains some mass, becoming heavier than earlier. That's incorrect. This was what I was trying to get through to you earlier. Let \gamma = 1/sqrt(1 - v2/c2), b = v/c, m = mass as measured when v << c. m is also called the [/b]proper mass[/b] or rest mass. I prefer the term proper mass since the "rest mass"is for particles at rest and a photon can never be at rest. Let M = gm be the inertial mass of a particle. It can be shown that the momentum of a point particle is 1) p = Mv = gmv The energy is 2a) E = Mc2 = gmc2 or 2b) E = Mc2 = gE0 where E0 is called the proper energy of the particle Eq. (2) can be rewritten to as 3) g = E/Mc2 = E/gmc2 Substitue Eq.(3) into Eq. (1) to obtain 4) p = mvg = mv(E/Mc2) = mv(E/gmc2) = vE/c2 Multiply Eq(4) through by c to obtain 5) pc = Ev/c This can be rewritten as 6) pc/E = v/c A luxon is a particle that always travels at the speed of light. For such a particle v = c. Substitute into Eq. (6) to obtain 7) pc/E = 1 ===> E = pc Now substitute Eq. (2) into Eq. (1) to obtain 8) p = Ev/c2 Multiply through by c to obtain 9) pc = Ev/c = Eb Square both sides to get 10) E2b2 = (pc)2 Subtract E2 from both sides 11) E2b2 - E2 = (pc)2 - E2 or 12) E2 - E2b2 = E2 - (pc)2 Factor E2 out of the left side to get 13) E2(1 - b2)2 = E2 - (pc)2 Note that g2 = 1/(1-b2) ==> (1 - b2) = 1/g2 14) E2/g2)2 = E2 - (pc)2 Note that E2/g2 = m2c4. We finally have 15a) E2 - (pc)2 = m2c4 = (mc2)2 or 15b) E2 - (pc)2 = (E02[/sup])2 Recall that for a photon v = c. If we take the limit in Eq. (15) for v --> c we find 16) m = 0 This is what it means for a photon to have zero proper mass. People often get ride of the "proper" and say "photons have zero mass". Recall the expression for momentum in Eq. (1), p = Mv. Let v = c to get p = Mc. Therefore M = p/c. Substitute p = E/c to get M = E/c[su]2[/sup]. For a photon E = hf where h = Planck's constant and f is the frequency of the photon. We now have M = hf/c2 which means that a photon has inertial mass. So the two expressions for the mass of a luxon is Proper Mass: m = 0 Inetial Mass: M = p/c = M = E/c[su]2[/sup] = hf/c2 Connection between E, p and m: E2 - (pc)2 = (mc2)2 Hopefully that should clear eveything up.
juanrga Posted May 28, 2012 Posted May 28, 2012 That's incorrect. This was what I was trying to get through to you earlier. Let \gamma = 1/sqrt(1 - v2/c2), b = v/c, m = mass as measured when v << c. m is also called the [/b]proper mass[/b] or rest mass. I prefer the term proper mass since the "rest mass"is for particles at rest and a photon can never be at rest. Let M = gm be the inertial mass of a particle. It can be shown that the momentum of a point particle is 1) p = Mv = gmv This is not the momentum of an electron in an electromagnetic field. Neither the momentum of Mercury near the Sun (due to general relativistic effects). The energy is 2a) E = Mc2 = gmc2 or 2b) E = Mc2 = gE0 More of the same. 16) m = 0 This is what it means for a photon to have zero proper mass. People often get ride of the "proper" and say "photons have zero mass". Recall the expression for momentum in Eq. (1), p = Mv. Let v = c to get p = Mc. Therefore M = p/c. Substitute p = E/c to get M = E/c[su]2[/sup]. For a photon E = hf where h = Planck's constant and f is the frequency of the photon. We now have M = hf/c2 which means that a photon has inertial mass. So the two expressions for the mass of a luxon is Proper Mass: m = 0 Inetial Mass: M = p/c = M = E/c[su]2[/sup] = hf/c2 Connection between E, p and m: E2 - (pc)2 = (mc2)2 Hopefully that should clear eveything up. The term "proper mass" for m is not broadly used because it is confusing, and because there is no utility in the definition of two or three different concepts of mass. One is enough and this is named just "mass". I do not know any application/utility of the concept of inertial mass M and both particle physics and general relativity use m. For instance four-momentum is defined using m in modern textbooks on general relativity.
pmb Posted May 28, 2012 Author Posted May 28, 2012 (edited) The concept of matter given by ajb and myself is the usual in the modern Standard Model. Yes. You can say it, but can you prove it? Edited May 28, 2012 by pmb
juanrga Posted May 28, 2012 Posted May 28, 2012 (edited) Yes. You can say it, but can you prove it? Many textbooks give the Hamiltonian for matter plus radiation. CERN website also explain to broad audiences why an electron is a matter particle, but a photon is not a matter particle http://public.web.ce...rdmodel-en.html Edited May 28, 2012 by juanrga
pmb Posted May 28, 2012 Author Posted May 28, 2012 (edited) juanrga - Please stop pulling this thread off topic. Nobody cares how many people use what. Edited May 28, 2012 by pmb
juanrga Posted May 28, 2012 Posted May 28, 2012 (edited) juanrga - Please stop pulling this thread off topic. Nobody cares how many people use what. You asked me in #20 if I could prove that the mainstream concept of matter used by ajb and me was the usual in the Standard Model. And I proved it in #21 with a link to CERN page devoted to the Standard Model. What is your problem? Edited May 28, 2012 by juanrga
pmb Posted May 28, 2012 Author Posted May 28, 2012 (edited) What is your problem? Oh man! Everybody thinks they're so unique as if they're the first one to point out the obvious, i.e. what a newbie in SR learns in the first week of relativity. What you think you're telling me for the first time I learned a quarter of a century ago. So why do you think you're different. Better yet, what makes you think I'm so ignorant? My problem is that this thread is getting off topic and you are responsible for doing it. Please stop it! As far as you so-called proof goes, it's not a proof at all. There's no logic to it which could be considered a proof. You really cant prover your assertion because its far from being true. All your doing is showing your igorance in these concepts. All you ended up doing was to show what some people think. That's far from being an assertion. But then again you weren't rigidly clear on that either. You state only Many textbooks give the Hamiltonian for matter plus radiation. CERN website also explain to broad audiences why an electron is a matter particle, but a photon is not a matter particle http://public.web.ce...rdmodel-en.html Which tells us only what "many" (whatever that means) textbooks say the Hamiltonian for matter plus radiation but you didn't say what that has to do with the terminology. And what was your textbook example supposed to do? You once again thought that you could argue a point without including the relavent information. E.g. quote one of those textbooks which you claim supports your assertion. I.e. show this Hamiltonian. Mving on, all your link shows is something I already know, that particle physicists use the term "mass" to refer to "proper mass". If I were a particle physicist then that's what I'd use in print too. All that tells us is what particle physicists use. It says nothing about the relativistic mechanics of continuous media. Nor does it say anything about active, passive and inertial mass as used in GR, astronomy and cosmology. That's no proof at all. Then there are relativists who study relativity for itself, i.e. study its useage in all concievebale applications. As far as textbooks go, relativity texts, SR and GR, the ones that use inertial mass to mean m = p/v = relativistic mass in at lese one derivation are Gravitation by Misner, Thorne and Wheeler, W.H. Freeman and Company Introducing Einstein's Relativity by Ray D'Inverno, Clarendon Press Relativity DeMystified by David McMahon, McGraw-Hill Basic Relativity by Richard A. Mould, Springer Relativity: Special, General and Cosmological by Wolfgang Rindler, Oxford University Press A First Course in General Relativity by Bernard F. Schutz, Cambridge University Press A Short Course in General Relativity - Second Edition by J. Foster and J.D Nightingale, Springer Special Relativity by A.P. French, The MIT Introductory Physics Series General Theory of Relativity by P.A.M. Dirac, Princeton University Press Introduction to the Theory of Relativity by Peter Gabriel Bergman, Dover Pub. Space-Time Structure by Erwin Schrodinger, Cambridge University Press Special Relativity by Albert Shadowitz, Dover Pub. The Meaning of Relativity - Fifth Edition by Albert Einstein, Princeton University Press Theory of Relativity by Wolfgang Pauli, Dover Pub. Relativity, Thermodynamics and Cosmology by Richard C. Tolman, Dover Pub. The Theory of Relativity by C.M. Möller, Oxford at the Clarendon Press The ones that were published recently (i.e. after 1990) are Introducing Einstein's Relativity by Ray D'Inverno, Clarendon Press Relativity DeMystified by David McMahon, McGraw-Hill Basic Relativity by Richard A. Mould, Springer Relativity: Special, General and Cosmological by Wolfgang Rindler, Oxford University Press A First Course in General Relativity by Bernard F. Schutz, Cambridge University Press A Short Course in General Relativity - Second Edition by J. Foster and J.D Nightingale, Springer Examples from the American Journal of Physics Apparatus to measure relativistic mass increase, John W. Luetzelschwab, Am. J. Phys. 71(9), 878, Sept. (2003). Relativistic mass increase at slow speeds, Gerald Gabrielse, Am. J. Phys. 63(6), 568 (1995). In defense of relativistic mass, T. R. Sandin, Am. J. Phys. 59(11) 1032 (1991). A simple relativistic paradox about electrostatic energy, Wolfgang Rindler and Jack Denur, Am. J. Phys. 56(9), Sept. (1988). An elementary development of mass-energy equivalence, Daniel J. Steck, Frank Rioux, Am. J. Phys. 51(5), May (1983). So please stop trying to pass of your misguided beliefs on me. I've heard them all before and have gotten into debates such as these thousands of times over the last 14 years. You're no different. Edited May 28, 2012 by pmb -2
D H Posted May 28, 2012 Posted May 28, 2012 No, Pete. juan is not dragging this thread off topic. The standard model does not view photons and other force carriers as matter. Matter is stuff that has mass and occupies space.
juanrga Posted May 28, 2012 Posted May 28, 2012 (edited) No Pete. This thread main topic is if photons are or are not matter as the OP asked. At least three posters including myself explained that photons are not usually considered matter. I added that our definition of matter agrees with the usual in the Standard Model where photons are not considered matter. I added a link to CERN website explaining why photons are not considered matter after you asked me explicitly in #20 if I could prove my statements. Pete, you are who is pulling this thread off-topic with several large posts about the definition of mass. I will not discuss more with you about mass neither about general relativity here (This is a thread under the Quantum Theory subforum!). I only want to add that modern textbooks in general relativity define the four momentum as [math]P^\mu = m U^\mu[/math] and that this m is the same m used by particle physicists. No other concept of mass is needed in general relativity. Indeed, general relativists and cosmologists agree on that photons are massless m=0. I recommend you start with A No-Nonsense Introduction to General Relativity. Edited May 28, 2012 by juanrga
pmb Posted May 28, 2012 Author Posted May 28, 2012 (edited) No, Pete. juan is not dragging this thread off topic. I disagree. The OPs question has been answered. You, juan and myself all agreed that whether a photon is matter depends on the definition of matter. Let that rest as is. After I posted post #16 the answer was all there for the OP to see. All that needed to be said had been said. juan the added some bogus stuff. I.e. I wrote p = gamma m v. juan then replied This is not the momentum of an electron in an electromagnetic field. Neither the momentum of Mercury near the Sun (due to general relativistic effects). juan never backed up his claim. That's just poor form. Without him being clear I'd wager that he's thinking that I was forgetting the momentum due to the the electrons field. If so then he is sorely misinformed. Let m = the proper mass of an electron. Let m0 = bare mass of the electron, dm = electromagnetic mass of electrons field. Then m = m0 + dm. The proper mass of an electron has the mass contribution of the electrons field built right into it. That's pretty well established. A bit later through my derivation he posts "More of the same." I find that just plain rude and condescending. The term "proper mass" for m...blah blah blah I don't care. Never have and I never will. Sell it somwhere else juan. I'd wager I know this subject much much much more that you do. I get that feeling from the arguments you've provided so far. Edited May 28, 2012 by pmb
michel123456 Posted May 29, 2012 Posted May 29, 2012 (edited) juanrga, on 28 May 2012 - 11:55 AM, said: Precisely photons travel to the speed of light because have not mass (m=0). As you said above Einstein showed that a massive particle with nonzero mass cannot travel at the speed of light because its mass would become infinite (which means that you need an infinite energy to accelerate one electron up to c, and this is impossible). I have a problem with that. You all must know I like diagrams: have you ever put this statement into diagram? You get a mass curve increasing with velocity, reaching the infinite a C For zero mass, the object should be at zero mass (tautology): that's the origin, the point A. But no, science tells us that for no mass, the "thing" (the photon) is closer to point B, actually infinitely far up out of the diagram, upon the C line, up, up,up to the moon. But not even, the photon is not infinitely up out of the diagram, the photon is down the C line, at the intersection with the horizontal zero-mass line, at point C. And certainly not at the zero-zero origin point. Which makes no sense to me. Edited May 29, 2012 by michel123456
studiot Posted May 29, 2012 Posted May 29, 2012 Thank you michel for posting the diagram. I fully appreciate juan's statement that the mass becomes asymptotic to a vertical line through C. However it provides no information whatsoever for what happens with v > c. No one has ever produced an experiment or observation to test if the relationship changes beyond c or remains the same. Now consider the tangent function. This has the same form up to a fixed value ie is asymptotic to a vertical line through 90. However if we proceed through 90 lo and behold the line which has disappeared out of the top of the paper is seen coming back in from the bottom. This behaviour could not have been predicted from the behaviour at x<90.
pmb Posted May 29, 2012 Author Posted May 29, 2012 For zero mass, the object should be at zero mass (tautology): that's the origin, the point A. By definition, the inertial mass of a particle is defined by m = m0/sqrt(1 - v2/c2) The derivation is here - http://home.comcast.net/~peter.m.brown/sr/inertial_mass.htm When v = 0, m = m0, not zero. That's why m0 is refered to as the particle's proper mass, or rest mass. You have A located at the wrong value of m. You have it a zero whereas the correct position is at m = m0. But no, science tells us that for no mass, the "thing" (the photon) is closer to point B, actually infinitely far up out of the diagram, upon the C line, up, up,up to the moon. The relationship m = m0/sqrt(1 - v2/c2) is derived under the the assumption that the speed of the particle can change and have any value 0 <= v < c. Therefore that derivation cannot be applied to a photon. In that diagram you are plotting inertial mass versus speed. You are not plotting proper mass vs speed. The derivation for the value of the proper mass of a photon is given by Post #3. At the end you simply take the limit in Eq. (15) for v --> c we find m[sub[0[/sub] = 0. But not even, the photon is not infinitely up out of the diagram, the photon is down the C line, at the intersection with the horizontal zero-mass line, at point C. And certainly not at the zero-zero origin point. Which makes no sense to me. Not to worry. You are simply misinterpretting your diagram. You are plotting m vs v, not m0 vs v. The prope mass m0 of a particle is independant of speed. I highly recommend that you take a very close read of post #3. It should clear a lot of this up. Note: There are zero errors in post #3.
pmb Posted May 30, 2012 Author Posted May 30, 2012 (edited) standard model does not view photons and other force carriers as matter. Matter is stuff that has mass and occupies space. I'm not interested in the Standard Model. The standard model is not the owner of the soul of the photon. If one were to claim that matter is that which is stuff that had mass and occupies space then that depends on how the term mass is defined. To that end we have already seen how experts in the field of SR/GR (not just people who apply the field to get their work done) decide to define the term mass. There are different views, contrary to what people claim. As I've explained before this list is a list of sr/gr texts which define mass such that photons have mass Introducing Einstein's Relativity by Ray D'Inverno, Clarendon Press Basic Relativity by Richard A. Mould, Springer Relativity: Special, General and Cosmological by Wolfgang Rindler, Oxford University Press The exact quotes are here - http://home.comcast.net/~peter.m.brown/ref/relativistic_mass/relativistic_mass.htm Alan H. Guth does the same thing and he's no slouch! Take notice how Schutz uses the term "inertial mass" onm [age 94 i is GR text (as well as his "Gravity From the Ground Up" text). Those are just to name a few. There is also that new sr/gr text I got written by Hans Stephani,k published not that long ago in 2004 he uses the term "inertial mass" as Schutz does. juan used a logical fallicy above (which is against the rules) when he asserted that because one web page, the CERN page, uses the term matter such that photons don't have matter is the end all and be all in definitions in the particle physicist community. Since we've all said our piece on this matter thing, let's let it go. We could argue this forever in the hopes to convert the other to their beliefs. If that's the case then so be it. But please leave me out of it. Thanks Edited May 30, 2012 by pmb
juanrga Posted May 30, 2012 Posted May 30, 2012 (edited) I disagree. The OPs question has been answered. You, juan and myself all agreed that whether a photon is matter depends on the definition of matter. Let that rest as is. After I posted post #16 the answer was all there for the OP to see. All that needed to be said had been said. juan the added some bogus stuff. I.e. I wrote p = gamma m v. juan then replied This is not the momentum of an electron in an electromagnetic field. Neither the momentum of Mercury near the Sun (due to general relativistic effects). juan never backed up his claim. That's just poor form. Without him being clear I'd wager that he's thinking that I was forgetting the momentum due to the the electrons field. If so then he is sorely misinformed. Let m = the proper mass of an electron. Let m0 = bare mass of the electron, dm = electromagnetic mass of electrons field. Then m = m0 + dm. The proper mass of an electron has the mass contribution of the electrons field built right into it. That's pretty well established. You substitution m --> m = m0 + dm changes nothing because is m --> m. Your momentum p=m gamma v is not the momentum of an electron in an electromagnetic field because its momentum receives a correction due to interaction with the field. Neither your p is the momentum of Mercury planet near the Sun because its momentum p receives a corrections due to deviation of spacetime from flatness. I have a problem with that. You all must know I like diagrams: have you ever put this statement into diagram? You get a mass curve increasing with velocity, reaching the infinite a C For zero mass, the object should be at zero mass (tautology): that's the origin, the point A. But no, science tells us that for no mass, the "thing" (the photon) is closer to point B, actually infinitely far up out of the diagram, upon the C line, up, up,up to the moon. But not even, the photon is not infinitely up out of the diagram, the photon is down the C line, at the intersection with the horizontal zero-mass line, at point C. And certainly not at the zero-zero origin point. Which makes no sense to me. You mixed what I said about a massless particle such as the photon with what I said about a massive particle such as the electron. Moreover, you are mixing the modern concept of mass (which I like) with the old concept of "relativistic mass" (used by the poster whom I replied). This "relativistic mass" (which I dislike) is what others call "inertial mass". The term "relativistic mass" has acquired a bad fame among most academics and is virtually not used in research. juan used a logical fallicy above (which is against the rules) when he asserted that because one web page, the CERN page, uses the term matter such that photons don't have matter is the end all and be all in definitions in the particle physicist community. Not even close to reality. In the other thread, I started posting that there are different definitions of matter. Next, I added that one of those definitions is accepted by the immense majority of physicists and chemists and can be found in many textbooks. I gave a link to astrophysics page (it is in the other thread) that agrees. I gave a link to CERN page that agrees and I gave a link to an introductory textbook on general relativity by a cosmologist that agrees (see the link above). At least three posters more used the same definition of matter that myself when replied to you and none is (I am rather sure) a particle physicist... The definiton of mass that I use is standard among physicists, chemists, biologists, engineers... The m in the above book in general relativity is the same m that I use. This is the same m that one finds in thousands of books from thermodynamics to quantum field theory. Edited May 30, 2012 by juanrga
D H Posted May 30, 2012 Posted May 30, 2012 I'm not interested in the Standard Model. The standard model is not the owner of the soul of the photon. Yes, it is. Photons are a subject of quantum mechanics, not relativity. Both special and general relativity are classical (non-quantum) theories. One can legitimately talk about photons in the context of special relativity because quantum electrodynamics has bridged the gap between special relativity and quantum mechanics. Talking about photons in the context of general relativity is not quite so legitimate because there is no theory (yet) that bridges general relativity and quantum mechanics. 1
michel123456 Posted May 30, 2012 Posted May 30, 2012 (edited) @pmb By definition, the inertial mass of a particle is defined by m = m0/sqrt(1 - v2/c2) The derivation is here - http://home.comcast.net/~peter.m.brown/sr/inertial_mass.htm When v = 0, m = m0, not zero. That's why m0 is refered to as the particle's proper mass, or rest mass. You have A located at the wrong value of m. You have it a zero whereas the correct position is at m = m0. The relationship m = m0/sqrt(1 - v2/c2) is derived under the the assumption that the speed of the particle can change and have any value 0 <= v < c. Therefore that derivation cannot be applied to a photon. In that diagram you are plotting inertial mass versus speed. You are not plotting proper mass vs speed. The derivation for the value of the proper mass of a photon is given by Post #3. At the end you simply take the limit in Eq. (15) for v --> c we find m[sub[0[/sub] = 0. Not to worry. You are simply misinterpretting your diagram. You are plotting m vs v, not m0 vs v. The prope mass m0 of a particle is independant of speed.I highly recommend that you take a very close read of post #3. It should clear a lot of this up. Note: There are zero errors in post #3. Bolded mine. "The prope mass m0 of a particle is independant of speed." except for m0= zero. ------------- edit from post #3 you wrote 2a) E = Mc2 = gmc2 then Eq. (2) can be rewritten to as 3) g = E/Mc2 = E/gmc2 Are you sure? Isn't it g=E/mc2 ? ------------------------ @Juanrga You substitution m --> m = m0 + dm changes nothing because is m --> m. Your momentum p=m gamma v is not the momentum of an electron in an electromagnetic field because its momentum receives a correction due to interaction with the field. Neither your p is the momentum of Mercury planet near the Sun because its momentum p receives a corrections due to deviation of spacetime from flatness. You mixed what I said about a massless particle such as the photon with what I said about a massive particle such as the electron. Moreover, you are mixing the modern concept of mass (which I like) with the old concept of "relativistic mass" (used by the poster whom I replied). This "relativistic mass" (which I dislike) is what others call "inertial mass". The term "relativistic mass" has acquired a bad fame among most academics and is virtually not used in research. Not even close to reality. In the other thread, I started posting that there are different definitions of matter. Next, I added that one of those definitions is accepted by the immense majority of physicists and chemists and can be found in many textbooks. I gave a link to astrophysics page (it is in the other thread) that agrees. I gave a link to CERN page that agrees and I gave a link to an introductory textbook on general relativity by a cosmologist that agrees (see the link above). At least three posters more used the same definition of matter that myself when replied to you and none is (I am rather sure) a particle physicist... The definiton of mass that I use is standard among physicists, chemists, biologists, engineers... The m in the above book in general relativity is the same m that I use. This is the same m that one finds in thousands of books from thermodynamics to quantum field theory. bolded mine. You are perfectly correct, I mixed things. But in order to do so there must be 2 different things to mix. isn't it? Edited May 30, 2012 by michel123456
pmb Posted May 30, 2012 Author Posted May 30, 2012 (edited) Yes, it is. Photons are a subject of quantum mechanics, .. This is getting carried away. One last time and then let it go, okay? I mean we have to stop somewhere and I hate debating about definitions. Its something a teenager would do. When the photon is used in relativity its treated as a classical particle. It's not treated as a quantum object. This also means it has a classical trajectory and the Heisenberg Undertainty Relation is not used in non-relativistic QM. In that context its just another luxon (a particle that travels at v = c). And as I've already explained many times, in relativity its ofen useful to think of light as having mass. I won't repeat what I've already provided. So let's drop the debate/ We all know what the other guy thinks. There is absolutley no reason to continue this debate on a definition. @pmb Bolded mine. "The proper mass m0 of a particle is independant of speed." except for m0= zero. The velocity never changes so we can't speak of the photons mass at differrent speeds. So the relationship m = gm0 doesn't apply to photons so we can't speak of independant of speed ------------- edit from post #3 you wrote then Are you sure? Isn't it g=E/mc2 ? ------------------------ Correct! Thank you very much. I must have made a mistake translating from paper to text. You substitution m --> m = m0 + dm changes nothing because is m --> m. Your momentum p=m gamma v is not the momentum of an electron in an electromagnetic field because its momentum receives a correction due to interaction with the field. blah blah blah... Wrong. Read Classical Charged Particles - Third Edition by Fritz Rohrlich, World Scientific Pub. Co. 2006. I won't be reading your posts again unless circumstances require it and I don't see that happening. You just have too many errors. Ever time you attempt to correct me you add an error or two in response. I don't have the time or the will to keep doing that. It's just plain silly. And remember, it makes absolutely no difference whether you think its an error or not (since you ususally reject the corretions given you). With regards to me responding to you, it only matters whether I think its an error or not. Good bye juan. I hope you take an open mind and read that book. It will do you a lot of good. It will teach you what Poincare mass is (See Feynman lectures to learn the meaning of that term). Edited May 30, 2012 by pmb
juanrga Posted May 30, 2012 Posted May 30, 2012 (edited) When the photon is used in relativity its treated as a classical particle. In that context its just another luxon. That means it has a classical trajectory and the Heisenberg Undertainty Relation is not used in non-relativistic QM. First as quantum theory shows a photon is not a localizable particle. The classical trajectories studied in relativity correspond to a kind of average position and speed of that collection of photons that we name "light signal". Second, the Heisenberg Uncertainty Relation is a fundamental property of non-relativistic QM; it is not something that you can use or ignore at your own choice. Wrong. I recommend that you learn the physics on this subject from Classical Charged Particles - Third Edition by Fritz Rohrlich, World Scientific Pub. Co. 2006. It's a superb text for this subject. This is a text about non-quntum mechanics relativity. You really should read it and learn the subject matter from Rohrlich, a superb physicist. You'll learn what Poincare stress is and how it pertains to the inertia of the electron. Oh yeah! You don't believe that stress has anything to do with inertia! LOL!! By the way. I don't see a reason to read or espond to any more of your posts. There's just too many errors in it and its too insulting to bother with. Take my advice and learn how to play with others. Rohrlich's classic text is a rather good source (although outdated in some parts) and he agrees 100% with what I said of course. I know very well that Poincaré stresses are, but it seems that you do not know that the Poincaré speculative model of the electron is inconsistent and was abandoned about 100 years ago. Edited May 30, 2012 by juanrga
michel123456 Posted May 30, 2012 Posted May 30, 2012 This is getting carried away. It's as if people are ignoring everthing else I said about light/photon having mass and how its applied in other areas. There's no reason to assume that anything else I said will be read since it appears that everything else I said seems to be ignored. Nothing I've said has been directly commented on so I see nothing else to talk about on this point beyond this post. When the photon is used in relativity its treated as a classical particle. It's not treated as a quantum object. This also means it has a classical trajectory and the Heisenberg Undertainty Relation is not used in non-relativistic QM. In that context its just another luxon. And in relativity its ofen useful to think of light as having mass. Even Feynman said that because light has energy it has gravitational mass and is thus deflected in a gravitational fiel. Radiation is a source of gravity. It contrubutes to inertia. It plays a role in cosmology in the radiation dominated era. The luxon is assigned mass by m = E/c2. That relationship is found in the lecture notes of someone who I respect more than anybody else who is alive in physics today. So let's drop the debate and let people use what they want withut being harassed. This constant badgering trying to force their views on others is bad juju. I find it irritating. Isn't that enough for all to let this debate die? We all know what the other guy things. There is zero reason to continue other than to harras in an attempt to force one to chanhing their viewpoint. The velocity never changes so we can't speak of the photons mass at differrent speeds. So the relationship m = gm0 doesn't apply to photons so we can't speak of independant of speed Correct! Thank you very much. I must have made a mistake translating from paper to text. Wrong. I recommend that you read Classical Charged Particles - Third Edition by Fritz Rohrlich, World Scientific Pub. Co. 2006. It's a superb text for this subject. You really should read it and learn the subject matter from Rohrlich, a superb physicist. You'll learn what Poincare stress is and how it pertains to the inertia of the electron. Oh yeah! You don't believe that stress has anything to do with inertia! Oh well. Just try to keep an open mind. At least while you're reading the book. By the way. I don't see a reason to read or espond to any more of your posts. They're just too offensive and error filled to read. bolded mine wrong excuse(*) (**). pmb just reached the bottom of the abyss of my unconsideration. Don't worry you are not alone there. (*)quoting (...) Note: There are zero errors in post #3. and (**) 3) g = E/Mc2 = E/gmc2 Substitue Eq.(3) into Eq. (1) to obtain bolded mine No. Bolded part does not look to me as an error of translation from paper to text. And no one goes from Eq(3) to (4) when there is an error in Eq(3).
pmb Posted May 30, 2012 Author Posted May 30, 2012 bolded mine wrong excuse(*) (**). pmb just reached the bottom of the abyss of my unconsideration. Don't worry you are not alone there. I don't understand. What do you mean by that? I had deleted a lot of the things in your post. I was having a hard time restraining my irritation at juan's poor argument skills, i.e. all claims, no valid proofs (invalid proofs, sure). But I didn't want to let my anger get in my way and I don't want to give juan more attention that he deserves. No. Bolded part does not look to me as an error of translation from paper to text. And no one goes from Eq(3) to (4) when there is an error in Eq(3). Well, that is actually the case. I still have the papers where I derived it and the paper contains the right derivations and equations. All you need do is ask whether I made a translation or not and I'll be glad to tell you. Only I have the ability to answer that question. I have no need to lie, that's for sure.
juanrga Posted May 30, 2012 Posted May 30, 2012 (edited) Pmb, I note that you have modified your original post #20 after my response was posted in #21. E.g. you have deleted the term "Poincaré stress" from your post #20 after I criticized it. But your original words are still quoted in my #21. The term "Poincaré stress" is also quoted in #22 because michael also replied to you before you edited your post. I am sorry about all that. You have added more material to #20, which was not in the original post when I (or michael) replied to you. For instance now you add that a luxon is a particle that travels at v = c and that in relativity it is often useful to think of light as having mass. However a luxon is defined as a massless particle (http://en.wikipedia.org/wiki/Luxon) and, regarding light, I already gave to you a link to an introduction to general relativity where light is considered to be massless without any problem. Edited May 30, 2012 by juanrga
swansont Posted May 31, 2012 Posted May 31, 2012 but it seems that you do not know that ! Moderator Note juanrga, you really need to stop doing this. If you disagree with a point someone has made, you are free to correct it. But you are not free to get personal about it. Do not repsond to this modnote in the thread
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