pmb Posted May 31, 2012 Author Share Posted May 31, 2012 (edited) Yes, it is. Photons are a subject of quantum mechanics, not relativity Sorry about getting back to you on this. Yes. When photons are used in relativity the quantum effects are ignored. The photon can be assigned a position, velocity and momentum, simultaneously. In relativity a photon moves on a classical trajectory. In essense in SR and GR the quantum effects are assumed to be insignificant for the purposes which it is used. When QED is brought in then one is no longer talking about classical mechanics. A relavent point was suggested by the late Phillip Morrison in Taylor and Wheeler's relavent text Exploring Black Holes. In the front matter it says Philip Morrison made several suggestions and convinced us not to invoke that weird quantum particle, the photon, in a treatment of the classical theory of relativity (except in some exercises). Taylor and I had several conversations about this for the above reasons, i.e. people seem to get confused when you use a quantum particle in a classical theory. That's why I prefer the term Luxon so that there will be no confusion. Nobody tries to analyze the quantum mechanics of a luxon. lol! In fact just because a particle is a luxon (a classical particle) it doesn't mean that its a photon. A photon has a spin associated with it. A luxon, again, a classical particle, need not have a spin. Pete Edited May 31, 2012 by pmb Link to comment Share on other sites More sharing options...

D H Posted May 31, 2012 Share Posted May 31, 2012 This is getting carried away. One last time and then let it go, okay? So what's this post #26, then? Follow your own advice in post #20: Let it go. I too don't want to get into ontology/etymology wars. They're silly and useless for the most part. Link to comment Share on other sites More sharing options...

pmb Posted May 31, 2012 Author Share Posted May 31, 2012 (edited) So what's this post #26, then? Follow your own advice in post #20: Let it go. I too don't want to get into ontology/etymology wars. They're silly and useless for the most part. This will end here. In the following post I use the symbol "m" to mean inertial mass (aka relativisic mass) defined by m = p/v In a previous post I told juan that I won't be reading your posts again unless circumstances require it and I don't see that happening. I was wrong. It’s very difficult to not accidently read posts which are in a thread when you’re reading the thread. Circumstance requires that I post in order for me to be right with my conscious. The circumstance is that it appears, very much so, that juan is not being honest in his responses and is utilizing a strawman to argue some of his claims. This means that juan is using a logical fallacy of the strawman which is a violation of the forum rules. 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. From that comment its obvious that you either didn't understand it or didn't read it. Rohrlich wouldn’t put out a new edition and keep parts which are outdated. No author would. (1) First off I find it extremely difficult to believe that you've ever heard of Rohrlich's text, never mind have it right at your side and know as fact what Rohrlich's stand on Poincare stress is. You never demonstrated that you even knew what the ineria of stress is, never mind Poincare stress. Feynman's text explains the same thing about a classical electron in terms of classical mechanics. You have demonstrated a tendancy not to learn about the inertia of stress. When I asked you about an article on the subject I could send you you said no. That's an important article for someone who doesn't understand the inertia of stress. You wouldn't even consider it when you failed at your attempt at solving the mass density of the magnetic field. The resolution to understanding the solution is to undestand the inertia of stress. (2) Another reason that I don't trust you is because of your total dismissal of that text. Fine, don't read it. Who cares? But don't claim that you have. In this case you claimed that Rohlich's text agrees with you 100%. All you did was to make a claim that he agrees with you and, as usual, never proved it. You're in the habit of never backing up what you claim. Do you know how irritating that is? You have this bad habit of assuming that merely stating what you believe is considered poof. Example: You claimed that, for an electron, p = mv is wrong and then didn't even back up your claim. You later made an attempt and said something about it and it was then that you explained about the electron being in an EM field. I.e. 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. Here you're responding with a strawman again. I never said that the electron was in an electromagnetic field. That's what a strawman is. I say one thing and then you claim I said another thing and attempt to prove that other thing wrong. The second part which I underlined is just plain nonsense as anyone can tell. However under any circumstances p = mv, is valid. Now that you said that it's clear that you don't kow what you're talking about very clearly. What you've demonstrated that you're thinking about is that when I said mv = momentum you mistook it as the canonical momentum of the electron. What has changed is that the mechanical momentum would be the same with or without the EM field but the canonicalmomentum is a function of the presence of the magntic field. E.g. if there is an electromagnetic field then the canonical momentum of an electron in such a field is given by Mechanical Momentum = p = mv Canonical Momentum = P_{i} = @L/@v_{i} = mv/sqrt(1 – v^2/c^2) + (e/c) A_{i} (3) The name of the text is Classical Charged Particles. This is not about quantum mechanics of an charged particle. Poincare stress applies here. You ignored that fact. Big mistake. (4) You already demonstrated that you are unaware of the inertia of stress. For that reason I know you didn’t read Rohrlich’s text since you have to know about the inertia of stress in order to follow his derivations. (5) Time to tell the audience the definition of Poincare Stress. From Oxford Dictionary of Physics: Poincaré Stresses - Nonelectric forces postulated to give stability to a model of the electron. Because of the difficulties in regarding an electron as a point charge it is possible to postulate that the electron is a charge distribution with a nonzero radius. However, an electric charge distribution alone is unstable. In 1906 Henri Poincaré postulated unknown nonelectric forces, now called Poincaré stresses, to give stability to the electron. Considerations such as these are now thought to be irrelevant, as it is accepted that an electron should be described by quantum electrodynamics rather than classical field theory. Notice the part where it says Considerations such as these are now thought to be irrelevant, as it is accepted that an electron should be described by quantum electrodynamics rather than classical field theory. What you ignored here is that the name of Rohrlich's text “Classical Charged Particles” (note the “classical” part) tells you that the subject of Poincare stress is classical and that the subject matter of the author's text is classical mechanics. (6) Poincare stresses apply to the momentum of a classical electron. Stress contributes to inertia as I've tried to explain to you many times, but for which you keep ignoring. (7) By definition, the momentum and mass of a particle is given by p = mv. That is to say that m is defined such that for particles which interact only by contact that mv is a conserved quantity. Momentum p is then defined as p = mv. (8) You claim that Rohrlich agrees with you, yet, again, you provide no proof and expect us to take your word on it. Sorry. I don't trust you at this point. Either you had the text at hand and chose not to quote or you didn't have the text at hand and are claiming you knew it from memory. (9) I'm quite different from you. I prove what I assert. From page 16 Rohrlich writes in the last paragaph An ingnious solution was suggested by Poincare. The Coulomb repulsion, which is responsible for the instability of the electron, can be compensated by nonelectromagnetic forces, a kind of negative pressure. These forces must be postulated ad hoc, to be sure, but the mount of pressure neccesary to provide stability also exactl compensates W_{self}/3 in (Eq. 2.27) ... gives p = { (4/3)W_{self}/c^{2} - W_{coh}/c^{2} } = (W_{self}/c^{2}) v because W_{coh} = (1/3)W_{self}/c^{2}. This is consistent with theLorentz tansformations. Th Poincare chesive forces have a double purpose: they make the electron stable and at the same time brings the theory in accord with the special theory of relativity. And that shoots down your claim about not being used in over a hundred years. I see it used quite often. I read an article not too long ago by David Griffiths on the topic. (10) In this thread you claimed 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. Wow! You are sooo wrong that its safe to say that your claim is just total nonsense. The term proper mass is a widely known term in the field of relativiy. It's found in most SR/GR texts. Modern ones too. You just keep thinking that it doesn't appear in QFT or QM then it doesn't exist. Boy! How wrong you are. I doubt you even know QFT. Moden Usage: Rindlers SR/GR text defines it as does. D'Inverno's as well. And those texts were published in the last decade. And again - nobody cares what you use I can't fathom why you keep bringing it up when you've beat that horse to death already. (11) Modern usage: From Mass renormalization in classical electrodynamics, David J. Griffiths and Russell E. Owen, Am. J. Phys. 51(12), December 1983 Abstract - The electromagnetic mass of a charged object can be calculated by three independent methods: (1) m_{u} = u/c^{2}, where u is the field energy of the object at rest, (2) m_{p} = p/v, where p is the field momentum when the particle is moving at speed v, and (3) m_{s} = F/a, where F is the self-force when the object has acceleration z. In the context of a simple dumbbell model we demonstrate that m_{p} = m_{s}, but these in general exceed m_{u}. And I was curious as to how it's applied now so I asked Griffith himself who returned my e-mail saying Well, personally I think they are an absolutely essential part of the story. Some people have tried to build a consistent theory without including Poincare stress, but I think this is doomed to failure, since SOME force is required to hold a charged particle together. On the other hand, the detailed nature of this force is irrelevant, so I don't see much in the recent literature about it. If I was going to be blind and wipe my memory of this subject cleen, then I'd take Rohrlich and Griffiths over you, who can't seem to provide a logical proof of anything people have asked of you. All you do is show one instance of something and expect people to accept that as proof that it holds in all cases. That to me is nonsense. (12) Then there is your constant use of a strawmen (which the board keeps ignoring for some reason). For some reason you got it into your head that you had to force your beliefs on me regarding what the term "mass" is. You argue your point that everyone who uses the stanard model uses what you do. Well big deal. Nobody cares. I don't even disagree with it. Although I challenged to you prove it because I kew you couldn't. It's just fun to watch. When it comes to particle physics its best to stick with proper mass. And you insist that everyone in relativity uses it and for some reason you'll keep at it, badering me by keep repeating your claim until you get what you want - to force me to believe what you believe. You pointed to a GR book which uses only proper mass, as if that was proof of something to me. You can't logically point to a single, or few, examples of what you said and then expect someone that everone uses it. I just told you I got a new text, published 2004, which uses inertial mass. 13) Basically you keep wanting to prove to me that everyone in the Standard Model uses proper mass rather than inertial mass. I never said otherwise. When you keep claiming that you've proved theory X and never argued X then you just used a strawman. That's a logical fallicy, and that's prohibited on the forum. 14) Otherwise I said that people in GR and cosmology tend to use inertial mass and gave you several examples to back up what I sai and you promptly ignored it for no reason other than you were unable to admit you were wrong. If you ca't admit your mistake then just let it go and stop using a strawman in an attempt to save face. This is the reason I said I won't read your posts. There are far too many errors to correct. It was dumb of me to assume I wouldn't read them because they're mixed in with all the other ones. Oh well. I might have to read part of them in the passing out of the corner of my eye but I'll have to live with the challenge of restraining myself from all the errors you make from misconceptions to logical fallacies. Edited May 31, 2012 by pmb Link to comment Share on other sites More sharing options...

swansont Posted May 31, 2012 Share Posted May 31, 2012 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. ! Moderator Note pmb, this kind of commentary is inappropriate and counterproductive. Everyone, let's stick to the physics and dispense with the sniping. Do no not respond to this. Your collective appeals have run out. Link to comment Share on other sites More sharing options...

pmb Posted May 31, 2012 Author Share Posted May 31, 2012 (edited) swansont = Please delete my post. Note: See proper mass at http://en.wikipedia.org/wiki/Invariant_mass Edited May 31, 2012 by pmb Link to comment Share on other sites More sharing options...

michel123456 Posted June 1, 2012 Share Posted June 1, 2012 (edited) Note: There are zero errors in post #3. That's incorrect. This was what I was trying to get through to you earlier. Let \gamma = 1/sqrt(1 - v^{2}/c^{2}), 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 = Mc^{2} = gmc^{2} or 2b) E = Mc^{2} = gE_{0} where E_{0} is called the proper energy of the particle Eq. (2) can be rewritten to as 3) g = E/Mc^{2} = E/gmc^{2} Substitue Eq.(3) into Eq. (1) to obtain 4) p = mvg = mv(E/Mc^{2}) = mv(E/gmc^{2}) = vE/c^{2} 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/c^{2} Multiply through by c to obtain 9) pc = Ev/c = Eb Square both sides to get 10) E^{2}b^{2} = (pc)^{2} Subtract E^{2} from both sides 11) E^{2}b^{2} - E^{2} = (pc)^{2} - E^{2} or 12) E^{2} - E^{2}b^{2} = E^{2} - (pc)^{2} Factor E^{2} out of the left side to get 13) E^{2}(1 - b^{2})^{2} = E^{2} - (pc)^{2} Note that g^{2} = 1/(1-b^{2}) ==> (1 - b^{2}) = 1/g^{2} 14) E^{2}/g^{2})^{2} = E^{2} - (pc)^{2} Note that E^{2}/g^{2} = m^{2}c^{4}. We finally have 15a) E^{2} - (pc)^{2} = m^{2}c^{4} = (mc^{2})^{2} or 15b) E^{2} - (pc)^{2} = (E_{0}2[/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/c^{2} 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/c^{2} Connection between E, p and m: E^{2} - (pc)^{2} = (mc^{2})_{2} Hopefully that should clear eveything up. No comment. Edited June 1, 2012 by michel123456 Link to comment Share on other sites More sharing options...

pmb Posted June 1, 2012 Author Share Posted June 1, 2012 (edited) No comment. When I wote "there are no errors" I assumed Ityped everything in write. You never had to make such a big deal out of it. I already said you were right michel. I admitted that I made a mistake the first time you pointed it out. You're now getting carried away. To much at this point with no gain. Moving on: I wanted to get back to the concept of bare mass that I touched on earlier. Again I want to warn everyone that this is about classical EM and the classical electron/charge. Somewhere in this thread or the original one I explained that the contribution of the Coulomb field to the mass of a charged particle is expressed as follows. From Rohrlich's text top of page 137, and using Rohrlich's notation m = m_{bare}_{ + m}_{Coul} _{ } _{The first term is defined as the }_{bare mass}_{ of the electron, i.e. the mass the particle would have if there it was uncxharged. I originally expressed as dm since that's how its often expressed. Rohrlich uses different notation. m is what is known as the proper mass of the electron.} Edited June 1, 2012 by pmb Link to comment Share on other sites More sharing options...

juanrga Posted June 1, 2012 Share Posted June 1, 2012 (edited) This will end here. In the following post I use the symbol "m" to mean inertial mass (aka relativisic mass) defined by m = p/v In a previous post I told juan that I was wrong. It's very difficult to not accidently read posts which are in a thread when you're reading the thread. Circumstance requires that I post in order for me to be right with my conscious. The circumstance is that it appears, very much so, that juan is not being honest in his responses and is utilizing a strawman to argue some of his claims. This means that juan is using a logical fallacy of the strawman which is a violation of the forum rules. From that comment its obvious that you either didn't understand it or didn't read it. Rohrlich wouldn't put out a new edition and keep parts which are outdated. No author would. Evidently a new edition does not guarantee that everything in that edition is free of error or up-to-date. But the important part of my comment was that Rohrlich's is a good text and that he agrees with me. I add snapshots of relevant parts of the textbook cited by Pmb, including a relevant paragraph where Rohrlich criticizes Pmb definition of mass. Rohrlich uses a much more correct, well-defined, and consistent concept of mass, which is the same that I use. (1) First off I find it extremely difficult to believe that you've ever heard of Rohrlich's text, never mind have it right at your side and know as fact what Rohrlich's stand on Poincare stress is. You never demonstrated that you even knew what the ineria of stress is, never mind Poincare stress. Feynman's text explains the same thing about a classical electron in terms of classical mechanics. You have demonstrated a tendancy not to learn about the inertia of stress. When I asked you about an article on the subject I could send you you said no. That's an important article for someone who doesn't understand the inertia of stress. You wouldn't even consider it when you failed at your attempt at solving the mass density of the magnetic field. The resolution to understanding the solution is to undestand the inertia of stress. After saying you thanks, I explained to you why I do not need that outdated article. (2) Another reason that I don't trust you is because of your total dismissal of that text. Fine, don't read it. Who cares? But don't claim that you have. In this case you claimed that Rohlich's text agrees with you 100%. All you did was to make a claim that he agrees with you and, as usual, never proved it. You're in the habit of never backing up what you claim. Do you know how irritating that is? You have this bad habit of assuming that merely stating what you believe is considered poof. Example: You claimed that, for an electron, p = mv is wrong and then didn't even back up your claim. You later made an attempt and said something about it and it was then that you explained about the electron being in an EM field. I.e. Here you're responding with a strawman again. I never said that the electron was in an electromagnetic field. That's what a strawman is. I say one thing and then you claim I said another thing and attempt to prove that other thing wrong. The second part which I underlined is just plain nonsense as anyone can tell. However under any circumstances p = mv, is valid. Now that you said that it's clear that you don't kow what you're talking about very clearly. What you've demonstrated that you're thinking about is that when I said mv = momentum you mistook it as the canonical momentum of the electron. What has changed is that the mechanical momentum would be the same with or without the EM field but the canonicalmomentum is a function of the presence of the magntic field. E.g. if there is an electromagnetic field then the canonical momentum of an electron in such a field is given by Mechanical Momentum = p = mv Canonical Momentum = P_{i} = @L/@v_{i} = mv/sqrt(1 – v^2/c^2) + (e/c) A_{i} First, your above expressions are mutually inconsistent. As you write at the very start of your message you use the symbol "m" to mean inertial mass (aka relativisic mass) This means that your P_{i} is wrong unless you change m to mean rest-mass. But if you change m to mean "rest-mass" then your above p is lacking the [math]\gamma[/math] term. No matter what is your choice for m, both equations are mutually inconsistent. Second, Rohrlich denotes the canonical (four)momentum by the symbol [math]\Pi^\mu[/math] Third, Rorhlich denotes the four-momentum by [math]p^\mu[/math] The momentum p is the spatial component of the four-momentum [math]p^\mu[/math] [math]\mathbf{p} = m\mathbf{v} + (e/c)\mathbf{A}[/math] It must be remarked that Rorhlich is using a tau-parametrization [math]\mathbf{v} = \mathbf{v}(\tau)[/math] instead of the more usual t-parametrization, the relationship is [math]\mathbf{v}(\tau) = \gamma\mathbf{v}(t)[/math]. Other authors use a different notation. The term [math]mv^\mu[/math] is usually named the kinetic (four)momentum. The term "mechanical momentum" is a misnomer, although can be still found in the literature. Rohrlich also uses the term kinetic momentum as I do As I said above the kinetic momentum coincides with the momentum p only in absence of electromagnetic fields [math]\mathbf{p} - (e/c)\mathbf{A} = m\mathbf{v} [/math] Fourth, Rohrlich agrees with me on that the addition of [math]\gamma[/math] to the definition of mass is useless and only found in older literature. He uses the concept of "rest mass" (which is simply named mass in modern literature), Again both Rohrlich and me are in full agreement: (3) The name of the text is Classical Charged Particles. This is not about quantum mechanics of an charged particle. Poincare stress applies here. You ignored that fact. Big mistake. (4) You already demonstrated that you are unaware of the inertia of stress. For that reason I know you didn't read Rohrlich's text since you have to know about the inertia of stress in order to follow his derivations. (5) Time to tell the audience the definition of Poincare Stress. From Oxford Dictionary of Physics: Poincaré Stresses - Notice the part where it says Considerations such as these are now thought to be irrelevant, as it is accepted that an electron should be described by quantum electrodynamics rather than classical field theory. What you ignored here is that the name of Rohrlich's text "Classical Charged Particles" (note the "classical" part) tells you that the subject of Poincare stress is classical and that the subject matter of the author's text is classical mechanics. (6) Poincare stresses apply to the momentum of a classical electron. Stress contributes to inertia as I've tried to explain to you many times, but for which you keep ignoring. (7) By definition, the momentum and mass of a particle is given by p = mv. That is to say that m is defined such that for particles which interact only by contact that mv is a conserved quantity. Momentum p is then defined as p = mv. (8) You claim that Rohrlich agrees with you, yet, again, you provide no proof and expect us to take your word on it. Sorry. I don't trust you at this point. Either you had the text at hand and chose not to quote or you didn't have the text at hand and are claiming you knew it from memory. (9) I'm quite different from you. I prove what I assert. From page 16 Rohrlich writes in the last paragaph And that shoots down your claim about not being used in over a hundred years. I see it used quite often. I read an article not too long ago by David Griffiths on the topic. (10) In this thread you claimed Wow! You are sooo wrong that its safe to say that your claim is just total nonsense. The term proper mass is a widely known term in the field of relativiy. It's found in most SR/GR texts. Modern ones too. You just keep thinking that it doesn't appear in QFT or QM then it doesn't exist. Boy! How wrong you are. I doubt you even know QFT. Moden Usage: Rindlers SR/GR text defines it as does. D'Inverno's as well. And those texts were published in the last decade. And again - nobody cares what you use I can't fathom why you keep bringing it up when you've beat that horse to death already. (11) Modern usage: From Mass renormalization in classical electrodynamics, David J. Griffiths and Russell E. Owen, Am. J. Phys. 51(12), December 1983 And I was curious as to how it's applied now so I asked Griffith himself who returned my e-mail saying If I was going to be blind and wipe my memory of this subject cleen, then I'd take Rohrlich and Griffiths over you, who can't seem to provide a logical proof of anything people have asked of you. All you do is show one instance of something and expect people to accept that as proof that it holds in all cases. That to me is nonsense. (12) Then there is your constant use of a strawmen (which the board keeps ignoring for some reason). For some reason you got it into your head that you had to force your beliefs on me regarding what the term "mass" is. You argue your point that everyone who uses the stanard model uses what you do. Well big deal. Nobody cares. I don't even disagree with it. Although I challenged to you prove it because I kew you couldn't. It's just fun to watch. When it comes to particle physics its best to stick with proper mass. And you insist that everyone in relativity uses it and for some reason you'll keep at it, badering me by keep repeating your claim until you get what you want - to force me to believe what you believe. You pointed to a GR book which uses only proper mass, as if that was proof of something to me. You can't logically point to a single, or few, examples of what you said and then expect someone that everone uses it. I just told you I got a new text, published 2004, which uses inertial mass. 13) Basically you keep wanting to prove to me that everyone in the Standard Model uses proper mass rather than inertial mass. I never said otherwise. When you keep claiming that you've proved theory X and never argued X then you just used a strawman. That's a logical fallicy, and that's prohibited on the forum. 14) Otherwise I said that people in GR and cosmology tend to use inertial mass and gave you several examples to back up what I sai and you promptly ignored it for no reason other than you were unable to admit you were wrong. If you ca't admit your mistake then just let it go and stop using a strawman in an attempt to save face. This is the reason I said I won't read your posts. There are far too many errors to correct. It was dumb of me to assume I wouldn't read them because they're mixed in with all the other ones. Oh well. I might have to read part of them in the passing out of the corner of my eye but I'll have to live with the challenge of restraining myself from all the errors you make from misconceptions to logical fallacies. I already wrote that Poincaré stresses model is an completely inconsistent and outdated model which was abandoned more than 100 years ago. After showing how Rohrlich is using the same concept of mass that I am using, let me cite the next article by Okun, where the concept of relativistic mass used by Pmb is rejected http://www.physicsto...isAuthorized=no Or just check Taylor & Wheeler (Spacetime Physics, 2nd Ed.): The concept of 'relativistic mass' is subject to misunderstanding. That's why we don't use it. First, it applies the name mass - belonging to the magnitude of a 4-vector - to a very different concept, the time component of a 4-vector. Second, it makes increase of energy of an object with velocity or momentum appear to be connected with some change in internal structure of the object. In reality, the increase of energy with velocity originates not in the object but in the geometric properties of space-time itself. That's incorrect. This was what I was trying to get through to you earlier. Let \gamma = 1/sqrt(1 - v^{2}/c^{2}), 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 = Mc^{2} = gmc^{2} or 2b) E = Mc^{2} = gE_{0} where E_{0} is called the proper energy of the particle Eq. (2) can be rewritten to as 3) g = E/Mc^{2} = E/gmc^{2} Substitue Eq.(3) into Eq. (1) to obtain 4) p = mvg = mv(E/Mc^{2}) = mv(E/gmc^{2}) = vE/c^{2} 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/c^{2} Multiply through by c to obtain 9) pc = Ev/c = Eb Square both sides to get 10) E^{2}b^{2} = (pc)^{2} Subtract E^{2} from both sides 11) E^{2}b^{2} - E^{2} = (pc)^{2} - E^{2} or 12) E^{2} - E^{2}b^{2} = E^{2} - (pc)^{2} Factor E^{2} out of the left side to get 13) E^{2}(1 - b^{2})^{2} = E^{2} - (pc)^{2} Note that g^{2} = 1/(1-b^{2}) ==> (1 - b^{2}) = 1/g^{2} 14) E^{2}/g^{2})^{2} = E^{2} - (pc)^{2} Note that E^{2}/g^{2} = m^{2}c^{4}. We finally have 15a) E^{2} - (pc)^{2} = m^{2}c^{4} = (mc^{2})^{2} or 15b) E^{2} - (pc)^{2} = (E_{0}2[/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/c^{2} 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/c^{2} Connection between E, p and m: E^{2} - (pc)^{2} = (mc^{2})_{2} Hopefully that should clear eveything up. After stating why m is a better concept of mass. Let me simply add this letter by Einstein about the subject: It is not good to introduce the concept of the mass of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the 'rest mass' m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion. — Albert Einstein in letter to Lincoln Barnett, 19 June 1948 (quote from L. B. Okun (1989), p. 42) Edited June 1, 2012 by juanrga Link to comment Share on other sites More sharing options...

pmb Posted June 1, 2012 Author Share Posted June 1, 2012 (edited) I am rewriting an earlier part of this thread because michel really wanted me to. Since he was very polite when he asked in a PM to me I thought I’d do it just for michel’s sake! qft1234. This was what I was trying to get through to you earlier. Let g = 1/sqrt(1 - v^{2}/c^{2}), b = v/c, m = inertial mass for v << c. m is also called the proper mass or rest mass) of the particle. I prefer the term proper mass since the term "rest mass" cannot literally be applied to particles at rest and a photon can never be at rest. Let M = gm be the inertial mass of a particle for any speed. It can be shown that the momentum of a classical point particle is (see http://home.comcast.net/~peter.m.brown/sr/inertial_mass.htm) 1) p = Mv = gmv The energy is 2a) E = Mc^{2} = gmc^{2} or 2b) E = Mc^{2} = gE_{0} where E_{0} is called the proper energy of the particle Eq. (2a) can be rewritten to as 3) g = E/mc^{2} Substitue Eq.(3) into Eq. (1) to obtain 4) p = mvg = mv(E/mc^{2}) = mv(E/mc^{2}) = vE/c^{2} 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 get (this is really a limiting process) 7) pc/E = 1 ===> E = pc Solve Eq. (2) for M to get M = E/c^{2} and then plug this result into Eq. (1) to get 8) p = (E/c^{2})v Multiply through by c to obtain 9) pc = Ev/c = Eb Square both sides to get 10) E^{2}b^{2} = (pc)^{2} Now subtract E^{2} from both sides and rearrange to get 11) E^{2} - E^{2}b^{2} = E^{2} - (pc)^{2} Factor E^{2} out of the left side to get 12) E^{2}(1 - b^{2})^{2} = E^{2} - (pc)^{2} Note that g^{2} = 1/(1-b^{2}) ==> (1 - b^{2}) = 1/g^{2} 13) E^{2}/g^{2})^{2} = E^{2} - (pc)^{2} Note that E^{2}/g^{2} = m^{2}c^{4}. We finally have 14a) E^{2} - (pc)^{2} = m^{2}c^{4} = (mc^{2})^{2} or 14b) E^{2} - (pc)^{2} = (E_{0}2[/sup])^{2} Recall that for a luxon v = c. If we take the limit in Eq. (14) for v --> c we find 16) m = 0 This is what it means for a luxon to have zero proper mass. Of course this results holds for photons. People often get rid of the adjective "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^{2}. For a photon E = hf where h = Planck's constant and f is the frequency of the photon. We now have M = hf/c^{2} which means that a photon has inertial mass. So the two expressions for the mass of a luxon is Proper Mass: m = 0 Inertial Mass: M = p/c = M = E/c^{2} = hf/c^{2} Connection between E, p and m: E^{2} - (pc)^{2} = (mc^{2})_{2} After stating why m is a better concept of mass. Let me simply add this letter by Einstein about the subject: Sorry juan, but a I already explained, I can't keep up with all the mistakes you keep making. See you after my sabatical. I'll see you but that'll be about it. Glad to see you got that book from te library and chose to learn from it. Can this really be true or were you just seeing how wrong you were! Hmmmmm. Edited June 1, 2012 by pmb Link to comment Share on other sites More sharing options...

juanrga Posted June 1, 2012 Share Posted June 1, 2012 (edited) Sorry juan, but a I already explained, I can't keep up with all the mistakes you keep making. See you after my sabatical. I'll see you but that'll be about it. Glad to see you got that book from te library and chose to learn from it. Can this really be true or were you just seeing how wrong you were! Hmmmmm. Sorry Pete, but you are also completely wrong about this. I knew Rohrlich's work before you cited it here. Edited June 1, 2012 by juanrga -1 Link to comment Share on other sites More sharing options...

pmb Posted June 6, 2012 Author Share Posted June 6, 2012 (edited) juan - I suggested to you several posts back that you read Rohlichs text. You gave no hint whatsoever that you even heard of it. Only several days later, perhaps after you got a copy from the library did you acknowledge that you had it. Then you claim I was wrong? Nope. I don't buy it. There is nothing about the entire situation that would make me believe you. IT goes against your track record. It'd be easier fopr everyont if you just told the truth instead of worrying about making a mistake. Nobody cares if you make a mistake. What people do care about is when you make a mistake and then lie to cover it up You claimed that the momentum I was talking about was canonical momentum, which as anyone can see, is a wrong assertion. Nobody should ever have assumed that when someone writes p = mv that its supposed to mean canonical momentum. That's just bad juju. 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). This is the most perfect example of a strawman that I've ever seem. Bad juju!!! Nobody was talking about electrons juan. It was you who brought it up, not I. That action is a logical fallacy known as a strawman. In any case the expression p = mv always holds true, by definition mind you!!!! This is the mechanical momentum of the particle. It should never be confused with canonical momentum such as you did. The terms in the expression are defined as follows (using bold symbols this time for precission). For a charged particle of charge e p = 3-momentum v = 3-velocity m = inertial mass p = mv = mechanical momentum P = p + (e/c)A = canonical momentum where P_{j} = @L/@v_{j} Never confuse P with p since that'd be a very serious mistake. Had you read my paper on the subject of mass in relativity then you'd have known all this and not made this mistake. And I wrote that paper a very long time ago so don't try to take credit for anything I speak of regarding canonical momentum and its relationship to mechanical momentum. It's all right here http://arxiv.org/abs/0709.0687 Please stop passing off your mistakes as someone elses Edited June 6, 2012 by pmb -1 Link to comment Share on other sites More sharing options...

imatfaal Posted June 7, 2012 Share Posted June 7, 2012 ! Moderator Note pmb. Further to Hyper-V's warning hereAccusations of lying, untrustworthiness, and immaturity are completely unacceptable in a science discussion forum. Any further repetitions of this form of post - or any other breaches of forum rules - will lead to suspension. I note that you have retracted some of your comments in your latest edit - Many Thanks for that; but please ensure that you do fully comply with forum rules in future.Do not respond to this modnote in the thread. You can report this post or PM me (or any other member of staff) if you believe this message is inaccurate or unjustified. Link to comment Share on other sites More sharing options...

juanrga Posted June 7, 2012 Share Posted June 7, 2012 juan - I suggested to you several posts back that you read Rohlichs text. You gave no hint whatsoever that you even heard of it. Only several days later, perhaps after you got a copy from the library did you acknowledge that you had it. Then you claim I was wrong? Nope. I don't buy it. There is nothing about the entire situation that would make me believe you. IT goes against your track record. It'd be easier fopr everyont if you just told the truth instead of worrying about making a mistake. Nobody cares if you make a mistake. What people do care about is when you make a mistake and then lie to cover it up You claimed that the momentum I was talking about was canonical momentum, which as anyone can see, is a wrong assertion. Nobody should ever have assumed that when someone writes p = mv that its supposed to mean canonical momentum. That's just bad juju. This is the most perfect example of a strawman that I've ever seem. Bad juju!!! Nobody was talking about electrons juan. It was you who brought it up, not I. That action is a logical fallacy known as a strawman. In any case the expression p = mv always holds true, by definition mind you!!!! This is the mechanical momentum of the particle. It should never be confused with canonical momentum such as you did. The terms in the expression are defined as follows (using bold symbols this time for precission). For a charged particle of charge e p = 3-momentum v = 3-velocity m = inertial mass p = mv = mechanical momentum P = p + (e/c)A = canonical momentum where P_{j} = @L/@v_{j} Never confuse P with p since that'd be a very serious mistake. Had you read my paper on the subject of mass in relativity then you'd have known all this and not made this mistake. And I wrote that paper a very long time ago so don't try to take credit for anything I speak of regarding canonical momentum and its relationship to mechanical momentum. It's all right here http://arxiv.org/abs/0709.0687 Please stop passing off your mistakes as someone elses You continue repeating the same mistakes that I reported in post #33 above, except that you have now corrected your expression for P_{j}after that I noticed above that your previous expression for P_{j} was wrong if by m you mean "inertial mass". Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 9, 2012 Share Posted June 9, 2012 That have zero mass in what sense? 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. What is mass? It is a quantity of inertia, gravitational mass. It is the presence of an object moving below the speed of light. What is an object which moves at lightspeed? It is the lack of all of the above. The fact that a particle with mass would gain an infinite amount of energy means that this is an ''unrealistic'' proposition. No particle with matter gains an infinite amount of energy, so as a limit, it works well as an explanation why particles with mass can never move exactly at the speed of light. The assertion that Einstein made was not wrong as far as we know it, because the more you accelerate a particle to the speed of light, the more energy is required... more energy in the visible universe, which is why we cannot get a particle with mass to that speed. Maybe some might find my newest post here a little interesting. It attempts to explain mass as a charge of the field. http://www.scienceforums.net/topic/66985-a-forgotten-theory-of-mass/page__p__683191#entry683191 Charge is simply the coefficients of the Lie Algebra. But as my post states, mass has also very close connections to the electromagnetic phenomena. Link to comment Share on other sites More sharing options...

pmb Posted June 13, 2012 Author Share Posted June 13, 2012 You continue repeating the same mistakes that I reported in post #33 above, except that you have now corrected your expression for P_{j}after that I noticed above that your previous expression for P_{j} was wrong if by m you mean "inertial mass". The mistakes are all yours, and that includes your confusion of canonical momentum with mechanical momentum. Nobody else in this forum would make such a mistake. Simply start a thread and pose the question and they'll enlighten you What is mass? The inertial mass m of a body is that property which resists changes in momentum. The inertial mass of a body is defined as p = mv where v = 3-velocity and 3 = 3-momentum = mechanical momentum. The active gravitational mass of a body is that which as as the source of gravity. The passive gravitational mass of a body is that quantity on which gravity acts. Thanks for the link. I'll take a peak. Link to comment Share on other sites More sharing options...

juanrga Posted June 13, 2012 Share Posted June 13, 2012 (edited) You continue repeating the same mistakes that I reported in post #33 above, except that you have now corrected your expression for P_{j}after that I noticed above that your previous expression for P_{j} was wrong if by m you mean "inertial mass". The mistakes are all yours, and that includes your confusion of canonical momentum with mechanical momentum. Nobody else in this forum would make such a mistake. Simply start a thread and pose the question and they'll enlighten you It is fair to emphasize that now you write "mechanical momentum" only after I corrected your early posts. In your early posts you never wrote "mechanical momentum" but you wrote "momentum" (an example is your #3) you only wrote "the momentum of a point particle is". If you are now suggesting that where you wrote "momentum", you did really mean "mechanical momentum", then part of the criticism that I did in my response #4 and in other places vanishes, evidently. However, other corrections to your mistakes remain, including that "kinetic momentum" is a better term than "mechanical momentum". Edited June 13, 2012 by juanrga Link to comment Share on other sites More sharing options...

pmb Posted June 14, 2012 Author Share Posted June 14, 2012 ...(snipped erroneous claims)... I recommend once again that you simply admit your mistakes and move on with life. Even if you just admit them to yourself. You've got to let this bone go. I'll explain it yet once more. In relativity when one sees "p = mv" it is understood that the p here is mechanical momentum. It's universally accepted to be that. That expression is the definition of what the p is. Canonical momentum has a different symbol and a different definition and is made explicit what the symbols mean. In QM the 'p' is always assumed to be canonical momentum. We know that due to the context. One never needs to say that p = mv is the mechanical momentum since when p = mv appears its understood. That's universally true with the exception of QM but even there its made clear. All QM texts will state that p is canonical momentum. So accept your mistake and move on. It's a small mistake which you're making into a mountain. I can see how you resist admitting your mistakes though. Try it. You might find it refreshing. Link to comment Share on other sites More sharing options...

juanrga Posted June 15, 2012 Share Posted June 15, 2012 (edited) I recommend once again that you simply admit your mistakes and move on with life. Even if you just admit them to yourself. You've got to let this bone go. I'll explain it yet once more. In relativity when one sees "p = mv" it is understood that the p here is mechanical momentum. It's universally accepted to be that. That expression is the definition of what the p is. Canonical momentum has a different symbol and a different definition and is made explicit what the symbols mean. In QM the 'p' is always assumed to be canonical momentum. We know that due to the context. One never needs to say that p = mv is the mechanical momentum since when p = mv appears its understood. That's universally true with the exception of QM but even there its made clear. All QM texts will state that p is canonical momentum. So accept your mistake and move on. It's a small mistake which you're making into a mountain. I can see how you resist admitting your mistakes though. Try it. You might find it refreshing. Initially you said us that p was "momentum". After receiving a pair of critical replies you changed your discourse to p is "mechanical momentum" and said us that "canonical momentum" was denoted by P, not by p. Some further critical replies latter you are starting to admit now that p can denote "canonical momentum"... in QM literature. If I cite the page 460 of Wald textbook on general relativity (where he defines "momentum" as [math]\partial L / \partial \dot{q}[/math]), would you agree that some general relativists use the term momentum as I use? Or will you maintain that we confound momentum with canonical momentum? And if I cite Feynman textbook on QED, where he uses p for momentum ([math]\partial L / \partial \dot{q}[/math]) and [math]\pi[/math] for kinetic momentum [math]\pi \equiv (p - eA)[/math], would you admit that some particle physicists use the same terminology and notation that I use? Or will you maintain that we do not know what canonical momentum is? Edited June 15, 2012 by juanrga Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 18, 2012 Share Posted June 18, 2012 (edited) The kinetic momentum [math]\pi[/math] can just be written [math]\frac{\partial \mathcal{L}}{\partial \dot{\phi}}[/math]. This is the Canonical momentum to [math]\phi[/math]. The ordinary momentum term is just freely switched for [math]\pi[/math] even though they don't exactly describe the the same things. For instance, [math]\sum_i p_i(q) \rightarrow \int \pi(x) dx[/math]. I don't call it kinetic momentum, I tend to call it a field momentum. Edited June 18, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

pmb Posted June 23, 2012 Author Share Posted June 23, 2012 (edited) The kinetic momentum [math]\pi[/math] can just be written [math]\frac{\partial \mathcal{L}}{\partial \dot{\phi}}[/math]. This is the Canonical momentum to [math]\phi[/math]. The ordinary momentum term is just freely switched for [math]\pi[/math] even though they don't exactly describe the the same things. For instance, [math]\sum_i p_i(q) \rightarrow \int \pi(x) dx[/math]. I don't call it kinetic momentum, I tend to call it a field momentum. I've seen that term used in Jackson's text where it's also used to refer to ordinary momentum. I found an article on this topic. See Momentum conservation and the vector potential of moving charges, Grant R. Fowles, Am. J. Phys.. 48(8), Sept. 1980 The specific problem of the interaction between two freely moving charges has been treated in this journal in an excellant artilce by Breitenberger. His analysis is a simplified version of an earlier for formulation by Darwin in which it is demonstrated that conservation of linear momentum pi]does[/i] hold when, instead of the ordinary particle momentum mv + eA where A is the vector potential. The author explains that eA can be identified with momentum of the joint electromagnetic field of the moving chages. Edited June 23, 2012 by pmb Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 23, 2012 Share Posted June 23, 2012 (edited) Pmb, where was the thread you posted your work on mass? Was it this one, I had been looking for it because I had a question. In your paper, you define the gravitational charge component in a gravitational 3-force equation. I believe it had the form [math]f^k = m\Gamma_{ij}^{k}v^i v^j[/math] I just wanted to know, how one would derive this equation. Presumably the connection is playing your usual role of the gravitational field yes? I see there is a dependance on the velocity. The [math]\Gamma_{ij}^{k}v^i v^j[/math] part looks a little bit like it comes from the geodesic equation of motion which involves usually a term of second derivatives. The equation I had in mind was this one [math]\frac{\partial^2 x^{\mu}}{\partial s^2} + \Gamma^{\mu}_{ij} \frac{\partial x^i}{\partial s} \frac{\partial x^j}{\partial s}=0[/math] Ok, they look quite different but if you rearrange it [math] \Gamma^{\mu}_{ij} \frac{\partial x^i}{\partial s} \frac{\partial x^j}{\partial s} = \frac{\partial^2 x^{\mu}}{\partial s^2}[/math] The left hand side of the equation does look a little like [math]\Gamma_{ij}^{k}v^i v^j[/math] So I am just wondering how the force equation ''gets'' the terms it has. I'm especially interested in the dependence of the velocity and how this comes about. Pmb, where was the thread you posted your work on mass? Was it this one, I had been looking for it because I had a question. In your paper, you define the gravitational charge component in a gravitational 3-force equation. I believe it had the form [math]f^k = \Gamma_{ij}^{k}v^i v^j[/math] I just wanted to know, how one would derive this equation. Presumably the connection is playing your usual role of the gravitational field yes? I see there is a dependance on the velocity. The [math]\Gamma_{ij}^{k}v^i v^j[/math] part looks a little bit like it comes from the geodesic equation of motion which involves usually a term of second derivatives. The equation I had in mind was this one [math]\frac{\partial^2 x^{\mu}}{\partial s^2} + \Gamma^{\mu}_{ij} \frac{\partial x^i}{\partial s} \frac{\partial x^j}{\partial s}=0[/math] Ok, they look quite different but if you rearrange it [math] \Gamma^{\mu}_{ij} \frac{\partial x^i}{\partial s} \frac{\partial x^j}{\partial s} = \frac{\partial^2 x^{\mu}}{\partial s^2}[/math] The left hand side of the equation does look a little like [math]\Gamma_{ij}^{k}v^i v^j[/math] So I am just wondering how the force equation ''gets'' the terms it has. I'm especially interested in the dependence of the velocity and how this comes about. I'm actually starting to think its definitely from the geodesic equation. The s is just 'proper time'. ct = x and c = x/t. [math] \Gamma^{\mu}_{ij} \frac{\partial x^i}{\partial s} \frac{\partial x^j}{\partial s} = \Gamma^{\mu}_{ij} v^i v^j[/math] I think that works out. This adds up to the same dimensions as you find in that force equation. But then I start thinking about the dimensions of the equation. [math]m\Gamma^{\mu}_{ij} v^i v^j[/math] How does this equate to a 3-force? If we just take it at face value, the part described as [math]\Gamma^{\mu}_{ij} v^i v^j[/math] would have to have dimensions of acceleration. Edited June 23, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

juanrga Posted June 23, 2012 Share Posted June 23, 2012 (edited) I found an article on this topic. See Momentum conservation and the vector potential of moving charges, Grant R. Fowles, Am. J. Phys.. 48(8), Sept. 1980 The author explains that eA can be identified with momentum of the joint electromagnetic field of the moving chages. It was already commented before that one cannot universally claim that eA is a field-like momentum. Since you insist on repeating the claim I will add more info. If one restricts himself to field electrodynamics, sure that eA is related to the field, but when one works with Wheeler-Feynman electrodynamics or with more advanced and recent formulations of electrodynamics the term eA is not related to any field because A is given as a functional of particles path. Pmb, where was the thread you posted your work on mass? Was it this one, I had been looking for it because I had a question. In your paper, you define the gravitational charge component in a gravitational 3-force equation. I believe it had the form [math]f^k = m\Gamma_{ij}^{k}v^i v^j[/math] I just wanted to know, how one would derive this equation. Presumably the connection is playing your usual role of the gravitational field yes? I see there is a dependance on the velocity. The [math]\Gamma_{ij}^{k}v^i v^j[/math] part looks a little bit like it comes from the geodesic equation of motion which involves usually a term of second derivatives. The equation I had in mind was this one [math]\frac{\partial^2 x^{\mu}}{\partial s^2} + \Gamma^{\mu}_{ij} \frac{\partial x^i}{\partial s} \frac{\partial x^j}{\partial s}=0[/math] Ok, they look quite different but if you rearrange it [math] \Gamma^{\mu}_{ij} \frac{\partial x^i}{\partial s} \frac{\partial x^j}{\partial s} = \frac{\partial^2 x^{\mu}}{\partial s^2}[/math] The left hand side of the equation does look a little like [math]\Gamma_{ij}^{k}v^i v^j[/math] So I am just wondering how the force equation ''gets'' the terms it has. I'm especially interested in the dependence of the velocity and how this comes about. I'm actually starting to think its definitely from the geodesic equation. The s is just 'proper time'. ct = x and c = x/t. [math] \Gamma^{\mu}_{ij} \frac{\partial x^i}{\partial s} \frac{\partial x^j}{\partial s} = \Gamma^{\mu}_{ij} v^i v^j[/math] I think that works out. This adds up to the same dimensions as you find in that force equation. But then I start thinking about the dimensions of the equation. [math]m\Gamma^{\mu}_{ij} v^i v^j[/math] How does this equate to a 3-force? If we just take it at face value, the part described as [math]\Gamma^{\mu}_{ij} v^i v^j[/math] would have to have dimensions of acceleration. In the first place, the geodesic equation of motion does not use partial derivative but total derivatives [math]d/ds[/math]. In the second place, [math]s[/math] in the geodesic equation of motion cannot be proper time for massless particles. In the third place, pmb's [math]v^i[/math] cannot be derivative with respect to s, unless he is now using another notation than that he used before here when he tried to define kinetic momentum (e.g. in #3). In the fourth place, [math]m\Gamma_{ij}^{k}v^i v^j[/math] is not a gravitational force because it includes a Christoffel symbol! Gravitation is not a force in general relativity. In the fifth place, you are lacking a minus sign in one of the equations. Edited June 23, 2012 by juanrga Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 23, 2012 Share Posted June 23, 2012 (edited) It was already commented before that one cannot universally claim that eA is a field-like momentum. since you insist on repeating the claim I will add more info. If one restricts himself to field electrodynamics, sure that eA is related to the field, but when one works with Wheeler-Feynman electrodynamics or with more advanced and recent formulations of electrodynamics the term eA is not related to any field because A is given as a functional of particles path. In the first place, the geodesic equation of motion does not includes partial derivative but total ones [math]d/ds[/math]. In the second place, [math]s[/math] in the geodesic equation of motion cannot be proper time for massless particles. In the third place, pmb's [math]v^i[/math] cannot be derivative with respect to s, unless he is now using another notation than that he used before here when he tried to define kinetic momentum (e.g. in #3). In the fourth place, [math]m\Gamma_{ij}^{k}v^i v^j[/math] is not a gravitational force because it includes a Christoffel symbol! Gravitation is not a force in general relativity. Well, his paper defined it as the gravitational 3-force. Don't shoot the messenger, this is why I asked the questions I did. I would say, should it be surprising that a force can have a christoffel symbol in it? I mean... after all, the Christoffel Symbol (is the gravitational field) in GR. (more) I have the geodesic equation written down somewhere, I was working by memory... but if you say so. The equation however will not describe massless particles, since the equation has M defined as a gravitational charge, or passive mass in other words. So yes, s would be the proper time interval. Here it is. You where right, it is not partial derivatives, but it does use proper time. http://en.wikipedia.org/wiki/Geodesic_(general_relativity) Edited June 23, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

juanrga Posted June 23, 2012 Share Posted June 23, 2012 In the first place, the geodesic equation of motion does not use partial derivative but total derivatives [math]d/ds[/math]. In the second place, [math]s[/math] in the geodesic equation of motion cannot be proper time for massless particles. In the third place, pmb's [math]v^i[/math] cannot be derivative with respect to s, unless he is now using another notation than that he used before here when he tried to define kinetic momentum (e.g. in #3). In the fourth place, [math]m\Gamma_{ij}^{k}v^i v^j[/math] is not a gravitational force because it includes a Christoffel symbol! Gravitation is not a force in general relativity. Well, his paper defined it as the gravitational 3-force. Don't shoot the messenger, this is why I asked the questions I did. I would say, should it be surprising that a force can have a christoffel symbol in it? I mean... after all, the Christoffel Symbol (is the gravitational field) in GR. (more) I have the geodesic equation written down somewhere, I was working by memory... but if you say so. The equation however will not describe massless particles, since the equation has M defined as a gravitational charge, or passive mass in other words. So yes, s would be the proper time interval. Here it is. You where right, it is not partial derivatives, but it does use proper time. http://en.wikipedia....ral_relativity) No shooting here, only a polite and impersonal remark about the non-existence of gravitational forces in GR. The christoffel symbols are not gravitational fields. The christoffel have no true physical meaning. The mass of the particle does not appear in the geodesic equation of motion by virtue of the equivalence principle. And no, s is not proper time if the particle is massless. The geodesic equation is the same but s is not proper time then. Link to comment Share on other sites More sharing options...

pmb Posted June 24, 2012 Author Share Posted June 24, 2012 Pmb, where was the thread you posted your work on mass? Was it this one, I had been looking for it because I had a question. I don't recall the thread. The paper I wrote about the concept of mass is found at http://arxiv.org/abs/0709.0687 In your paper, you define the gravitational charge component in a gravitational 3-force equation. I believe it had the form [math]f^k = m\Gamma_{ij}^{k}v^i v^j[/math] I just wanted to know, how one would derive this equation. What you have as a superscript should be a subscript. The derivation is at http://home.comcast.net/~peter.m.brown/gr/grav_force.htm The same derivation can be found in Basic Relativity by Richard A. Mould. The author was motivated by Moller's text so I assume you can also find that derivation there as well. Presumably the connection is playing your usual role of the gravitational field yes? Yes. To be precise it was Einstein who made this identification, not I. I see there is a dependance on the velocity. The [math]\Gamma_{ij}^{k}v^i v^j[/math] part looks a little bit like it comes from the geodesic equation of motion which involves usually a term of second derivatives. All forces in relavity are velocity dependant, e.g. like the Lorentz force on a charged partilce. The equation I had in mind was this one...So I am just wondering how the force equation ''gets'' the terms it has. I'm especially interested in the dependence of the velocity and how this comes about. You'll see when you follow the derivation in the URL I posted. This adds up to the same dimensions as you find in that force equation. But then I start thinking about the dimensions of the equation. [math]m\Gamma^{\mu}_{ij} v^i v^j[/math] How does this equate to a 3-force? If we just take it at face value, the part described as [math]\Gamma^{\mu}_{ij} v^i v^j[/math] would have to have dimensions of acceleration. The expression is supposed to be in Newtons, not m/s^{2}. I'm sure that if you were to write the expression out in all its detail and I'm sure that you the dimensions are correct. Meanwill I'd double check it for myself when I have the time. Link to comment Share on other sites More sharing options...

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