gib65 Posted July 2, 2008 Posted July 2, 2008 Are photons the only particles that don't exert a gravitational force on other particles?
Gilded Posted July 2, 2008 Posted July 2, 2008 No indeed, mostly because they do exert a gravitational force on other particles. Gravitons don't, but they're hypothetical particles.
ecoli Posted July 2, 2008 Posted July 2, 2008 a massless particle exerts gravitational forces? What in the world is going on here?
gib65 Posted July 2, 2008 Author Posted July 2, 2008 a massless particle exerts gravitational forces? What in the world is going on here? Yeah, that's exactly what I want to know. So if you took two photons and fired them at the same time in the same direction (say an inch apart) they would eventually pull themselves together under their gravitational influence?
timo Posted July 3, 2008 Posted July 3, 2008 The problem with your question is that in any widely-accepted theory that has photons (i.e. quantum theories of the electromagnetic field) there is no handle for gravitation. If you come from pure Relativity, then energy and momentum density contribute to a term that creates curvature of space (the energy-momentum tensor) - although for a universe with only a single photon you could probably make that term vanish by taking the limit to a state where the photon has zero energy. A more simple answer that you are probably looking for: The sources of gravitation are energy and momentum. A photon has both. That said, I'd like to re-emphasize that photons are the result of quantum mechanics (a special quantized base of the electromagnetic field) and that we do not have a theory of quantum gravity. While it would violate many of the guidlines in modern physics I am not even sure that we have a convinving (=experimental) evidence against the statement that gravity only works by/on gluons, for example.
Mr Skeptic Posted July 3, 2008 Posted July 3, 2008 So does this mean that "relativistic mass" really is mass, despite the particle physicists not liking it?
Pete Posted July 3, 2008 Posted July 3, 2008 (edited) Are photons the only particles that don't exert a gravitational force on other particles? Photons have both active and passive gravitational mass. That means that they generate gravitational fields and are effected by gravitational fields. Mass is the source of gravity. The stress-energy-momentum (SRM) tensor provides a complete description of gravity. It is not enough to associate the components of the SEM tensor as sources because that is insufficient. Pressure is also a source of gravity. The active gravitational mass density rho of a relativistic fluid is a function of the energy density u and its pressure p. i.e. (let c = 1) rho = u + 3p Yeah, that's exactly what I want to know. So if you took two photons and fired them at the same time in the same direction (say an inch apart) they would eventually pull themselves together under their gravitational influence? No. Photons moving parallel to each other generate a gravitational field which doesn't exert a gravitational force on photons moving in the same direction (the gravitational force is velocity dependant). If they are moving in opposite directions then there will be a force. So does this mean that "relativistic mass" really is mass, despite the particle physicists not liking it? Yes. Note that the only thing particle physicists work with are particles. As such it makes no difference what you call mass. But in general it does make a difference and physicists who work with relativity consider more than just particles and as such proper mass (aka rest mass) is not even meaningfull in the most general circumstances. Pete Edited July 3, 2008 by Pete
gib65 Posted July 3, 2008 Author Posted July 3, 2008 Wow, that's interesting. Completely throws my misconceptions for a spin. So then are there any particles in the universe that don't exert gravity (gravitons notwithstanding)?
Mr Skeptic Posted July 3, 2008 Posted July 3, 2008 So then are there any particles in the universe that don't exert gravity (gravitons notwithstanding)? Maybe virtual particles?
timo Posted July 3, 2008 Posted July 3, 2008 So does this mean that "relativistic mass" really is mass, despite the particle physicists not liking it? It's more like relativistic mass really just being another word for the energy, despite the "everything is relative" crowd not liking it.
Pete Posted July 3, 2008 Posted July 3, 2008 It's more like relativistic mass really just being another word for the energy, despite the "everything is relative" crowd not liking it. The relativistic mass of an object is defined as the ratio of the magnitude momentum of that object to the object's speed. Inertial energy (defined as the sum of kinetic energy and rest energy) is defined otherwise. For that reason relativistic mass it not another name for energy. In fact energy includes potential energy of position which is not included in relativistic mass. Relativistic mass is not even proportional to inertial energy in general. I.e. if one were to calculate the general expression for mass density and compare it to the general expression for inertial energy density they'd see that they are quite different. Pete
iNow Posted July 3, 2008 Posted July 3, 2008 As I understand it, though, relativistic mass is not really used anymore to any great extent, and is generally frowned upon. When someone says "mass," it can generally be assumed that they mean rest mass unless otherwise indicated. It's not like if we don't say "assuming a spherical earth" in a post talking about earth that people won't know whether we're discussing a flat earth or spherical one...
Pete Posted July 3, 2008 Posted July 3, 2008 As I understand it, though, relativistic mass is not really used anymore to any great extent, and is generally frowned upon. I don't know about whether its not used to a great extent or not (because I'm unsure as to the precise meaning of that phrase). That's difficult to say. I myself have researched this subject and found that its not possible to define mass to refer to what is otherwise called rest mass (which I myself refer to as proper mass for various reasons). The mass = relativistic mass is completely consistent, logical and works in all possible circumstances. I've explained my reasons in a paper I wrote on the subject. For those who would like to take a look at it then you can find it online at - http://arxiv.org/abs/0709.0687 There have been many articles written on this topic. One is online at http://arxiv.org/abs/physics/0504110. The article is called On the Abuse and Use of Relativistic Mass, by Gary Oas, last revised 21 Oct 2005. In this article the author makes a very relevant observation, i.e. The discussion up to this point has been intended to demonstrate that the concept of relativistic mass is problematic. For those that already understand this point we would like to point out that this notion is still prevalent in the literature on relativity. To this end the results of an extensive, though not exhaustive, literature search are now discussed. The results of the authors research shows that 75% of the literature in the survey employs the concept of relativistic mass. Scientific journals were left out of the survey so as to keep the survey of a manageable size. When someone says "mass," it can generally be assumed that they mean rest mass unless otherwise indicated. I'm not sure what is meant by "generally be assumed." Each instance must be evaluated individually. One can't merely assume that it means rest mass. I'm sure you didn't mean that though. I believe that you meant that if one were to take a survey then you meant that in most cases the term would mean rest mass. However since most of the literature in the survey employs relativistic mass, and since authors who do so simply call it mass, then your assumption is generally not true. Especially since a great deal of the relativity literature was published several decades ago. Some of the most important relativity texts of all use the term "mass" to mean relativistic mass and they didn’t even mention the term "relativistic mass" at all. Case in point; Gravitation by Misner, Thorne and Wheeler. I examined that text for such usage and came across two important examples. The first had to do with the symmetry of the stress-energy-momentum tensor. The second was in their statement of the source of gravity. In those two, very important, cases they used the term mass unqualified to mean relativistic mass. I myself did a short survey and have recorded the results in this web page http://www.geocities.com/physics_world/ref/relativistic_mass.htm Pete
timo Posted July 3, 2008 Posted July 3, 2008 The relativistic mass of an object is defined as the ratio of the magnitude momentum of that object to the object's speed. I see three quirks in that statement: 1) It's supposedly not what the majority of people mean when they say "relativistic mass". I'd think most people actually do mean the energy and define it via m=E/c². I have not heard your definition before. To be honest: I am not convinced that many others have, either. 2) Taking your statement literally your RE wasn't even properly defined. I assume that for the v=0 case one has to think of a limit, though. 3) What would the so-defined quantity be good for?
Pete Posted July 5, 2008 Posted July 5, 2008 (edited) I see three quirks in that statement:1) It's supposedly not what the majority of people mean when they say "relativistic mass". I'd think most people actually do mean the energy and define it via m=E/c². I have not heard your definition before. To be honest: I am not convinced that many others have, either. I agree. It is true that the majority of people believe that relativistic mass is defined as m = E/c2. That is unfortunate. When momentum is defined as p = mv where m is a proportionality factor the m is called the mass of the particle. When it is defined as such people often refer to it as relativistic mass. The term is well defined in the relativity literatare as the m in p = mv. For example; if you have the text Special Relativity, by A.P. French then take a look at the footnote on page 16 which reads By inertial mass we mean the ratio of linear momentum to velocity. If you have The Feynman Lectures on Physics - Vol. I then turn to page 16-6 which reads We thus write the momentum vector as a certain coefficient times the vector velocity; p = mvv We put the subscript v on the coefficient to remind us that it is a function of velocity., and we shall agree to call this coefficient mv the "mass." Similarly, from The Theory of Relativity, by C. Moller, page 65 Therefore, to a material particle moving with the velocity u relative to a system of inertia S, we shall assign a momentum vector p proportional to u p = mu The proportionality factor m is called the mass of the particle. Those are just three examples. It can be traced back to Richard Tolman who seems to be the first physicist to employ this concept. Herman Weyl defined mass in this way. 2) Taking your statement literally your RE wasn't even properly defined. In what sense? I assume that for the v=0 case one has to think of a limit, though. Absolutely. A similar notion is employed if one were to defined inertial mass as the ratio m = F/a. One takes the limit a -> 0. 3) What would the so-defined quantity be good for? The answer to this question is identical to the question What is mass good for? An important fact in relativity is that stress/pressure contributes to inertia, i.e. the momentum of a body is a function of both energy and stress. This fact would not be reflected if one were to define inertial mass (aka relativistic mass) as the ratio E/c2. This definition holds in all conceivable cases. Using this definition it becomes obvious that pressure is a source of gravity. That's how one obtains the expression rho = (u + 3p) for the active gravitational mass of a body. Strangely enough the inertial mass density is rho = (u + p) I haven't figured out why there is a difference yet. One of the puzzles I am seeking to find a solution to. Does anybody know why there is a difference? It is important to keep in mind that m = p/v is the definition of relativistic mass whereas m = \gamma m0 is an equality. Once mass is defined by m = p/v the equality can be derived by applying the principle of conservation of momentum to two particles which undergo an elastic collision. It can then be shown that, under certain circumstances, the relativistic mass is related to total inertial energy by E = mc2. If the system under consideration is a particle then that relation will hold. If the body is an isolated system then it will hold in that case too. However if a body is under stress then E = mc2 will not hold. However the mass is still related to the momentum as m = p/v. It will always hold in fact because its a definition and not an equality. I imagine that what I said above regarding the fact that E = mc2 is not always true will come as a surprise to most people. For that reason I think it'd be a good idea if I quote a special relativity text. From Introduction to Special Relativity, by Wolfgang Rindler, page 150 It would seem at first sight that \rho and \rho0 should be related by the equation \rho = \gamma(u)2\rho0, where one \gamma is due to length contraction affecting what is a unit volume in the rest frame, and the other is due to mass increase according to formula (26.3). But that simple formula is valid only in certain special cases, e.g. for single particles and for systems of free particles[..]. It is not generally valid for constrained systems. That this is true is readily seen by examining the components of the stress-energy-momentum tensor. In a frame of reference which is moving relative to a body the momentum density will be a function of the x-component of stress. This is also the case in the new relativity FAQ regarding the relativistic mass of the photon which is at http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html Pete Edited July 5, 2008 by Pete multiple post merged
swansont Posted July 5, 2008 Posted July 5, 2008 Those are just three examples. It can be traced back to Richard Tolman who seems to be the first physicist to employ this concept. Herman Weyl defined mass in this way. Just want to note that in your three examples, none of them called the term "relativistic mass"
Pete Posted July 5, 2008 Posted July 5, 2008 (edited) Just want to note that in your three examples, none of them called the term "relativistic mass"The title of the section in which this is found is entititled Relativist mass. Pete Edited July 5, 2008 by Pete
pioneer Posted July 5, 2008 Posted July 5, 2008 (edited) If you look at SR, there are three equations, one each for time, distance and mass. The photon has zero mass, so it can travel at C. But it still has SR affected parameters in distance and time, reflected by wavelength and frequency. This is the massless momentum. This space-time aspect of energy has a connection to GR. If we add it all together, photons do not exert Newtonian Gravity due to zero rest mass, but can contribute to GR. This difference becomes more apparent only at very high mass which generates heat energy-mass but not any additional rest mass. For example, if we had a cloud of space gas with mass M, as it contracts the gravity work will add energy to the system. If this energy-mass was going into Newtonian mass we would see mass increasing as it contracted. All we see is the rest mass conserved. The energy only adds to GR so Newtonian breaks down. Edited July 5, 2008 by pioneer
Pete Posted July 5, 2008 Posted July 5, 2008 If you look at SR, there are three equations, one each for time, distance and mass. /quote] I don't understand this comment. What are these three equations that you are talking about? The photon has zero mass, so it can travel at C. But it still has SR affected parameters in distance and time, reflected by wavelength and frequency. This is the massless momentum. To be precise; a photon has zero proper mass and a well-defined, non-zero inertial mass (as well as a non-zero active gravitational mass and non-zero passive gravitational mass) which I refer to simply as mass. This space-time aspect of energy has a connection to GR. If we add it all together, photons do not exert Newtonian Gravity due to zero rest mass, but can contribute to GR. I don't understand your reason for saying this. What exactly do you mean when you say photons do no exert Newtonian Gravity? If you mean that photons don't exert a gravittional forces on other particles then that is clearly wrong. Of course I'm assuming that a photon can be treated as a localized EM field that propagates through space with speed c. You do understand that a beam of light generates a gravittional field, don't you, Pete
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