juanrga Posted May 28, 2012 Share Posted May 28, 2012 (edited) I prefer the mine which in c=1 units and convention summation (i.e. sum over index i) is [math]\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2} = \sqrt{(T^{00})^2 - \sum_{ij} \delta_{ij}T^{i0}T^{j0}}[/math] This also gives the energy in the general case when the electric field is not zero and coincides with the fundamental definition used in particle physics when the above is applied to matter in a quantum context. I forgot to ask before, why do you think that energy flux would contribute to the energy density? In my expression [math]T^{00}[/math], [math]T^{i0}[/math], and [math]T^{j0}[/math] are densities (recall I am using c=1). The energy density consists of two terms: an 'inertial' term (given by mass) plus a 'kinetic' term (associated to motion). This is the same structure that energy has for particles as electrons. When there is not electric field the mass density of the EM field is [math]\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2} = T^{00}[/math] which coincides with yours (when reintroducing c). Edited May 28, 2012 by juanrga Link to comment Share on other sites More sharing options...

pmb Posted May 28, 2012 Author Share Posted May 28, 2012 (edited) In my expression [math]T^{00}[/math], [math]T^{i0}[/math], and [math]T^{j0}[/math] are densities (recall I am using c=1). The energy density consists of two terms: an 'inertial' term (given by mass) plus a 'kinetic' term (associated to motion). This is the same structure that energy has for particles as electrons. When there is not electric field the mass density of the EM field is [math]\rho \equiv \sqrt{(T^{00})^2 - (T^{i0})^2} = T^{00}[/math] which coincides with yours (when reintroducing c). What you're doing is to use the expression [math]E^2 - (pc)^2 = m^2c^4[/math] which was derived for isolated particles. Your application to a density turns out to be invalid in general and in particular it doesn't work in this case. The reason you don't recognize your mistake is because you are operating under the assmption that what holds for an isolated single particle will also hold for a continuos medium like a magnetic field. By this I mean that you assume that inertial mass density equals mass density (c=1). That relationship doesn't hold in this scenario. Try transforming to another frame and see if your attempt yields the same result since you're trying to get an invariant mass density. Try again. This time go back to the definition of mass instead of an equality that was derived for particles. Edited May 28, 2012 by pmb Link to comment Share on other sites More sharing options...

juanrga Posted May 28, 2012 Share Posted May 28, 2012 (edited) What you're doing is to use the expression [math]E^2 - (pc)^2 = m^2c^4[/math] which was derived for isolated particles. Your application to a density turns out to be invalid in general and in particular it doesn't work in this case. The reason you don't recognize your mistake is because you are operating under the assmption that what holds for an isolated single particle will also hold for a continuos medium like a magnetic field. By this I mean that you assume that inertial mass density equals mass density (c=1). That relationship doesn't hold in this scenario. Try transforming to another frame and see if your attempt yields the same result since you're trying to get an invariant mass density. Try again. This time go back to the definition of mass instead of an equality that was derived for particles. For interacting particles one merely substitutes the total energy and momentum by the kinetic energy and kinetic momentum [math]K^2 - (\pi c)^2 = m^2c^4[/math] For fields there is no kinetic vs. non-kinetic distinction, only the total energy and momentum are defined locally for fields. Therefore don't worry about that. I have not assumed what you say. What I did was to apply the transition to the continuum limit as is done in mechanics to derive the equations of continuum systems. In fact a field is merely a collection of particles (harmonic oscillators). I do not know what you mean by "you assume that inertial mass density equals mass density", because I do not know exactly what you mean by inertial mass (I only can guess). I do not know why you think that I am trying to obtain an "invariant mass density". Where I said or even suggested that a density is an invariant? Nowhere I said such one thing about the density. Edited May 28, 2012 by juanrga Link to comment Share on other sites More sharing options...

pmb Posted May 28, 2012 Author Share Posted May 28, 2012 (edited) Before I continue I'd like to make one thing clear - I have no intention of ever getting into a debate regarding a definition of a term used in physics ever again on the internet. That's my goal and I might slip accidently. But I get all of you, please don't try to drag me into a discussion about a definition. I hate them and they serve no purpose. I will introduce a term to somone who has never heard of it but that'll be the extent. I might discuss it e-mail. People have much less to prove in PMs. That's why I don't mind PMs. For interacting particles one merely substitutes the total energy and momentum by the kinetic energy and kinetic momentum [math]K^2 - (\pi c)^2 = m^2c^4[/math] You didn't define what K is. If you mean E then you're using a new symbol that is rarely if ever never used to express inertial energy. And that relationship works only for particles. It doesn't hold for continuous media. For fields there is no kinetic vs. non-kinetic distinction, ... I have no idea what you mean by that. Please explain. ...only the total energy and momentum are defined locally for fields. You're forgetting that a field is an example of a continuous medium. The mathematical object that used to find the mass of something is the stress-energy-momentum tensor. So far you haven't taken the stress in the magnetic field into account. Are you aware that stress has inertia? I do not know what you mean by "you assume that inertial mass density equals mass density", because I do not know exactly what you mean by inertial mass (I only can guess). You don't know what inertial mass is? The inertial mass m of a particle is the m in p = mv. The proper mass M of a particle is the M in p = Mv/sqrt(1-v^2/c^2). m and M are related by m = M/sqrt(1-v^2/c^2). Sometimes people use m for proper mass and M for inertial mass. The people who prefer to use proper mass instead of inertial mass also prefer to use the term "inertial mass" to refer to proper mass. I choose inertial mass to mean the m in p = mv. I do not know why you think that I am trying to obtain an "invariant mass density". Where I said or even suggested that a density is an invariant? Nowhere I said such one thing about the density. You have a habit sometimes of just tossing out equations and expecting everyone to know what its supposed to mean. This is one time that you can't do that. Especially since you wrote an expression that has never written in relatvity (because it's wrong for your use). You wrote rho = sqrt (energy density^2 - momentum densoity ^2) without explaining why. The M in M = sqrt(E^2 - P^2) is often called the invariant mass of a particle. You never offered an eplanation of why you were doing what you were doing. One can't just toss out equations and expect people to know why you're doing what you're doing. Especially in this case. Tell you what. I don't think that it's a good idea for me to show you the solution until you learn the physics of the inertial of stress. A nice text that does this is by either Rindler or Mould. Hit the library and learn about the mechanics of continuous media. I really don't expect that you'd accept the solution because I don't think you'd accept the definitions used in the relativity community. Or you could get a copy of the article The inertia of stress, Rodrigo Medina, Am. J. Phys. 74(11), November 2006 Abstract - We present a simple example in which the importance of the inertial effects is evident. The system is an insulating solid narrow disk whose faces are uniformly charged with equal charges of equal magnitude and opposite signs. The motion of the system in two different directions is considered. It is shown how the contribution of energy and momentum of the stress that develops inside the solid to balance the electromagnetic forces have to be added to the electromagnetic contributions to obtain the results predicted by the relativistic equivalence of mass and energy. To others - I'll send it to you in e-mail. I've already sent it to one person. Edited May 28, 2012 by pmb Link to comment Share on other sites More sharing options...

juanrga Posted May 28, 2012 Share Posted May 28, 2012 (edited) You didn't define what K is. If you mean E then you're using a new symbol that is rarely if ever never used to express inertial energy. And that relationship works only for particles. It doesn't hold for continuous media. I said that K is the kinetic energy. The symbol K is also standard. Kinetic energy and total energy are two different concepts. I have no idea what you mean by that. Please explain. For matter one can split the total energy and the total momentum into a kinetic plus an interaction term as correspond to the structure of the Hamiltonian. There is no such split for electromagnetic radiation. You're forgetting that a field is an example of a continuous medium. It is difficult to believe because in the message that you are replying I wrote: What I did was to apply the transition to the continuum limit as is done in mechanics to derive the equations of continuum systems The mathematical object that used to find the mass of something is the stress-energy-momentum tensor. So far you haven't taken the stress in the magnetic field into account. Are you aware that stress has inertia? Are you aware that my definition of mass uses components of [math]T^{\mu\nu}[/math]? I have not used the stress components of the tensor, because they do not define the mass of a system. I am using a standard concept of mass which is denoted by m (this is not your m below). You don't know what inertial mass is? The inertial mass m of a particle is the m in p = mv. The proper mass M of a particle is the M in p = Mv/sqrt(1-v^2/c^2). m and M are related by m = M/sqrt(1-v^2/c^2). Sometimes people use m for proper mass and M for inertial mass. The people who prefer to use proper mass instead of inertial mass also prefer to use the term "inertial mass" to refer to proper mass. I choose inertial mass to mean the m in p = mv. I wrote something different: I do not know exactly what you mean by inertial mass (I only can guess). It is only now that you give a precise definition of what you mean by inertial mass. You have a habit sometimes of just tossing out equations and expecting everyone to know what its supposed to mean. This is one time that you can't do that. Especially since you wrote an expression that has never written in relatvity (because it's wrong for your use). You wrote rho = sqrt (energy density^2 - momentum densoity ^2) without explaining why. The M in M = sqrt(E^2 - P^2) is often called the invariant mass of a particle. You never offered an eplanation of why you were doing what you were doing. One can't just toss out equations and expect people to know why you're doing what you're doing. Especially in this case. Anyone can ask for clarifications, but attributing to me stuff that I never said is something that I am not going to accept. Tell you what. I don't think that it's a good idea for me to show you the solution until you learn the physics of the inertial of stress. A nice text that does this is by either Rindler or Mould. Hit the library and learn about the mechanics of continuous media. I really don't expect that you'd accept the solution because I don't think you'd accept the definitions used in the relativity community. Or you could get a copy of the article The inertia of stress, Rodrigo Medina, Am. J. Phys. 74(11), November 2006 To others - I'll send it to you in e-mail. I've already sent it to one person. Thank you very much for your advice and the article, although I doubt that it can be useful for me. Time ago I abandoned the continuum limit approximation and I almost always work with more accurate and fundamental descriptions of nature. Precisely in one or two weeks I will plan to start to write a paper on a generalized energy-momentum-stress tensor [math]\Theta^{\mu\nu}[/math] which includes gravitational effects that are not included in the ordinary [math]T^{\mu\nu}[/math] used in general relativity. But thanks in any case. Edited May 28, 2012 by juanrga Link to comment Share on other sites More sharing options...

pmb Posted May 28, 2012 Author Share Posted May 28, 2012 Thank you very much for your advice and the article, although I doubt that it can be useful for me. And that's why you won't get the answer. You're just one of those people who, when they don't get the answer right they attempt to change physics so that the do get it right. Link to comment Share on other sites More sharing options...

pmb Posted May 29, 2012 Author Share Posted May 29, 2012 I decided not to post that. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 8, 2012 Share Posted June 8, 2012 (edited) People often ask me about when proper mass can't work as good as relativistic mass or even better. I wrote an entire paper to answer this question but people don't read it carefully enough. Probably because they believe that no matter what the paper says they've already made up their mind because the already thought about it carefully alread a long time ago. And that is a good reason. I'd probably do the same thing - Too much reason with no real expectations of changing what they think in any sape or form So to take a shot at perhaps clarifying why physicists hold on to the notion of relativist mass. So I've decided to create a challenge for everyone. I have a SR text by Hans C. Ohanian. One of the homework problems is to find the mass density of a magnetic field. That's my challenge to you all. Solve this introductory level SR problem. Let's make it as simple as possible and assume that the magnetic field be uniform. Find the mass density of a the magnetic field. Use whatever definition of mass that you see fit. Good luck. So... you want someone here, to show how one would find the mass density of an electromagnetic field... For a non-dispersive material, the energy density of a magnetic field is [math]\rho = \frac{B \cdot B}{2\mu}[/math] I did find this paper which helps determine a mass density along magnetic field points. http://www-pw.physics.uiowa.edu/~dag/publications/2001_DeterminingTheMassDensityAlongMagneticFieldLinesFromToroidalEigenfrequenciesPolynomialExpansionAppliedToCRRESData_JGR.pdf Edited June 8, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

pmb Posted June 13, 2012 Author Share Posted June 13, 2012 (edited) So... you want someone here, to show how one would find the mass density of an electromagnetic field... No. I didn't want anything. I was merely making a challenge. I think I decided not to post the result in open forum. If you'd like to see the answer please send me a PM and I'll send it to you. Thanks for the link. When I get a printer I'll prinit it out and read it. I suspect that is wrong though. I can't concentrate reading involved documents online. Sitting too long causes pain and the pain distracts me. By the way, you gave the value of the energy density for a magnetic field, not the mass density. Please take note that if the answer was just mass density = energy density/c[sup2[/sup] then I'd never have asked. That would be trivial and not worth discussing about. The answer is also in this paper that I wrote - http://arxiv.org/abs/0709.0687 Edited June 13, 2012 by pmb Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 13, 2012 Share Posted June 13, 2012 (edited) No. I didn't want anything. I was merely making a challenge. I think I decided not to post the result in open forum. If you'd like to see the answer please send me a PM and I'll send it to you. Thanks for the link. When I get a printer I'll prinit it out and read it. I suspect that is wrong though. I can't concentrate reading involved documents online. Sitting too long causes pain and the pain distracts me. By the way, you gave the value of the energy density for a magnetic field, not the mass density. Please take note that if the answer was just mass density = energy density/c[sup2[/sup] then I'd never have asked. That would be trivial and not worth discussing about. The answer is also in this paper that I wrote - http://arxiv.org/abs/0709.0687 Can you define [math]\Box[/math] this more clearly, you say Minkowski defined it as the mechanical mass, but I have never seen this before - I believe this symbol is more generally used to define the d'Alembertian. Edited June 13, 2012 by Aethelwulf Link to comment Share on other sites More sharing options...

pmb Posted June 13, 2012 Author Share Posted June 13, 2012 Can you define [math]\Box[/math] this more clearly, you say Minkowski defined it as the mechanical mass, but I have never seen this before - I believe this symbol is more generally used to define the d'Alembertian. Yes, it's the d'Alembertian operator. If/when you see that in my paper its because there was an error translating it from an MS Word file to a PDF file. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 13, 2012 Share Posted June 13, 2012 So, are you saying in your paper, its not meant to be the operator, because it was an error when it was error in the translation? Link to comment Share on other sites More sharing options...

pmb Posted June 13, 2012 Author Share Posted June 13, 2012 So, are you saying in your paper, its not meant to be the operator, because it was an error when it was error in the translation? Yes. Link to comment Share on other sites More sharing options...

Aethelwulf Posted June 13, 2012 Share Posted June 13, 2012 Can I ask what it is meant to be then? thank you. Link to comment Share on other sites More sharing options...

pmb Posted June 13, 2012 Author Share Posted June 13, 2012 (edited) Can I ask what it is meant to be then? thank you. I think that it appears anywhere a greek symbol appears inline. Edited June 13, 2012 by pmb Link to comment Share on other sites More sharing options...

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