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Natural Science Forum / Physics / Relativity / December 2006Relativistic Mass - Part 1,865,234
Yep. Back to relativistic mass again. During the time I've been gone from the newsgroup I've continued to learn as much as I could about the concept of mass in relativity. I don't wish to rehash what we've gone over in the past but to cover aspects that haven't been covered anywhere else on the web save a forum or two I frequented. In a rather new SR text I have the author asks a question regarding what the energy density and the mass density is of a given magnetic field. I recall a past conversation where someone disagreed with me regarding my answer. That person held the opinion that the mass density is that mass density as measured in the rest frame. I.e. go to the rest frame of the magnetic field and calculate the energy density. Then divide through by c^2 to get the mass density (i.e. rest mass density). Seems simple doesn't it? However if you take a good look at this you'll see that its not as simple as we thought it was. The problem is that we first have to determine what is to be the rest frame for this magnetic field. I choose to think of it in terms of a bar magnet and the rest frame of a bar magnetic would then be the rest frame of the field. Since we are asked to evaluate the energy density then we are given a problem in which we are examining a non-closed system. An infinitesimal volume element of the field cannot be said to be isolated. Therefore there cannot be a 4-velocity associated with the field density. The relation m^2 = E^2 - p^2 (c=1, m = rest mass) is not valid in this situation either. So we must of necessity go to the definition of mass. Definition of proper mass = m Let p = 3-momentum and v = 3-velocity. Choose M such that p = Mv is a conserved quantity. It is then shown that M is a function of speed and equal to M = gamma*m. Therefore m is defined as m = lim (v->0) M(v). Now we go back to the magnetic field problem. We eveluate the momentum density of a small element of the field and multiply the result by the small volume of the element. We then let v -> 0 and we then have the proper mass. So what's the problem you ask? The problem is that your answer will depend on the direction of motion you've chosen to measure things in. And that is true for all velocities even those near zero. Definition of relativitist mass = M = gamma*m I worked this out and made a web page at http://www.geocities.com/physics_world/sr/mass_mag_field.htm I was inspired to work with this due to an article in the American Journal of Physics by Rindelr and Denur as well as Rindler's intro to SR text. References included in above web page. The reason for this post is to bring to light the wider application of mass in physics. In this case its the mass of continuos media. Something we've never discussed before. This is also related to what Schutz refers to as the inertia of pressure in his books. Last month there was an article in AJP called "The inertia of pressure." I e-mailed the author of that text about this and he was glad I did since he found it pretty cool too. :) Best wishes Pete > Yep. Back to relativistic mass again. During the time I've been gone from > the newsgroup I've continued to learn as much as I could about the concept > of mass in relativity. I don't wish to rehash what we've gone over in the > past but to cover aspects that haven't been covered anywhere else on the web > save a forum or two I frequented. There is no reason to revisit this topic. Special relativiy is defined by the Poincare group. The Poincare group has 2 Casimir operators. One of these is called m^2, which then defines the mass as m = |sqrt(m^2)|. *snip* > The problem is that we first have to determine what is to be the rest frame > for this magnetic field. I choose to think of it in terms of a bar magnet > and the rest frame of a bar magnetic would then be the rest frame of the > field. You have now left the realm of special relativity and entered the realm of electromagnetism. > Since we are asked to evaluate the energy density then we are given a > problem in which we are examining a non-closed system. An infinitesimal > volume element of the field cannot be said to be isolated. Therefore there > cannot be a 4-velocity associated with the field density. That happens to apply to electromagnetism but is not intrinsic to relativity. The field known as the weak interaction consists of massive particles, which can obviously have a four velocity. Special relativity is NOT a theory of electromagnetism. Electromagnetic theory is a relativistic field theory. > The relation m^2 = > E^2 - p^2 (c=1, m = rest mass) is not valid in this situation either. Since the relation, E^2 - p^2 = m^2 comes from integrating the Hamiltonian density over all space, that statement doesn't make sense in the above context. > So we must of necessity go to the definition of mass. In special relativity, that definition comes straight from Poincare invariance. The Poincare group has 10 generators (or Killing vectors, if you prefer to think of it those terms). > Definition of proper mass = m > > Let p = 3-momentum and v = 3-velocity. Choose M such that p = Mv is a > conserved quantity. Then, you have only 9 of the ten generators of the Poincare group and you have not given the relations between those three momenta and the other six generators of the Lorentz group. The result is that you haven't restricted the possible theories to only special relativity. For example, the Galilean group also fits if you don't require invariance under Galilean boosts. In that case, p^2 is not only conserved, it is a constant under changes in velocity. > It is then shown that M is a function of speed and equal > to M = gamma*m. And, if p^2 is a constant, as noted above, M is a function of speed which is NOT equal to gamma m. In special relativity, gamma m is called the energy, so why call it the relativistic mass? You are attempting to do what Lorentz tried to do - find an equivalent interpretation of relativity which retains classical conceots that are inconsistent with special relativity. >> Yep. Back to relativistic mass again. During the time I've been gone from >> the newsgroup I've continued to learn as much as I could about the [quoted text clipped - 5 lines] > > There is no reason to revisit this topic. I didn't start this thread because I thought there was a need. In fact threads aren't always started because of needs. In this case the thread was started because I wanted to share a thought related to continuous media. >> The problem is that we first have to determine what is to be the rest >> frame [quoted text clipped - 4 lines] > You have now left the realm of special relativity and entered the > realm of electromagnetism. Electromagnetism is inherently relativistic. In any case this is the beginning of the topic and not the entire topic. The subject matter is E = mc^2 etc. and electromagnetism is being used as an application for clarity. The subject matter is the relativistic treatment of continuous media. >> Since we are asked to evaluate the energy density then we are given a >> problem in which we are examining a non-closed system. An infinitesimal [quoted text clipped - 4 lines] > That happens to apply to electromagnetism but is not intrinsic to > relativity. It applies to systems. This is about what systems and not what theories it applies to. > Since the relation, E^2 - p^2 = m^2 comes from integrating the > Hamiltonian > density over all space, that statement doesn't make sense in the above > context. E^2 - p^2 = m^2 is derived in many ways. In any case nobody said that E^2 - p^2 = m^2 could be used for anything else but complete systems. The subject matter of this example came from a text which asked a question. The author of the text was trying to apply the definition of mass he defined in the text to the EM field energy density. By his definition it doesn't work as he thought it would, i.e. mass density can't reasonably be applied as he defined it. In that web page it is demonstrate that E^2 - p^2 = m^2 will not work in that case. In fact this thread is about the mass of non-closed systems and its definition. >> It is then shown that M is a function of speed and equal >> to M = gamma*m. [quoted text clipped - 3 lines] > called > the energy, so why call it the relativistic mass? Because they are not always equal. M is not always proportional to energy. In cases such as that of stressed bodies M will not be proportional to the energy of the body. E = Mc^2 only applies in certain circumstances, i.e. when the object is a particle or when it is totally isolated. In the case of a stressed rod it will not hold. However it is still meaningful to speak of the mass of a body even when it is under stress. However there are those who wou,d choose not to define mass for anything but the total mass of a closed system. I take it you're one of those people? > You ... Me? When I create something new which can't be find in the relativity literature then I'll say so. But this topic is found in the relativity literature so I cannot be given credit for the subject matter. So "You" must be replaced with "They". > ...are attempting > to do what Lorentz tried to do - find an equivalent interpretation > of relativity which retains classical conceots that are inconsistent > with special relativity. Nope. I'm simply spelling out the relativistic mechanics of continuous media. What I'm doing is nothing that can't be found in the relativity literature such as the American Journal of Physics or Rindler's SR text. The reason being that if a relativist like the author of that text I spoke of can make an error such as asking for the mass density of a magnetic field and give the wrong answer then its a subjecct worthy of starting a thread for. This is in hopes that others will be aware of the physics and not make the same mistake. Regards Pete > "bergeron" <badd_xi2@yahoo.com> wrote in message > >> The problem is that we first have to determine what is to be the rest [quoted text clipped - 7 lines] > > Electromagnetism is inherently relativistic. Perhaps you missed the point. Relativity is not an electromagnetic theory. Electromagnetism is a relativistic theory. Put another way, relativity is a theory about the geometry of spacetime. Electromagnetism is a theory about one of many possible theories that are consistent with that geometry. See the difference? > In any case this is the > beginning of the topic and not the entire topic. The subject matter is E = > mc^2 etc. and electromagnetism is being used as an application for clarity. > The subject matter is the relativistic treatment of continuous media. Then you need to use field theory. > >> Since we are asked to evaluate the energy density then we are given a > >> problem in which we are examining a non-closed system. An infinitesimal [quoted text clipped - 6 lines] > > It applies to systems. I beg to differ. The radiation fields corresponding to the nuclear force, for example, are mesons. Mesons are massive and therefore have a well defined four-velocity. Same for the weak interaction radiation fields consisting of Ws and Zs. > This is about what systems and not what theories it > applies to. [quoted text clipped - 5 lines] > > E^2 - p^2 = m^2 is derived in many ways. It can be made plausible to students in an introductory course in many ways by identifying those quantities with their Newtonian analogs without questioning the meaning of the Newtonian analogs. However, today, physicists recognize the necessity of identifying quantities with their origin in a theory rather than invite the conflicts which arise from ill-defined quantities. The geometry of special relativity is sufficient to define energy, momentum and mass (as well as other kinematic quantities), just as Galilean relativity is sufficient to derive Newtonian mechanics. If you are trying to define mass differently than that which is derived directly from the spacetime geometry of relativity, either you get the same result or you will eventually arrive at an inconsistency of your own making. There is no third option. > In any case nobody said that E^2 - p^2 = m^2 could be used for anything else > but complete systems. So, do incomplete systems violate the laws of physics? > The subject > matter of this example came from a text which asked a question. The author > of the text was trying to apply the definition of mass he defined in the > text to the EM field energy density. Energy density is energy per unit volume. That is comparable to mass density, not mass. > By his definition it doesn't work as he > thought it would, i.e. mass density can't reasonably be applied as he > defined it. Then, perhaps his definition is at fault. > In that web page it is demonstrate that E^2 - p^2 = m^2 will not > work in that case. In fact this thread is about the mass of non-closed > systems and its definition. In other words, if you introduce energy and don't properly account for it, it becomes mass? *snip* > > And, if p^2 is a constant, as noted above, M is a function of speed > > which is NOT equal to gamma m. In special relativity, gamma m is [quoted text clipped - 4 lines] > In cases such as that of stressed bodies M will not be proportional to the > energy of the body. Sure it will, unless you ignore the physics which underlies the meaning of ``stressed.'' > E = Mc^2 only applies in certain circumstances, i.e. > when the object is a particle or when it is totally isolated. In the case of > a stressed rod it will not hold. However it is still meaningful to speak of > the mass of a body even when it is under stress. Sure, and it doesn't require the rather weird argument you are trying to put forth to do so. > However there are those > who wou,d choose not to define mass for anything but the total mass of a > closed system. I take it you're one of those people? I define mass as one if the Casimir operators of the Poincare group, since that is what is dictated by the Poincare invariance of special relativity. If your definition is not consistent with that, then your definition is also not consistent with relativity. Or, is it your contention that electromagnetism is not Poincare invariant? a constant, as noted above, M is a function of speed > which is NOT equal to gamma m. In special relativity, gamma m is > called > the energy, so why call it the relativistic mass? Actually even for a particle the relativistic mass of a particle will not in general be equal to its energy. E.g. the quantity which is conserved for a particle in free-fall in a conservative field is the time-component of the momentum 1-form, not the time-component of the momentum 4-vector. I.e. it is P_0 which has the right to be called "energy" rather than M = P^0. When one defines a quantity like 4-momentum one must demand that the meaning of the components is not coordinate dependant. If one chooses a coordinate system which corresponds to a non-inertial frame (and where the gravitational potentials are constant) then it is P_0 which is the energy and not P^0. This is a very good reason to have a seperate name for these quantities. The best reason is that they refer to two very different quanties wich totally different meanings. Because they are poportional to each other is no excuse to do away with one of them. Pete <snip> > Definition of proper mass = m > found it pretty cool too. :) > Best wishes > Pete Hi Pete, I think that Physicists generally prefer to talk of invariant mass rather than relativistic mass because the former is independant of velocity wheras the latter tends to infinity at close to c which is a bit of a bugger. If this is partonisingly simple then forgive me - I'm not a physicist and had to work through this stuff myself as you seem to be doing. The way I think of it is m0 = sqrt(E^2/c^4 - p^2/c^2) (invariant) mr = E/c^2 (relativistic from the famous e=mc^2) (where E is en energy & p is momentum) Obviously a quick bit of algebra and you can form the velocity dependant relationship mr = m0 /sqrt(1 - v^2/c^2) > <snip> >> [quoted text clipped - 4 lines] > > Hi Pete, Hi! > I think that Physicists generally prefer to talk of invariant mass ... I understand why they do that. This thead is not about that. Its about the relativistic mechanics of continuos media, specifically mass density. The purpose of this thread is to make aware to others the problems in attempting to define mass denisty. This has little to do with that whole debate ... I hope. > rather than relativistic mass because the former is independant of > velocity wheras the latter tends to infinity at close to c which is a [quoted text clipped - 3 lines] > The way I think of it is > m0 = sqrt(E^2/c^4 - p^2/c^2) (invariant) This expression makes no sense to me. > mr = E/c^2 (relativistic from the famous e=mc^2) > (where E is en energy & p is momentum) Only in certain circumstances. > Obviously a quick bit of algebra and you can form the velocity > dependant relationship > mr = m0 /sqrt(1 - v^2/c^2) Regards Pete > This has little to do with that whole debate ... I hope. My fault - no nuisance intended :-) > > The way I think of it is > > m0 = sqrt(E^2/c^4 - p^2/c^2) (invariant) > > This expression makes no sense to me. I just checked it (it is right, thank God, otherwise even more blushes). You derive it from simply from the basic Einstein energy term: E = (Mo.c^2)/(SQRT(1-(v^2/c^2)) This can be played with by substituting relativistic momentum for relativistic mass (ie. p = γp0 = mrel * v).and it gives the above equation. Obviously, as you quite rightly point out, only when v=0 (and therefore p=0) do you get e=mc^2 (Are there any methods for getting symbols into postings so formulae can be read easier? On MBoards you can often switch a font to good effect..I'm not really familiar with Usenet (not used it for a couple of decades).) Regards (and sorry for sidetracking your thread) Chris Pmb wrote: > This has little to do with that whole debate ... I hope. My fault - no nuisance intended :-) > > The way I think of it is > > m0 = sqrt(E^2/c^4 - p^2/c^2) (invariant) > > This expression makes no sense to me. I just checked it (it is right, thank God, otherwise even more blushes). You're correct. I screwed up. > You derive it from simply from the basic Einstein energy term: > E = (Mo.c^2)/(SQRT(1-(v^2/c^2)) This can be played with by substituting relativistic momentum for relativistic mass (ie. p = ?p0 = mrel * v).and it gives the above equation. Obviously, as you quite rightly point out, only when v=0 (and therefore p=0) do you get e=mc^2 When v = 0 then E_0 = m_0 c^2 where m_0 is proper mass. If you wish to use m to refer to proper mass then the expression E = mc^2 is incorrect since E is the energy of the particle and not the rest energy. It is the rest energy that is proportional to proper mass. But you probably know all this anyway. >(Are there any methods for getting symbols into postings so formulae can be read easier? I wish! :) >On MBoards you can often switch a font to good effect..I'm not really familiar with Usenet (not used it for a couple of decades).) Regards (and sorry for sidetracking your thread) Chris Thanks Chris Pete > When v = 0 then E_0 = m_0 c^2 where m_0 is proper mass. If you wish to use m > to refer to proper mass then the expression E = mc^2 is incorrect since E is > the energy of the particle and not the rest energy. It is the rest energy > that is proportional to proper mass. But you probably know all this anyway. Do you hear that scraping sound ? That's my fingernails clawing at the cliff as I try to avoid falling! I'm struggling to try and bring my math up to a level which is at least not humiliating, and so, no, I don't know a lot of this :-) (I'm just struggling through Penrose's Road to Reality at the moment so in a few years I might be able to disguise my ignorance better - hopefully before I get to retirement age :-) Best wishes Chris PS - I promise not to interrupt again off=topic. >> When v = 0 then E_0 = m_0 c^2 where m_0 is proper mass. If you wish to >> use m [quoted text clipped - 5 lines] > Do you hear that scraping sound ? That's my fingernails clawing at the > cliff as I try to avoid falling! LOL! Don't worry. We've all been there. And I think its safe to say that this happens to almost everyone when they attempt to understand some physics they haven't worked with in years. E.g. right now I'm going over Ohanian's EM text cover to cover to refresh myself on the subject matter and to prepare myself for Jackson's text. So I'm in there struggling with you Chris! :) > I'm struggling to try and bring my math up to a level which is at least > not humiliating, and so, no, I don't know a lot of this :-) Feel free to contact me in e-mail. I'd be most happy to help you to the best of my abilities. I'm disabled at the moment so I have all the time in the world to help you. It would also be benificial to me too since explaining physics to someone else helps the person explaining to understand the physics better. > (I'm just struggling through Penrose's Road to Reality at the moment so > in a few years I might be able to disguise my ignorance better - [quoted text clipped - 3 lines] > Chris > PS - I promise not to interrupt again off=topic. Nonsense. Please post whatever you wish. I personally don't mind. For help please e-mail me at peter102560NOSPAM at comcast.net. Just yank out the NOSPAM and replace "at" with "@" and that's my e-mail address. Best regards Pete > Nonsense. Please post whatever you wish. I personally don't mind. For help > please e-mail me at peter102560NOSPAM at comcast.net. Just yank out the [quoted text clipped - 3 lines] > > Pete You are a gentleman and a scholar sir. I will very possibly be taking you up on that kind offer fairly soon :-) Best regards Chris >> When v = 0 then E_0 = m_0 c^2 where m_0 is proper mass. If you wish to >> use m [quoted text clipped - 14 lines] > Chris > PS - I promise not to interrupt again off=topic. By the way, I wish to end my participation in this thread. The subject title is wrong and is drawing the wrong attention. I started a new thread called "Mass density" to replace this one. Thanks. Pete > I think that Physicists generally prefer to talk of invariant mass > rather than relativistic mass because the former is independant of > velocity While it is true that mass is independent of velocity, the real reason physicists use mass rather than "relativistic mass" is that mass is what appears in the Lagrangian, not "relativistic mass" -- this makes mass fundamental and "relativistic mass" derived. Another reason is the lesson learned in intermediate mechanics: describing systems via invariants often/usually results in much simpler equations than using quantities that are not invariant. Another reason is that the whole notion of "mass" involves "amount of stuff", and that must be intrinsic to an object; mass is intrinsic while "relativistic mass" is not. AFAICT the only use of "relativistic mass" is in discussing the relationship between SR and Newtonian mechanics. As it is inherently tied to an inertial frame and the Lorentz transform, it is useless in GR except in such a limit. Mass is the norm of an object's 4-momentum and remains well defined in GR. In the phrase "invariant mass", the first word is redundant, and I omit it. Pmb keeps bringing this subject up through ignorance and obstinacy -- please write down a Lagrangian using "relativistic mass". Or any other fundamental equations. Tom Roberts {a few insults} Sorry Tom but I do not respond to people who are insulting me. I thought you could avoid insulting me but I guess you were wrong. Please consider yoursefl plonked. > {a few insults} > > Sorry Tom but I do not respond to people who are insulting me. I thought > you could avoid insulting me but I guess you were wrong. Correction. I meant to say "I thought you could avoid insulting me but I guess I was wrong." Pete Due to the response in this thread it appears as if the purpose of it was lost. I assume that was caused by the title of this thread. I started this thread due to some physics that I've been working across lately. The subject has never been discussed in this forum in the past from what I have seen. The subject is mass density. I am therefore no longer reading this thread. I started another to replace it and am hoping people don't over focus on relativist mass and will direct their attention to the actual problem at hand, the homework problem which appeared in Ohanian's SR text regarding the mass density of a magnetic field. Best wishes Pete > While it is true that mass is independent of velocity, the real reason > physicists use mass rather than "relativistic mass" is that mass is what > appears in the Lagrangian, Please show us this Lagrangian with this rest mass. > not "relativistic mass" -- this makes mass > fundamental and "relativistic mass" derived. Relativistic mass should be interpreted as the observed mass where the observed mass becomes observer dependent. As an object is observed to gain speed, its observed mass and observed kinetic energy increases. Relativistic mass should also apply to potential energy in which potential energy reduces the observed mass. > Another reason is the > lesson learned in intermediate mechanics: describing systems via > invariants often/usually results in much simpler equations than using > quantities that are not invariant. Another reason is that the whole > notion of "mass" involves "amount of stuff", and that must be intrinsic > to an object; mass is intrinsic while "relativistic mass" is not. Rest mass is intrinsic to an object. Relativistic mass is how an observer would observe this intrinsic mass. <shrug> > AFAICT the only use of "relativistic mass" is in discussing the > relationship between SR and Newtonian mechanics. As it is inherently [quoted text clipped - 4 lines] > In the phrase "invariant mass", the first word is redundant, > and I omit it. You (plural) choose not to identify relativistic mass as the observed mass where the potential energy is a reduction in the observed mass itself. This is done to avoid the early embarrassment of claiming energy conservation not the general case in GR. Properly derived through the calculus of variations with a valid Lagrangian, one can trivially show how the observed energy and the observed mass as a function of speed and the curvature in spacetime. In this case, energy conservation becomes a universally fundamental principle as solid as the rock of Gibraltar. > Pmb keeps bringing this subject up through ignorance and obstinacy -- > please write down a Lagrangian using "relativistic mass". Or any other > fundamental equations. And I couldn't wait to see you writing down this Lagrangian with this mass you have been bragging about. <shrug> >> While it is true that mass is independent of velocity, the real reason >> physicists use mass rather than "relativistic mass" is that mass is what >> appears in the Lagrangian, > > Please show us this Lagrangian with this rest mass. There are two such Lagrangians but they are tightly related to each other but both of which will do the job. The relativistic but non-covariant Lagrangian is in this page http://www.geocities.com/physics_world/sr/relativistic_energy.htm See Eq. (15). I had the covariant Lagrangian in one of those pages but I can't find it now. As an example consider the free-particle Lagrangian L defined by (m_0 = proper mass) L = -m_0*c^2*sqrt(U^aU_a) If one wanted to then you can replace m_0 by m_0 = m/gamma (m = rel-mass) to get L = -(m/gamma)*c^2*sqrt(U^aU_a) Seems silly to do this though since its not in its simplest form. Proper mass has its place in relativity and that is where ever an equation which is expressed in terms of tensors is found. The components of these equations have a specific meaning and that is where coordinate dependant physical quantities appear. >> not "relativistic mass" -- this makes mass >> fundamental and "relativistic mass" derived. [quoted text clipped - 4 lines] > Relativistic mass should also apply to potential energy in which > potential energy reduces the observed mass. I assume you are referring to the internal potential energy of an object and not the potential energy of position. Is that correct? >> Another reason is the >> lesson learned in intermediate mechanics: describing systems via [quoted text clipped - 31 lines] > And I couldn't wait to see you writing down this Lagrangian with this > mass you have been bragging about. <shrug> Good luck with that. Pete > http://www.geocities.com/physics_world/sr/relativistic_energy.htm Defining (K = - m0 c^2 sqrt(1 - B^2)), the kinetic energy, is invalid. If L(q1, q2, q3...) is a function of q1, q2, q3..., and q1(t), q2(t), q3(t)... are functions of t, then we have dL/dt = SUM((@L/@q_i) (dq_i/dt)) Where ** I = 1, 2, 3... ** @ = Partial derviative Your equation (1) is also invalid. I don't think you understand how your equation (2), the Euler-Lagrange equation, is derived. Your application of the Euler-Lagrange equation violates the very essence of what defining it as the Euler-Lagrange equation. > See Eq. (15). I had the covariant Lagrangian in one of those pages but I > can't find it now. Are you not the original author? > As an example consider the free-particle Lagrangian L defined by (m_0 = > proper mass) > > L = -m_0*c^2*sqrt(U^aU_a) Show me what qualifies it as a Lagrangian. Claim is cheap. > If one wanted to then you can replace m_0 by m_0 = m/gamma (m = rel-mass) to > get > > L = -(m/gamma)*c^2*sqrt(U^aU_a) Same as above. > Seems silly to do this though since its not in its simplest form. Proper > mass has its place in relativity and that is where ever an equation which is > expressed in terms of tensors is found. The components of these equations > have a specific meaning and that is where coordinate dependant physical > quantities appear. Does proper mass = rest mass? Whether the answer is yes or no, I still cannot comprehend what you are trying to say. > I assume you are referring to the internal potential energy of an object and > not the potential energy of position. Is that correct? I cannot answer your question because I do not understand what you mean by the internal potential energy versus the potential energy of position. Call me ignorant if you like, but I only understand one and only one concept of the potential energy. > > And I couldn't wait to see you writing down this Lagrangian with this > > mass you have been bragging about. <shrug> > > Good luck with that. Thanks anyway. I am not the one to do that. Dr. Robert is challenged to do so. May lady luck smile on him this time for a change. However, there is a very good chance he will not answer to the fair challenge. > The energy-momentum tensor fully describes the mass of any body under any > situation. Does the energy momentum tensor fully describe the mass of the gravitating body or the mass of the object under the gravitational influence of this gravitating body? >> http://www.geocities.com/physics_world/sr/relativistic_energy.htm > > Defining (K = - m0 c^2 sqrt(1 - B^2)), the kinetic energy, is > invalid. Thanks. I didn't notice that error until now. Much appreciated. > If L(q1, q2, q3...) is a function of q1, q2, q3..., and q1(t), q2(t), > q3(t)... are functions of t, then we have [quoted text clipped - 7 lines] > > Your equation (1) is also invalid. Only if your "if" is correct which it isn't. The most general form of the Lagrangian is as a function of q1, q2, ..., dq1/dt, dq2/dt ... and t. As such equation 1 is correct. > I don't think you understand how your equation (2), the > Euler-Lagrange equation, is derived. I know very well how it is derived, thanks. > Your .... Mine? I didn't invent these definitions. They are standard and can be found in a decent mechanics text. >> See Eq. (15). I had the covariant Lagrangian in one of those pages but I >> can't find it now. > > Are you not the original author? Yes I am the author. However I made so many pages over the last several years that I can't remember what each one contains and where it is found. >> As an example consider the free-particle Lagrangian L defined by (m_0 = >> proper mass) >> >> L = -m_0*c^2*sqrt(U^aU_a) > > Show me what qualifies it as a Lagrangian. Claim is cheap. When substituted into the covariant Euler-Lagrange equations it yields the correct equation of motion >> Seems silly to do this though since its not in its simplest form. Proper >> mass has its place in relativity and that is where ever an equation which [quoted text clipped - 4 lines] > > Does proper mass = rest mass? No. > Whether the answer is yes or no, I still > cannot comprehend what you are trying to say. What are you unclear on? >> I assume you are referring to the internal potential energy of an object >> and [quoted text clipped - 20 lines] > gravitating body or the mass of the object under the gravitational > influence of this gravitating body? Yes. Pete >> http://www.geocities.com/physics_world/sr/relativistic_energy.htm > > Defining (K = - m0 c^2 sqrt(1 - B^2)), the kinetic energy, is > invalid. Hold on a sec. I just recall what that K is. It does **not** represent kinetic energy. It merely represents a term in the Lagrangian. If you were to look at the Lagrangian for a charged particle moving in an EM field then the Lagrangian would have that part (i.e. the K as defined above) followed (in the relativistic/non-covariant form) by a term consisting of the potential energy and another term involving velocity and the magnetic potential. Pete > Are you not the original author? I found the web page which contains the covariant Lagrangian. Its at http://www.geocities.com/physics_world/em/relativistic_charge.htm See Section 4. The Lagrangian is given in Eq. (32). Its labeled as capital gamma so as to distinguish it from the relativistic non-covariant Lagrangian in that page. >> As an example consider the free-particle Lagrangian L defined by (m_0 = >> proper mass) >> >> L = -m_0*c^2*sqrt(U^aU_a) > > Show me what qualifies it as a Lagrangian. Claim is cheap. Okay. See the above URL. In that web page (in section 4) the covariant Lagrangian is plugged into the covariant Euler-Lagrange equations and the result is the covariant equations of motion Eq. (41). > Does proper mass = rest mass? Whether the answer is yes or no, I still > cannot comprehend what you are trying to say. [quoted text clipped - 7 lines] > position. Call me ignorant if you like, but I only understand one and > only one concept of the potential energy. Internal potential energy - The energy which exists due to the mutual interaction of two particles. E.g. the magnitude of the potential energy of two point charges of magnitude Q and q which are a distance D apart is U = Qq/r (in cgs units). The potential energy of position is that funtion which, when the negative gradiant is taken of it, the force results. I.e. F = -grad U. This value, U, as you may recall for a charged particle in an electric field is U = q*Phi where Phi is the electric potential where E (electric field) = -grad Phi. >> > And I couldn't wait to see you writing down this Lagrangian with this >> > mass you have been bragging about. <shrug> >> >> Good luck with that. > > Thanks anyway. I am not the one to do that. Dr. Robert ... Why do you refer to him as "Dr."?? Best regards Pete [ snip ] > > Thanks anyway. I am not the one to do that. Dr. Robert ... > > Why do you refer to him as "Dr."?? Presumably Tom Roberts has a PhD in physics, and so may be properly addressed as "Dr. Roberts". I don't recall whether Tom has ever actually stated that he has a PhD but given that he's a professional physicist at a major lab it's a very reasonable guess that he does. In general, it seems like only the cranks who probably have no real degrees talk a lot about the degrees they supposedly have. Now, as to why K-W would call Tom "Dr." in that response .. that's a different question! KW's posts generally consist of errors, falsehoods, and snide comments. This isn't an error and it isn't a falsehood, so it must be snide. It appears to be a dig at both you and Roberts at once -- sort of like he's saying "DOCTOR Roberts disagrees with you, nyah nyah!! And besides he says he's a DOCTOR but he doesn't understand this either, so there!" > Best regards > > Pete > [ snip ] >> > [quoted text clipped - 17 lines] > with you, nyah nyah!! And besides he says he's a DOCTOR but he doesn't > understand this either, so there!" Nice analysis ;-) You could also have said that Koobee Wublee is a downright troll. Dirk Vdm > [ snip ] >> > [quoted text clipped - 17 lines] > with you, nyah nyah!! And besides he says he's a DOCTOR but he doesn't > understand this either, so there!" The fact that people disagree with me on newsgroups means nothing to me so I guess they're wasting their time. Its kind of cute the way that K-W uses "<shrug>" with Tom. He uses it so often its a bit funny when someone is doing it to him. Pete | [ snip ] | > > [quoted text clipped - 3 lines] | | Presumably Tom Roberts has a PhD in physics, Hahahahaha! Where from, University of Antarctica? Outback of Australia? The Moon? Famous quotes by Roberts: | SR is strictly valid only in a flat Lorentzian manifold with the | topology of R^4. This of course is a very poor model of the world we [quoted text clipped - 8 lines] | some: | the binary pulsars, and observations of accretion disks near black holes" | Imagine a train leaving one city at 12:00 and arriving in a city 60 | miles to its west at 12:01. Do you really think that train traveled | 3,600 miles per hour? Of course not! This example used two _different_ | coordinate systems for "time", the two timezones of those two cities. To | obtain the speed you _must_ use a single coordinate system; then you'll | realize it traveled just under 60 miles per hour. and so may be properly | addressed as "Dr. Roberts". I don't recall whether Tom has ever | actually stated that he has a PhD but given that he's a professional | physicist at a major lab it's a very reasonable guess that he does. Hahahahahaha! He got fired from Lucent Technologies! He's even more deranged than you or Hammond. | In general, it seems like only the cranks who probably have no real | degrees talk a lot about the degrees they supposedly have. That's Roberts! | Now, as to why K-W would call Tom "Dr." in that response .. that's a | different question! KW's posts generally consist of errors, | falsehoods, and snide comments. HAHAHAHAHA! How the f.ck would you know, snide rabbit hugger? > Tom Roberts has a PhD in physics, and so may be properly > addressed as "Dr. Roberts". I don't recall whether Tom has ever > actually stated that he has a PhD but given that he's a professional > physicist at a major lab it's a very reasonable guess that he does. I would at least like to find out when he received his doctorate degree and on what thesis. Then, if possible, I like to find out the institution. Aren't these supposed to be public information? >From this information, I can get a more accurate profile of him. Right now, I picture him to be over 70, possibly over 75 or even 80. > In general, it seems like only the cranks who probably have no real > degrees talk a lot about the degrees they supposedly have. It is not fair for you to label anyone who does not agree with you on seeing the emperor's clothes a crank. > Now, as to why K-W would call Tom "Dr." in that response .. that's a > different question! KW's posts generally consist of errors, [quoted text clipped - 3 lines] > with you, nyah nyah!! And besides he says he's a DOCTOR but he doesn't > understand this either, so there!" You are totally wrong. Among all the posters, I do find Dr. Robert's remarks the most interesting to engage in discussions. I have to admit my posts do contain some grammatical errors and spelling errors in which Dr. Robert's posts are impeccably clean in terms of these errors. I am no literary giant. However, the subject matter should be clean. Yes, my posts are labeled abrasive by Dr. Roberts and others. Just because I address the very core beliefs in their education, that should not warrant my posts abrasive. These highly educated individuals need to understand the opposing point of view. When they are not addressing the core of my posts but rather the outer skin, I know my message has finally gotten through. Aren't you supposed to have plonked me after hurting so much by my snide comments? > Presumably Tom Roberts has a PhD in physics, and so may be properly > addressed as "Dr. Roberts". No need to presume. I received a Ph.D. in physics in 1975 from the University of Illinois in Urbana. > Now, as to why K-W would call Tom "Dr." in that response .. that's a > different question! KW's posts generally consist of errors, > falsehoods, and snide comments. Yes. I think he's trying to be snide, and failing at even that. Tom Roberts > http://www.geocities.com/physics_world/em/relativistic_charge.htm > > See Section 4. The Lagrangian is given in Eq. (32). Its labeled as capital > gamma so as to distinguish it from the relativistic non-covariant Lagrangian > in that page. How did you derive or put together equation (1)? What is a relativistic non-covariant Lagrangian? Is there a non-relativistic non-covariant Lagrangian? Is there a relativistic covariant Lagrangian? Is there a non-relativistic covariant Lagrangian? If yes, please give examples. > Okay. See the above URL. In that web page (in section 4) the covariant > Lagrangian is plugged into the covariant Euler-Lagrange equations and the > result is the covariant equations of motion Eq. (41). I only know of one concept to the Euler-Lagrange equations. So far, the covariant version escapes me. > Internal potential energy - The energy which exists due to the mutual > interaction of two particles. E.g. the magnitude of the potential energy of [quoted text clipped - 4 lines] > field is U = q*Phi where Phi is the electric potential where E (electric > field) = -grad Phi. It sounds like they are the same. "Koobee Wublee" aka Australopithetuc Afarensis aka <koobee.wublee@gmail.com> wrote in message news:1167243626.674997.124850@f1g2000cwa.googlegroups.com... >> http://www.geocities.com/physics_world/em/relativistic_charge.htm >> [quoted text clipped - 20 lines] > I only know of one concept to the Euler-Lagrange equations. So far, > the covariant version escapes me. There seems to be a lot that escapes you. What's that smell? http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LonelyTop.html http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SmellHere.html http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/TwoMetrics.html http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/DiffGeoAero.html http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/FlatSphere.html http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LorentzTale.html http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/SRBogus.html http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/ReasonLaws.html http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewLagrangian.html http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LosingIt.html http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/AerospaceRelativity.html http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/NewPotential.html http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/WhatWrong.html Dirk Vdm > "Koobee Wublee" > aka Australopithetuc Afarensis [quoted text clipped - 42 lines] > > Dirk Vdm I did not know that they migrated from Australia to Irvive, California. Looks like they did :-) >> http://www.geocities.com/physics_world/em/relativistic_charge.htm >> [quoted text clipped - 5 lines] > > How did you derive or put together equation (1)? I guess it came from being bored while studying. > What is a relativistic non-covariant Lagrangian? Let c = 1. Then L = K - U where K = -mc^2sqrt(1 - v^2/c^2) > Is there a non-relativistic non-covariant Lagrangian? Yes. For the charged particle its L = 1/2 mv^2 - V where V = potential energy > Is there a relativistic covariant Lagrangian? Yes. Its L = mc^2 sqrt(1 - v^2/c^2) - V > Is there a non-relativistic covariant Lagrangian? Not that I know of >> Okay. See the above URL. In that web page (in section 4) the covariant >> Lagrangian is plugged into the covariant Euler-Lagrange equations and the [quoted text clipped - 14 lines] >> field is U = q*Phi where Phi is the electric potential where E (electric >> field) = -grad Phi.nti > It sounds like they are the same, No. They are not the same to my level Pete > > While it is true that mass is independent of velocity, the real reason > > physicists use mass rather than "relativistic mass" is that mass is what > > appears in the Lagrangian, > > Please show us this Lagrangian with this rest mass. Here is one: L = Psibar(d_slash - m)Psi Any questions? >> While it is true that mass is independent of velocity, the real reason >> physicists use mass rather than "relativistic mass" is that mass is what >> appears in the Lagrangian, > > Please show us this Lagrangian with this rest mass. For a scalar field phi with mass m, the usual Lagrangian (density) is: L = g^uv (d_u phi) (d_v phi) - m^2 phi^2 As this must be invariant, one cannot write anything similar at all using "relativistic mass". Similar remarks hold for other types of fields.... > Relativistic mass should be interpreted as the observed mass where the > observed mass becomes observer dependent. Sure. But of course it is identical to the energy of the object, and "energy" is a much better name than "relativistic mass", because it carries more of the original meanings. > Rest mass is intrinsic to an object. Relativistic mass is how an > observer would observe this intrinsic mass. Sure, except the last is more like energy than "mass". It is primarily the inappropriate PUN on the word "mass" that I object to. Tom Roberts "Pmb" <peter102560_nospam@comcast.net> wrote in message ... In case I din't make my point clear, this thread is about "mass density" and not mass and how asking for mass density can lead one to error. To be precise its about a homework problem in an SR text I have. The problem is discussed in this web page http://www.geocities.com/physics_world/sr/mass_mag_field.htm Whether you prefer proper mass or rel-mass it makes no difference since the problem remains. Actually I think I labeled this thread incorrectly. This isn't about relativistic mass. Sorry. The paradox by Rindler/Denur is a good example of being careful. See http://www.geocities.com/physics_world/sr/rd_paradox.htm The physics of this subject matter can be found in "Introduction to Special Relativity," Wolfgang Rindler, Oxford University Press (1982). See chapter on continuous media. Pete > http://www.geocities.com/physics_world/sr/mass_mag_field.htm The underlying problem is that this usage of "mass" is just plain different from all other usages. And IMHO just plain wrong. This usage is unrelated to the norm of the 4-velocity, because the electromagnetic field is not a 4-vector. The actual quantity that is important in GR is the energy-momentum tensor. There is no general method to take an ARBITRARY e-m tensor and from it obtain a mass, like there is for an arbitrary object's 4-momentum (take its norm). There is such a method for an object: one can transform to the rest frame of the object, and in that frame T_00 is the only nonzero component, and from that one can obtain a mass (integrate over the volume of the object). Now consider electromagnetic radiation. The key difference from the bullet is that for electromagnetism the e-m tensor is traceless. So there is no frame in which T_00 is the only nonzero component, and the above procedure to determine the mass density fails. Note this is true for both static electromagnetic fields and for radiation; it is intimately related to the fact that electromagnetic radiation travels with speed c, and to the form of Maxwell's equations. The above link says "In the present case of a magnetic field the speed v of the element of matter [includes field] is identical to the speed of the source that generates the magnetic field." This is a rather big confusion on the part of the author -- one cannot really ascribe any speed to an electromagnetic field, for precisely the reason above -- there is no frame in which the "field is at rest" (in such a frame T_00 would be the only nonzero component, and there simply is no such frame). The underlying mistake is attempting to discuss an "infinitesimal volume" of the field as if it were an "object". It is sensible to consider the ENTIRE SYSTEM (solenoid + field), but it simply does not make sense to attempt to separate out an infinitesimal region and consider it an isolated "object". I suppose all the formulas the author writes make some sort of sense, but they do not yield anything that is of any importance or use -- this is just an isolated textbook-like exercise in mathematical masturbation. Yes, for electromagnets we often discuss the TOTAL field energy, and that is useful for various engineering tasks (e.g. determining how much quench protection is required for a superconducting solenoid). But the "mass" or "speed" of an infinitesimal region is useless. Tom Roberts > > http://www.geocities.com/physics_world/sr/mass_mag_field.htm > > The underlying problem is that this usage of "mass" is just plain > different from all other usages. And IMHO just plain wrong. This usage > is unrelated to the norm of the 4-velocity, because the electromagnetic > field is not a 4-vector. not considered you mean becus there is no t? then it must be a 5-vector, i tell you is not a 3-vector > The actual quantity that is important in GR is the energy-momentum > tensor. There is no general method to take an ARBITRARY e-m tensor and [quoted text clipped - 37 lines] > > Tom Roberts > > http://www.geocities.com/physics_world/sr/mass_mag_field.htm > [quoted text clipped - 34 lines] > make sense to attempt to separate out an infinitesimal region and > consider it an isolated "object". but many of such of infinitisimal adiacent control volumes are correlated, so it make sense > I suppose all the formulas the author writes make some sort of sense, > but they do not yield anything that is of any importance or use -- this [quoted text clipped - 5 lines] > > Tom Roberts > The actual quantity that is important in GR is the energy-momentum > tensor. How is this? In most applications such as in free space, the energy-momentum tensor is zero. The only way to work it backwards to calculate the gravitating mass or energy is through the integration constants. By choosing the constant to be of energy related, we have the gravitating energy; by choosing the constant to be of mass related, we have the gravitating mass. The gravitating mass means only the rest mass with its energy of the rest mass while energy involves speed and others. So, energy and the energy of the rest mass just do not mix well. How do you justify to define mass as only the rest mass and tossing out relativistic mass? The energy-momentum tensor is not as divine as you have pictured it to be. There is not a single halo on top of this tensor. <shrug> > I suppose all the formulas the author writes make some sort of sense, > but they do not yield anything that is of any importance or use -- this [quoted text clipped - 3 lines] > quench protection is required for a superconducting solenoid). But the > "mass" or "speed" of an infinitesimal region is useless. They are just another way to look at a problem. Your (plural) way has so many silly consequences. An example is waiting for you to address above. >> The actual quantity that is important in GR is the energy-momentum >> tensor. > > How is this? In most applications such as in free space, the > energy-momentum tensor is zero. The energy-momentum tensor fully describes the mass of any body under any situation. Pete >> The actual quantity that is important in GR is the energy-momentum >> tensor. > > How is this? Because it is the invariant quantity that includes both the energy and momentum of an object or density. There is no other _invariant_ quantity that does so (except for various contractions of this tensor). [Note that in the case of a pointlike object, the e-m tensor is completely determined by its position and 4-momentum, so the number of degrees of freedom is reduced.] And, of course, the field equation is: G + L g = T Where T is the energy-momentum tensor. This is the central equation of GR. > In most applications such as in free space, the > energy-momentum tensor is zero. Sure. That does not diminish its importance -- in "most applications" there are some regions of the manifold with non-vanishing e-m tensor. > The only way to work it backwards to > calculate the gravitating mass or energy is through the integration > constants. Hmmm. The integration constants usually represent boundary conditions, not matter within the manifold. After all, one must solve the field equation throughout the manifold, including non-vacuum regions. [I suspect you are thinking of a handful of vacuum manifolds, such as the Schwarzschild manifold, for which the universe is completely empty but there is an integration constant that behaves as if it were a mass. These are all quite special, and are by no means the entire story.] > By choosing the constant to be of energy related, we have > the gravitating energy; by choosing the constant to be of mass related, > we have the gravitating mass. The gravitating mass means only the rest > mass with its energy of the rest mass while energy involves speed and > others. So, energy and the energy of the rest mass just do not mix > well. You are quite confused about "constants". One must solve the field equation throughout the manifold, and "constants" cannot represent that in the non-vacuum regions. Moreover, as I said, one must use the e-m tensor for that, not just mass or "relativistic mass" or energy. Note that for pointlike particles the e-m tensor is completely determined by their positions and 4-momenta -- mass appears in the latter, but "relativistic mass" is nowhere to be found. For non-interacting dust particles the situation is the same. For more complicated mass distributions other quantities enter into the e-m tensor (e.g. pressure).... > How do you justify to define mass as only the rest mass and > tossing out relativistic mass? Read my earlier posts. I don't "toss out" the "relativistic mass", I merely point out it is not of fundamental importance, is derived from mass, is observer dependent, and does not represent anything physical. Tom Roberts > > Please show us this Lagrangian with this rest mass. > [quoted text clipped - 5 lines] > using "relativistic mass". Similar remarks hold for other types of > fields.... How do you get this? > > Relativistic mass should be interpreted as the observed mass where the > > observed mass becomes observer dependent. > > Sure. But of course it is identical to the energy of the object, and > "energy" is a much better name than "relativistic mass", because it > carries more of the original meanings. Thus, you do agree that energy is merely an observed quantity depending on which observer is doing the observation. > > Rest mass is intrinsic to an object. Relativistic mass is how an > > observer would observe this intrinsic mass. > > Sure, except the last is more like energy than "mass". It is primarily > the inappropriate PUN on the word "mass" that I object to. Your objection is only skin deep. Mathematically, (m = E / c^2). Thus, observed mass and observed energy are both the same except by a constant (1 / c^2). Also, your objections do not render others wrong in this subject. > > How is this [energy-momentum tensor important n GR]? > [quoted text clipped - 9 lines] > G + L g = T > Where T is the energy-momentum tensor. This is the central equation of GR. Are you sure that T is an invariant quantity? Earlier you are hinting the energy is an observed quantity. Since T is the energy-momentum tensor, it must be observer dependent! What T is is described as follows. T = Constant rho g_ij Where ** rho = Rest mass density, or energy density ** g_ij = The metric (interpretation matrix to coordinate measurement) rho can also be charge density as well. There is nothing mathematically stopping what rho can be except through the content it is describing. If charge, it could mean the difference of + and - charge as addressed by Weber. The founders of GR chose energy instead. Yes, energy is an observed quantity! > > In most applications such as in free space, the > > energy-momentum tensor is zero. > > Sure. That does not diminish its importance -- in "most applications" > there are some regions of the manifold with non-vanishing e-m tensor. Importance is just your opinion. For all practical applications, the energy-momentum tensor being observer dependent is zero. <shrug> Thus, in order for this tensor to be observer independent, it must be a function of intrinsic substance such as the rest mass itself. Then, if you are thinking what I am thinking, you deserve credit. The cosmological constant pulled out of Einstein's *ss should indicate the rest mass density of vacuum instead of the energy density of vacuum. Does the rest mass density of vacuum sound silly? > > The only way to work it backwards to > > calculate the gravitating mass or energy is through the integration [quoted text clipped - 9 lines] > behaves as if it were a mass. These are all quite special, > and are by no means the entire story.] The integration constant associated with the field equations is not a true integration constant. If you understand how the field equations are derived you should not have misunderstood what I said. Thus, your suspicion is very groundless. <shrug> > > By choosing the constant to be of energy related, we have > > the gravitating energy; by choosing the constant to be of mass related, [quoted text clipped - 6 lines] > equation throughout the manifold, and "constants" cannot represent that > in the non-vacuum regions. The "integration" constant was pulled out of his *ss by Hilbert. Hilbert designed the following Lagrangian which became the basis where the field equations are derived. L = (constant g^ij R_ij + rho) sqrt(-det(g^ij))) Where ** L = Einstein-Hilbert Lagrangian ** g^ij = inverted matrix of g_ij, the metric ** R_ij = Ricci tensor ** rho = Mass Lagrangian also put together by Hilbert Do you not see there are at least 2 degrees of freedom to design your universe? Or you can be like Einstein who basically modified the Lagrangian into the following to add in the effect of the Cosmological constant. <yawn> L = (constant g^ij R_ij + rho + rho0) sqrt(-det(g^ij))) ** rho0 = Cosmological constant (sign may be inverted) > Moreover, as I said, one must use the e-m tensor for that, not just mass > or "relativistic mass" or energy. In vacuum, the e-m tensor is zero. So, I still do not understand how you can distinguish rest mass, observed mass, or observed energy from zero? > Note that for pointlike particles the e-m tensor is completely > determined by their positions and 4-momenta -- mass appears in the > latter, but "relativistic mass" is nowhere to be found. For > non-interacting dust particles the situation is the same. For more > complicated mass distributions other quantities enter into the e-m > tensor (e.g. pressure).... Momentum, however, is dependent on the relativistic mass which means the momentum is observer dependent as well! You are designing a universe based on what an observer observes. 'The Lathe of Heaven' is only a science fiction. You are defending your silly religion called GR with shields made out of toilet papers. <shrug> > > How do you justify to define mass as only the rest mass and > > tossing out relativistic mass? > > Read my earlier posts. I don't "toss out" the "relativistic mass", I > merely point out it is not of fundamental importance, is derived from > mass, is observer dependent, and does not represent anything physical. Yes, I did. You contradict yourselves many times over in the past hundred years. <shrug> >> > Please show us this Lagrangian with this rest mass. >> [quoted text clipped - 17 lines] > Thus, you do agree that energy is merely an observed quantity depending > on which observer is doing the observation. That is quite true. But it can't be said that relativistic mass is the same as energy. E.g. the relativistic mass of a particle is proportional to the time component of the particle's 4-momentum whereas the particle's energy is proportional to the time component of the 1-form dual to the 4-momentum. I.e. (let c = 1) P^0 = M and P_0 = E. These quantities have different values in general, e.g. when the particle is in free fall in a static g-field. >> > Rest mass is intrinsic to an object. Relativistic mass is how an >> > observer would observe this intrinsic mass. [quoted text clipped - 34 lines] > ** rho = Rest mass density, or energy density > ** g_ij = The metric (interpretation matrix to coordinate measurement) That is not how the stress-energy-momentum tensor T is defined. > rho can also be charge density as well. There is nothing > mathematically stopping what rho can be except through the content it [quoted text clipped - 101 lines] > Yes, I did. You contradict yourselves many times over in the past > hundred years. <shrug> True. Pete >>> Please show us this Lagrangian with this rest mass. >> For a scalar field phi with mass m, the usual Lagrangian (density) is: [quoted text clipped - 4 lines] > > How do you get this? General knowledge of modern physics. You can look it up. > Thus, you do agree that energy is merely an observed quantity depending > on which observer is doing the observation. Of course. Always has been, always will be. >>> Rest mass is intrinsic to an object. Relativistic mass is how an >>> observer would observe this intrinsic mass. [quoted text clipped - 4 lines] > Thus, observed mass and observed energy are both the same except by a > constant (1 / c^2). Not at all. They have a completely different character and usage. The energy of an object is the time component of its 4-momentum (and is thus dependent on coordinates); its mass is the norm of its 4-momentum (and is thus independent of coordinates). > Also, your objections do not render others wrong in this subject. Sure. But the people around here who advocate "relativistic mass" display no significant reason to do so. >> And, of course, the field equation is: >> G + L g = T >> Where T is the energy-momentum tensor. This is the central equation of GR. > > Are you sure that T is an invariant quantity? Of course it is! it is a _tensor_. All tensors are invariant. > Earlier you are hinting > the energy is an observed quantity. Since T is the energy-momentum > tensor, it must be observer dependent! Nonsense. Energy is NOT the energy-momentum tensor. Nor is energy a component of the e-m tensor. The relationship is more subtle. > T = Constant rho g_ij This is just plain wrong. Not even close. For instance in the instantaneous rest frame of a pointlike object, the energy-momentum tensor has just one non-zero component, T_tt. The metric, however, has 4 nonzero components. > Importance is just your opinion. For all practical applications, the > energy-momentum tensor being observer dependent is zero. <shrug> You are confused. No tensor is "observer dependent". > [... further nonsense omitted] You _REALLY_ need to learn what GR actually says, rather than making nonsense like this up. Tom Roberts > For a scalar field phi with mass m, the usual Lagrangian (density) is: > L = g^uv (d_u phi) (d_v phi) - m^2 phi^2 > > > How do you get this? > > General knowledge of modern physics. You can look it up. I did, and I did not find anything you have claimed. What justifies it to be a Lagrangian? > > Thus, you do agree that energy is merely an observed quantity depending > > on which observer is doing the observation. > > Of course. Always has been, always will be. Excellent, at least we can agree on both energy and mass are observed quantities. I know the New Year is still days away, but I am going to open that bottle of Champaign to celebrate. > > Your objection is only skin deep. Mathematically, (m = E / c^2). > > Thus, observed mass and observed energy are both the same except by a [quoted text clipped - 4 lines] > dependent on coordinates); its mass is the norm of its 4-momentum (and > is thus independent of coordinates). The energy is how you choose to present it in that silly 4-momentum stuff. Despite the silliness, it is mathematically valid. Thus, I am only expressing my own mild objection. However, do you deny (E = m c^2) or (m = E / c^2)? If the answer is no which I predict that is what you are going to answer, then what you are saying is very silly because (mass = energy). If you are talking about the intrinsic mass or the rest mass, it still does not make any sense because the rest mass is just a number independent of velocity, curvature in spacetime, or whatever. It is noway related to the norm of a tensor. <shrug> > > Also, your objections do not render others wrong in this subject. > > Sure. But the people around here who advocate "relativistic mass" > display no significant reason to do so. How do you know? Relativity implies observed observation. Relativistic mass implies observed mass in which you have fully agreed. So, please stop talking down on the folks with less credential on paper than you do. > And, of course, the field equation is: > G + L g = T [quoted text clipped - 3 lines] > > Of course it is! it is a _tensor_. All tensors are invariant. The tensors applied in GR are merely matrices in disguise. By redefining matrices as tensors do not make them immune from the rules of mathematics. <shrug> > > Earlier you are hinting > > the energy is an observed quantity. Since T is the energy-momentum > > tensor, it must be observer dependent! > > Nonsense. Energy is NOT the energy-momentum tensor. Nor is energy a > component of the e-m tensor. The relationship is more subtle. There is nothing divine about the energy-momentum tensor either. In fact, according to Einstein-Hilbert Lagrangian, the derivative of it with respect to each of the elemement of the metric matrix must be a function of the mass Lagrangian which is very simply a constant times rho (mass, energy, or whatever density per volume) times this element of the metric matrix. The founding fathers of GR chose rho as the energy density without much thought about energy being observer dependent, and now their descendants are paying the price. However, rho can be time dependent. That is why the momentum is in there. However, you cannot avoid that the destiny of energy and momentum both involved in this tensor. When talking about energy and momentum, these are observer dependent quantities. Therefore, the e-m tensor matrix must be observed dependent. As a sheep, you cannot become a wolf by wearing wolf's skin. > > T = Constant rho g_ijThis is just plain wrong. Not even close. > > For instance in the instantaneous rest frame of a pointlike object, the > energy-momentum tensor has just one non-zero component, T_tt. The > metric, however, has 4 nonzero components. Nonsense, (T = 0) in vacuum. > > Importance is just your opinion. For all practical applications, the > > energy-momentum tensor being observer dependent is zero. <shrug> > > You are confused. No tensor is "observer dependent". No, I am not confused. You are acting irresponsibly by claiming this tensor as invariant while the components of this tensor are energy and momentum which you have agreed to be observer dependent already. <shrug> > > [... further nonsense omitted] > > You _REALLY_ need to learn what GR actually says, rather than making > nonsense like this up. You really need to go through how the field equations are derived. I guarantee you that you will be utterly surprised and stop licking off from your GR teachers' nonsensical regurgitations. <disgusting> >> For a scalar field phi with mass m, the usual Lagrangian (density) is: >> L = g^uv (d_u phi) (d_v phi) - m^2 phi^2 [quoted text clipped - 43 lines] > So, please stop talking down on the folks with less credential on > paper than you do. What Tom claims is untrue. I myself have posted several reasons for relativistic mass. Tom seems to ignore things which he disagrees with in this case. However he has jumped to false conclusions many times. E.g. he keeps asserting that I'm an advocate for relativistic mass when actually I have no opinion either way. I merely note the differences when the subject comes up but he seems to take that as me pushing relativistic mass on others. >> And, of course, the field equation is: >> G + L g = T [quoted text clipped - 8 lines] > redefining matrices as tensors do not make them immune from the rules > of mathematics. <shrug> Tensors are quite different objects than matrices. They have different definitions altogether. One merely represents a tensor in matrix notation. If I give you a matrix there is no way to determine whether it represents a tensor or not. >> > Earlier you are hinting >> > the energy is an observed quantity. Since T is the energy-momentum [quoted text clipped - 35 lines] > momentum which you have agreed to be observer dependent already. > <shrug> The term "invariant" has many uses, the meaning of which depends on the context. When some people say that a tensor is "invariant" they typically mean that its definition is not coordinate dependant. E.g. a tensor can be defined as a multilinear map of vectors and 1-forms to the set of real numbers. Such a map does not depend on a coordinate system - hence the use of the term invariant. I, personally, don't use the term as such since it can lead to confusion. >> > [... further nonsense omitted] >> [quoted text clipped - 4 lines] > guarantee you that you will be utterly surprised and stop licking off > from your GR teachers' nonsensical regurgitations. <disgusting> Yipes. That's pretty gross! :) Pete > > For a scalar field phi with mass m, the usual Lagrangian (density) is: > > L = g^uv (d_u phi) (d_v phi) - m^2 phi^2 [quoted text clipped - 4 lines] > > I did, and I did not find anything you have claimed. Then, you did not try very hard, since the Lagrangian above is the relativistic wave equation, also know as the Klein-Gordon equation. > What justifies it to be a Lagrangian? WTF? It appears that you are rather unfamiliar with the concept of a Lagrangian. There is no prescription for ``deriving'' a Lagrangian, although understanding some physics is very helpful in eliminating the losers from among the possible choices. The justification is that it correctly describes experimental data - the same criteria used for centuries to justify any other equations being offered as physics. Is your criteria different? > > > Thus, you do agree that energy is merely an observed quantity depending > > > on which observer is doing the observation. [quoted text clipped - 16 lines] > The energy is how you choose to present it in that silly 4-momentum > stuff. Despite the silliness, it is mathematically valid. I guess that from your standpoint as the One True Disciple of Riemann, the more mundane considerations like using quantities with well-defined physical meanings must appear silly. > Thus, I am > only expressing my own mild objection. However, do you deny (E = m > c^2) or (m = E / c^2)? You omitted half of the rhs. The correct expression is E^2 = (pc)^2 + (mc^2)^2. > > > For a scalar field phi with mass m, the usual Lagrangian (density) is: > > > L = g^uv (d_u phi) (d_v phi) - m^2 phi^2 [quoted text clipped - 6 lines] > is the relativistic wave equation, also know as the Klein-Gordon > equation. I don't see any resemblance to the following. http://en.wikipedia.org/wiki/Klein-Gordon_equation > > What justifies it to be a Lagrangian? > > WTF? WTF? > It appears that you are rather unfamiliar with the concept > of a Lagrangian. Wrong. > There is no prescription for ``deriving'' a > Lagrangian, although understanding some physics is very helpful > in eliminating the losers from among the possible choices. Wrong, again. Just because the classical Lagrangian is (L = K + U) where (L = Lagrangian), (K = Kinetic energy), (U = Potential energy), that does not mean it is not derived. I choose to define (Energy = K - U) where (U >= 0). Get over with my definition. Apparently, you do not understand how the Euler-Lagrange equations are derived. <shrug> > The justification is that it correctly describes experimental > data - the same criteria used for centuries to justify any other > equations being offered as physics. You got lucky. <shrug> Your luck would eventually run out. We are discussing physics not voodoo experimentation in case if you have not noticed. > Is your criteria different? Yes, a Lagrangian has to satisfy all the conditions imposed on the derivation of the Euler-Lagrange equations. > > > > Thus, you do agree that energy is merely an observed quantity depending > > > > on which observer is doing the observation. [quoted text clipped - 20 lines] > Riemann, the more mundane considerations like using quantities > with well-defined physical meanings must appear silly. The 4-momentum was put together as a mathematical tool. I don't mean the tool is silly. I meant the way the 4-momentum is regarded as a physical entity and being worshiped as one is rather silly. Again, it is merely a mathematical tool. <shrug> > > Thus, I am only expressing my own mild objection. However, > > do you deny (E = m c^2) or (m = E / c^2)? > > You omitted half of the rhs. The correct expression is > E^2 = (pc)^2 + (mc^2)^2. It depends on how you define what m is. You are correct if you define m as m0, the intrinsic mass or the rest mass. I am correct if I define m as the observed mass or the relativistic mass. You also need to get over with that. <shrug> It is good to see you that you have balls to confront me. In the past few days, you have been nabbing at me with a silly sentence or two like a shy school boy. <shrug> > > > > For a scalar field phi with mass m, the usual Lagrangian (density) is: > > > > L = g^uv (d_u phi) (d_v phi) - m^2 phi^2 [quoted text clipped - 10 lines] > > http://en.wikipedia.org/wiki/Klein-Gordon_equation Please tell me that you are kidding. Even if you didn't know how to get from the Lagrangian to the Hamiltonian shown throughout most of the Wikipedia page, the expression at the bottom of the page under ``Action'' minus the integral sign is the same as the one above. > > > What justifies it to be a Lagrangian? > > [quoted text clipped - 6 lines] > > Wrong. |
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