Severian 410 Posted July 20, 2012 Share Posted July 20, 2012 It is perfectly consistent to define mass as the frame dependent "relativistic mass" if you like. But it is a bit silly for 3 reasons: 1. All professional physicists mean "rest mass" when they say mass. 2. Rest mass is frame invariant, so a property of the particle, whereas relativistic mass is frame dependent, so dependent on the observer. 3. If m is used to denote relativistic mass, we have momentum p=mv. This is a waste of notation since p and v are now always proportional to one another. It is much more convenient to keep a non-linear relation between them, i.e. [math]p=\gamma m v[/math] with [math]\gamma = 1/\sqrt{1-v^2/c^2}.[/math] 3 Link to post Share on other sites

pmb 12 Posted July 20, 2012 Share Posted July 20, 2012 It is perfectly consistent to define mass as the frame dependent "relativistic mass" if you like. But it is a bit silly for 3 reasons: It's hardly silly. Many relativity textbooks us it. Percentage wise about 67% of textbooks published between 1970 and now use it. I see no valid reason to think it silly when that many physicists use it 1. All professional physicists mean "rest mass" when they say mass. That's incorrect. I was speaking to a well-known cosmologist/particle physicist yesterday and he told me that in cosmology physicists use the term "mass" to refer to E/c^{2}. H'e's very very well-known in his field so I trust him. Plus I have several relativity texts which prove otherwise. The list inlcludes such texts by such authors as Misner Thorne and Wheeler (page 141), Mould, Rindler, D'Inverno, Schutz. Stephani. You can see a short list here with the relevant quotes http://home.comcast.net/~peter.m.brown/ref/relativistic_mass/relativistic_mass.htm 3. If m is used to denote relativistic mass, we have momentum p=mv. This is a waste of notation since p and v are now always proportional to one another. It is much more convenient to keep a non-linear relation between them, i.e. [math]p=\gamma m v[/math] with [math]\gamma = 1/\sqrt{1-v^2/c^2}.[/math] That relation doesn't always hold. On example is when the body is under stress. [math]p=\gamma m v[/math] is inlid under such a case but p = mv is still valid. Another example is an extended body which is emitting radiation. In such a case the mass per unit length of the rod can be uniformly decreasing. In an inertial frame moving parallel to the rod the mass per unit length is not uniform. The proper time in this case has no meaning so that [math]p = m\frac{dt}{d/tau}[/math] has no meaning. However p = mv still has meaning. For details please see http://home.comcast.net/~peter.m.brown/sr/invariant_mass.htm See last section entitled An Incorrect Application of Invariant Mass Link to post Share on other sites

juanrga 106 Posted July 20, 2012 Share Posted July 20, 2012 (edited) It is perfectly consistent to define mass as the frame dependent "relativistic mass" if you like. But it is a bit silly for 3 reasons: 1. All professional physicists mean "rest mass" when they say mass. 2. Rest mass is frame invariant, so a property of the particle, whereas relativistic mass is frame dependent, so dependent on the observer. 3. If m is used to denote relativistic mass, we have momentum p=mv. This is a waste of notation since p and v are now always proportional to one another. It is much more convenient to keep a non-linear relation between them, i.e. [math]p=\gamma m v[/math] with [math]\gamma = 1/\sqrt{1-v^2/c^2}.[/math] There are more reasons. Some were given in a specific thread about mass. Edited July 20, 2012 by juanrga Link to post Share on other sites

Severian 410 Posted July 20, 2012 Author Share Posted July 20, 2012 It's hardly silly. Many relativity textbooks us it. Percentage wise about 67% of textbooks published between 1970 and now use it. I see no valid reason to think it silly when that many physicists use it. [snip] Plus I have several relativity texts which prove otherwise. The list inlcludes such texts by such authors as Misner Thorne and Wheeler (page 141), Mould, Rindler, D'Inverno, Schutz. Stephani. You can see a short list here with the relevant quotes http://home.comcast....vistic_mass.htm Do you have any source to back up your 67%? I find that hard to believe. The link you include has a handful of texts. Also, the very fact that this thread exists proves that it is silly to use relativistic mass in text books, because it confuses the students. That's incorrect. I was speaking to a well-known cosmologist/particle physicist yesterday and he told me that in cosmology physicists use the term "mass" to refer to E/c^{2}. H'e's very very well-known in his field so I trust him. I also find that hard to believe. Who was this person? The only reason I can think of for doing this would be if their energies are so large that rest mass becomes negligible, allowing them to use the term "mass" for something else. Still seems a bit daft though. If you use the term "mass" to refer to E/c^2, why not refer to it as "energy"? They are even identical in Planck units. That relation doesn't always hold. On example is when the body is under stress. [math]p=\gamma m v[/math] is inlid under such a case but p = mv is still valid. Another example is an extended body which is emitting radiation. In such a case the mass per unit length of the rod can be uniformly decreasing. In an inertial frame moving parallel to the rod the mass per unit length is not uniform. The proper time in this case has no meaning so that [math]p = m\frac{dt}{d/tau}[/math] has no meaning. However p = mv still has meaning. For details please see http://home.comcast....ariant_mass.htm See last section entitled An Incorrect Application of Invariant Mass Well, it also breaks down for something moving at the speed of light. My point was that having a definition of mass that provides no extra benefit is pointless. p=mv is not useful. Lev Okun agrees with me. Link to post Share on other sites

imatfaal 2481 Posted July 20, 2012 Share Posted July 20, 2012 I have opened this thread to split off an argument from a thread about Mass of Light. The initial exchanges were as follows I prefer the term proper mass ove rest mass anyday. Even though the proper mass of photons is zero we still have 1) Inertial mass (aka relativistic mass) of photon is non-zero 2) Passive gravitational mass of light is non-zero 2) Active gravitational mass of light is non-zero Ummmm ........ he's here to learn physics. Hence his questions. Modern physics uses a fundamental concept of mass which is invariant and denoted by m. For a photon m=0; as is well-known photons are "massless particles". "Proper mass" is misleading, it could seem that m is the mass of a object only when measured in the proper frame, which is not true. m is also the mass of the object in the laboratory frame or in any other frame. 1 Link to post Share on other sites

juanrga 106 Posted July 20, 2012 Share Posted July 20, 2012 (edited) It's hardly silly. Many relativity textbooks us it. Percentage wise about 67% of textbooks published between 1970 and now use it. I see no valid reason to think it silly when that many physicists use it Older textbooks from the 60s--70s use relativistic mass. They also use other outdated concepts/notations today abandoned. Most modern and/or advanced textbooks do not use relativistic mass: Weinberg, Wald, Feynman, Caroll... Some introductory textbooks use relativistic mass as a trick to teach students some effects of relativity over Newtonian expressions. However, no serious treatment of relativity gives "relativistic mass" any relevance and it plays no role in fundamental physics or research. You can see a short list here with the relevant quotes http://home.comcast....vistic_mass.htm Which proves the above point. You are mostly citing introductory textbooks: "Introducing Einstein's Relativity", "A First Course in General Relativity", "A Short Course in General Relativity"... The only textbook which is not an introduction is that by Misner, Wheeler and Thorne, but I have a copy of this old book and I know that the authors use "rest mass" in many expressions since the firsts pages In fact, Wheeler and Taylor write in their textbook Spacetime Physics against relativistic mass: Ouch! The concept of 'relativistic mass' is subject to misunderstanding. That's why we don't use it. First, it applies the name mass - belonging to the magnitude of a 4-vector - to a very different concept, the time component of a 4-vector. Second, it makes increase of energy of an object with velocity or momentum appear to be connected with some change in internal structure of the object. In reality, the increase of energy with velocity originates not in the object but in the geometric properties of space-time itself." Not only the concept of relativistic mass gives misunderstandings to students, but its not used in research! It would be completely eliminated from literature. I also noticed that you mention a page in your own site where you give supposed links to supposed lecture notes online, but the links that I tried give three "404 not found" errors, one "bad request" error, and the two remaining links that I tried are very very elementary introductions to relativity [again relativistic mass is a pedagogical trick to introduce some relativistic effects to students who only know Newtonian physics] http://aether.lbl.go...imation/sr.html http://galileo.phys....s_increase.html I have just checked another supposed link that you give but I obtain another "404 Document not found". Wow! You also mention Einstein as support for your use of relativistic mass, but it is well-known that Einstein wrote: It is not good to introduce the concept of the mass M = m/(1-v2/c2)^{1/2} of a body for which no clear definition can be given. It is better to introduce no other mass than 'the rest mass' m. Instead of introducing M, it is better to mention the expression for the momentum and energy of a body in motion. That relation doesn't always hold. On example is when the body is under stress. [math]p=\gamma m v[/math] is inlid under such a case but p = mv is still valid. Another example is an extended body which is emitting radiation. In such a case the mass per unit length of the rod can be uniformly decreasing. In an inertial frame moving parallel to the rod the mass per unit length is not uniform. The proper time in this case has no meaning so that [math]p = m\frac{dt}{d/tau}[/math] This does not even has the correct dimensions (units). Edited July 20, 2012 by juanrga 1 Link to post Share on other sites

PaulS1950 23 Posted July 20, 2012 Share Posted July 20, 2012 The only reason a photon has energy or momentum is because of its velocity. "At rest" there is nothing - the photon can't exist. Therefore it is meaningless to have a discussion about the mass of an object that doesn't exist. Of course it has zero mass - it isn't there! Since its velocity is a function of its existence then we have to calculate its, energy, momentum and MASS in that reference frame. What is the mass of a photon traveling at C? What is the mass of a billion photons travelling at C? The momentum is a billion times higher than one... If a photon has energy and momentum then it MUST have mass (even if it is a result of its velocity which is a function of its existence). If M=0 then MC^2 =0 it cannot have kinetic energy (it does) and MC=0 it cannot have momentum (it does), it follows logically and mathmatically that a photon must have mass. M cannot be zero. Paul Link to post Share on other sites

juanrga 106 Posted July 20, 2012 Share Posted July 20, 2012 What is the mass of a photon traveling at C? Zero. What is the mass of a billion photons travelling at C? The momentum is a billion times higher than one... Zero. If a photon has energy and momentum then it MUST have mass (even if it is a result of its velocity which is a function of its existence). If M=0 then MC^2 =0 it cannot have kinetic energy (it does) and MC=0 it cannot have momentum (it does), it follows logically and mathmatically that a photon must have mass. M cannot be zero. No. The special relativistic expression for the energy is [math]E^2= m^2c^4 + \mathbf{p}^2 c^2[/math]. For massless particles [math]m = 0[/math] this reduces to [math]E = p c[/math]. Massless particles have both energy and momentum. http://en.wikipedia.org/wiki/Photons#Physical_properties Link to post Share on other sites

PaulS1950 23 Posted July 20, 2012 Share Posted July 20, 2012 (edited) We all know that you can't use zero when multiplying or dividing - it isn't proper math because zero is valueless. If p=0 then it has no meaning. cp = 0 but you don't get zero because p = x>0. If you multiply infinity by zero you still get zero (no infinity's) What is the value of p in E=pc? it has to be >0 Edited July 20, 2012 by PaulS1950 Link to post Share on other sites

juanrga 106 Posted July 20, 2012 Share Posted July 20, 2012 What is the value of p in E=pc? it has to be >0 For a photon [math]p=h\nu/c[/math]. The frecuency [math]\nu[/math] of photon is not zero; therefore its momentum cannot be zero. Link to post Share on other sites

PaulS1950 23 Posted July 20, 2012 Share Posted July 20, 2012 The frequency adds mass? I believe that mass times velocity is momentum. Link to post Share on other sites

juanrga 106 Posted July 20, 2012 Share Posted July 20, 2012 (edited) The frequency adds mass? A photon is massless http://en.wikipedia....ical_properties I believe that mass times velocity is momentum. [math]\mathbf{p}=m\mathbf{v}[/math] is only valid for massive particles at low velocities [math]v\ll c[/math]. Edited July 20, 2012 by juanrga Link to post Share on other sites

PaulS1950 23 Posted July 20, 2012 Share Posted July 20, 2012 How much mass would you have if you did not exist? How much mass would you have at the velocity C? A photon has mass because it is at velocity C. Taking it out of that reference frame is the same as my first question in this post. The second question is as meaningless as an at rest photon. Neither can happen. Link to post Share on other sites

swansont 7598 Posted July 20, 2012 Share Posted July 20, 2012 The frequency adds mass? I believe that mass times velocity is momentum. That's a Newtonian relation. Relativity supersedes that. 1 Link to post Share on other sites

alpha2cen 18 Posted July 20, 2012 Share Posted July 20, 2012 (edited) The frequency adds mass? I believe that mass times velocity is momentum. Mass definition problem? Which one is mass? 1)E^{2}=m^{2}C^{4}+p^{2}C^{2} m=1/C^{2}.(E^{2}-p^{2}C^{2}) 2)P=mv m=P/v P; momentum 3)E=(1/2)rmv^{2} m=2E/(rv^{2}) 4)F=ma m=F/a 5)F_{g}=mg m=F_{g}/g 6) another equation m=.. Edited July 21, 2012 by alpha2cen Link to post Share on other sites

juanrga 106 Posted July 21, 2012 Share Posted July 21, 2012 How much mass would you have if you did not exist? Being a massive body. I cannot exist if I do not have mass. How much mass would you have at the velocity C? Massive bodies cannot reach the speed of light, only massless particles can travel at c. Link to post Share on other sites

alpha2cen 18 Posted July 21, 2012 Share Posted July 21, 2012 Mass definition problem? Which one is mass? 1)E^{2}=m^{2}C^{4}+p^{2}C^{2} m=1/C^{2}.(E^{2}-p^{2}C^{2}) 2)P=mv m=P/v P; momentum 3)E=(1/2)rmv^{2} m=2E/(rv^{2}) 4)F=ma m=F/a 5)F_{g}=mg m=F_{g}/g 6) another equation m=.. About the mass definition. Which one is the more exact definition between the energy basis and the force basis? Link to post Share on other sites

Severian 410 Posted July 24, 2012 Author Share Posted July 24, 2012 What is the mass of a billion photons travelling at C? The momentum is a billion times higher than one... As juanrga said, zero, but this is in part because photons don't interact with each other. If these were gluons instead, then they would have a mass (from their interaction). Link to post Share on other sites

PaulS1950 23 Posted August 2, 2012 Share Posted August 2, 2012 [math] E=mc^2 + pc[/math] from [math] E^2 = m^2c^4 + p^2c^2[/math] by finding the square root of both sides of the equation. so [math] m=E/(c^2+pc) [/math] Since the photon exists only at the speed of light it is as improper to find the "rest" mass as it is to find your personal mass at c...... Both are meaningless. The only mass for a photon that has any meaning is the relativistic mass. Paul Link to post Share on other sites

Severian 410 Posted August 2, 2012 Author Share Posted August 2, 2012 (edited) [math] E=mc^2 + pc[/math] from [math] E^2 = m^2c^4 + p^2c^2[/math] by finding the square root of both sides of the equation. so [math] m=E/(c^2+pc) [/math] Since the photon exists only at the speed of light it is as improper to find the "rest" mass as it is to find your personal mass at c...... Both are meaningless. The only mass for a photon that has any meaning is the relativistic mass. Paul Firstly, your maths is wrong. Secondly, the rest mass of the photon is zero, so [math]E=pc[/math]. Thirdly, "zero" does have meaning. Edited August 2, 2012 by Severian Link to post Share on other sites

PaulS1950 23 Posted August 2, 2012 Share Posted August 2, 2012 Firstly, your maths is wrong. Secondly, the rest mass of the photon is zero, so [math]E=pc[/math]. Thirdly, "zero" does have meaning. Since the photon can't exist at rest you are right; the REST mass of a photon is zero. If you don't exist your mass is also zero. A photon, in its natural frame, has mass because of its velocity (at least). Paul Link to post Share on other sites

juanrga 106 Posted August 2, 2012 Share Posted August 2, 2012 I would add that "rest mass" is an old name which can be misleading. In modern literature we say invariant mass or simply mass. E.g. we say that the mass of a photon is zero. Link to post Share on other sites

Autumn_Man 0 Posted August 3, 2012 Share Posted August 3, 2012 [math] E=mc^2 + pc[/math] from [math] E^2 = m^2c^4 + p^2c^2[/math] by finding the square root of both sides of the equation. so [math] m=E/(c^2+pc) [/math] Since the photon exists only at the speed of light it is as improper to find the "rest" mass as it is to find your personal mass at c...... Both are meaningless. The only mass for a photon that has any meaning is the relativistic mass. Paul A better term than rest mass is proper mass. It has no conotations about being measured in a rest frame. It's an intrinsic property of a particle and can be calculated using energy ann momentm by the relation (letting m = proper mass) E^2 - (pc)^2 = m^2 c^4. Solve for m and you have the proper mass of any particle as determined from any frame of reference. Link to post Share on other sites

juanrga 106 Posted August 4, 2012 Share Posted August 4, 2012 A better term than rest mass is proper mass. It has no conotations about being measured in a rest frame. But proper mass is also a misleading name. Students could believe that proper mass is the mass in the proper frame, when m is the mass in any frame. This is the reason why we use "invariant mass" or simply "mass" to refer to m. Link to post Share on other sites

Autumn_Man 0 Posted August 4, 2012 Share Posted August 4, 2012 But proper mass is also a misleading name. Only to people who don't know what the term proper means. In this context the term proper means intrinsic. Proper mass is defined as the m in E^2 - (pc)^2 = m^2c^4. Until students learn what the terms mean who cares what mistakes they make. Mistakes are cleared up when the student learns the physics. Bad students believe all sorts of weird things This is the reason why we use "invariant mass" or simply "mass" to refer to m. "we" is what is found in the literature and "we" understand and use the term proper mass or rest mass too when that is the subject Link to post Share on other sites

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