DrRocket Posted March 2, 2011 Share Posted March 2, 2011 Also, mass is invariant — it doesn't increase with speed.[/modtip] That depends on what one means by "mass", in special relativity. The current fashion, apparently inspired by elementary particle physics, is to equate "mass" ([math]m[/math]) with "rest mass" ([math]m_0[/math]), wich indeed is invariant, essentially by definition. But there is also the very useful concept of "relativistic mass", [math]m = \gamma m_0 [/math]. [math] E=mc^2[/math] is correct (admitting that the zero-rest-mass case must be handled separately) and easily remembered if one interprets [math]m[/math] as relativistic mass. If one works only with rest mass one is stuck with [math]E^2 = m_0^2c^4 +p^2c^2[/math]. This latter equation applies to the zero-rest-mass case, but also raises the question of what one means by momentum, [math]p[/math]. In the case [math]m_o \ne 0[/math], [math]p=mv[/math] where [math]m = \gamma m_0[/math] so one still has to deal with relativistic mass, at least implicitly. There is also the concept of "invariant mass", which is the relativistic mass of a system of particles, measured in center-of-mass coordinates so that the total momentum is zero, so that the mass is [math] E/c^2[/math]. This corresponds to the mass that would be measured by a labratory balance for a macroscopic object -- a hot bucket of water in principle weighs more than a cold one. If instead one were to cling to "mass" as being "rest mass" in the macroscopic setting, then the mass measurement of a laboratory balance would not equal the sum of the masses of the molecules of which an object is composed -- creating all sorts of confusion. My position is to consciously adopt whatever convention is appropriate for the problem at hand, keeping in mind that different authors use different conventions in different situations. "Mass" is not author-invariant. http://en.wikipedia....cial_relativity The situation in general relativity is even more murky: http://en.wikipedia....eral_relativity http://matheuscmss.w...s-applications/ Link to comment Share on other sites More sharing options...

swansont Posted March 3, 2011 Share Posted March 3, 2011 That depends on what one means by "mass", in special relativity. The current fashion, apparently inspired by elementary particle physics, is to equate "mass" ([math]m[/math]) with "rest mass" ([math]m_0[/math]), wich indeed is invariant, essentially by definition. But there is also the very useful concept of "relativistic mass", [math]m = \gamma m_0 [/math]. "Usefulness" is subjective. Relativistic mass is just a proxy for total energy, so why not use total energy? [math]E^2 = m_0^2c^4 +p^2c^2[/math] is the equation you start with. [math]E=mc^2[/math] is only valid at rest; if you use it in general, you're making up a new equation, and then have to worry about what you mean by mass. It's not an ad hominem. What you wrote was dismissed because it was wrong. Why can't I think about physics qualitatively without understanding all the math? This has been explained at some length in other threads, but the fact that ask it again is emblematic of the issue here. You do not appear to be willing to accept explanations, because you continually question the validity of them when they don't fit some preconceived notion you have. That would be easier to accept if those questions came from a basis of understanding some physics. It isn't the job of anyone here to spoon-feed you. You can go to a foreign country and get by with hand gestures and pointing and a couple of travel-book phrases, but if you don't speak the language you can't have a conversation. If you want to discuss physics at a non-superficial level, you need to learn the language. 1 Link to comment Share on other sites More sharing options...

DrRocket Posted March 3, 2011 Author Share Posted March 3, 2011 (edited) "Usefulness" is subjective. Relativistic mass is just a proxy for total energy, so why not use total energy? [math]E^2 = m_0^2c^4 +p^2c^2[/math] is the equation you start with. [math]E=mc^2[/math] is only valid at rest; if you use it in general, you're making up a new equation, and then have to worry about what you mean by mass. Yes, you have to worry about what you mean by mass. That was the point. No, I am not making up a new equation, Einstein made up [math]E=mc^2[/math] (expressed a bit differently in the first paper, but clearly involving variable mass). http://www.fourmilab...tein/E_mc2/www/ If you use [math]E^2 = m_0^2c^4 +p^2c^2[/math] you have to come to grips with what you mean by momentum, and that will take you right back to "relativistic mass". Using total energy is fine. But you need to know what it is. Since what it is is [math]\gamma m_0 c^2[/math] for a body of positive rest mass you are right back to relativistic mass. Best to remain flexible and use the approach most convenient for the problem at hand. I find the recent tendancy to be dogmatic and demand that mass be rest mass rather amusing, and limiting. Edited March 3, 2011 by DrRocket Link to comment Share on other sites More sharing options...

swansont Posted March 3, 2011 Share Posted March 3, 2011 Yes, you have to worry about what you mean by mass. That was the point. No, I am not making up a new equation, Einstein made up [math]E=mc^2[/math] (expressed a bit differently in the first paper, but clearly involving variable mass). http://www.fourmilab...tein/E_mc2/www/ Einstein assumed the system to be at rest. When he concludes that the emission of energy L reduces mass by L/c^2, one must acknowledge that L — by his terminology — is the rest frame energy of the photon. The equation is not valid in any other frame. Link to comment Share on other sites More sharing options...

DrRocket Posted March 3, 2011 Author Share Posted March 3, 2011 Einstein assumed the system to be at rest. When he concludes that the emission of energy L reduces mass by L/c^2, one must acknowledge that L — by his terminology — is the rest frame energy of the photon. The equation is not valid in any other frame. That is simply wrong. [math] E=mc^2[/math] is valid in any inertial frame if [math]m[/math] is relativistic mass. From the Einstein paper : "The mass of a body is a measure of its energy-content; if the energy changes by L, the mass changes in the same sense by L/9 × 10^{20}, the energy being measured in ergs, and the mass in grammes." The reference to variable mass makes it clear that Einstein was considering what is now called "relativistic mass". BTWThere is no such thing as "the rest frame of the photon" in special relativity -- that would invoke singularities in Lorentz transformations that one cannot live with. You can do this either way, there is no need to be dogmatic. [math]E = \sqrt{m_0^2 c^4 + p^2c^2}[/math] [math]= \sqrt{m_0^2 c^4 + \gamma ^2 m_0^2 v^2 c^2}[/math] [math] = \sqrt { m_0^2c^4(1 + \gamma^2 \frac {v^2}{c^2}})[/math] [math]= \sqrt{\gamma^2 m_0^2 c^4}[/math] [math]= mc^2[/math] so long as [math]m_0 \ne 0[/math] The notion of relativistic mass goes back to Einstein, and was taught as a matter of course for many years. The convention to regard mass as "rest mass" is relatively recent and not universal. In either case we are talking about a convention, and not a question of "right" or "wrong" Regarding mass as rest mass is very convenient in quantum field theories (where m_0 can easily be 0), and less so for macroscopic problems in special relativity. In general relativity "mass' is even more murky, and nobody knows what to do about quantum gravity. Link to comment Share on other sites More sharing options...

swansont Posted March 3, 2011 Share Posted March 3, 2011 That is simply wrong. [math] E=mc^2[/math] is valid in any inertial frame if [math]m[/math] is relativistic mass. No, your claim was that relativistic mass shows up in the paper, and it doesn't. E=mc^2 appears in the context of being true in what Einstein defines as the rest frame. From the Einstein paper : "The mass of a body is a measure of its energy-content; if the energy changes by L, the mass changes in the same sense by L/9 × 10^{20}, the energy being measured in ergs, and the mass in grammes." The reference to variable mass makes it clear that Einstein was considering what is now called "relativistic mass". BTWThere is no such thing as "the rest frame of the photon" in special relativity -- that would invoke singularities in Lorentz transformations that one cannot live with. I didn't say rest frame of the photon. I said the rest frame of the system, i.e. what Einstein defined in the paper, the rest frame of the massive object. The energy of the photon is not frame invariant; the frame in which the photon energy is L is the rest frame of the massive object. You can do this either way, there is no need to be dogmatic. Nothing dogmatic about this. It's a matter of what you can derive. [math]E = \sqrt{m_0^2 c^4 + p^2c^2}[/math] [math]= \sqrt{m_0^2 c^4 + \gamma ^2 m_0^2 v^2 c^2}[/math] [math] = \sqrt { m_0^2c^4(1 + \gamma^2 \frac {v^2}{c^2}})[/math] [math]= \sqrt{\gamma^2 m_0^2 c^4}[/math] [math]= mc^2[/math] so long as [math]m_0 \ne 0[/math] The notion of relativistic mass goes back to Einstein, and was taught as a matter of course for many years. The convention to regard mass as "rest mass" is relatively recent and not universal. In either case we are talking about a convention, and not a question of "right" or "wrong" Regarding mass as rest mass is very convenient in quantum field theories (where m_0 can easily be 0), and less so for macroscopic problems in special relativity. In general relativity "mass' is even more murky, and nobody knows what to do about quantum gravity. Lev Okun has written a few papers documenting that this is not the case. Here's one http://arxiv.org/abs/0808.0437 Link to comment Share on other sites More sharing options...

DrRocket Posted March 3, 2011 Author Share Posted March 3, 2011 No, your claim was that relativistic mass shows up in the paper, and it doesn't. E=mc^2 appears in the context of being true in what Einstein defines as the rest frame. Wrong. Read the paper. Note the reference to variable mass, noted in my earlier post, which makes no sense if one is in the rest frame of the object. It was your claim that [math]E=mc^2[/math] applies only in the rest frame: "Usefulness" is subjective. Relativistic mass is just a proxy for total energy, so why not use total energy? [math]E^2 = m_0^2c^4 +p^2c^2[/math] is the equation you start with. [math]E=mc^2[/math] is only valid at rest; if you use it in general, you're making up a new equation, and then have to worry about what you mean by mass. It is quite clear that this is not true, since as you have asserted [math] E^2=m_0^2c^4 + p^2c^2[/math] is valid in any inertial frame and as shown previously and below it is eqiuivalent to [math]E=mc^2[/math] You do, of course have to worry about what you mean by mass, and [math]m[/math] is "relativistic mass". But you have to worry about what you mean by mass in any case. If you cling to "mass = rest mass" in all cases then what you measure on a laboratory scale is not mass. What you measure on a laboratory scale is in fact the relativistic mass in a reference frame in which the net momentum of the system of molecules is 0, what is called "invariant mass". Invariant mass is not the sum of the rest masses of the individual molecules. So, using a dogmatic "mass = rest mass" approach results in a macroscopic definition of mass that is not additive -- the mass of a system, as measured in the conventional manner, is not the sum of the masses of its constituent parts. You can adopt that stance, but it is, kindly, unconventional, and contrary to normal useage at the macroscopic scale. It requires some new definition of mass for the old reliable [math] F=ma[/math] and [math]F=\dfrac {E}{c^2}a[/math] just doesn't satisfy. Replacing the more rigorous [math]F=\dfrac {dmv}{dt}[/math] with [math]F= \dfrac {d (\frac{E}{c^2})v}{dt}[/math] is equally unappealing. I didn't say rest frame of the photon. I said the rest frame of the system, i.e. what Einstein defined in the paper, the rest frame of the massive object. The energy of the photon is not frame invariant; the frame in which the photon energy is L is the rest frame of the massive object. Einstein assumed the system to be at rest. When he concludes that the emission of energy L reduces mass by L/c^2, one must acknowledge that L — by his terminology — is the rest frame energy of the photon. The equation is not valid in any other frame. I misinterpreted what you meant by "rest frame energy of the photon. But the statement that "The equation is not valid in any other frame" is incorrect, as shown below. Nothing dogmatic about this. It's a matter of what you can derive. Yes, it is a matter of what you can derive. And [math]E=mc^2[/math] is derivable from and equivalent to [math] E^2 = m_0^2c^4 + p^2c^2[/math] and I showed you the derivation. I'll repeat it here. [math]E = \sqrt{m_0^2 c^4 + p^2c^2}[/math] [math]= \sqrt{m_0^2 c^4 + \gamma ^2 m_0^2 v^2 c^2}[/math] [math] = \sqrt { m_0^2c^4(1 + \gamma^2 \frac {v^2}{c^2}})[/math] [math]= \sqrt{\gamma^2 m_0^2 c^4}[/math] [math]= mc^2[/math] so long as [math]m_0 \ne 0[/math] So, "mass=rest mass" in all situations is in fact dogmatic. Better to fit your definition to the problem at hand, being clear of the definition used. It has the advantage that you don't have to give up [math] F=\dfrac {dp}{dt}[/math] which is rather useful for macroscopic dynamics problems; you just have to accept [math]m[/math] as relativistic mass. Lev Okun has written a few papers documenting that this is not the case. Here's one http://arxiv.org/abs/0808.0437 I don't care about Lev Okun's view of history. The mathematics is right here. There is nothing radical in what I am telling you. This is in any number of texts on relativity. If you accept [math]E^2 = m_0^2 c^4 + p^2c^2[/math] you must also logically accept [math]E=mc^2[/math] for they are the same equation (for [math]m_0 \ne 0 [/math]). The bottom line is that "mass" has several different meanings, all of which are legitimate in mainstream physics. Link to comment Share on other sites More sharing options...

ajb Posted March 4, 2011 Share Posted March 4, 2011 The bottom line is that "mass" has several different meanings, all of which are legitimate in mainstream physics. Sure. Now when one says "a particle of mass m" then one is referring to the rest mass. The reason we like the rest mass is that is is a Casimir of the Lorentz group: [math]P_{\mu}P^{\mu} = E^{2}-p^{2} = m^{2}[/math]. In general relativity, the mass of a space-time is essentially the same as the energy; "E = m" in appropriate units. As energy is directly related to invariance under time, in GR the notion of energy becomes much more involved and generally impossible to define. For composite systems "relativistic mass" could be a useful notion, but I don't see how it is really any different to the total energy (in a specified frame), as stated by swansont. (The same goes for mass of a space-time in GR.) Link to comment Share on other sites More sharing options...

Schrödinger's hat Posted March 4, 2011 Share Posted March 4, 2011 My problem with rest mass for composite systems is where do you stop? You're going to have energy bound up in bonds of various kinds as well as thermal, pressure etc. You'd need to go to all the effort of figuring out how much that is (yes, I know it's usually <<1%, but it's the principle that irritates me) rather than weighing it. Most of the time, for a composite system we don't really have rest mass and energy. Instead we have some of the energy, and some more of the energy. To me this means that neither has any right to conceptual primacy. On the other hand, calling both of them mass is annoying to the highest degree as well. Maybe we need a new word for total energy when you put it in a box, and stop the box? Rest mass is often used for this, but then when you open the box and only photons fly out you suddenly have to rename it. </bugbear since 1st year when we were asked a question about mass after being told about relativistic mass, but before anyone had mentioned rest mass> Link to comment Share on other sites More sharing options...

DrRocket Posted March 4, 2011 Author Share Posted March 4, 2011 My problem with rest mass for composite systems is where do you stop? You're going to have energy bound up in bonds of various kinds as well as thermal, pressure etc. You'd need to go to all the effort of figuring out how much that is (yes, I know it's usually <<1%, but it's the principle that irritates me) rather than weighing it. Most of the time, for a composite system we don't really have rest mass and energy. Instead we have some of the energy, and some more of the energy. To me this means that neither has any right to conceptual primacy. On the other hand, calling both of them mass is annoying to the highest degree as well. Maybe we need a new word for total energy when you put it in a box, and stop the box? Rest mass is often used for this, but then when you open the box and only photons fly out you suddenly have to rename it. </bugbear since 1st year when we were asked a question about mass after being told about relativistic mass, but before anyone had mentioned rest mass> I suspect that the problem goes way back to very early science classes where everyone was exposed to the littany differentiating between mass and weight (no arguement there) and being inculcated with the notion that mass is fundamental. Mass is not quite so fundamental as folks were told in those early days. It just is not that easy. What you measure on a labratory scale is relativistic mass, in the version of "invariant mass" taken in a reference frame of zero momentum. The mass that shows up in general relativity is also relativistic mass, but momemtum flux also shows up in the stress-energy tensor, and curvature is independent of any net linear velocity. So there is not a direct and easy correspondence between just rest mass or just relativistic mass either. GR is very subtle. In quantum field theories it is natural to consider mass as a charactristic of a particle and rest mass is the natural concept. What is really fundamental is the perspective that mass and energy are the same thing, and what is conserved is neither (rest) mass nor energy but mass/energy as a single entity. I suspect that the rigid "mass is rest mass" stance that is common these days comes from the particle physics influence. There are a lot more particle physicists than relativists. Perhaps you are right and we need a different word, or words. We are somewhat prisoners of our language, and "mass' is with us. It just is one of those words with several definitions. It is not a big problem as long as one is clear on the definition used in any given situation. Misner, Thorne and Wheeler attempt to break the language straight-jacket by using words like momeneregy, mass/energy, and of course spacetime. That does sometimes help, but their terminology is not widely used. Similar terminology issues exist in mathematics -- compact vs the French (compact + Hausdorf), Riemannian vs Pseudo-Riemannian, etc. -- but mathematicians seem more accepting of adopting local author-dependent conventions, perhaps because mathematicians are very careful about the definitions that are used in any specific theorem. For composite systems "relativistic mass" could be a useful notion, but I don't see how it is really any different to the total energy (in a specified frame), as stated by swansont. (The same goes for mass of a space-time in GR.) It isn't any different. The terminology does reinforce that mass and energy are the same thing, two sides of a single coin. Link to comment Share on other sites More sharing options...

swansont Posted March 4, 2011 Share Posted March 4, 2011 Wrong. Read the paper. Note the reference to variable mass, noted in my earlier post, which makes no sense if one is in the rest frame of the object. I did read the paper. "mass" is mentioned four times, all at the end: If a body gives off the energy L in the form of radiation, its mass diminishes by L/c². The fact that the energy withdrawn from the body becomes energy of radiation evidently makes no difference, so that we are led to the more general conclusion that The mass of a body is a measure of its energy-content; if the energy changes by L, the mass changes in the same sense by L/9 × 1020, the energy being measured in ergs, and the mass in grammes. In all instances of the use of the term, it is clearly in the reference frame where the mass is at rest, because this is the frame where the photon energy is L. So he is not referring to relativistic mass. Variable mass is not the same as relativistic mass; the mass in this case varies because they system releases energy. This is the same concept behind binding energy and mass defect of a nucleus, which is mentioned by Einstein (without the terminology) as a test for the concept. You have noted this yourself, that invariant mass is a separate concept from relativistic mass. It was your claim that [math]E=mc^2[/math] applies only in the rest frame: It is quite clear that this is not true, since as you have asserted [math] E^2=m_0^2c^4 + p^2c^2[/math] is valid in any inertial frame and as shown previously and below it is eqiuivalent to [math]E=mc^2[/math] "Equivalent to" is not the same as "equals." [math]E^2=m^2c^4 + p^2c^2[/math] and [math]E=mc^2[/math] cannot both be true. You only get to relativistic mass by redefining what you mean by mass. The fact that you used the former equation to derive the latter just proves the point I made in post #2. You do, of course have to worry about what you mean by mass, and [math]m[/math] is "relativistic mass". But you have to worry about what you mean by mass in any case. If you cling to "mass = rest mass" in all cases then what you measure on a laboratory scale is not mass. What you measure on a laboratory scale is in fact the relativistic mass in a reference frame in which the net momentum of the system of molecules is 0, what is called "invariant mass". Invariant mass is not the sum of the rest masses of the individual molecules. So, using a dogmatic "mass = rest mass" approach results in a macroscopic definition of mass that is not additive -- the mass of a system, as measured in the conventional manner, is not the sum of the masses of its constituent parts. You can adopt that stance, but it is, kindly, unconventional, and contrary to normal useage at the macroscopic scale. It requires some new definition of mass for the old reliable [math] F=ma[/math] and [math]F=\dfrac {E}{c^2}a[/math] just doesn't satisfy. Replacing the more rigorous [math]F=\dfrac {dmv}{dt}[/math] with [math]F= \dfrac {d (\frac{E}{c^2})v}{dt}[/math] is equally unappealing. Since I did not say anything about rest mass (I said it was invariant), most of this is moot. I don't care about Lev Okun's view of history. Is he wrong about the history? I only brought it up because you made a claim that Einstein was a proponent of relativistic mass. Do you retract it? Link to comment Share on other sites More sharing options...

DrRocket Posted March 5, 2011 Author Share Posted March 5, 2011 I only brought it up because you made a claim that Einstein was a proponent of relativistic mass. Do you retract it? Einstein, at least in his in later years was not a proponent of relativistic mass. But the notion comes directly from his special theory and from [math]E=mc^2[/math] which is perfectly valid, as I proved above. I don'rhink you are interpreting the Einstein paper properly (see end of this post). You are being dogmatic. Unnecessarily.. But it is equally clear that you will not be convinced that there are several different definitions of mass. All are valid, and all are useful in the proper context. You can certainly continue with little penalty, as one can formulate a valid description of physics using only one single concept. But that formulation will be very clumsy in some situations. Since you are wedded to "mass is rest mas" you should, to be consistent to your principles, reject any and all mass measurements made with a labotatory balance or a bathroom scale. You will need to chamge [math]F=ma[/math] to [math]F=\frac {E}{c^2}a[/math] and [math]F=G{m_1m_2}{r^2}[/math] to [math]F=G \dfrac {\frac {E_1}{c^2} \frac {E_2}{c^2}}{r^2}[/math]. You will need to revise what you mean by mass in general relativity, as the mass entry in the stress-energy tensor is relativistic mass, not rest mass, and it is relativistic mass that figures in the momentum flux entries as well. Denying relativistic mass and insisting that "m" always be rest mass will make reading much of the literature at best confusing. There are tons of texts that use the concept of relativistic mass, and to good effect. ALL books on classical mechsanics use" m" to denote "invariant mass", relativistic mass in the center of mass frame. That allows them to measure mass on a labratory scale. Do you propose to revise Goldstein to match your standard ? You can certainly do what you propose. It is most certainly correct. But it is hard to do with a straight face. Addendum Now, with regard to the Einstein paper we have (saving me some typing) this bit from Wiki, with which I agree "Einstein considered a body at rest with mass M. If the body is examined in a frame moving with nonrelativistic velocity v, it is no longer at rest and in the moving frame it has momentum P = Mv.Einstein supposed the body emits two pulses of light to the left and to the right, each carrying an equal amount of energy E/2. In its rest frame, the object remains at rest after the emission since the two beams are equal in strength and carry opposite momentum. But if the same process is considered in a frame moving with velocity v to the left, the pulse moving to the left will be redshifted while the pulse moving to the right will be blue shifted. The blue light carries more momentum than the red light, so that the momentum of the light in the moving frame is not balanced: the light is carrying some net momentum to the right. The object has not changed its velocity before or after the emission. Yet in this frame it has lost some right-momentum to the light. The only way it could have lost momentum is by losing mass. This also solves Poincaré's radiation paradox, discussed above. The velocity is small, so the right-moving light is blueshifted by an amount equal to the nonrelativistic Doppler shift factor 1 − v/c. The momentum of the light is its energy divided by c, and it is increased by a factor of v/c. So the right-moving light is carrying an extra momentum ΔP given by: The left-moving light carries a little less momentum, by the same amount ΔP. So the total right-momentum in the light is twice ΔP. This is the right-momentum that the object lost. The momentum of the object in the moving frame after the emission is reduced by this amount: So the change in the object's mass is equal to the total energy lost divided by c^{2}. Since any emission of energy can be carried out by a two step process, where first the energy is emitted as light and then the light is converted to some other form of energy, any emission of energy is accompanied by a loss of mass. Similarly, by considering absorption, a gain in energy is accompanied by a gain in mass. Einstein concludes that the mass of a body is a measure of its energy content." Link to comment Share on other sites More sharing options...

swansont Posted March 5, 2011 Share Posted March 5, 2011 Einstein, at least in his in later years was not a proponent of relativistic mass. But the notion comes directly from his special theory and from [math]E=mc^2[/math] which is perfectly valid, as I proved above. I don'rhink you are interpreting the Einstein paper properly (see end of this post). You haven't yet demonstrated that he was a proponent of relativistic mass in his younger years. You are being dogmatic. Unnecessarily.. Here's the problem with that (and related accusations of impropriety). If they're wrong, they merely serve as a distraction and are often used as an ad hominem. Dogma is a doctrine that is proclaimed as true without proof or justification. But I am providing proof/justification. Ergo, it is not dogma, and I am not being dogmatic. It's really that simple. But it is equally clear that you will not be convinced that there are several different definitions of mass. All are valid, and all are useful in the proper context. But that is neither my objection nor my argument. My objection is that relativistic mass is a defined term, not a derived term. It doesn't show up by itself in any of these equations. You have to define it, for convenience, and only as a proxy for total energy. The drawback that I see is that the first thing lost in discussion, especially with physics neophytes, is the context, and then you have people trying to equate relativistic mass with invariant mass, and that causes problems. You can certainly continue with little penalty, as one can formulate a valid description of physics using only one single concept. But that formulation will be very clumsy in some situations. Since you are wedded to "mass is rest mas" you should, to be consistent to your principles, reject any and all mass measurements made with a labotatory balance or a bathroom scale. You will need to chamge [math]F=ma[/math] to [math]F=\frac {E}{c^2}a[/math] and [math]F=G{m_1m_2}{r^2}[/math] to [math]F=G \dfrac {\frac {E_1}{c^2} \frac {E_2}{c^2}}{r^2}[/math]. <Sigh> I will point out again that I mentioned invariant mass, not rest mass. Saying I am wedded to mass is rest mass is a mischaracterization (a straw man). Denying relativistic mass and insisting that "m" always be rest mass will make reading much of the literature at best confusing. There are tons of texts that use the concept of relativistic mass, and to good effect. ALL books on classical mechsanics use" m" to denote "invariant mass", relativistic mass in the center of mass frame. That allows them to measure mass on a labratory scale. Do you propose to revise Goldstein to match your standard ? Again, that is not necessary, as it was not my position to begin with. Addendum Now, with regard to the Einstein paper we have (saving me some typing) this bit from Wiki, with which I agree "Einstein considered a body at rest with mass M. If the body is examined in a frame moving with nonrelativistic velocity v, it is no longer at rest and in the moving frame it has momentum P = Mv.Einstein supposed the body emits two pulses of light to the left and to the right, each carrying an equal amount of energy E/2. In its rest frame, the object remains at rest after the emission since the two beams are equal in strength and carry opposite momentum. But if the same process is considered in a frame moving with velocity v to the left, the pulse moving to the left will be redshifted while the pulse moving to the right will be blue shifted. The blue light carries more momentum than the red light, so that the momentum of the light in the moving frame is not balanced: the light is carrying some net momentum to the right. The object has not changed its velocity before or after the emission. Yet in this frame it has lost some right-momentum to the light. The only way it could have lost momentum is by losing mass. This also solves Poincaré's radiation paradox, discussed above. The velocity is small, so the right-moving light is blueshifted by an amount equal to the nonrelativistic Doppler shift factor 1 − v/c. The momentum of the light is its energy divided by c, and it is increased by a factor of v/c. So the right-moving light is carrying an extra momentum ΔP given by: The left-moving light carries a little less momentum, by the same amount ΔP. So the total right-momentum in the light is twice ΔP. This is the right-momentum that the object lost. The momentum of the object in the moving frame after the emission is reduced by this amount: So the change in the object's mass is equal to the total energy lost divided by c^{2}. Since any emission of energy can be carried out by a two step process, where first the energy is emitted as light and then the light is converted to some other form of energy, any emission of energy is accompanied by a loss of mass. Similarly, by considering absorption, a gain in energy is accompanied by a gain in mass. Einstein concludes that the mass of a body is a measure of its energy content." You agree with this. The part, at the outset, that says "Einstein considered a body at rest with mass M." Where does relativistic mass make its appearance? Link to comment Share on other sites More sharing options...

DrRocket Posted March 5, 2011 Author Share Posted March 5, 2011 (edited) You haven't yet demonstrated that he was a proponent of relativistic mass in his younger years. Whether he was a proponent has not been the real issue. The question arose when I correctly stated that the equation [math] E=mc^2[/math] is valid in any inertial reference frame if "m" is taken as relativistic mass. Variability of mass with energy, the equivalence of mass and energy, comes straight from Einstein's work. But see the observation at the end of this post. Einstein initially used "none of the above". But that is neither my objection nor my argument. My objection is that relativistic mass is a defined term, not a derived term. It doesn't show up by itself in any of these equations. You have to define it, for convenience, and only as a proxy for total energy. The drawback that I see is that the first thing lost in discussion, especially with physics neophytes, is the context, and then you have people trying to equate relativistic mass with invariant mass, and that causes problems. I don't follow this at all. None of the quantities, mass (any version) or energy are defined in terms of more fundamental ideas or derived in special relativity. Relativistic mass in the form of [math] m = E/c^2 [/math], valid in any reference frame is as much derived as is [math]m_0=E/c^2[/math], which is valid only in the rest frame of the particle. Even in Newtonian mechanics [math]F=ma[/math] serves only to define either force or mass in terms of the other, but no independent definition of either is extant. Mass is simply taken as a primitive. In special relativity we have three versions of mass and any of the two are definable in term of the third. <Sigh> I will point out again that I mentioned invariant mass' date=' not rest mass. Saying I am [i']wedded to mass is rest mass[/i] is a mischaracterization (a straw man). This then seems to me to be an inconsistent position. Rest mass ([math]m_0[/math]) requires referral to a special reference frame -- the rest frame. But as you originally stated, does not change with motion. It has the appealing property of being the scalar factor relating 4-velocity to 4-momentum, or equivalently it is the Minkowski norm of the 4-momentum vector (in units in which c=1). In this sense rest mass really is invariant in the sense of Lorentz invariance. Relativistic mass ([math]m = \gamma m_0[/math]) is frame-dependent and very definitely does change with motion. It is not invariant in any sense. Invariant mass is most commonly applied to a system of particles. In the case of a single particle it coincides with rest mass. In the case of many particles (as with the molecules in a macroscopic system) invariant mass is the sum of the relativistic masses of the particles plus any potential energy from interactions,, in the reference frame in which the total momemtum of the system is 0 -- the center-of-mass frame. So to deal with invariant mass you must first accept relativistic mass and then go further and consider a system-specific reference frame. Invariant mass is actually the most conceptually difficult of the three ideas. Invariant mass varies with thermal motion and so does not conform with your statement that mass does not vary with motion. One needs a bit of care with invariant mass. this is because in general invariant mass[math](A \cup B) \ne [/math] invariant mass [math] (A) + [/math] invariant mass[math](B) [/math]. Consider two bodies moving at relativistic speed along paths at right angles. Then individually their invariant masses are their rest masses, but the invariant mass of the system is not just the sum of the rest masses. The center of mass is moving along a 45 degree line in our "fixed" frame and the "invariant mass" is referred to a frame with that point as the origin. For large systems in thermal motion this problem goes away. Again' date=' that is not necessary, as it was not my position to begin with.[/quote'] I have become a bit confused as to what your position is. More so because I know that you understand relativity. You agree with this. The part' date=' at the outset, that says "Einstein considered [b']a body at rest[/b] with mass M." Where does relativistic mass make its appearance? As with any quantitative analysis using special relativity, one must select a reference frame. Einstein selected the rest frame of the body as the most convenient one in which to write his equations. What is revealing is his statement, "If a body gives off the energy L in the form of radiation, its mass diminishes by L/c². The fact that the energy withdrawn from the body becomes energy of radiation evidently makes no difference, so that we are led to the more general conclusion that The mass of a body is a measure of its energy-content; if the energy changes by L, the mass changes in the same sense by L/9 × 10^{20}, the energy being measured in ergs, and the mass in grammes." -- bold added In more usual notation L=E and 9 × 10^{20}= c^2 Hence what this equation states is quite simply [math]E=mc^2[/math] Changing of mass momentumin the rest frame makes limited sense (only when electromagnetic energy is emitted or absorbed), since as you have said, rest mass is invariant and rest mass and "invariant mass" are the same thing for the monolithic body that Einstein was addressing. Einstein's statement in greater generality would seem to apply to kinetic as well as electromagnetic energy. Remember that in 1905 when Einstein wrote the paper the atomic hypothesis was not universally accepted, so the equivalence of "invariant mass" with the mass of Newton was not so clear. I don't know if "invariant mass" was even a concept in use at the time. I will admit, however, that others interpret m in that same paper to be rest mass. I don't ascribe to that interpretation because of the words in bold above. With regard to what Einstein used in the early years, I have done a little research. The answer is a bit ugly. Relativistic mass was introduced by Tolman in 1932. Einstein and Lorentz, prior to that, used the notions of longitudinal and transverse mass -- [math]m_T= \gamma m_0[/math], [math]m_L= \gamma ^3 m_0[/math]. Dynamics in this formulation is repulsive. With relativistic mass one has simply [math] F= \frac {dp}{dt} = \frac {d(mv)}{dt}[/math] just as with Newton, simple and elegant. http://en.wikipedia....cial_relativity My only point in bringing up Einstein was to counter the statement that [math]E=mc^2[/math] is any sort of "new equation" or invalid. It is neither. Using Einstein's early writings from a technical perspective is probably not a good idea. Any good physicist or physics student today probably understands relativity better than did Einstein. They should. Einstein's genius was in developing relativity from scratch. Modern scholars have the benefit of over a century of work on the subject by a boat load of geniuses. You are free to stick to one and only one version of mass if you wish. I will continue to use whatever is most convenient for me in the application at hand, being clear as to what concept is being used. I doubt that you will see me using longitudinal or transverse mass (Ohanian in his book says that Einstein himself made mistakes with those concepts). In my first course involving relativity, we used relativistic mass, and I seem to have survived without permanent damage. Edited March 6, 2011 by DrRocket Link to comment Share on other sites More sharing options...

swansont Posted March 6, 2011 Share Posted March 6, 2011 Whether he was a proponent has not been the real issue. The question arose when I correctly stated that the equation [math] E=mc^2[/math] is valid in any inertial reference frame if "m" is taken as relativistic mass. Variability of mass with energy, the equivalence of mass and energy, comes straight from Einstein's work. Variability of mass is not the same as relativistic mass. You can't use one to justify using the other. You claimed relativistic mass showed up in the link, as one justification for using it. It's not there. I don't follow this at all. None of the quantities, mass (any version) or energy are defined in terms of more fundamental ideas or derived in special relativity. Relativistic mass in the form of [math] m = E/c^2 [/math], valid in any reference frame is as much derived as is [math]m_0=E/c^2[/math], which is valid only in the rest frame of the particle. Even in Newtonian mechanics [math]F=ma[/math] serves only to define either force or mass in terms of the other, but no independent definition of either is extant. Mass is simply taken as a primitive. In special relativity we have three versions of mass and any of the two are definable in term of the third. You said it yourself. What we use, and are used to, is invariant mass. Rest mass is a subset of that — a special case for a single particle. Relativistic mass is not. It is a redefinition of mass. This then seems to me to be an inconsistent position. Rest mass ([math]m_0[/math]) requires referral to a special reference frame -- the rest frame. But as you originally stated, does not change with motion. It has the appealing property of being the scalar factor relating 4-velocity to 4-momentum, or equivalently it is the Minkowski norm of the 4-momentum vector (in units in which c=1). In this sense rest mass really is invariant in the sense of Lorentz invariance. Relativistic mass ([math]m = \gamma m_0[/math]) is frame-dependent and very definitely does change with motion. It is not invariant in any sense. And we like having Lorentz invariance, no? Link to comment Share on other sites More sharing options...

Schrödinger's hat Posted March 6, 2011 Share Posted March 6, 2011 (edited) I don't see why this is worth getting so worked up about, there's only one vaguely real/consistent thing, which is momentum-energy. Ambiguity in these terms clearly came from somewhere, otherwise there would be no argument. I think the most useful thing to do at this point is start thinking of what you can do whenever you talk about these concepts to reduce this. There are three useful ways I know of to break momentum-energy down. 1) The scalar product of momentum-energy of a system/object of interest. I don't think we usually care why it's there, just how much there is. 1 photon has none of this, 2 photons can have some if they aren't in the same direction. [math] \left| \sum_i P_i \right| [/math] 2) The scalar product of momentum energy of a single elementary particle, or the sum of these quantities for each individual particle. This is the only time that rest mass is truly meaningful and useful, although for composite systems rest mass is a good approximation. Any time you're dealing with anything as or more complicated than a proton you don't actually have rest mass. [math] \sum_i \left| P_i \right| [/math] 3) Time component of momentum-energy in a given frame. [math]\sum_i \gamma_0\cdot P_i [/math] Edit: I suppose transverse mass would be a fourth. Quite useful, although it seems like a silly concept to call mass in any reasonable sense. The only issue comes up because we used to call them all the same thing. If you put something on a set of scales you're always measuring 1, which is degenerate with 3 because you have to hold it still If you're using non lorentz invariant equations you're using/measuring 3 If you're doing theoretical work, or particle or nuclear physics you will probably use 2, or sometimes 1. When 2 is appropriate, it is usually degenerate with 1 They all get called mass for a number of reasons I can see: People are lazy and just shorten relativistic mass to mass People are lazy and just shorten rest mass to mass People aren't using/don't care about lorentz invariance They decided to redefine mass as one of them, but in no relativity text I have seen -- caveat: I do not read many papers -- do people seem to distinguish between 1, 2 and 3 correctly, they either call 1 and 3 relativistic mass and 2 rest mass, or 1 and 2 rest mass, and 3 relativistic mass. Sometimes energy/relativistic energy is thrown in here too, to add confusion. Suggested solution: 3 shall be known as total energy/relativistic energy -- could do with something better here 2 shall be known as rest mass, most situations in which it is useful/will be used it is degenerate with 1 1 shall be known as mass or rest energy These definitions are as close to consistent with everything I have read as I can manage without inventing new words As the term relativistic mass is used for both 1 and 3 very frequently it should be avoided because it causes confusion and ambiguity. Further thoughts/edit: With relativistic mass one has simply just as with Newton, simple and elegant. How often is force a useful concept in a relativistic setting? This seems a bit like trying to keep one foot in each pond. Again, I'm inexperienced, but surely a (2?)-form/rotor of some kind is much more useful than something projected onto a frame, at which point you'd be using a vector for momentum anyway? Oop: Didn't read all of swansont's post. Invariant mass==1 Edited March 6, 2011 by Schrödinger's hat Link to comment Share on other sites More sharing options...

DrRocket Posted March 6, 2011 Author Share Posted March 6, 2011 Variability of mass is not the same as relativistic mass. You can't use one to justify using the other. You claimed relativistic mass showed up in the link, as one justification for using it. It's not there. You said it yourself. What we use, and are used to, is invariant mass. Rest mass is a subset of that — a special case for a single particle. Relativistic mass is not. It is a redefinition of mass. And we like having Lorentz invariance, no? As far as I can tell you are arguing semantics, not physics. None of "rest mass", "invariant mass" or "invariant mass" are "re-definitions" in any meaningful sense. All go back over three-quarters of a century. All are valid. All have a long history of use. All are recognized terms, with well-defined meanings. The whole point is that "mass" without further qualification is ambiguous in the context of special relativity. If you insist that one be solely entitled to the term "mass" then you can do that, but recognize that in so doing you are being arbitrary. I have seen other physicists argue as vociferously for "rest mass" as you argue for "invariant mass". You will have difficulty communicating with such a pereson since you will be using one word for two distinct concepts. Just out of curiousity, how do you relate "mass" to "inertia" ? I personally try to avoid the term and simply stick to equations of dynamics with the type of mass being clearly specified; e.g. [math] F=\frac {d(mv)}{dt}[/math] where [math] m = \gamma m_0[/math] Link to comment Share on other sites More sharing options...

csmyth3025 Posted March 25, 2011 Share Posted March 25, 2011 (edited) Hello DrRocket! Fancy meeting you here! This thread has been very interesting but, as you know, my ability to appreciate the nuances of the arguments presented here is limited - to say the least. To simplify things (for me), let me take a question posted elsewhere in another thread and pose it here in the hope that the answer might help me differentiate the various definitions you good folks have been debating. If I take a 1 kg mass, pick it up off the floor and place it on a shelf 2 m high, what - if anything - changes. From what I've read here I'm guessing that the rest mass and the invariant mass remain the same. Of course I could be wrong about that. That reduces the question to: Does the relativistic mass or equivalent energy content (whichever term you prefer) increase, decrease, or remain the same relative to the Earth? It's my understanding that these last two terms are always relative to something else. Best Regards Chris Edited to correct spelling error Edited March 25, 2011 by csmyth3025 Link to comment Share on other sites More sharing options...

DrRocket Posted March 25, 2011 Author Share Posted March 25, 2011 (edited) Hello DrRocket! Fancy meeting you here! This thread has been very interesting but, as you know, my ability to appreciate the nuances of the arguments presented here is limited - to say the least. To simplify things (for me), let me take a question posted elsewhere in another thread and pose it here in the hope that the answer might help me differentiate the various definitions you good folks have been debating. If I take a 1 kg mass, pick it up off the floor and place it on a shelf 2 m high, what - if anything - changes. From what I've read here I'm guessing that the rest mass and the invariant mass remain the same. Of course I could be wrong about that. That reduces the question to: Does the relativistic mass or equivalent energy content (whichever term you prefer) increase, decrease, or remain the same relative to the Earth? It's my understanding that these last two terms are always relative to something else. Best Regards Chris Edited to correct spelling error This entire discussion was predicated on special relativity. That implies no gravity. Mass in general relativity is a whole new can of worms. http://en.wikipedia....eral_relativity Edited March 25, 2011 by DrRocket Link to comment Share on other sites More sharing options...

csmyth3025 Posted March 25, 2011 Share Posted March 25, 2011 (edited) This entire discussion was predicated on special relativity. That implies no gravity. Mass in general relativity is a whole new can of worms. http://en.wikipedia....eral_relativity Thanks for the link. It's obvious that the answer to my question is much more complicated than the deceptively simple statement of the question itself. This, of course, leads to another question (and, perhaps, another thread): Can this question be answered in any meaningful way? Chris Edited to correct spelling error Edited March 25, 2011 by csmyth3025 Link to comment Share on other sites More sharing options...

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