Talk:Mass/Archive 2

The strong equivalence principle

In another wikipedia's page on Equivalence Principle, the strong equivalence principle is stated as:

"The gravitational motion of a small test body depends only on its initial position in spacetime and velocity, and not on its constitution. and The outcome of any local experiment, whether gravitational or not, in a laboratory moving in an inertial frame of reference is independent of the velocity of the laboratory, or its location in spacetime. "

But here it is said that strong equivalence principle means "the equivalence of inertial and gravitational mass".

I can't see the direct relationship between these two seemingly unrelated concepts. Perhaps this is a disagreement. If it's not, I suggest that someone could write a few more lines on the equivalence of these two defintion of "strong equivalence principle".

measurement of mass

after digesting the topic page and this discussion page, i'm wondering what components of a macroscopic object are included in determining the mass/energy of said object. emf spectrum output? gravitational effect on emf spectrum passing by? gravitational waves generated by its angular momentum? rotational speed? what about the emf field that leads to solar flares? the so-called solar wind? or is that part of the emf spectrum output?

there is the adage, "when you scream in space, no one can hear you." even in space, it's not entirely vacuous. there is the quantum fluctuations of particles existing and then not. can sound waves be generated by a macroscopic body orbiting a galaxy?

do gravitational waves behave the way light-waves do, reinforcing or mitigating each other? if yes, will it mess with the measurement?

Mass of compound objects

I've had to go through this section and fix it, in the same way as the section on mass in special relativity. Boy, it looks like this whole mass=energy thing still confuses people!

Look, kinetic energy of systems contributes to their REST mass. Similarly, a pair of photons headed in opposite directions has a mass, AS a system. Even though neither of them have mass, when considered SEPERATELY. It's only the NON invarient part of kinetic energy (what you see with a single particle) which has no rest mass. As soon as you're dealing with systems in the COM frame, where momentum sums to zero, then mass is separately conserved, because it IS the total relativistic energy. This leads to much confusion, which I've tried to remove from both of these articles. Sbharris 22:06, 14 April 2006 (UTC)

Yes, it still confuses people, and rightly so! It's a confusing subject!

For example, I disagree with your claim here; or at best I find it misleading. I infer from what you say that the inertia in the center of mass frame of two moving objects of mass m is 2mγ. However, the inertia of this system is not a scalar, its acceleration depends on the direction of the applied force. This makes it difficult to talk about any kind of invariant mass of this system. If you consider a large composite system composed of many particles with momentum isotropically distributed, then this issue disappears in the average, so you can say that the mass of a box of many photons is the sum of their energies, but not for two photons.

With this in mind, you final comment about conservation of mass holding in the center of mass frame also becomes suspicious. If I have two particles of mass m collide and stick together, then I'm left with a single particle of mass 2mγ. Mass has not been conserved. Anyway, the only way that claim could have been true is if you take "mass" to mean relativistic mass. -lethe ^{talk +} 22:23, 14 April 2006 (UTC)

You're certainly giving me a hard time for a few misspellings of the word invariant.

Look, the invariant mass of systems is correctly defined in the short wiki invariant mass. And not by me. However, the equation says it all, and you should go there and read. If you're a mathematician, that's the end of our discussion. It isn't "difficult" to talk about the invariant mass of systems-- particle physicists do it all the time.

The problem is not the definition, the problem is you trying to apply the definition of invariant mass to systems of moving objects. -lethe ^{talk +} 03:23, 16 April 2006 (UTC)

Why is that a problem?? The invariant mass is most commonly used in systems of "moving objects" and in fact usually objects traveling nearly at c. The invariant mass (which is the square root of the Mandelstam variable s) is often calculated by high energy physicists from the momenta of components of a particle decay (a particle which is usually traveling at high velocity in the lab frame), to obtain the mass = invariant mass= rest mass of the original particle. But there are a vast number of other ways this constant can be used. Here's a primer on relativistic kinematics with several examples. Note they all calculate invariant masses for unbound high energy systems, for example proton-antiproton creation from proton-on-proton collisions. http://www.phys.ufl.edu/~korytov/phz5354/note_C03_kinematics.pdf

>See the equation, apply it to your system of photons or your colliding objects, and observe that invariant mass is conserved.

You are incorrect on this point. Invariant mass is not conserved. -lethe ^{talk +} 03:23, 16 April 2006 (UTC)

Of course it is! It must be, since it's merely the length of the summed energy-momentum 4-vector, the conservation of which, is the basic SR treatment of the conservation of momentum and energy. It's not only conserved, but (as with all 4-vectors) it's invariant, so all observers see it as the same quantity.

Since in the COM frame the summed momenta of a system are zero, obviously invariant mass of the system is equal to relativistic mass in this frame. Thus, if relativistic mass (or total relativistic energy) is conserved in the COM frame (and we're all agreed that it is) then invariant mass is also conserved THERE. But since the invariant mass of objects (or systems of objects) is observer-independent (that's what we MEAN by invariant here), that means that if it's conserved in any one frame, it's conserved in all of them. QED.

What makes the COM frame so useful is that in this frame the invariant mass is the same as the total energy. Total energy may not be (generally is not) conserved in jumping from frame to frame (and neither is momentum), but since invariant mass is the same in all frames, if you find the energy and mass in the COM frame, at least you know that's the mass in the lab frame also (or whatever frame you're interested in), even if it isn't any longer proportional to the energy. That allows you to *calculate* the total energy, in any frame. That's so useful, that high energy physicists often transform to the COM to calculate total reaction mass (or the Mandelstam equiv). See examples above.

But even if you're not a high energy physicist, the COM frame is useful because it's then usually the lab frame. In any case, the mass *of systems* (both bound and unbound) is conserved in any given frame, even in relativity. And here I mean invariant mass-- that which causes gravitation. It would be endlessly difficult if it weren't so, because otherwise gravity would jiggle every time some reaction changed "matter" (ie, fermions) to "energy" (kinetic energy or radiation).

>If you want to define mass differently for this section (as something other than invariant mass) then go right ahead.

It is not me, but you who has to take a different definition of mass. Relativistic mass is conserved, this is another way of stating that energy is conserved. Invariant mass is not conserved, and I already gave you an example to show that it's not.

Your example discounted and ignored the mass associated with the kinetic energies of your moving objects. You can't do that. Kinetic energy has mass (invariant mass!) in systems, and your example presented a system where the kinetic energy would indeed have mass, even before the objects struck and the energy was transformed to heat.

If you want to see how this works, run your example in reverse: start with a composite object (initially at rest) which springs apart into two pieces. We follow from the initial rest frame, which is the COM frame. Does mass of the system decrease when the spring releases? No. Kinetic energy of the fragments contributes now to system mass, just as it does with a heated gas. Put a box around the fragments and let them bounce off the walls like gas molecules, if you don't believe it. Weigh the box.

The reason you have made this mistake is that you think the invariant mass of a composite system is its total energy.  But this is not correct. -lethe talk + 03:23, 16 April 2006 (UTC)

No. They are only the same in the COM frame. But that's the rest frame for common objects, so that's the basic reason that total energy gives the "rest mass" for common objects.

However, as noted above, invariant mass is not only conserved, but invariant also. Energy is conserved, but it is not invariant. In some ways the invariant mass of a system (which takes into account both energy and momentum) is an even more basic property of a system than either its total energy, or its 3-momentum (both of which are separately conserved, but neither of which is invariant).

But that's not the way relativists do it. And that's not the way the wiki on mass does it, when it defines mass as the scalar length of the energy-momentum 4-vector (which is the definition in relativity).

Ahh... yes, this gets to the bottom of why your definition is wrong: mass is defined to be the Casimir invariant of the Poincaré group. The invariant length of momentum. A theorem of Schur guarantees that this number is in fact an invariant for all irreducible systems. But of course, two particles do not constitute an irreducible system. It has two invariant reps: the COM frame and the displacement vector between the two particles. -lethe ^{talk +} 03:23, 16 April 2006 (UTC)

Not sure what you're talking about regarding "two invariant representations of a system of two particles." If you mean invariant under Lorentz transformations, there is only one quantity for a system of the type we seek. It is the scalar length of the sum of the particle energy-momentum 4-vectors. This can be calculated for any system in any frame, and it gives the same answer. "The displacement vector between the particles"? That's doesn't sound like an inertial frame. What kind of observer sees that? Representations in physics should be, well, physical.

>Of course, that applies to systems also, but the article implies that, when it notes that heat contribues to masses of objects. Which of course it does.

Yes, heat contributes. Kinetic energy contributes in any large composite system where momenta are distributed evenly, all but the COM rep become trivial. -lethe ^{talk +} 03:23, 16 April 2006 (UTC)

So does kinetic energy.

This is your mistake. -lethe ^{talk +} 03:23, 16 April 2006 (UTC)

You're going to have to explain that. Kinetic energy contributes in large composite systems, but not in 2-particle systems? Are you proposing some kind of complexity cut-off?

If you take this as your definition of mass, then the mass is no longer a measure of the inertia of the system.

It isn't intended to be. The only invariant inertia is the one that's measured when the body is at rest (or in the COM frame for a system).

> Remember, moving bodies have different transverse inertias than longitudinal inertia. I submit that any definition of mass which is not inertia, is not a good definition. -lethe ^{talk +} 03:23, 16 April 2006 (UTC)

I submit that your definition of inertia isn't a good definition. You're after something that is invariant, and your definition of inertia isn't. Bodies have different transverse and longitudinal LENGTHS also, as seen by different observers. So what? That only means that length isn't invariant, and isn't as interesting, for that reason. Neither is inertia, if you insist on measuring it from a non-rest frame.

>I left the section on inertia alone, but I dislike the discussion because "inertia," like energy, is a varying and not useful quantity,

I disagree with you. Inertia, as measured in the rest frame, is an invariant quantity for elementary systems.

You're not talking about the "rest frame" when you go on about "transverse and longitudinal inertias"-- you're talking about the moving frame which gives those ideas meaning. IOW, those apply by definition only to inertias measured from moving frames. You can't have it both ways. We going in circles.

In physics "invariant" is short for "Lorentz invariant", i.e., does not vary under Lorentz space-time transformations. I have no idea if inertias for two-particle systems are invariant, but since they are directly connected to system momenta, I doubt it. However, the same is not true of energy-momentum 4-vectors.

Just as Schur's lemma guarantees you it has to be. Schur's lemma fails for composite systems, and therefore the inertia as measured in the rest frame of a composite system is not invariant. The physics and the mathematics line up perfectly. -lethe ^{talk +} 03:23, 16 April 2006 (UTC)

If you say so. The inertia of composite systems may not be an invariant (may not even have a good single definition), but I guarantee you that the invariant mass is invariant and does have a good and single definition.

You're making some fairly controversial edits. You're doing it to many articles. You have provided no references, you have no consensus to support you, your spelling is bad, and your latex is broken. I think you should try a little harder to ensure the quality of your edits, this may protect you from being reverted.

<Sigh> There must be some Latin phase for argumentum ad hominum subspecies "better speller." And another which must translate as "I'm right because I can use the Wiki editor better". I recognize argumentum ad populam, but it doesn't work too well in science.

If you want a general reference to the use of 4-vectors in SR, you can try Taylor and Wheeler's Spacetime Physics, which I'm sure is on your shelf. You can also take a look at the reference to the usenet physics FAQ on photon mass (which is already referenced in the article on mass) to get some idea about the mass of photons in a box; that alone is a good illustration that massless objects may add mass to systems (total system mass is NOT the sum of masses of system objects, but I never said it was). The reference I gave above on kinematics should illustrate that particle physicists use the invariant mass of systems of rapidly moving decay products to back-calculate the rest mass of parent particles. Your problem is that you don't really believe that corresponds with something you can weigh on a scale, which is to say, mass as we usually measure it. Your problem.

As it stands, I'm still trying to decide whether to revert you at the 10 other articles about mass to which you added this (in my opinion incorrect) idea about the rest mass of two moving particles. I stand by my decision to revert your edits. -lethe ^{talk +} 03:23, 16 April 2006 (UTC)

Well, you're wrong. And the longer you take with your position, the sillier you'll look!

Look, I understand your position: if we define invariant mass to be the total energy in the rest frame, then invariant mass is conserved. You've discovered a convoluted way of saying "energy is conserved even in the rest frame".

No, more than that. Mass is conserved and in every frame. I'm talking about mass as we weigh it on the scale, for compound objects. But also this includes a generalized definition of mass for unbound systems, a definition which also presents a quantity which does not change in reactions, and is also observer-independent. This quantity is conserved. In the rest frame and also all other frames (if by "mass" we mean invariant mass).

Consider now the way the wiki article (and some other texts, I will admit) present the idea. "Mass" is not conserved, they say. You see, if we have a reaction (chem or nuclear) and it gets hot and we let heat leak away, the whole thing will weigh less! (or have less inertia, or less gravity). Golly. And if we put the heat back in, it would weigh a little more! And how much more? Why, just enough to get back to how much it weighed to begin with. Hmmm. So if you measure a system's mass and you let a little of the mass escape out of your system, you'll have less mass after it escapes. And what you let out-- lo!-- is just what you found you were missing. I agree it's a convoluted idea. Can you think of a short "conservation" presentation of it? Because you don't seem to like mine.

>Now here is my position: this definition of invariant mass is wrong.

No, here you're treading on new physics. It's the definition for invariant mass of systems (sum length of energy-momentum 4-vector) used in the wiki and also all the texts, including Taylor and Wheeler. You may not like it, and may want to use something else for "mass." But this is what physicists use to calculate rest mass of particle systems, so if you think it's not up to the job, I think you're wrong. If you say it's not the same as composite system inertia, I can't even say if you're wrong. But if you're right about the inertia of composite systems not being invariant I think it's perverse of you to want to use it as a definition of mass, and then complain that by your definition, mass isn't conserved. Why, then, don't you use a definition of mass which IS conserved? Sbharris 04:05, 17 April 2006 (UTC)

But look, we don't have to argue the relative merits of our definitions. If you can show me some sources that attest that "invariant mass is conserved", I will bow out of this discussion, and you can revert to your version. -lethe ^{talk +} 03:23, 16 April 2006 (UTC)

PS, I would like to continue this conversation with you, and get this matter straightened out. Can I ask you to try to keep your responses to a paragraph or too? I don't think anyone else will be interested in joining this conversation if they have to read through your thousand word essay above. I'm going to open a new section with my summary of our two positions and ask for input. -lethe ^{talk +} 03:27, 16 April 2006 (UTC)

invariant mass of composite systems: please comment

This is a summary of a very long conversation above, shortened to facilitate your comments.

Sbharris has made substantial edits to several articles (mass, conservation of mass, mass in special relativity, energy) which I consider contentious. I have reverted his edits on this page, and am considering other reverts, while Harris thinks the articles are wrong and the edits were important. I seek the input of others. I summarize our positions as I see them.

Harris puts forth the view that invariant mass of a composite system should be defined as the total energy in the rest frame (center of momentum frame) of the system. This definition fits naturally with the equation $m^{2}=E^{2}-p^{2}$, where you expect to get the total invariant mass is equal to the total energy in the rest frame. In this view, since invariant mass is defined to be the total energy, it is always conserved.

It is my view that invariant mass should be defined is the inertia of a system in the rest frame, and that since there is no single invariant inertia of a composite system, the invariant mass of the system is the sum of the invariant masses of the elementary constituents. The failure of the equation $m^{2}=E^{2}-p^{2}$ is a consequence of the fact that composite systems to not transform irreducibly and Schur's lemma does not apply, so the Casimir P² need not act as a scalar. With this definition, mass is not conserved; it can be converted into energy and vice versa.

I think that the idea that conservation of mass should hold in special relativity is misleading and nonstandard and that Harris's stuff should be reverted in all the articles, but I think I'd like to get some feedback on this before I do anything drastic. Thank you. -lethe ^{talk +} 03:44, 16 April 2006 (UTC)

...there is no single invariant inertia of a composite system... This is news to me, unless we're talking GR or something. I really don't understand Schur's lemma or Casimir invariant though, so maybe I'm out of my depth here... —Keenan Pepper 07:59, 16 April 2006 (UTC)

OK, like Sbharris, I'm guilty of dreaming up my own nonstandard calculations in order to justify my position. Certainly you're right; that there is no invariant inertia is not something I took from a textbook. It's something I came up with for this argument. But you've heard that the inertia of a moving object is not a scalar? That is, the inertia of the object under a transverse force is m, while the inertia of an object under a longitudinal force is mγ. This is all I mean when I say "no single invariant inertia". There is no frame in which both objects are at rest , so there is no way to treat both objects as a single mass.

I dreamed up these nonstandard arguments to explain why the seemingly correct calculation of Sbharris is actually wrong. But if you don't want to comment on my refutation of his calculation, you could still help me to arrive at this conclusion: according to the standard definitions in textbook relativity, conservation of mass is violated, and redefining the terms so that conservation of mass is respected is not a task for a wikipedia article. -lethe ^{talk +} 14:41, 16 April 2006 (UTC)

I agree with Sbharris that it would be good to include some discussion of masses of systems, but several changes in the way of presentation are necessary. First, the discussion of these things should be more concise. Second, we shouldn't say that the mass is conserved: usually such a statement would mean that the sum of the masses of the products is equal to the sum of the masses of the reactants, which is not the case here (even though I agree with Sbharris that the mass of the system as a whole is conserved, I think such a statement is misleading, and also not very useful since it only means that the total energy and momentum are conserved), so in my opinion the articles should keep saying that mass is not conserved. On the other hand, I don't understand why Lethe disagrees to talk about the inertial mass of a composite system. For example, proton is composed of quarks. Does it mean that one cannot talk about the inertial mass of a proton (at least when the applied force is much weaker than the strong force that keeps the proton together)? Yevgeny Kats 14:57, 16 April 2006 (UTC)

Well, I won't defend my position about composite systems having invariant mass. I will just agree that the notion that conservation of mass holds in relativity is not correct, and we should excise it. -lethe ^{talk +} 04:55, 17 April 2006 (UTC)

Harris comments: I think lethe has his main problems with the idea of the inertial mass of unbound systems.

As for the sum of masses of reactants not being equal to the sum of masses of the products, we need to specify how and where these are to be measured. If we change inertial frames to look at products, it's hardly surprising that our masses change. Also, if we fail to maintain a closed system, and let heat or light leak away. The problem with statements that mass isn't conserved is that they really have more to do with how mass is measured (i.e., under standard conditions and temps) rather than the reality of what happens in perfectly closed systems. The short substance of physical law is that mass is actually conserved, unless you let it escape. The new message of relativity is that mass can escape from systems in tricky ways you maybe never thought of, such as light and heat. But if you prevent that, you'll see that mass is neither created or destroyed. If the system is closed, you can't get mass out, no matter what happens. Mass then doesn't change. Very simple.

And yes, sometimes mass is hard to locate, such as with the system mass associated with kinetic energies. That mass seems to be located in the space-time fabric somewhere, as Taylor and Wheeler remind us. But we can't ignore such things, because most of what we think of as "ponderable matter" is really kinetic energy. If you believe what the high energy guys say about the rest mass of quarks, then the hadron masses are mostly NOT due to rest quark mass (i.e., fermionic), but due to other stuff. Thus, when you shake a solid object like a cup, most of the "inertia" is NOT quarks and leptons (ie, fermions), but instead fermion kinetic energy plus various bosonic fields (virtual gluons and pions and photons and such). So the idea that what we measure as "mass" is total energy in the COM frame is pretty important.

I'm getting resistance to this from people who seem to figure that mass is stuff, and they want to think of the stuff as certain little bits to be added up. The Wiki on mass defined it as a quantitation of "matter" and then went on to define "matter" as fermions (ie, quarks and electrons). I edited this to say that mass is actually matter (if by "matter" you mean fermions) PLUS energy (which takes care of all the kinetics and the force fields). Actually, it's mostly energy. But somebody took out my edit. I can see that standard model physics hasn't quite penetrated, yet.Sbharris 04:05, 17 April 2006 (UTC)

I fail to see what any of this has to do with the standard model. I think the removal of your edits is in concord with standard model physics. -lethe ^{talk +} 04:27, 17 April 2006 (UTC)

Be glad to explain. It goes like this: Back in the bad old days before the standard model, physicists thought that nucleons were basic particles, and so matter was mostly made up of "particles". Physicists knew that nucleons attracted each other strongly and that there was a mass defect in all nuclei with respect to hydrogen, but it wasn't large (less than 1%) and in any case it was a mass defect, or missing mass, so nobody gave much thought to its mass contribution. The protons and neutrons in nuclei were a bit lighter than free ones, but not by much. The mass of objects was still seen as composed of fermionic particles, with the mass contributions from fields (like the nuclear fields, which were inferred) seen as negative contributions, not positive contributions.

With the standard model all this changes. Hadrons are seen to be made of quarks, and the bare quark masses (see up quark and down quark) now thought, from modern experiments using standard model analysis, to be small (1.5 to 8 Mev). That means that 99+% of ordinary matter (the stuff we're made of) is not particles (fermions) at all, but rather quark kinetic energies and gluon fields. Of course, all fields can be thought of as consisting of particles, but the bosons, particularly the virtual ones, have always been seen as more "energy" and less "matter." That could be explained, but, the wiki articles on matter and mass, which rely on the idea of matter as basically fermions, and have no good place for the huge contribution of energy to ordinary object "rest mass", just end up confusing this issue more.Sbharris 05:16, 17 April 2006 (UTC)

The binding energy of a nucleon contributes to its rest mass. If this isn't mentioned in the article, it needs to be, but this is a consequence of the special theory of relativity, not of the standard model. -lethe ^{talk +} 05:31, 17 April 2006 (UTC)

Until the standard model, nucleons HAD no "binding energy," as they were not known to be (or thought to be) composite particles. Binding energy of nucleons to each other contributes to nuclear rest mass in a negative (and relatively minor) way, as noted.

Here's the standard model as I learned it: fundamental particles live in irreps of PoincarexSU(3)xSU(2)xU(1). momentum squared is a casimir of this group, and so acts as a scalar on irreps. Composite particles are tensor reps, and so the Casimir does not act on a scalar. Does any part of this contradict the formulation of the standard model that you learned? -lethe ^{talk +} 05:39, 17 April 2006 (UTC)

Sounds way overcomplicated or oversimplified, or both. "Physics is to math as sex is to masturbation." I'm quoting a physicist, of course. Anyway, understand relativistic kinematics before applications of Lie algebras. Walk before run. Sbharris 06:58, 17 April 2006 (UTC)

So you don't find any problem with that description, except that it's overcomplicated? Being overcomplicated doesn't make something wrong. Perhaps it's the case that you don't understand Lie algebras and Casimir operators. If that's so, I find it odd that you assume that I don't understand the standard model or basic relativistic mechanics. As a matter of fact, you cannot understand the standard model without knowing Lie algebras and such. But let me try to explain and translate it for you: when you put two spins together, you don't get a new spin with with the sum of the angular momenta. You get a new system whose spin is in the range |j₁–j₂| ≤ j ≤ j₁ + j₂. The combined system can be decomposed with Clebsh-Gordan coefficients into lower spin systems. For example, two spin-1/2 particles reduce to a singlet (spin 0) and a triplet (spin 1). The same thing happens with momentum: put two particles together and you get a new system which reduces into two reps, one with the total mass and rest frame, and one with the reduced mass and displacement vector. A singlet state or a triplet state are characterized by a single spin quantum number and behave simply under rotations, but a combination of a singlet and a triplet state do not. It behaves like a tensor under rotations. Similarly, a two particle system does not behave like a single mass under the Poincare group. -lethe ^{talk +} 07:22, 17 April 2006 (UTC)

As I said last time, we seek invariants for transformations between different inertial observers at the same spacial location, which is mere Lorentz invariance. Invariant mass is one such Lorentz scalar, and yes it applies to systems of 2 particles, such as two gamma rays or a particle-antiparticle pair. That what the concept of invariant mass was invented to do, unlike inertia, which isn't even Lorentz invariant for ONE particle, let alone two. In any case, when you implicitly bring up the case of particles that attract or repel each other (a requirement for the very concept of reduced mass, which does not appear with non interacting systems) you're introducing unnecessary problems. FYI, in that case the mass appears "reduced" due to somebody's transformation to an accelerated frame located at one of the particles the force acts upon. Mass appears reduced (or can be treated as if reduced) because acceleration is relative in that case, and thus acceleration appears reduced IF you forget to use your full acceleration relative to some non-accelerated frame. But new fields appear for accelerated observers. If one of these cancels out some mass effect, that doesn't mean the mass is actually gone or not observed. It only means that somebody forgot to observe properly. (Shuttle astronauts don't feel the gravity of the Earth. That doesn't mean the mass of the Earth has disappeared somehow for them, and thus was not conserved. If they look out the window they notice the truth)

1. Reduced mass does not require an interaction, you're wrong about that. Your comments about accelerated frames leave me scratching my head. Perhaps you don't know what reduced mass is?

Hey, you're the one using the term in a non-standard manner. Look at the Wiki or google it.

Reduced mass is how you pass to a irreducible represenation of the translation group.

Reduced mass is an algebraic device for solving the two body problem exactly, by translating to the coordinate system of one (accelerated) body. However, there needs to be a force between the bodies, or else there's no point in calculating this variable; it doesn't help you solve the problem. The very origin of the term, and the math behind it, presumes an interaction.

You're wrong about this. Reduced mass is used to turn any two body problem with into a one body problem. Why is an interaction needed? -lethe ^{talk +} 20:42, 18 April 2006 (UTC)

More exactly the "reduced mass" was invented to solve the classic "two-body problem" in celestical mechanics (but also it occurs in classic treatments of scattering ala Rutherford, etc). This involves both bound and unbound states of two particles/bodies. To answer your question, the interaction is "needed" for the problem in sense that without an interaction, there is no "problem" to be solved in the first place! Or at very most, it's the trivial "problem" of seeing what the linear equations of motion of two particles look like from various origins and frames. That certainly doesn't require use of the reduced mass, and isn't helped by by the concept, in the slightest. Anyway, you can look at the Wikis for reduced mass and two-body problem. No, there's no explicit statement in the Wiki that the "problem" involves an interaction between bodies due to a single distance-dependent force. But there should be, perhaps, since that fact is implicit in the equations. Perhaps it ought to be brought to the reader's attention in the text, for the benefit of people who have difficulty seeing the implication of stuff in equations :). Perhaps you'd like to do the job? If I did it, I'd probably get it reverted.... Sbharris 23:42, 18 April 2006 (UTC)

By the way, it's not as if I introduced pairs of particles to complicate the picture. It is you who insists on introducing a description in terms of pairs of particles. I'm simply trying to show you why your description is wrong. Now, you say inertia isn't Lorentz invariant, and that's true. Inertia in the rest frame is invariant however.

But that's a meaningless statement. Lorentz invariance *means* invariance under Lorentz transformation, when passing from one inertial frame to another (while keeping the same coordinate origin). How do you talk about Lorentz invariance in ONE frame (like a rest frame) when the whole point is invariance between an infinite number of frames in relative motion?

The situation is exactly the same as yours: you define the invariant mass to be the total energy in the rest frame. Total energy is not a Lorentz invariant, but total energy in the rest frame is. Therefore if you want to argue that inertia is a bad measure, then you've also killed your own definition. -lethe ^{talk +} 04:46, 18 April 2006 (UTC)

Still missing the point. Invariant mass is indeed total energy in the rest frame, but even though total energy will change if you boost to another frame, invariant mass will not, so you can use the original value of mass IN ANOTHER FRAME, but not the original total energy. You can use the value of invariant mass from ANY frame as they're all the same; I merely point out that it's a handy fact that in one particular frame (the COM frame) that value happens to be equal to the total relativistic energy (which is something you can often easily calculate in particle physics, for systems). This causes confusion, since in that frame people sometimes wonder why the total relativistic energy determines the object's "rest" mass (why does a heated object or box of radiation weigh more, etc). They think it's insecure to use total relativistic energy/c^2 for mass because it's frame/observer-dependent. And indeed it is. Which is why it's NOT used to caculate mass except in this one special frame. For all others, however, you use the energy-momentum 4-vector length, the invarant mass, which will have two components, not just one. And this is the measure which causes gravitation, etc. Which is nice because we don't want the total gravitational field of an object to be frame/observer dependant.

Similar comments apply to your introduction of spin in an example-- spin comes with complex effects which don't really illustrate our topic. Lie groups and their representations and their Casimir operators are a big topic, but far broader than physics, and way overtechnical here. Though no doubt to the man with a hammer, the world tends to look like a nail.Sbharris 04:28, 18 April 2006 (UTC)

You're wrong about this. Lie algebras and their representations are exactly the theory that tells you how to understand what an elementary particle is, and what particles are not elementary. It tells you which systems have well-defined quantum numbers like spin and mass. Your dismissal of the subject makes dialogue with you difficult. -lethe ^{talk +} 04:48, 18 April 2006 (UTC)

It tells you NOTHING about what particles are elementary! That's an experimental result. Physicists pick groups to correspond with the particles they find. When a group is nearly filled it makes "predictions", but if you go over that and find more, you must pick larger groups.

Of course it's an experimental result, and the language of group theory is what we use to describe to describe the result. Two electrons does not transform irreducibly under Poincaré, so two electrons do not constitute a fundamental particle. A pion does transform irreducibly under Poincaré, so it might be a fundamental particle, and indeed, it is treated so in Yukawa's theory. However, a pion doesn't transform irreducibly under the standard model symmetry group, so it cannot be viewed as a fundamental particle in that theory, and of course this group was constructed with experimental input. Nevertheless, for a given model, you can easily check when something is not a fundamental particle: it doesn't transform irreducibly. Two particles will not usually transform irreducibly, and therefore some things which can be said about fundamental particles cannot be said about them. Now please, stop arguing with me whether group theory is a good language to use to talk about particle theory, and provide me one reference which supports your claim that mass is conserved. -lethe ^{talk +} 19:55, 18 April 2006 (UTC)

Okay, read http://www2.slac.stanford.edu/vvc/faqs/faq5.html This is a FAQ from SLAC. The question is whether mass is conserved in high energy particle event and the answer is "no." Qualified to mean that "mass" as measured if you add up rest masses before and after, is not conserved. Which of course means if you drain away kinetic energy, change reference frames, and do a lot of other no-no stuff. But how can you expect mass to be conserved if you don't close your system? But then, the article goes sheepishly on to say that if you put a box around the whole system and don't let any mass or energy in or out, mass is conserved though all processes. So in other words, it is conserved if you keep track of it. Which is my point, and what they should have said at the beginning. And the way the Wiki should read. And what I've been trying to do in fixing it. But it's heavy lifting!Sbharris 03:44, 19 April 2006 (UTC)

All right, I'm just a freshman physics student here, so I probably don't know what I'm talking about, but I have a question that's been bothering me. Say you have two particles that are attracted to each other and they're vibrating back and forth in one direction very fast. Now according to special relativity, they should both be harder to push (i.e. the same force causes less acceleration) in the direction of vibration, because the change in momentum is greater for both of them. On the other hand, the total momentum is zero in the rest frame, and the change in momentum should be the same no matter which direction you push the two particle system. So if you look at it as two particles, it should be harder to push in one direction, but if you put them in a box and consider that as one object, there's no reason why it should be harder to push in any direction. What is the resolution of this apparent paradox? —Keenan Pepper 04:58, 18 April 2006 (UTC)

Ah, I was looking for the word isotropy. What I mean to say is, is the inertia of this system isotropic or anisotropic? —Keenan Pepper 05:08, 18 April 2006 (UTC)

The inertia of a moving object is not a scalar. How much acceleration is exhibited depends on which direction you push. If you push an object in the direction of its motion, the inertia is mγ, while if you push it in a perpendicular direction, it exhibits an inertia of m. The same thing applies to two particles, whether they're moving linearly, vibrating, in a box or no. The only way to get a scalar mass is to consider momentum distributed equally in all directions. I suppose you could accomplish this with a single particle vibrating and rotating just right. More easily, this occurs with a system with many particles which are thermal. Then, no matter which way you push the particle, you see the same inertia. To sum up: yes, the inertia of a moving particle is anisotropic, and there is no paradox. -lethe ^{talk +} 19:55, 18 April 2006 (UTC)

Wow, I guess I was assuming all this stuff that just isn't true in relativity. For example, the acceleration is not even necessarily parallel to the force. The damn thing might not even go the same direction you push it! So "inertia" as the ratio of force to acceleration isn't even well defined. —Keenan Pepper 03:55, 19 April 2006 (UTC)

"For a macroscopic object, the rest energy mc^2 includes the thermal energy"

This is in conflict with the definition of mass given earlier in the article; the article does not define mass as energy divided by c^2. The definition of mass given in the article, i.e. the invariant rest energy divided by c^2, cannot be generalized to macroscopic objects in a relativisically correct way. You can only define the mass of macroscopic objects this way in the classical limit.

Count Iblis 14:28, 16 April 2006 (UTC)

The main definition in the article is this: mass is a measure of inertia. Using the theory of special relativity, you can show that the the inertia increases with thermal energy. Therefore, I see no incongruity. You say classical limit, but it is my impression that it is actually the thermodynamic limit that allows the calculation of the mass of this macroscopic energy, something which my discussion above touches upon. -lethe ^{talk +} 14:56, 16 April 2006 (UTC)

Yes, you can indeed define this in the thermodynamic limit. But then, macroscopic objects don't need to be in thermal equilibrium, so you can't use this to define a (scalar) relativistic mass. Macroscopic objects are, in general, described by the energy-momentum tensor.

Count Iblis 16:34, 16 April 2006 (UTC)

The point is that compound objects are not made of things at rest, so it doesn't work to use "rest energy" (as for particles) as any part of the definition of mass of objects. Objects contain more than rest energies (as for example, kinetic energy, pressure, various kinds of other fields and stresses). Please note the energy momentum tensor for something simple like a container of thin ideal monatomic gas (neglecting pressure contribution) is just the invariant mass (summed energy-momentum 4-vector) of the particle system. Since we're working in the center of mass/momentum frame, this is the relativistic energy/c^2, and it happens to be invariant for systems, even though relativistic energy per se, sometimes isn't. What happens if you move an object (or go to a moving frame) is the energy increases but the momentum goes to something non-zero, and the two effects cancel, so again you're left with invariant mass, which is total relativistic energy in the rest or COM frame. But it has nothing to do with "rest energy" of particles. Sbharris 04:05, 17 April 2006 (UTC)

I agree with this. I think that the article should mention that mass is a redundant concept in relativity. You just have energy and you can define an energy in the rest frame which is an invariant, because it is the norm of the 4-momentum vector.Count Iblis 01:30, 19 April 2006 (UTC)

Well, it's NOT redundant if you're then going to use two kinds of energy. Why not call one of these kinds of energy by the name it had before Einstein? In relativity there's 1) the kind of energy which is invariant and then there's 2) the kind that isn't. For the kind that is, I think "mass" (appropriately rescaled as a scalor) is a good name. Not least because this is already what we measure with balances and scales and whatnot, and stick into our equations for gravity. So we're familiar with the concept, and already have a name for it. Taylor and Wheeler and most relativists agree with me on this. By contrast, for the kind of energy which isn't invariant, but is still conserved in SR, we can still call that "energy or relativistic energy." And we can try to forget "relativistic mass," because it really is redundant if we already have a relativistic energy.

Of course, this is more or less the convention already used in the article. I tried to make no changes there. My problems with the article have more to do with the article's insistance on connection of mass to inertia, which isn't a good relativistic idea. The concept of mass as a conserved and invariant quantity survives SR just fine; it is the concept of inertia that doesn't. This is no reason to throw the baby out with the bathwater. It is better to simply say that Einstein killed the idea of absolute inertia along with the ideas of absolute time and space. But mass survives relativity as a concept, just as charge does. And for much the same basic reasons: we need it as a source of an invariant field. Sbharris 02:56, 19 April 2006 (UTC)

I agree with you that mass is not redundant. Invariants are useful, and you'd have a hard time writing the standard model lagrangrian without using the mass of your particles. Now, you've read that SLAC FAQ page, and you know that mass is not conserved, because mass means the sum of the rest masses of the particles.

If you interpret mass to mean the energy, then you have a conservation law, and this is the mass that you measure if you put the system in a box. But the standard usage of the term mass is the sum of the rest masses, and if you want to edit the article to describe the alternate interpretation in which mass is conserved, you have to do so vary carefully, mentioning that you're using a nonstandard description. I do not think it's a bad idea to mention this, but it has to be done just right. -lethe ^{talk +} 03:58, 19 April 2006 (UTC)

I'll see what I can do. Yes, we finally agree that we can do it by pointing out strongly that conservation of mass "by summation of rest masses" has to go, as a concept in relativity, but conservation of mass in systems is still good.Sbharris 18:00, 19 April 2006 (UTC)

Some Comments

The article doesn't mention mass as an inverse correlation length in field theories. This is the way mass manifests itself in fundamental physics and also in statistical physics.

Also, the article completely fails to mention that mass is not an independent physical quantity from energy in relativistic physics. If you take the classical limit c --> infinity you can distil a mass as an independent physical quantity from energy. You can, of course, still define mass as the rest energy divided by c^2, but you could then just as well omit the c^2 and call it rest energy.

But as noted above, the rest energy has special properties which cause us to want to give it a special name. And available for this is the name it already historically had, which is "mass." Unlike the relativistic energy, the rest energy is invariant mass/c^2, which means that it's the same quantity for all observers in all frames. And since this is what we're already used to calling "mass" in Newtonian physics, let's just keep doing THAT. In contrast, it's the idea of trying to keep the Newtonian idea of mass as dp/dt, even in relativity, which is a bad one, and should rightly be dropped. Ditto for the Newtonian idea of mass as defined by inertia. Just let them both go, folks. Some things are outdated by Einstein, and others survive. Sbharris 03:16, 19 April 2006 (UTC)

It should be noted that c is only dimensionful because we have chosen to measure time in different units than spatial lengts. But that is just convention and has nothing to do with physics. The argument that c is not dimensionless and that therefore mass is different from energy is thus a circular argument. You can put c = 1 and measure time and space in the same units and consider mass and energy to be the same thing. You won't ever encounter any inconsistencies if you do that. Count Iblis 01:47, 19 April 2006 (UTC)

I'm not claiming mass is "different" from energy. But mass is a historical concept which we might as well simply keep, rather than renaming it "rest energy." Why not call everything energy? Because as explained, the rest energy is special because it's an invariant quantity which determines gravitation, whereas other energies aren't. So it deserves a special term. "Rest energy" is two words. "Mass" is one word, and already has a history and is familiar as the source of gravity. So there you are. Why insist on making things more complicated and using different words, in cases where the old ones serve perfectly well? Sbharris 19:09, 20 April 2006 (UTC)

Why do you say rest energy determines gravity? The source of the gravitational field is the stress tensor, which includes kinetic energy, as well as momentum. -lethe ^{talk +} 20:37, 20 April 2006 (UTC)

Actually, that's the source of gravity (curvature) at a point, but if you want the gravitational mass of a system, or extended object, you're going to have to integrate that tensor (which represents essentially an energy density) through the appropriate spacial volume. Which you must be very careful how you do, since different observers in different frames see different volumes. However, not to worry, since this is a tensor, and in relativity these things transform between frames using the Minkowski metric and all observers end up seeing the same integrated total energy. And how much energy is that, you ask? Why, the simplest representation involves the energy density and (static) pressures which observers in the COM frame see, where all the off-diagonal tensor components zero out for perfect fluids which have no viscosity, etc. That first component T00 is the total energy density in the COM frame, and that includes all the rest mass and kinetic energy and whatnot energies in the system COM frame. For an ideal gas the other components in the COM frame are pressures, which reduce to zero for small system of particles like the one we're discussing. End result is the system invariant mass (Minkowski norm of the summed 4-momenta) in the COM frame. In other words, what I've been saying. Which is to say, the mass you'd weigh on a scale if you had the whole thing confined in a box. I hope I don't have to reference this. If you have Misner, Thorne and Wheeler you can look at chapters 5 and 22.

By the way, I was annoyed enough at the waffling of the SLAC FAQ that I wrote them, and they were kind enough to reply back, and this is what they said about mass of systems:

You are right! What you convince me is that there is no end to the possible confusions that arise if I try to avoid talking about frame --because when I say energy gravitates I am implicitly assuming the we are in the center of mass frame of the system in question --otherwise, as you point out, it is just not a true statement.

So for the FAQ's reader, here's what we agree:

Mass is a relativistically (frame) invariant quantity well defined for any system, and it does not change due to any process within that system. (This is why both Steve and I hate the silly trick of text books defining "relativistic mass =gamma m c^2" -- this is actually the energy of a moving particle of invariant mass m. )

In the center of mass frame for the system the total energy of the system is E=m c^2

The mass of a system is NOT in general the sum of the masses of any set of subsytems that make up the system. It includes contributions from interaction energies between subsytems, and motion of the subsystems.

The mass of a molecule is not the sum of the masses of the atoms it contains.(an example of the above statement)

In chemical reactions the invariant mass of the total system of many molecules, as defined above, does not change. However the sum of the masses of the molecules that exist in the sytsem is different before and after the reaction. In nuclear processes (fission or fusion) that same statements are true, replacing "molecules" by "nuclei".

I hope this finally ends the discussion. I think MTW's Gravitation and the SLAC physics FAQ writers are on my side.Sbharris 20:28, 24 April 2006 (UTC)

Count Iblis, even if we used natural units, c would be 1 planck length per planck time, not just the number 1. It would not be dimensionless, it would have units of (planck length)(planck time)⁻¹. —Keenan Pepper 00:17, 21 April 2006 (UTC)

Count Iblis is correct, c is actually equal to the dimensionless number 1 in Planck units, not some dimensionful number whose value is 1. The Planck length is actually equal to the Planck time in Planck units, so the unit you quote actually vanishes. The property of c=1, h/2π=1, and G=1 is the definition of Planck units. -lethe ^{talk +} 01:04, 21 April 2006 (UTC)

That doesn't make sense. A dimensionless number has the same value regardless of the measurement units used to calculate it. (That's straight from dimensionless number.) The value of c is different depending on the choice of units. —Keenan Pepper 01:28, 21 April 2006 (UTC)

Well I guess to understand this, consider natural units, instead of Planck units. Natural units are defined by the equations g=h/2π=1. Since there are three degrees of freedom in the three choices of fundamental units (in SI) and we've only imposed two choices with these two equations, there is still a single choice left to make here, and people usually take MeV to be it. Light travels 1 MeV^–1 every MeV^–1. Thus, the speed of light is 1. If you change units (say, from MeV to Joules), the speed of light will still be 1. Therefore it's dimensionless. In Planck units, there are no choices of units, all three degrees of freedom are eaten by the equations c=G=hbar=1, therefore there are no changes of units which can preserve the three equations. If there are no changes of coordinates, then actually every number stays the same. Every physical quantity is dimensionless in Planck units. That's how they're usually interpreted. -lethe ^{talk +} 01:48, 21 April 2006 (UTC)

Let me explain it another way: the number of allowable fundamental units depends on the number of dimensionful constants you're willing to have in your equations. If you're willing to have a dimensionful constant in E=mc², then you're allowed to measure distance and time with different units (years and light years, if you like, then the speed of light will be a dimensionful 1 lightyear per year). The fewer dimensionful constants you're willing to carry around, the fewer fundamental units you're allowed to have. Each constraint you put on a dimensionful constant eliminates one unit. The notion of changing your units is usually not allowed to change the number of units. To do so entails changing the form of the equations of physics, which is a no-no. Thus, with natural units, the only way to change units is by changing the energy unit, which has no effect on the value of the speed of light. In Planck units, all units have been eliminated and all quantities are dimensionless, since there are no changes of units left (except the identity operation). -lethe ^{talk +} 01:55, 21 April 2006 (UTC)

All right, your point of view still seems really weird to me but I think I understand it at least. So you're saying the meter and the second are simply different units for measuring the same kind of quantity (spacetime interval) and the figure 299 792 458 m/s is simply the conversion factor between them, like 12 inches/foot? —Keenan Pepper 02:36, 21 April 2006 (UTC)

Yep, that's it in a nutshell. The number 299 792 458 m/s and the number 12 inches/ foot are both dimensionful. However, you can choose to measure short lengths and long lengths in the same unit, which eliminates one dimensionful constant from your description (the dimensionful conversion factor then becomes the dimensionless number 1). This is equivalent to defining your units by the equation s=l, where s is a variable holding the length of something short that you measure, and l is a variable holding the length of something long that you measure. Similarly, you can choose to measure spatial distance and time intervals with the same units , in which case the speed of light becomes the dimensionless number 1. This point of view is weird, and takes a while to get used to, but I think getting on top of it gives you a better understanding of units, and I assure you that it is quite standard in theoretical physics. -lethe ^{talk +} 03:12, 21 April 2006 (UTC)

Sbharris, yes I see your point now. In high energy physics we do indeed need the term mass, e.g. like in case of "massive fields" etc. It would be cumbersome not to have a separate term for this.

Lethe, thanks for explaining "dimensionless c"! Michael Duff has written a few articles about this (probably because to his frustration that there are still a lot of physicists who still believe in fundamentaly incompatible dimensions/units). In one of his article on this topic he gave this example. You could define separate units for distances in the x, y and z directions. The formula for the space time interval then becomes:

ds^2 = c^2 dt^2 - dx^2 - c_{y}^2 dy^2 - c_{z}^2 dz^2

distances in the x, y and z-directions are assigned the dimension "xylophone", "yacht" and "zebra" respectively. the quantities c_{y} and c_{z} would then become dimensionful, xylophone/yacht and xylophone/zebra respectively. You could imagine that intelligent beings would have defined such units if they lived in a world where rotational symmetry was broken, e.g. due to very strong fields, confining then in a one dimensional world. The existence of y and z directions would have lead to physical effects that would seemingly have nothing to do with spatial dimensions. These effects would at first be described using effective laws, but because the connection with spatial distance would not be understood in the beginning, they would have defined units incompatible with distance to descibe the effects. If they later discovered special relativity then they would have arrived at the formula for ds^2 with the dimensionful constants in there.

Our situation is similar to that of the hypothetical 1 dimensional beings. We have defined our units for physical quantities before quantum physics and relativity were discovered. Without these theories you cannot relate mass, time and space to each other. If we do general relativity or quantum physics and use our old units, then we encounter dimensionful constants, precisely because the old units were defined to be incompatible while relativity and quantum physics tells us otherwise. The constants h-bar, c and G are the conversion factors. If we insist on using incompatible dimensions, they appear to be dimensionful.Count Iblis 13:59, 21 April 2006 (UTC)