Talk:Mass/Archive 3

Hammer lands before feather

Both objects accelerate at the same rate towards the moon, but the moon accelerates towards the hammer faster than it accelerates towards the feather; ergo, the hammer hits the ground first. This isn't an "unsourced claim"; it's physics. Anonymous 198736 03:46, 23 April 2006 (UTC)

I guess it's probably true. -lethe ^{talk +} 03:58, 23 April 2006 (UTC)

Love the Goya painting! :-) Anonymous 198736 04:05, 23 April 2006 (UTC)

Do you? I love it too, it's my favorite painting, and I have a print of it hanging in my kitchen. But I've lately been getting some harrassment from a crackpot who thinks that my preference for that painting indicates that I rape and murder children. -lethe ^{talk +} 04:40, 23 April 2006 (UTC)

Not that I know much about him, but I remember being blown away seeing some of his early work compared to his (boringly traditional) contemporaries. Subversive, and sometimes quietly so; this portrait he did of some Spanish noblewoman, around 1800 or so, in the Getty collection, I found particularly impressive. Anonymous 198736 05:48, 23 April 2006 (UTC)

de Goya, like Beethoven, was at the forefront of the Romantic movement. It's my impression that yes, their contemoraries' works seem pretty boring by comparison. -lethe ^{talk +} 09:26, 23 April 2006 (UTC)

WTF, the hammer and the feather are both pulling the moon in the same direction. Anyway, it doesn't matter if it's true, it matters if it's verifiable. —Keenan Pepper 04:08, 23 April 2006 (UTC)

Verifiable, what? Do we really need to verify that 2 + 2 = 4? Run the equations: F = ma, F = (G * m1 * m2) / r^2. Gravity is indiscriminate, you know, and gravity is postmodern. Even if you adopt a simplistic, spatially absolutist perspective towards the concept of "falling," to deny that even as hammer and feather fall towards the moon, so to does the moon fall towards them, would be to demonstrate astonishing lunocentrism. Anonymous 198736 04:14, 23 April 2006 (UTC)

What you're claiming is not as simple as 2 + 2 = 4, and needs a verifiable source. As I said, the two objects are pulling the moon in the same direction. The moon is not accelerating faster toward one than the other. —Keenan Pepper 04:23, 23 April 2006 (UTC)

If you drop two objects at the same time, then you're right, the fall at the same rate (and the planet accelerates according to the combined mass). But if you do two separate experiments, first dropping the heavy object and then dropping the light object, it seems to me that anonymous 19 is right: the heavier object will fall faster. As for whether this fact is interesting and deserves mention in the article, or whether it requires a source, I haven't made up my mind. -lethe ^{talk +} 04:40, 23 April 2006 (UTC)

Okay, you've got a point there. —Keenan Pepper 04:48, 23 April 2006 (UTC)

Actually, the hammer lands first even if you drop them simultaneously, since they can't both be dropped from the exact same place at the exact same time. The side of the moon closer to the hammer will accelerate faster than the side closer to the feather. This should be mentioned in the article, since the very possibility seems to have escaped least one person already, even to the point of denial. Anonymous 198736 05:01, 23 April 2006 (UTC)

The feather will land first!! (and please don't include this in the article). Let's do a calculation. The mass of the Earth is about 10²⁴ times bigger than the mass of the hammer. Therefore, the acceleration of the Earth is 10²⁴ times smaller than the acceleration of the hammer, so if the hammer falls through 1 m, the Earth will move 10^-24 m during the same time. Such short distances, which correspond to an energy of 100000 TeV, are not probed even by the most energetic particle accelerators! Let's now do another calculation. Suppose we're dropping the feather. The hammer, which is presumably on the ground, attracts the feather with its gravity. The additional acceleration of the feather is given by GM/r², where M is the mass of the hammer and r is the distance from the feather to the hammer. This gives roughly 10^{-10 m/s2, which is much bigger than the effect suggested by the Anonymous. Yevgeny Kats 05:22, 23 April 2006 (UTC)}

Of course, that's only if you drop the hammer first and the feather second, which just goes to show how arbitrary and pointless this all is. —Keenan Pepper 05:35, 23 April 2006 (UTC)

Well, the article as it currently stands is just plain wrong, so I'm going to go ahead and make the necessary corrections. I'll try not to make it overwordy. Please don't revert it again. Anonymous 198736 05:48, 23 April 2006 (UTC)

It's an important fact for the theoretical basis of general relativity that the instantaneous acceleration that a falling object undergoes be independent of the mass of the object. Two objects should undergo exactly the same acceleration. If it's not exact, then GR is broken. So I'm not too happy with the current wording "the hammer and the feather should fall at approximately the same rate". Nevertheless, I do agree with Anonymous 19 that we shouldn't keep statements that are wrong. Perhaps we can do better? -lethe ^{talk +} 06:16, 23 April 2006 (UTC)

Lethe, almost any statement in most articles about physics is approximate. If you fix every such statement you will never get out of it. In our case, the approximation is better than anything one could dream about (as we calculated above), so I suggest that we shouldn't confuse the readers with that. I completely agree with Keenan who said "how arbitrary and pointless this all is". Yevgeny Kats 15:38, 23 April 2006 (UTC)

I agree with you, that it's probably bad to have a slavish aversion from things that are only approximately true. In the current situation, I think the right solution is to just not talk about how long it takes for two objects to hit the ground, and instead only mention the acceleration. Right? -lethe ^{talk +} 23:46, 23 April 2006 (UTC)

Let me take another stab. Anonymous 198736 07:30, 23 April 2006 (UTC)

OK, I just changed what was basically "they will hit the ground at the same time" to "they will accelerate towards the ground at the same rate." Then stuck a full explanation in parentheses at the end of the paragraph. On reflection, this should probably be broken out into another section. Anonymous 198736 07:39, 23 April 2006 (UTC)

The Moon won't respond like a rigid object when the objects are released. The acceleration of the entire Moon as calculated above is just the average acceleration of all the parts of the Moon. What you need to calculate is the motion of the surface, and this doesn't depend on the gravitational response of the Moon at all! The most important effect is due to the elastic response. If you are standing on the moon then the surface is slightly depressed at the place you are standing. If you release the hammer your weight will become suddenly slightly less. The surface will respond to that by oscillating around the new equilibrium shape. Count Iblis 15:48, 23 April 2006 (UTC)

This funny effect seems to be only a matter of principle: if it happens, as suggested above, that the distance traveled by the ground is of the order (or less) than the radius of an electron, then no experiment can ever observe a difference. --Philipum 07:52, 24 April 2006 (UTC)

Since the theoretical experiment is carried out within Newton's framework, it should be noted in the article that even within this framework the objects do not land together. Yevgeny, please stop deliberately spreading ignorance by reverting the article to contain inaccurate statements. Anonymous 198736 04:26, 9 May 2006 (UTC)

Equivalence principle --> gravity is nonlinear

It is possible to explain that gravity must couple to itself in the article without using general relativity. You just consider the case of a gravitationally bound system, like e.g. the Earth Moon system. The total energy, E, is less than M_{moon} + M_{earth} due to gravitational binding. Equivalence principle says that this system should fall toward the Sun with the same acceleration as any other object. This means that the Sun's gravity couples to E which includes the gravitational binding energy of the Moon and the Earth. Count Iblis 12:26, 24 April 2006 (UTC)

The binding energy of Earth and Moon is positive-- they have have less mass then if they were free. But since they would fall toward the sun at the same rate no matter what mass they had, I fail to see why that means gravity couples to "E". Look at it this way: suppose gravitational binding didn't change the Earth-Moon system mass at all. It would still orbit the sun in the same way. The feather/hammer discussion is relevant here. The reduced mass might be smaller, but we'd have no reference to measure that. The energy is on the order of GmM/2r which is 4e28 J or so, which is 4e11 kg of mass, or 440 million tons. A cube of rock 1,700 ft on a side = small mountain or two big hills. How would we ever tell if it's missing or not? Answer: we couldn't. Sbharris 03:42, 25 April 2006 (UTC)

Hi Sbharris, the point here is that you do assume that the inertial mass is less due to gravitational binding and then you ask if gravity couples to the inertial mass of the gravitationally bound system or to the mass the system would have had if it wasn't gravitationally bound. Of course, the former is true. The point is that this means that gravity is nonlinear. The nice thing about this argument is that it doesn't use general relativity and is easily understood by lay people.

I've read some time ago that the effect of the gravitational binding as a source of gravity (which is automatically included in general relativity) was tested using lunar ranging data. If this effect were absent then general relativity would be invalid. This is not like adding a small mountain, rather it's more like adding the gravity of a mountain without adding the inertia of the mountain.

Count Iblis 12:37, 25 April 2006 (UTC)

The Apollo [1] laser lunar ranging experiments were essentially looking at gravitational binding mass of the Earth, not the Earth moon system. The Earth has significantly larger gravitational self-energy per kg with regard to the moon. This experiment looked to see if gravitational mass loss might be different from inertial mass loss in these circumstances (Earth vs. Moon, as "seen" by Sun), but the whole experiment only makes sense if you posit that the gravitational binding of the Earth would result in Earth mass loss (as expected-- it would go to heat and be radiated away), but somehow not lost Earth inertia loss. IOW the lost gravitational field would have to not gravitate, but keep its inertia, which would be truly bizarre. The lack of THAT really doesn't test GR very strongly, for if a gravity field didn't gravitate or have inertia EITHER, we'd never see that in this experiment.

BTW, as side note, the chemical binding energy of the Earth (with regard to free silicon and oxygen atoms), is more or less of the same magnitude as the gravitational binding energy. It's just that this quantity for chemistry goes up in proportion to M rather than M^2 for gravity, so chem binding energy/kg is the same for the Earth and moon (more or less). Sbharris 01:47, 28 April 2006 (UTC)

Thanks for the info! I agree with your points, but that only makes it easier to argue that gravity is indeed non-linear! Count Iblis 23:20, 6 May 2006 (UTC)

h=c=1

It is intersesting to note that when c=1,

$E=m$

which shows that rest energy and rest mass are equivalent.

Likewise, when h=c=1,

$E={\frac {1}{\lambda _{C}}}$

and

${\tilde {\nu }}=1/\lambda$

This shows that rest energy and rest mass are equivalent to the Compton wavenumber. GoldenBoar 20:44, 6 May 2006 (UTC)

That's indeed the case and this should be mentioned more prominently on this wiki page plus others on units etc. Small masses are usually expressed in inverse centimeters, see e.g. here Count Iblis 21:05, 6 May 2006 (UTC)

Within Particle Physics (where the smallest masses are dealt with) all masses are given in units of energy (eV, MeV, GeV etc)Jameskeates

Mass of a photon

I continue to dislike the way this article just dismisses the mass of many forms of energy, essentially by just refusing to use the initial reference frame or refusing to close the system. Which is what happens when you have an annihilation and state that the product photons are massless. Yes, individually they are, but if you're looking at them individually you haven't closed the system. You're redrawn it around each individual photon.

ONE photon has a mass (and a weight) in relativity, if you trap it in a box so that it bounces around, and put the box on a scale. The weight and mass turn out to be due to the gravitational Doppler shift. Here, I'll do the calc (excuse my editing):

Put one (well localized) gamma photon of ground energy E_o = hf_o in a perfectly gamma mirrored box (okay, it's Gedanken) of height x, in a gravitational field g. Let it bounce up and down, hitting the top and bottom of the box, each bounce after time t=2x/c. Each bounce transfers 2 times the photon momentum E/c. Now the force that the photon exerts on box top (or bottom) is dP/dt = Force = F = (2E/c)/(2x/c) = E/x.

The difference in photon force (ΔF) between the top and bottom of the box is the "weight" of the photon, as measured by the scale the box is on, and THAT is due to the Doppler change in energy, (ΔE)/x. Now ΔE is just h*(Δf) where f is the photon frequency. From Einstein's formula, the Δf from a gravitational shift through a constant field g and height x, is given by Δf = f_o[gx/c²]. Here f_o is the frequency at ground level. Substituting gives: Weight = ΔE/x = h(Δf)/x = h(f_o)[gx/c²]/x. Cancel x's and note that hf_o = E_o which is the energy of the the photon at the ground, and you have:

Weight = (E_o) g/c²

If you equate the weight as mg, then the "mass" you weigh for the photon is m = E_o/c². How about that? Photons have mass and weight, if you trap them and pay attention to your system boundaries.Sbharris 22:33, 6 May 2006 (UTC)

I think it would be best to give some simple examples in the article. We could take an empty box filled with radiation at temperature T. Then the mass is

${\frac {\pi ^{2}k_{\mbox{b}}^{4}}{15\hbar ^{3}c^{5}}}VT^{4}$. There is also a contribution from the zero point energy, but I'm not sure what the formula for that is. It isn't zero, because if you expand an empty box you have to perform work to overcome the Casimir force. Count Iblis 23:36, 6 May 2006 (UTC)

Dear Sbharris, you are talking about what is usually called energy rather than mass. And the statement is that gravity couples to energy(-momentum). This is the standard way of talking about it. Therefore, in my opinion, your example is irrelevant to the article Mass. Yevgeny Kats 05:34, 7 May 2006 (UTC)

They can talk about it any way you like, but the way you MEASURE the mass of an object is to weigh it. Or look at the motion of a free test particle in its vicinity. You can say gravity couples to "energy-momentum" but you don't know (and may have no way of ever knowing) how much of that gravitating object over there is "energy" and how much is "momentum". How much of the "mass" of a neutron or proton is "energy"?? A lot. We don't know how much. Gravity couples to all of it, though, and we CALL all of it "mass". Rest mass. Stuff. Most of atoms and "matter". That's been the normal use of the term in the language for centuries, and Einstein didn't change it.

Nearly all the stuff you want to dismiss from the article as mere "energy" and not "mass" has all of the characteristics we've always associated with "mass", including weight, inertia, and active and passive gravitational interaction potential. Moreover, you have no possible way of telling how much of the familiar objects you deal with, including your own body, is composed of your new proposed Platonically-purified essense of mass. So what good is this idea of yours going to do you? In any case, I can assure you that it's not my ideas, but yours, that are non-standard here (if I understand you correctly)Sbharris 19:04, 8 May 2006 (UTC)

I also agree that your recent edit, Sbharris, is not so good. I maintain my old position that including kinetic energy in the mass is nonstandard, and your program of inserting that into many articles should be reconsidered. I agree with Yevgeny's removal. -lethe ^{talk +} 05:40, 7 May 2006 (UTC)

Well, you can be proud of yourselves. In an attempt not to "confuse people" you've produced an article with two examples which are simply wrong. If two particles collide and stick, they don't gain in mass. That's equivalent to saying that if a sealed container of thin monatomic hydrogen or oxygen sat long enough for all the atoms to combine into gas molecules, that the thing would actually sit there and gain weight. So you all really believe that? Is that physics, in your worldview? In the same way, do you really think that everytime a positron means an electron, that the gravitational field of the earth decreases?Sbharris 19:23, 8 May 2006 (UTC)

Sbharris, please don't sound so bitter. Let me ask you to first consider this point: no one is wrong. It's only a question of definitions. A definition cannot be wrong, it can only be useful or not.

If the first part of your last sentence is true, the second part is meaningless and you wasted your time typing it. Perhaps you're talking about a very limited subset of definitions? Define "useful." In the real world we often bet with our lives that our technology is good and the bridge or the airplane or the ship doesn't go down. In practice that takes a certain amount of elasiticity out of terms like "useful" and "wrong." For myself, I don't hold to the social constructivist theory of physics; I usually ask to see the driver licenses of whose who claim to do so.

Well I don't know what social constructivism means at all, so I guess you have also wasted your time by typing that. -lethe ^{talk +} 01:44, 9 May 2006 (UTC)

Whether or not the mass of two particles increases after they collide and stick together is simply a matter of what you define mass to be. The task left to us is to establish which definition is standard. I contend that the standard definition is that mass does not include kinetic energy.

You already admitted that it did, when you admitted that the mass of a container of ideal gas increases when you heat it. That heat energy goes entirely into kinetic energy, and thus you weigh the kinetic energy. It therefore has mass, Q.E.D. Perhaps you are now backpedaling and claiming that the classical mass (and weight = mg) of a container of ideal gas does NOT increase with heat Q/c^2?

I'm not backpedaling. -lethe ^{talk +} 01:44, 9 May 2006 (UTC)

Then would you explain yourself? Does it or doesn't it?Sbharris 03:27, 10 May 2006 (UTC)

If your problem is with number of particles, you can do a little exercise and replace my photon bouncing up and down in a box on scales, with a single helium atom. At low temperatures where the atom sits on the box bottom and can't move, you weigh just the atom. Give the atom kinetic energy (either a classical or relativistic amount), and you'll find the momentum transfer difference between top and bottom of the box (which determines the increase in box weight) is exactly g * {kinetic energy}/c^2. That's what you see with the scale, a quantity which has been our measure of mass for as long as we've been measuring mass. So what new definition do you propose?

And what if the atom is moving side to side in the box? What is its mass then? -lethe ^{talk +} 01:44, 9 May 2006 (UTC)

Why, its "mass" (i.e., what you infer from what you weigh when you weight the increase in the box weight) is the particle's rest mass plus its kinetic energy. In the Newtonian picture the sideways bouncing particle applies a downward force on the sides of the box because it doesn't hit with exactly transverse (x-component only) velocity, but rather picks up mv_y downward momentum during mean journey time t. That v_y is gt, and if you divide the momentum transferred per hit by the time between hits to get the y momentum transfer rate, you get mg. Which is the force and thus the particle's weight. It's actually a little more complicated because the particle transfers twice that y momentum per hit, but on the other hand in doing so it gets a downward kick from the wall which gives it a total twice that much downward v, so those factors of 2 cancel, and it's still mg. Don't worry about it. In the full SR case, guess what, the momentums are all γmv's but the arguments are all the same and the momentum transfer and thus weight ends up γmg. Which you will note is g* the total relativistic energy/c^2. That is, rest plus kinetic energy is the "mass" which counts as weight/g. You can't win at this. The mean weight of the box in the COM frame, no matter what kind of total energy E you add to it, or in what form, is always g[E/c^2]. As an exercise you may wish to try it with a laterally bouncing photon. I suggest replacing gravity with a uniformly accelerated box at a = g, and look at what aberration does to the momentum transfer in the acceleration direction. Again you'll find a photon of energy E adds g (E/c^2) weight.

Mass changes when objects collide. This is the origin of the term "mass defect" in nuclear physics.

Yes, this dates from the days when people thought particles came in even lumps which all weighed the same. And also didn't realize that heat and radiation had mass, and they should thus expect mass loss if they let heat and radiation escape a system. It's rather as though people discovered a "mass defect" whenever autos crashed, forgetting to take into account the weight of the gasoline and radiator water that inevitably ran away down the gutters. But such defect is not a good argument for nonconservation of mass; just one for better closure of systems.

The concept of mass defect comes from after Einstein's publications. It comes from nuclear physics. Nuclear physicists understand very well that heat and radiation can change the mass. They built a bomb out of the principle.

Yes, so? Between Einstein's publications and the time of the bomb, they figured it out. But mass defect doesn't mean the mass has disappeared, merely that it has moved away.

The concept of a mass defect is taught in schools today.

Yes. It's taught by high school teachers who think that that the "missing mass" in the nucleus has "turned into" massless energy. It's also taught by other teachers in higher institutions, who generally know better. The reason we don't weigh that "missing" mass is that it's very light, and in any case has escaped our weighing system. But that's not the same as saying it has no mass, in all the sense we mean when we speak of everyday, ordinary mass. If a single observer keeps track of it, it still has all the mass it always did. It's just somewhere else, other than the nucleus. That's a very simple thing to say, and it's not a complicated concept. I agree that it's widely misunderstood at the secondary school level, but I'm not responsible for that.

This is one of the reasons I claim that rest mass of individual particles is the standard definition of mass, not rest mass plus kinetic energy.

Yes, but that's individual particles with no bound subcomponents. It's not the same case as systems of particles, or composites made of bound particles. For which kinetic energy and any other kind of energy "inside there," contributes to mass. I already suggested you read chapter 5 of Misner, Thorne and Wheeler as to the issue of gravitational mass and the "time-time" (0,0) component of the stress-energy tensor, which includes KE density in the COM frame. I'm not repeating myself again.

Whatever you may think of the concept of "mass defects", you must admit that the concept is ubiquitous in nuclear physics, which illustrates that this is simply a matter of what your definition of "mass" is. -lethe ^{talk +} 01:44, 9 May 2006 (UTC)

The fact that there's a mass defect spoken of, in no way is intended to mean that somebody thinks the mass has disappeared. It's just short for the fact that in this bound system, the sum of the mass of the system is not the sum of the rest masses of the particles which make it up. But we all know by now, that this is generally not the case for any system. The term originates from a time when this WAS a big deal. It isn't, any more.Sbharris 03:27, 10 May 2006 (UTC)

Well, once you've taught the physicists of the world why your definition of mass is better than theirs, and it starts appearing in textbooks, then we can bootstrap ourselves out of alchemy here on wikipedia. Until then, we have to do what the physicists do. -lethe ^{talk +} 01:44, 9 May 2006 (UTC)

Er, I AM using the definition of mass that the average physicist uses. It's not the sum of rest masses of elementary particles in systems ("systems" includes nearly everything in nature, from hadrons to teacups), and I never pretended it was, and people (starting with Einstein) have realized it shouldn't generally be, for more than a century. However, that's not our issue, for it still doesn't mean mass is not conserved-- merely that when mass is "missing" from a system, that means that it's either moved, or else it hasn't been properly observed (ie, the "missing" part is noted by something other than a single inertial observer, and instead is a rash of reports from various inertial frames). The idea of a "rest" mass of closed systems (from hadrons to teacups) does indeed include their internal component kinetic energy and whatnot. For ordinary objects, it includes also their heat content and many kinds of potentials. If you don't believe this, you've got the problem, because this is really, really basic physics. Sbharris 03:27, 10 May 2006 (UTC)

Well, I'm sorry you're so pessimistic about the Wikipedia process. Neither one of us seems to be able to convince the other. It is time we started providing sources. I'm going to pick up 3 nuclear physics books and post what they say. -lethe ^{talk +} 01:44, 9 May 2006 (UTC)

This should be amusing. But the concept is so basic that I fear it won't be where you're looking for it. At that level, it's already understood. Sbharris 03:27, 10 May 2006 (UTC)

Why not say "if the particles of an object are moving relative to each other, then the mass of that object is greater because of it"? Does that satisfy everyone? —Keenan Pepper 01:20, 9 May 2006 (UTC)

Sure, but you might as well be quantitative about it. The mass increases by E/c^2 where E is the total kinetic energy of the particles in the COM frame. The charm of Einstein's famous equation is it actually is true in the COM frame, and when talking about mass CHANGES. That's how he originally wrote it down, and we've gone though an entire cycle of saying it's really not a good relativistic equation, THEN that it's only trivially true by (bad) definition, and now (finally), when I want to resurrect the meaning Einstein originally gave it, I'm here being accused of being obscurant and revisionist. The irony. Sbharris 01:33, 9 May 2006 (UTC)

That might satisfy me, yes. It seems that whether the modern definition of mass allows us to include kinetic energy depends on whether we're talking about several independent objects or a single object made up of parts. In particular, phrases like "if two objects stick together after a collision between them ...However note that the mass of the entire system does not change in such an encounter, since kinetic energy contributes mass to such systems)" are as far as I can tell in direct contradiction of modern physics definitions and usage, and should not be allowed to stay. -lethe ^{talk +} 01:44, 9 May 2006 (UTC)

Machine translation?

To me the paragraph "The object of inertial mass lower than any significantly high number tend to have low responsiveness to a change of resistance to accelerated motion. We therefore conclude that objects of a high inertial mass tend to increase resistance force less rapidly while experiencing acceleration, so accelerate more rapidly." appears to have been translated into English from a foreign language. Do other people read it that way? Can anybody actually understand it (especially the first sentence). P Koil (talk) 15:53, 4 January 2008 (UTC) 15:55, 4 January 2008

I, too, fail to extract sense from this. The second sentence appears to claim that bodies with high mass accelerate faster (than bodies with lower mass?). Of course, this is false if we are to suppose everything else (including the acting force) being equal, and I cannot imagine any reasonable alternative boundary condition that would make it true.

The stilted style does not sound machine-translated to me, more like a crank with a non-standard perception of mechanics who tries to hide his basic incompetence behind random learned-sounding phrasing (such as "significantly", "tend to", or "more rapidly" instead of "faster").

This is the edit where the paragraph first appeared. I have removed it. –Henning Makholm 02:12, 6 January 2008 (UTC)

Not true.

> To date, no deviation from universality, and thus from Galilean equivalence, has ever been found, at least to the accuracy 1/10**12. More precise experimental efforts are still being carried out. <

This part of the article is not true. I am aware of experiments conducted in deep mineshafts with copper and steel balls, which showed quite significant (0.1-0.4%) difference based on material quality. These experiments, however, are DISPUTED (which is not the same as "has never been found"). This issue should be clarified! 82.131.210.162 (talk) 18:07, 9 January 2008 (UTC)

• And it takes brass balls to do fringe physics of this type, too. SBHarris 01:56, 1 February 2008 (UTC)

Citation please? --Slashme (talk) 05:45, 10 January 2008 (UTC)

Physics or chemistry?

Someone changed the opening paragraph a couple of days ago, saying mass is a concept of chemistry. Seems to be no discussing here on this talk page to categorize mass as a chemistry concept, so I changed back to physics. Also, the article's categorization doesn't deal with chemistry. Although mass is very important in chemistry, in my opinion it should be viewed as a physical property. Mårten Berglund (talk) 00:38, 20 March 2008 (UTC)

I agree. The current definition reads far too closely to the definition of amount of substance, which is definitely different from mass. Physchim62 (talk) 16:30, 14 December 2008 (UTC)

(Au sujet de la masse) Origin of Mass

The last link to http://unification.selfip.org:2008/ is innapropriate to wikipedia and should be removed. The concept of particle as "a magnetic spherical loop" developped in this "theory" would be regarded as bogus by the vast majority of physicists. If required I can loose my time translating parts of this for you, but I believe it is not worth. —Preceding unsigned comment added by Humanino (talk • contribs) 15:55, 18 August 2008 (UTC)

As nobody seem to comment, I will now remove this link. Humanino (talk) 19:06, 2 September 2008 (UTC)

Effect of eating on body mass

Why doesn't your mass increase by 100 g after you eat a 100g sandwich? —Preceding unsigned comment added by 71.17.75.192 (talk) 02:26, 15 December 2008 (UTC)

You're not a closed system, and you lose carbon when you breathe in oxygen and exhale CO2. Some of that sandwich is carbon. If you put a man in a bubble or sealed spaceship and he eats a 100 g sandwich, mass of the system never changes. SBHarris 00:50, 5 February 2009 (UTC)

The mass of the bubble or sealed spaceship (including its contents) system would not change, but the mass of the man-sandwhich system would still change. Brian Jason Drake 09:34, 22 September 2009 (UTC)