Talk:Invariant mass

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Top[edit]

Shouldn't this be a redirect to rest mass? Keenan Pepper 06:26, 24 Mar 2005 (UTC)

Yes, yes it should. -lethe talk + 03:03, 16 April 2006 (UTC)
No, no it shouldn't. "Rest mass" is used to refer to a single particle or system at rest. "Invariant mass" is used to refer to a system of particles in motion, usually decay products. The rest mass of a single particle is also its invariant mass, but the two terms deserve their own discussions. Strait 18:56, 16 May 2006 (UTC)
Rest Mass currently directs to well click rest mass and see. This article discusses invariant mass and relativistic mass and describes rest mass as the same as invariant mass. I think this is the page you should be looking at. Jameskeates 13:12, 23 August 2006 (UTC)
No, it should not. I am an undergraduate researcher at Rutgers University currently analyzing data from the Large Hadron Collider, and the Invariant Mass page has proven invaluable to me as I receive my "crash course" in high energy physics to begin working (no pun intended). Ordinarily I would not learn this material for another year or two, so this article is incredibly valuable to me. Furthermore, the article addresses invariant mass within the scope of particle physics, whereas the special relativity page encompasses too much information for this topic. Don't fix what isn't broken. 15:38, 19 July 2010

—Preceding unsigned comment added by 128.6.27.165 (talk) 19:57, 19 July 2010 (UTC)

You say "the special relativity page encompasses too much information", but I don't think people are suggesting merging this with mass in special relativity or any other page, just suggesting that "rest mass" may be a more common term in which case the title of the article should be changed from "Invariant mass" to "Rest mass". I think Strait's comment that "rest mass" and "invariant mass" have different meanings, and that rest mass refers only to a system at rest, is incorrect--relativity textbooks routinely talk about the rest mass of objects which are moving relative to the frame in which the analysis is being performed. The sentence in the current opening paragraph (which I am going to edit out), saying "If the system is one particle, the invariant mass may also be called the rest mass", also seems to introduce a false distinction, since both "invariant mass" and "rest mass" can be used to describe bound systems of multiple particles. If anyone thinks there is ever a context where rest mass and invariant mass could not be used interchangeably according to the accepted definitions, please point it out. Hypnosifl (talk) 19:11, 23 July 2010 (UTC)
Well, the obvious one is the one for which "invariant mass" is most often used in particle physics, which is systems of UNBOUND particles. In that case, "invariant mass" is perfectly well-defined and often used. However, "rest mass" is undefined and meaningless. SBHarris 00:08, 24 July 2010 (UTC)
Wouldn't even an unbound system have a center of mass, which would have its own well-defined rest frame? (the frame where the total momentum was zero) Even so, it may be that in practice physicists never actually use "rest mass" to describe the total energy (divided by c^2) of unbound systems in the center-of-mass frame, in which case the two terms do have a distinct meaning in practice even if it would be perfectly logical to define "rest mass" in this way. Hypnosifl (talk) 01:35, 24 July 2010 (UTC)
Yes, it would be perfectly logical, by extension. But in practice, nobody uses the term to refer to unbound systems. So, we have a sort of oddity of language, but not THAT odd. After all, the "rest frame" (and thus also COM frame) of bound systems is easier than finding the COM frame, which may not be obvious at all. Consider an assymetically expanding gas cloud: it has a COM frame and an invariant mass, but it's not at all obvious by inspection where that frame is. In most bound systems, by contrast, you just look at the motion of the center of mass, which is (usually) obvious by inspection. In solid objects it's the rest frame of the whole extended collection, and even in rotating objects it's the rest frame of the center of mass, around which it rotates. In a bottle of gas it's the frame of the bottle, and so on. Thus, rest mass and invariant mass are the same for single particles and extended objects and bound systems, but not for unbound systems, where the "rest mass" isn't defined. This is in part because the COM frame isn't obvious. But also it is in part because "invariant" mass can be used for systems involving photons which are never at rest. Physicists are comfortable talking about the invariant mass of a system of decay photons from the decay of a pi-zero (neutral pion) for example, but the "rest mass" of such a system sounds funny, since no part of it can be brought to rest, by any shift in reference frame. SBHarris 02:42, 24 July 2010 (UTC)

This page is not needed?[edit]

I think that this page is not needed and should be merged with Mass in special relativity. Comments? Timb66 10:17, 28 May 2007 (UTC)

I thought so myself when I first saw it. But mass in special relativity became an experiment in popular science. I think this page should retain its conciseness, and correct message according to what is established today, for the physics student.
In fact, I would support copying everything from mass in special relativity that is relevant to today's physics to here. Thanks. Edgerck 22:59, 28 May 2007 (UTC)

I don't agree with maintaining two pages on essentially the same subject just to allow two different points of view to be expressed. Let's work to improve mass in special relativity (which i agree needs work). Timb66 09:28, 29 May 2007 (UTC)


This page is just redundant. I think that "mass in special relativity" is inclusive of relativistic mass and invariant mass. Explanation is already given in "mass in special relativity". The term invariant mass rises because of the special relativity. Thljcl (talk) 14:38, 22 April 2008 (UTC)


Has the Newtonian concept been utterly cast out? "Quantity of matter" is passe? Truth be known, "mass" is a motor function of massless charge circulating in a Bergman heliocycloid, producing a scanning vector potential the interactions of which scanning vector potentials of different particles result in the phenomena associated with what has always been called "mass." But, what do I know about anything. ---Nosey, 20:45, Jan 4, 2009 (UTC) —Preceding unsigned comment added by 199.80.75.2 (talk)

This page is absolutely needed. Speaking as someone who works in experimental particle physics the invariant mass of a particle or collection of particles is a concept that I use every day and although related is a different thing than the concept of rest mass. Actually the concept or idea that a particles mass changes as it speeds up is an antiquated idea that is not used in modern physics. The majority of physicists today choose to think of mass as invariant. This is a very different concept than mass-energy equivalence but still related in the scope of relativity. You should fully feel free to interpret the physics in a way that is consistent and convenient to you but this page should be kept because this article covers directly the way the concept is used in experimental particle physics. --Josh, 17:19, 16/08/2009.

Comment[edit]

There is a problem of accuracy in this article: Since the mass of a particle is an invariant, there is NO EQUIVALENCE between MASS and ENERGY. The author of this article seems to have forgotten that... Considering a "varying" mass may change nothing to the answers to calculations, but physically, in a encyclopedical article that treats on physics it's absolutely necessary to be scientifically accurate so prevent confusion to (maybe non scientific) people who read it.

When you have a particle of mass m, you can say that the rest-energy of this particle is equal to mc^2. When you give this mass a velocity it will acquire momentum and its energy will be \gamma mc^2. (And its mass remains m !!) Take the example of a photon: It has no mass. But it has an energy (E=pc= h\nu where p is the momentum of the photon, h the Planck constant, \nu the frenquency). Then how could you say that mass is equivalent to energy?

By the way, the frequency \nu of the photon is the frequency the electrmagnetical wave would have, speaking classicaly, but the photon is never a wave. (The wave-particle "duality" is a confusion that often appears. Early in the 20th century it was not very clear to scientists, but now it's very clear) Electromagnetical waves are made of photons, they are not waves. But it is classicaly convenient to treat it as waves. Similarly, in quantum mechanics, the way the concept of wave functions appears should not lead to a wave-particle confusion: Theses fonctions behave mathematically like waves, but they only express, when lead to square, the probability the particles to go here or there. Tupac Yupanqui (talk) 15:04, 6 June 2009 (UTC)

Why don't you just go back to read Jackson or any other textbook about special relativity? What you said above can only show that you don't really understand the basic concept of invariant mass.--Suiseiseki (talk) 01:56, 16 July 2009 (UTC)
J.D. Jackson doesn't say anywhere that the mass of a particle would change when it acquires velocity! If you find where, show it to me, which chapter, which page!
Perhaps I've not made myself clear, my sentences are not perfect at all, sorry I'm not englishman nor american, but I hope it's understable??!?
I've no problem with the concept of invariant mass of a system of particules.
What poses a problem for me is, in this article: (quote:)

The motivation for defining rest energy is in the Special Theory of Relativity. According to that theory, the mass of a body changes in proportion to its kinetic energy, via:

dm=\frac{dE_k}{c^2},

This leads to Einstein's famous conclusion that energy and mass are manifestations of the same phenomena. Defining rest energy as above makes the mathematical expression of mass-energy equivalence more elegant, but is still arbitrary in the way it places energy on an absolute scale. See background for mass-energy equivalence.


The equation above is a non sense, since dm = 0 !
Correct relation between mass and energy of a free particle with non zero mass:
E = \gamma m c^2 = \sqrt{\dfrac{1}{1-\frac{v^2}{c^2}}} m c^2
It brings to
E = mc^2 when v=0
and to newtonian behaviour when \dfrac{v}{c}\ll 1 :
E_{Newton} = mc^2 + \frac{1}{2}mv^2


Tupac Yupanqui (talk) 17:58, 22 December 2009 (UTC)

Tupac, there are two definitions of mass being used here. One is the invariant mass, which for single particles is the same as the rest mass. The other definition is the relativistic mass, which is NOT invariant, but changes with reference frame, same as total energy. People are always getting confused by the two kinds of mass and the two kinds of energy. Setting c=1, the rest mass (invariant mass) is the same as the rest/invariant energy. And the total energy is the same as the relativistic mass. But these 4 are equal to each other only when total momentum is zero, which means only in the COM frame. It is only there for systems that total energy is equal to invariant mass (and for particles, obviously total energy is rest mass). Anyway, see mass in special relativity for a full discussion of these 4 quantities, only 2 of which are invariant. SBHarris 03:56, 23 December 2009 (UTC)

I thought that Einstein calculated that it was the incremental mass decrease that gave the E = Mc^2 energy value and not necessarily the rest mass. The conversion calculation in Kaplan was noted to be an incremental momentum calculation, which resulted in the delta mV being equal to delta Mc^2. But that mathematics involves the assumption of the existence of some amount plus an increment of the M value.WFPM (talk) 09:27, 7 April 2011 (UTC)
Yes, it's the incremental mass increase, which means mass or energy subtracted (or added) from the system. The loss or gain of Δm is not due to mas-energy CONVERSION, but an added or subtracted mass resulting from energy ADDITION or SUBTRACTION. It's not the same thing. In such cases, both mass and energy go off together. There is no conversion, but simply removal. Of both of them. SBHarris 19:26, 7 April 2011 (UTC)
Yes but! If the invariant or rest mass of an electron is given (as 9.1036+ x 10E^-28 grams), we are by the theory forced into the assumption that all of that mass is at rest! And comes the question as to whether the particular electron under consideration is completely at rest, or only relatively stationary in a particular position. And since most electrons with we deal are in an orbital or other condition of motion, how are we to know what value of variant relativity mass value that would occur at any velocity of excitation. Of course the binding energy value of 0.502 MEV assigned to the C velocity electron makes implicit the above rest mass value. But if the electron is in motion, as it usually is, how do we determine this relativity revised additional kinetic energy value. I hope that this explains the complexity of the physical motion process of the problem.WFPM (talk) 02:52, 2 May 2012 (UTC)
And what I am really worried about is how it can be calculated that an orbital electron can be assumed to be in a stable situation in one orbital situation and then instantaneously imagined to have moved to a second orbital situation in such a manner that a photon's worth of energy is emitted in the process, and presumably in one direction. Is that a correct statement of the theory of the photon emission process?WFPM (talk) 03:06, 2 May 2012 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── The equation for rest mass corrects for any velocity of motion of the particle, whether it's moving or the observer is moving. So the rest mass of the electron comes out the same number whether it is sitting still (E = 511 keV) or is moving at half the speed of light. In the latter case it will have a larger total energy and a larger momentum, and these simply subtract out to the same m_0.

(m_0c^2)^2=E^2-\|\mathbf{p}c\|^2\,
That sounds to me that you're saying that a moving mass can have a larger content of energy than it does when moving at a velocity of c! In other words that the kinetic energy addition can be positive to the rest mass energy? And I can't see where the observer's velocity has anything to do with the rest mass plus kinetic energy of a particle other than his individual measurement of it.

As to emission of a photon, it's a complicated thing. It involves transition from two stable states in which the electron has two different energies but each of which has no dipole moment. In between the states is a a transition state that has a transition dipole moment. It is the frequency of this changing electric dipole moment that generates the photon. As the atom goes from one stationary state to the other, it passes through this transition state. You can see that the process of stimulated emission relies on an outside photon of just the right frequency to encourage formation of this transition state, which radiates (as in lasers and masers). On the other hand, in spontaneous emission the photon which gets the process going, like a snowflake starting an avalanche, is a photon from vacuum fluctuations. It's all pretty cool. SBHarris 03:31, 2 May 2012 (UTC)

And the intermediate energy emission of an electron certainly is a complicated thing! First, we have to have a way for the electron to have some detachable matter that can be used to create the photon. Next we have to have a means of emission such as to meet the required electromagnetic radiation properties of the photon. And it has to be emitted from the electron with an initial velocity of that of light energy, regardless of the velocity (or should I say speed) of light! I've always thought that meant that it had to be emitted at right angles to its direction of motion? And that emission process doesn't prevent it from subsequently emitting a different quantity (frequency) of radiation at a later date from some internal reservoir of energy, plus it must have an internal accounting procedure that controls the total radiation energy production process. And now you want to worry about the process of spontaneity! And given the definition of the photon as "a bundle of radiation energy", which we can expand into "an accumulation of bundles of electrostatic charge and radiation energy", can you see a real physical entity in this concept that would allow a Z number of these entities to exist and be organized and controlled by the Z number of protons that are supposed to exist within the atomic nucleus?WFPM (talk) 19:14, 2 May 2012 (UTC)
Due to special relativity, light is emitted at the speed of light even from moving objects. That includes your car headlights, too. As for the rest of your comments, you seem to be assuming things that aren't necessary. Photons aren't conserved, but newly minted. There is some kind of internal accounting that makes for conservation of energy, but nobody really knows how it works at the quantum level. When a photon is made, there's obviously a radiation reaction force on the electron and this acts though a distance (as does the Z electric field of the nucleus, which acts on the electron which moves through a distance closer to it) and all these have to come out to be the same. SBHarris 21:42, 2 May 2012 (UTC)
In the illustrations I've seen, the forward speed of the light beam is not permitted to exceed c, so the emission of the beam gets crowded into a flatter and flatter sideways beam of light energy away from the source, which I thought was what the formula called for. And I'm comparing a photon to a match, in that it evidently contains some material that can be caused to "ignite" and emit radiation with some of its content. And my models result in a stack of alpha particles on top of each other within the structure, and so I don't see how the inner protons are able to use their charge characteristic to influence the activity of an individual orbital electron, since the charge field is a summary charge field, which may or may not be penetrated by the path of the electron. And I certainly appreciate your attention to my comments on this matter.WFPM (talk) 15:12, 3 May 2012 (UTC)
The shape of the wavefront from a moving source depends on what frame you look at it from. From the moving source you do get a flattened shape since the forward light speed cannot exceed c, even if the source is moving. From the view of the source, of course, every thing is symmetrical as though it were standing still. That's what makes relativity different from other wave-propagations (if you're standing still and the wind is blowing by you, the envelope of a sound "ping" you make is not spherical; but in relativity, it is always spherical in the frame of the emitter).

As for your other question, the whole idea of an electric "field" is that it's a field. An electron can "feel" the field of a buried proton, just as you can feel the gravity field due to the core of the Earth.SBHarris 17:48, 3 May 2012 (UTC)

Okay and say the nucleus acquires an additional deuteron and hangs an additional electron in the outer reaches of its "cloud" of electrons, and the summary field increases in charge and shrinks in volume, evidently due to the increased kinetic activity of the electrons, and that results in in the spectrum of radiation of one (or more?) electrons having a higher frequency and shorter wave length of radiation property characteristics. And that is due to the smaller incremental energy jumps required for the electrons to move from one level to another. And this is all due to the action of attractive exchange force between the protons and the presumably nearest by electrons in their vicinity. And this is a finely tuned system such that the minor difference in radiation frequency can be determined and related to the activity of a particularly designated electron of the system. I guess it's not like Einstein said "not malicious but just meticulous", but as an engineer, that sounds like a pretty complicated design.WFPM (talk) 19:05, 3 May 2012 (UTC) I note that that last statement gets me involved in spectroanalysis about which I know very little and will read about in the CRC handbook. But I wish that somebody that kind of knowledge would pay more attention to the indication of the real physical models.WFPM (talk) 16:55, 4 May 2012 (UTC)

Merge proposal[edit]

Someone proposed that Bare mass be merged into this article. However, as far as I can gather from a quick perusal, the two concepts are quite different. RockMagnetist (talk) 21:57, 24 January 2012 (UTC)

Someone? Where namely, and what to merge? Incnis Mrsi (talk) 21:01, 4 February 2012 (UTC)

Additivity[edit]

Had Mr. Sbharris read the link he removed, intensive and extensive properties#Extensive properties, he probably would realize what do I mean. Namely, that it is an extensive quantity but without (exact) additivity. Incnis Mrsi (talk) 21:01, 4 February 2012 (UTC)

Next topic: I do not agree with the new wording, that we have to prove that invariant mass exists. It just is equal to the norm (more precisely, to square root of scalar square) of 4-momentum vector. So, if we trust in existence of 4-momentum for physical systems, there should be no proofs of existence. It is a matter of arithmetics to determine whether invariant mass is positive, zero or maybe even imaginary, but it is well defined. The non-collinear photons example proves that mass of a system consisting of massless particles only may be indeed positive. The term massless is a bit confusing, but good physicists know that it means "zero invariant mass", not "invariant mass is undefined". There should be no proof that this quantity exists. Incnis Mrsi (talk) 21:01, 4 February 2012 (UTC)

We mean it exists for the 2 photon system, which means it is positive and not zero. "Zero mass" does not mean mass exists but is zero-- it means mass does not exist. That is what "zero" means-- non existence. Don't make this complicated. Your addition of the possibilties of "imaginary invariant mass" for faster-than-light particles don't help this article either, since no such thing has ever been seen, and it violates causality and therefore logic (insoluble paradoxes arrise where things both happen and do NOT happen, and this is logically impossible). Again, you're making things complicated without need. Normally, "invariant mass" is not defined if "length" of the defining vector is imaginary. People have tried to define it to get tachyons and non-causal relationships and tine travel into the past and so on, but that's science fiction. Show me communication with the past and I'll grant you imaginary mass particles. Otherwise, it's just mental masturbation (or science fantasy).

Finally, whether invariant mass is an example of an extensive property without additivity depends entirely on how defines "extensive property." More unnecessary complication, especially for the lede of an article. SBHarris 21:59, 4 February 2012 (UTC)

I did not say anything about superluminal motion, time travel and causality – read my theses carefully. I did not say anything about particles, velocities, and about motion at all – we do not have to suppose that it exists. We actually need only two things: the Minkowski space-time and the four-momentum. Then we may just write a definition:
m^2 = {\mathbf p}_\mu {\mathbf p}^\mu = {\mathbf p}_\mu {\mathbf p}_\nu \eta^{\mu\nu}
I do not know, why the article does not contain this formula. This would allow to demonstrate that so named "massless" particles are special case by their symmetries, but not for this definition of the invariant mass. But for objects which are believed to exist IRL, exclusion of negative energies (i.e. such 4-momenta as p = (−m, 0, 0, 0) is also a difficulty – negative rest masses may be theoretically even worse than imaginary ones. Incnis Mrsi (talk) 22:39, 4 February 2012 (UTC)

Invariant mass is usually what we weigh as "mass"[edit]

An author has removed this statement: Only when the system as a whole is at rest (its total momentum is zero), will invariant mass be equivalent to the familiar concept of rest mass.

I've corrected the misuse of the apostophe, but so far as I can tell, this statement is completely true. Since ordinary objects (like a tank of gas or a solid composed of vibrating atoms) are "systems", what we weigh as "mass" on a scale in the COM frame IS the "invariant mass" of the object-at-rest. The "rest mass" of a system, is the mass of the system when the system is at rest, and that's the mass of the system in its COM frame, where the scale is. It is indeed the invariant mass of bound systems that we're all familiar with, in everyday life, since we measure the mass of bound systems in their COM frame in everyday life, and that mass includes contributions from all energies that show up in that frame, in the system (like the energy of molecules in a hot gas, or the kinetic and potential energies of vibrating atoms). I think that deserves a mention, but it has been systematically removed. So, defend doing that, please, or I'll put it back. SBHarris 17:58, 25 June 2012 (UTC)

Japanese Version[edit]

I noticed there's a Japanese version of this page (http://ja.wikipedia.org/wiki/%E4%B8%8D%E5%A4%89%E8%B3%AA%E9%87%8F) which links to the English page, but the English page doesn't link to the Japanese page. Sorry, I don't know how to edit this as I'm just a Wikipedia reader.. — Preceding unsigned comment added by 58.7.92.17 (talk) 13:52, 14 January 2013 (UTC)

Done. — Quondum 15:48, 14 January 2013 (UTC)

Mandelstam variable s[edit]

Shouldnt there be a link (or at least mention somewhere) the relation to s, the mandelstam variable? — Preceding unsigned comment added by 146.50.70.176 (talk) 09:24, 18 April 2013 (UTC)

Edit needed - rest mass is NOT invariant mass[edit]

Rest mass redirects here. Since they are NOT the same, they either need to be carefully differentiated here, or the article needs to be split. This whole article needs a rewrite. I am NOT familiar with the concept of invariant mass, or I would do it myself. The article goes to great lengths to explain that the sum of the rest masses is not equal to the invariant mass. I am not sure how useful that tidbit is, since it fails to discriminate between the two concepts. There are a multitude of other problems with this article. It seems to be claiming (I had to read it several times to understand it probably is not) that invariant mass changes with velocity. It claims that the invariant mass IS the total energy of the system divided by c² as long as a certain frame of reference exists (note the only requirement is its EXISTENCE, not its use as the actual frame of reference). Since total energy IS a function of velocity (relative to the f.o.r.), invariant mass is not invariant if we take the sentence as it is written. Too many clauses, too many qualifications. Why not say simply that for individual fundamental particles, invariant mass is identical to rest mass (if I understand the concept) but that in systems of particles, the center of momentum frame is not the rest frame, the individual particles have momentum in that frame and so the invariant mass is the sum of both the rest mass of each constituent and their momentum (or energy). { I am not clear whether the invariant mass includes other sources of potential energy??} { I am also not aware of how the concept fits into General Relativity. The article seems to dabble with GR, but avoids direct discussion - most likely a wise choice} Seems this article agonizes over the existence of composite particles and systems. And is tripped up by them. Nowhere is it clear (if it is actually true, as I am assuming) that two observers of the same system (why does it have to be bound? Isn't the only requirement that it be measurable?) will agree on the invariant mass, while the measured mass will be relative?? That is, the invariant mass does not change with a different choice of reference frame, while a particle's or system's measured mass may change. I also take exception to the use of m₀ or M₀ to mean both rest mass and invariant mass in the same article. At least, it sure confused me.72.172.11.228 (talk) 01:11, 31 May 2013 (UTC)