Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Why do we have an elementary charge $e$ in physics but no elementary mass? Is an elementary mass ruled out by experiment or is an elementary mass forbidden by some theoretical reason?

share|improve this question
    
Related: physics.stackexchange.com/q/122/2451 –  Qmechanic Mar 9 '12 at 20:09

7 Answers 7

up vote 12 down vote accepted

Let me add two references to points already mentioned in this discussion:

Today, there is no reason known why the electric charge has to be quantized. It is true that the quantization follows from the existence of magnetic monopoles and the consistency of the quantized electromagnetic field, which was shown first by Dirac, you'll find a very nice exposition of this in

  • Gregory L. Naber: "Topology, geometry and gauge fields." (2 books, of the top off my head I don't know if the relevant part is in the first or the second one).

AFAIK there is no reason to believe that magnetic monopoles do exist, there is no experimental evidence and there is no compelling theoretical argument using a well established framework like QFT. There are of course more speculative ideas (Lubos mentioned those).

AFAIK there is no reason why mass should or should not be quantized (in QFT models this is an assumption/axiom that is put in by hand, even the positivity of the energy-momentum operator is an axiom in AQFT), but a mass gap is considered to be an essential feature of a full fledged rigorous theory of QCD, for reasons that are explained in the problem description of the Millenium Problem of the Clay Institute that you can find here:

share|improve this answer
    
word of note: if you're quoting an article or book, use the > operator to put it in blockquote format. –  jcolebrand Jan 31 '11 at 15:34
    
I deleted my answer, and added my vote to yours. I think we are in agreement and all the information is pretty much contained here. I am particularly in agreement with the tone: in my mind we don't know why charge is quantized, but we may have some solid ideas. We have no idea about the issue of mass unit either way, so it is really a question of our degree confidence in our convictions at this point. In particular, I find the arguments Lubos gives on quantization of masses inconclusive or irrelevant (but I also think it is unlikely elementary particle masses come in quantized units) –  user566 Jan 31 '11 at 15:42
    
@drachenstern: Thanks for the hint, should I use the blockquote format when I quote from a book or also when I list a book or an article? (In my answer I only list a book and an article, I don't quote them.) –  Tim van Beek Jan 31 '11 at 16:01
    
that is an excellent question and I have no idea honestly. I just couldn't tell if you were quoting or merely referencing within the body of that. I might've prefaced the bullet point with "Refer to" because it seemed like you were attempting to quote him, even tho you might've been paraphrasing. I have a feeling you know more about article citing than I do tho so I'm gonna let you go with your gut on this one. :s ;) –  jcolebrand Jan 31 '11 at 16:16

Charge comes from discrete symmetries and is countable and additive. Mass comes from continuous 4d space, is exchangeable with energy and, in quantum mechanical dimensions not linearly additive, thus not countable.

Suppose you have an elementary quantum of mass, $m_q$. In the world we know two such quanta would not end up as $2m_q$.

One would add the four vectors and take the measure in 4space, and square root it, to get the invariant mass of two of them, etc for higher numbers at will. Given a mass, you could never know/count of how many $m_q$ it is composed. It is a continuum. Whereas charge is simply additive and countable.

The only way an elementary particle rest masses could be a linear sum of $m_q$s is for there to be no binding energy, and experiments tell us the elementary particles are bound, if stable. If there were no binding energy then the composites would crumble into the constituent $m_q$ with the slightest scattering.

share|improve this answer
    
Hmmmm. You're going to argue that since all orbital angular momenta are multiples of h-bar it must be impossible to have one with zero angular momentum? –  Carl Brannen Jan 31 '11 at 5:52
    
No, I was trying to edit it. It is not angular momentum that is the problem, since it is a quantum number coming from the solution of equations so can be 0. It is charge itself, since the photon also has 0 charge but nobody uses it as an elementary charge. –  anna v Jan 31 '11 at 6:10
1  
Good answer. It's as simple as possible, but no simpler. –  Ben Crowell Aug 10 '11 at 22:06

Mass can't be quantized because the contribution of a particle to a system's mass is not a scalar, but a 0 component of a 4-vector, so if you have a system of quantized mass particles, their bound states would not obey mass quantization.

In semi-classical gravity, there is a simple reason that charge has to be quantized. If the proton had a charge infinitesimally bigger than the positron, you could make a black hole, throw in some protons, wait for an equal number of positrons to come out in the Hawking radiation, and then let the resulting wee-charged black hole decay while throwing back all the charged stuff that comes out. This would produce a small mass black hole with charge equal to any multiple of the difference, and it could not decay except by undoing the process of formation. This is obviously absurd, so the charge is either quantized or there are particles of arbitrarily small charge.

Further, the small charge particles can't be too heavy, since the polarizing field of the black hole with these wee-charges must be strong enough to polarize the horizon to emit them. If their mass is bigger than their charge, then they are net-attracted to the black hole, which causes a constipation for the black hole--- it can't get rid of its charge. So the wee charged particles must be lighter than their mass generically.

These types of arguments reproduce the simpler swampland constraints. That out universe is not in the swampland is the only real testable prediction that string theory has made so far (for example, it excludes models where the proton stability is guaranteed by a new unbroken gauge charge).

share|improve this answer

Mass is determined by how a particle interacts with the Higgs boson(s). Mass is also determined by the relativistic mass-energy equation $E^2=m^2c^4+p^2c^2$ or more simply $m=\sqrt{(E^2-(pc)^2)}/c^2$ The energy values are a continuum, so there is no discrete elementary mass unit. In General Relativity, things are more complicated. In Stationary spacetimes, for example, the gravitational potentials (metrics) are not functions of time and the ST has time translational symmetry, so energy is conserved-- but while the stress-energy tensor is Lorentz covariant, in a non-isolated system, the system exchanges energy-momentum with its environment and its "mass" is not invariant--once again, no elementary or fundamental mass. I hope this is what you meant, but maybe you just meant the Planck mass... Frank Wilczek spends quite awhile in his popular book "The Lightness of Being" trying to say what mass is and isn't. He does quite a good job in non-technical terms- http://www.amazon.com/Lightness-Being-Ether-Unification-Forces/dp/B004HEXSXG/ref=sr_1_1?s=books&ie=UTF8&qid=1296458301&sr=1-1

share|improve this answer
    
@lubos--yes, I am sort of saying the same thing that energy is continuous and hence mass is continuous. In GR, in non-isolated systems, rest mass is co-ordinate dependent. –  Gordon Jan 31 '11 at 7:23
1  
"The energy values are a continuum, so there is no discrete elementary mass unit." This doesn't make sense. If you could choose $E$ and $p$ independently, then certainly you could make any value of $m$ you liked. But they aren't independent. By your argument, a single electron could have any mass at all. –  Ben Crowell Aug 10 '11 at 21:57

Dear asmailer, the reason is simple and completely understood: the electric charge is the generator of a $U(1)$ symmetry which is compact and may be parameterized by an angle, $\phi$. So wave functions may only depend on the angle $\phi$ in a periodic way, $\exp(iQ\phi)$ where $Q$ is integer (or an integer multiple of $e/3$, if I look at the elementary $U(1)$ rescaled by a factor of three that also allows quarks).

On the other hand, the mass is nothing else than the energy measured in the rest frame. The energy generates translations in time - and time is noncompact. So the corresponding phase $\exp(Et/i\hbar)$ isn't constrained by any condition of periodicity. So the energy is continuous even in the rest frame.

In the other frames, the continuous character of the energy is even more obvious because the "already continuous" rest mass is multiplied by the Lorentz factor $1/\sqrt{1-v^2/c^2}$ which changes - and has to change - continuously as we vary the velocity; the latter is required by the principle of relativity. So the mass and energy are continuous, have to be continuous, and will always remain continuous.

You could continue to ask "why" and in fact, you could get even deeper answers. You could ask why time is not periodic - which was used for the continuity of energy in a particular frame. Well, time has to be "aperiodic" because a periodic time would cause the grandfather paradox and other bad things - closed time-like curves. Time is also unbounded in the future because we live in a space with the positive cosmological constant.

On the other hand, groups such as $U(1)$ have to be compact and are compact in any quantum theory of gravity. This was argued e.g. by Cumrun Vafa in his Swampland program. For $U(1)$, the situation is simpler: the electric charge has to be quantized because of the Dirac quantization rule and because of the existence of the magnetic monopoles which is also guaranteed in a consistent theory of quantum gravity as was explained in another question on this server.

share|improve this answer
6  
Lubos: Insisting that the Abelian group in the standard model is compact is a choice which is equivalent to saying charge is quantized. It does not explain that fact, it just encodes it. You are also correct that mass/energy cannot be quantized, by LI, but I don't see anything going wrong if the rest mass of all elementary particles ends up being a multiple of some basic unit. As I wrote, I don't see any advantage in it either. –  user566 Jan 31 '11 at 7:04
2  
For two other argument in your answer: time is non-compact so the generator of time translation does not have to be quantized, even in the rest frame. This does not prevent those masses from being quantized anyway, for some other reason currently unknown (certainly that does not imply CTCs or anything like that). I think we also put different weights on our certainty that monopoles exist. I would not go as far as saying this is a done deal, I think you underestimate the amount of theoretical uncertainty around the subject, but we can just agree to disagree on this. –  user566 Jan 31 '11 at 7:19
1  
@Moshe, suppose you have an elementary quantum of mass, m_q. In the world we know two such quanta would not end up as 2*m_q. One would add the four vectors and take the measure in 4space and square root it to get the mass of two of them, etc for higher numbers at will. Given a mass, you could never know/count of how many m_q it is composed. It is a continuum. Whereas charge is simply additive and countable. –  anna v Jan 31 '11 at 9:59
    
@Lubos: So you are basically saying that we have an elementary charge, because magnetic monopoles exist and we have no elementary mass, because we measure a positive cosmological constant? –  asmaier Jan 31 '11 at 10:22
    
@Moshe continued: the only way elementary particle rest masses could be composed by n*m_q is if there were no binding energy, and then they would crumble into m_qs at the slightest interaction. This is not observed. –  anna v Jan 31 '11 at 10:28

The coupling constant for gauge theories is dimensionless, such as the fine structure constant $\alpha~=~e^2/(4\pi\epsilon_0\hbar c)$ $\simeq~1/137$. Mass has naturalized units of reciprocal length. This makes the establishment of a charge more reasonable, and a unitless number is something which is benchmarked as being an absolute constant. In other words, if $\alpha$ changed it would be a pure numerical variation. Occasionally there are claims of this. A quantity which has an actual dimension in units is so in relationship to other quantities.

This is a question related to the problem of quantum gravity. The Planck mass $m_p~=~\sqrt{\hbar c/G}$ can be thought of as the fundamental unit of reciprocal length, and the gravitational constant $G$ has units of area. This area corresponds to the unit area of a black hole event horizon. For a Yang-Mill field theory the coupling constant functions in a field which is unitary. By contrast units of mass are related to this reciprocal length, which in turn is not just a unit involving gravitational modes, but also the degeneracy of modes which have an entropy --- or entanglement entropy.

So mass does not quantize in quite the elementary fashion we might expects with charge and other coupling parameters for interactions.

share|improve this answer
    
Nice answer! Brings in the physical constants. (+1) –  Carl Brannen Jan 31 '11 at 3:12
1  
The dimensionless nature of $\alpha$ doesn't imply that all charges must be in the ratios of integers. If the ratio of the charges of the electron and the proton was an irrational number, we would just have to pick one of them to use in defining the fine structure constant. –  Ben Crowell Aug 10 '11 at 22:01
1  
Your speculation about quantum gravity is incorrect and pointless. (1) There are answers that don't resort to quantum gravity, and that's preferable, since we don't have a theory of quantum gravity. (2) Your argument doesn't make any sense. It's just a bunch of impressive-sounding words strung together. (3) You seem to be assuming that length is quantized in any theory of quantum gravity, but there are counterexamples. In fact, both of the leading contenders for a theory of quantum gravity (string theory and LQG) are counterexamples. –  Ben Crowell Aug 10 '11 at 22:04

I think it's beause do not have a fundamental understanding of mass. If we did, maybe that fundamental unit would have some relationship (i.e. tiny fraction) with the Planck mass.

The current effort in that direction probably begins with understanding the Higgs. There are several competing theories of the Higgs. They don't even agree on the number of such particles. So in that sense, the ball is in the experimentalist court.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.