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.

In particular I am curious if the values of the rest masses of the electron, up/down quark, neutrino and the corresponding particles over the next two generations can be defined as constant or if there is an intrinsic uncertainty in the numbers. I suppose that since there is no (as of yet) mathematical theory that produces these masses we must instead rely on experimental results that will always be plagued by margins of error. But I wonder does it go deeper than this? In nature do such constant values exist or can they be "smeared" over a distribution like so many other observables prior to measurement? (Are we sampling a distribution albeit a very narrow one?) Does current theory say anything about this? (i/e they must be constant with no wiggle room vs. no comment)

I am somewhat familiar with on-shell and off-shell particles- but I must confess I'm not sure if this figures into my question. I'd like to say that as an example I'm talking about the rest mass of the electron as could be determined by charge and charge/mass ratios. But perhaps this very number is itself influenced by off-shell electron masses? Perhaps this makes no sense. Needless to say I'd appreciate any clarification.

share|improve this question

7 Answers 7

up vote 4 down vote accepted

They most certainly are not. You are right that there is no theory that explains masses (these are input as parameters) but note that our current theories used to explain e.g. LHC data (that is, quantum field theories) inevitably come with a scale attached: you need to describe upto what energies you do physics otherwise the theory just doesn't make sense [insert usual story here about renormalization and infinities often told to scare little children before their going to bed].

Now, this shouldn't come as such a surprise since there are new particles awaiting discovery just behind the corner, so claiming that we have a complete theory would be preposterous. Instead, what we claim is that we have a good theory that works upto some scale. Consequently, all of the parameters that are inserted by hand must depend on the scale. Again, this is because theories at different scales are potentially completely different (e.g. at the "present scale" there is no supersymmetry assumed while it is conceivable that at a little higher scale our theories will have to include it) and so the parameters of the theories that are used to connect the theory with experiment potentially have no relation to each other. This phenomenon is known as running of coupling constants or, briefly, the running coupling.

The moral is that all the rest masses and interaction "constants" depend on some scale. They shouldn't be thought as something inherently deep about the nature but just as fitting parameters that describe only effective masses and effective coupling. To illustrate why they are just effective: consider an electron in classical physics. We can measure its charge by usual methods. This value is the long-distance low-energy $e(E \to 0)$ limit of the scale dependent coupling $e(E)$. As you increase the energy and try to probe electron at shorter distances you will find that lots of others electron-positron pairs appear, screening the electron, and the charge that you will measure will be different due to these changed conditions (we talk about the polarization of vacuum).

Just for the sake of completeness: one could say that $E \to 0$ limit is the most important thing about couplings and that we should take that as definition. If so, then these long-distance couplings are indeed constants as one was used in classical physics. But this point of view is worthless in particle physics where people instead try to make $E$ as high as possible to obtain a theory valid at high scales (since this is what they need at LHC).

share|improve this answer
    
Why the downvote? –  Marek Jul 24 '11 at 15:54
    
Initially I too was concerned about the downvote. I read the link and followed up with reading on renormalization. This does not seem do be just a trick of math but rather it is modeling what happens at smaller and smaller distances. In which case I see we view mass as not constant at all but rather a scale related parameter. But is it not the masses at E-->0 that are adjusted during renormalization- and these are true constants? –  jaskey13 Jul 24 '11 at 22:28
    
@jaskey: yes, there are two (interrelated) effects present: that of renormalization (known as subtraction of infinities) and renormalization group (i.e. scale dependence). In the example with electron I mentioned, even at the $E \to 0$ limit we need to use renormalization since there is always screening of charge due to electron's own field and this turns outs to be infinite if computed naively. One uses renormalization to make it finite. But then if we change the scale $E$ we'll find that the (now finite) charge changes again (due to higher-energy scale effects). –  Marek Jul 24 '11 at 22:47

I guess I can think of three possible ways in which masses could be non-constant. (1) They could change due to quantum-mechanical fluctuations, (2) they could be slightly different for different particles at the same time, (3) or they could change over cosmological time intervals.

Number 1 seems to be what you had in mind, but I don't think it works. The standard picture is that for a particle of mass m, its momentum p and mass-energy E can fluctuate, but the fluctuations are always such that $m=\sqrt{E^2-p^2}$ (with c=1) stays the same.

Re #2, here's some good info: Are all electrons identical?

Re #3, one thing to watch out for is that it is impossible, even in principle, to tell whether a unitful fundamental constant is changing. The notion only makes sense when you talk about unitless constants: Duff, http://arxiv.org/abs/hep-th/0208093 However, it certainly does make sense to talk about changes in the unitless ratios of fundamental constants, such as the ratio of two masses or the fine structure constant.

There are claims by Webb et al. J.K. Webb et al., http://arxiv.org/abs/astro-ph/0012539v3 that the fine structure constant has changed over cosmological timescales. Chand et al., Astron. Astrophys. 417: 853, failed to reproduce the result, and IMO it's bogus. I'm not aware of any similar tests for the ratios of masses of fundamental particles. If you change the ratio of masses of the electron and proton, it will change the spectrum of hydrogen, but at least to first order, the change would just be a rescaling of energies, which would be indistinguishable from a tiny change in the Doppler shift.

Brans-Dicke gravity (Physical Review 124 (1961) 925, http://loyno.edu/~brans/ST-history/ ) has a scalar field that can be interpreted as either a local variation in inertia or a local variation in the gravitational constant G. This could in some sense be interpreted as meaning that, e.g., electrons at different locations in spacetime had different masses, but all particles would be affected in the same way, so there would be no effect on ratios of masses -- hence the ambiguity between interpreting it as a variation in inertia or a variation in G. B-D gravity has a unitless constant $\omega$, and the limit $\omega\rightarrow\infty$ corresponds to general relativity. Solar system tests constrain $\omega$ to be at least 40,000, so B-D gravity is basically dead these days.

share|improve this answer
    
Or (most relevantly), they could be a consequence of the renormalization group theory of QFT where it turns out they actually aren't constants at all. This is e.g. the famous unification of forces in GUT theories: coupling parameters of electroweak (this has two components $U(1)$ and $SU(2)$, and consequently is described by two parameters, but these components are not the same as EM and weak parameters we are used to in low-energy physics) and strong force depend on the scale and at GUT scale all three of them meet at the same point. –  Marek Jul 24 '11 at 6:27
    
About point #1. Since we find rest mass as the invariant of the energy/momentum four-vector- we similarly find an invariant length for the position/time four-vector. I see the position-momentum uncertainty relation and then the energy-time uncertainty relation. Somewhere (think it was Griffiths Q&M) I read the energy-time relation is a byproduct of extending the position-momentum relation into special relativity (3-->4 vectors) If this is correct then should there be an uncertainty relationship for the invariant length of a particles four vector and it's rest mass? Or can't one do that? Why? –  jaskey13 Jul 24 '11 at 22:18

if you think about it, any time you perform a measuremeant, you must use a device which is limited by quantum mechanical laws. Any device measuring the weight of a particle must record the weight via some state transition internal to the device... (e.g. the device must change in some way when the particle is placed near it so even if the particle is at rest when the observation is made, the device must have some transition in internal position and momenta to record an observation). So I would say that quantum mechanics does impose a limit on how close we can get to finding the actual rest mass of a particle. That limit is always loosely proportional to plancks constant or plancks constant times 1/2.

share|improve this answer
    
notice that the uncertainty doesn't come from the mass itself, but from the process of measuring, which by definition will involve some particle motion, which will in turn limit the uncertainty to the heisenberg limits. –  Timtam Jul 24 '11 at 5:06
    
Then by your answer you would contend that the rest masses are real constants but forever out of our reach due to our limited measuring process? Perhaps I may be stepping into philosophy here- but how can one say something is real (or constant) if it cannot be verified as such by observation? –  jaskey13 Jul 24 '11 at 22:55
    
I'm not saying it's constant, i'm saying there will always be some uncertainty due to quantum mechanics. –  Timtam Jul 25 '11 at 0:01
    
thus it is impossible to verify whether any observable is "truly" a constant. But to my mind getting the uncertainty to a factor of $10^-34 is good enough approximation that to me, i'll just go ahead and assume it's a constant... what is motivating this question by the way –  Timtam Jul 25 '11 at 0:04
    
I read text books but am not in a university- so when I cannot resolve something I ask it here. I feel that I have over and over again seen the assumption of rest mass as a constant for a given particle but seen no reason as to why this should be so- especially given the nature of the quantum world. I'd like to see theoretical motivation for constancy- but as of yet (and especially considering the array of answers) I see none –  jaskey13 Jul 25 '11 at 1:26

The rest masses of fundamental particles certainly are NOT constants !

I will do copy/past from pages 6-9 (of 20) of a recent document by Alfredo (Independent researcher)
the sufficient information to show how atomic properties can change, including mass, at a cosmological level.
Starting only from data and making no hypothesis he formally presents a dilation(scaling) model of the universe where the atoms are not invariant and physical laws hold, without contradiction with GR, and compares the model with FRW and $\Lambda$CDM models.

(the whole paper deserves your attention, it is very clear and it only uses basic physics, imo accessible to undergraduated students. He makes a full study on how we measure, units, local and field constants and laws, how can exist a variation and why we are not aware of such, and a lot more)

quoting Alfredo:
How the universe can be scaling

We have seen that if space expansion traces a scaling phenomenon, we should expect to detect varying field constants; we have now to find out why that is not observed. The first thing to do is to look up to the dimension functions of field and some other constants:

$$ \begin{array}{ccl} \left[G\right] & = & M^{-1}L^{3}T^{-2}\\ \left[\varepsilon\right] & = & M^{-1}Q^{2}L^{-3}T^{2}\\ \left[c\right] & = & LT^{-1}\\ \left[h\right] & = & ML^{2}T^{-1}\\ \left[\sigma\right] & = & M^{-3}L^{-8}T^{5}. \end{array} $$

The equations of field constants ($G,\varepsilon$ and c) display a peculiar characteristic: the summation of exponents of the dimension function of each field constant is zero! This is unexpected and does not happen with the other constants. It means that if all the four base units concerned change by the same factor,

$$ M=Q=L=T, $$ then the measuring units of field constants hold invariant, $[G]=[\varepsilon]=[c]=1$. To see the relevance of this, let us consider that the atomic units of mass, charge, length and time change all at the same rate in relation to the space units. In that case, because of the property shown above, the atomic units of the field constants hold invariant in relation to the space ones and, therefore, the field constants are invariant in both systems (they are invariant in space units by definition of these ones). The geometry of space would be scaling in atomic units while the value of field constants would hold invariant--- which is exactly what cosmic data seems to display.

The fact that the dimensions of field constants display null summation of exponents can just be a coincidence, but it is also the kind of indication we were looking for, a property embedded in physical laws. This is the only way we can consider a previously unknown fundamental property without conflicting with established physics.

We have now the fundamental understanding that can support a scaling (dilation) model of the universe and we will now proceed to the formal development of that model.
...
Hence, one of the systems of units is defined from matter properties, designated here by atomic system and identified by A ("A" from "atomic") and the other is the space system of units, identified by S ("S" from "space"); the later is such that space properties (geometry and field constants) remain invariant in it, which is required to qualify the S system as internally defined in relation to space. Thus, the conditions that define the S system are the following:
- The units of S are such that the S measures of field constants hold invariant;
- The length unit of S is such that the wavelength of a propagating radiation in vacuum is time invariant.
The base quantities are Mass (M), Charge (Q), Time (T), Length (L) and Temperature ($\theta$), and the ratio between A and S base units is denoted by $M_{AS},Q_{AS},T_{AS},L_{AS},\theta_{AS}$. Note that the ratio between the A and S units of any quantity or constant is therefore expressed by the respective dimension function;
...
Postulates

The model will be deducted not from hypotheses but from relevant observational results, which are stated as postulates:

  1. In atomic units (A), all local and field constants are time-independent.
  2. $L_{AS}\,$decreases with time.

The first postulate is not fully supported in experience, as we cannot state it with the required error margin; however, we have also no sound indication from observations that it might be otherwise. The second postulate represents the observed phenomenon of space expansion in atomic units, stated in this unusual way because it is presented as a function of $L_{AS}\,$, i.e., of the ratio between atomic and space length units and not the inverse, as usual.
...
S units, by definition, are such that (eq.1)

\begin{equation} \frac{dG_{S}}{dt_{S}}=\frac{d\varepsilon_{S}}{dt_{S}}=\frac{dc_{S}}{dt_{S}}=0. \end{equation} Since the field constants are time-invariant also in atomic units, as stated by postulate 1, and since the two systems of units are identical at $t=0$, then the values of these constants are the same in the two systems at whatever time moment: (eq.2) $$ \begin{array}{ccccc} G_{A} & = & G_{S} & = & G\\ \varepsilon_{A}^{\vphantom{l}^{\vphantom{L}}} & = & \varepsilon_{S} & = & \varepsilon\\ c_{A}^{\vphantom{l}^{\vphantom{L}}} & = & c_{S} & = & c. \end{array} $$
The relation between the S and A values of each constant is the one between the respective A units and S units, which is given by the dimension function,
therefore (eq.3)
$$ \begin{array}{ccccl} \dfrac{G_{S}}{G_{A}} & = & \left[G\right]_{AS} & = & {M_{AS}^{-1}}{L_{AS}^{3}}{T_{AS}^{-2}}=1\\ \dfrac{\varepsilon_{S}^{\vphantom{l}^{\vphantom{L}}}}{\varepsilon_{A}} & = & \left[\varepsilon\right]_{AS} & = & {M_{AS}^{-1}}{Q_{AS}^{2}}{L_{AS}^{-3}}{T_{AS}^{2}}=1\\ \dfrac{c_{S}^{\vphantom{l}^{\vphantom{L}}}}{c_{A}} & = & \left[c\right]_{AS} & = & L_{AS}{T_{AS}^{-1}}=1. \end{array} $$ This set of equations implies $M_{AS}=Q_{AS}=T_{AS}=L_{AS}$. By postulate 2, $L_{AS}$ is a time function, therefore the solution can be presented as: (eq.4)
$$ \begin{equation} M_{AS}(t)\,=Q_{AS}(t)\,=T_{AS}(t)\,=L_{AS}(t)\,.\, \end{equation} $$ Note that temperature is independent of this result.

The next step is to define this time function, which is the space scale factor law. As all the above four base quantities follow this function, it is convenient to identify it by a specific designation; in this work this scaling law is identified by the symbol $\mathcal{\alpha}$: (eq.5)
$$ \begin{equation} \alpha(t)\,=\, L_{AS}(t). \end{equation} $$ ...
The scaling law

To make no hypothesis on the cause of the expansion is to consider that expansion is due to a fundamental property; to consider otherwise would imply a specific hypothesis on a particular phenomenon driving the expansion. Therefore, for this model, the space expansion is due to a fundamental property, tracing a self-similar phenomenon. Likewise, as no hypothesis is made on how fundamental properties may vary with position on space and time, it is assumed that they do not depend on it. This implies that the scaling has a constant time rate in some physically relevant system of units, i.e., that the scaling law is exponential in such system of units. There are only two possibilities in the framework established for this model: either space expansion is exponential in A units ($L_{SA}(t_{A})=\alpha^{-1}(t_{A})$ is exponential) or matter evanesces exponentially in S units ($L_{AS}(t_{S})=\alpha(t_{S})$ is exponential). The former case does not fit observations; only the later case is possible.

The general expression for a scaling law exponential in S units is (eq.6) $$ \begin{equation} \alpha(t_{S})=k_{1}e^{k_{2}\cdot t_{s}}\,; \end{equation} $$ at the moment $t_{A}=t_{S}=0$ it is $\alpha(0)=L_{AS}(0)=1,$ so $k_{1}=1$; note now that (eq.7)

$$ \begin{equation} \frac{dt_{S}}{dt_{A}}=T_{AS}=\alpha\, \end{equation} $$ which shows that the variation of the measure of time is inversely proportional to the time unit; and that (eq.8) $$ \begin{equation} r_{A}=r_{S}{L_{AS}^{-1}}=r_{S}\cdot\alpha^{-1}, \end{equation} $$ where r is the distance to some point, or its length coordinate; as the rate of space expansion at t=0 is, by definition, the value of Hubble constant, represented by $H_{0}$, then (eq.9)
$$ \begin{equation} H_{0}=\left(\frac{1}{r_{A}}\frac{dr_{A}}{dt_{A}}\right)_{0}=-k_{2}, \end{equation} $$ therefore (eq.10)
$$ \begin{equation} \alpha(t_{S})=e^{-H_{0}\cdot t_{S}}. \end{equation} $$

Hubble constant is the present space expansion rate for an atomic observer and is the matter evanescence rate (negative) for a space observer.

share|improve this answer
1  
If you decided that masses depend on time, then there is no need to prove that masses depend on time ;-) –  Vladimir Kalitvianski Jul 25 '11 at 9:02
    
@Vladimir I see no logic in your sentence. It's a fallacy. I will try to rephrase it in a logical way. "If you decided to show that masses can depend on time, you must use a referential 'above' the particles, then you need to prove that 'physical laws'(eq.3) are still valid, as the author did using dimensional analysis. –  Helder Velez Jul 25 '11 at 13:45
    
Sorry this is utterly ridiculous! –  Columbia Jul 25 '11 at 14:29
    
@Columbia When I was a little boy I needed 40 steps to cross the road in front of my home. Now I measure it to be 20 steps and we have a consensus, among all that grew with me, that the road is shrinking, backed with careful measures. My special world is made of equal kids and when I was told that our height and weight had changed I reply utterly ridiculous!. Mass/length atom invariance is a hidden claim of BBT (it is unavoidable and it was never proved, and not been explicit makes it worse). Columbia please read the argument carefully and try to find a flaw. It should be utterly simple. –  Helder Velez Jul 25 '11 at 15:20

Marek, you know well that the total wave function of any system is a product of wave functions of independent degrees of freedom (or "particles") of this system. Each particle wave function may be in the ground or excited state and is determined with some specific to it constants, including mass. This fact does not depend on the state of other wave functions in this product so if a heavy particle wave function remains in its ground state (not yet born particle), it does not influence the other constants (masses of already born particles). When we discover a new heavy particle and use its excited wave function, its presence does not influence the previous particle masses. Masses in such a construction are scale and energy independent. For example the photon mass is the same in presence of electron and in absence of it, and vice versa.

The energy in an experiment is not a scale of the theory but a given energy of the system, here somebody fooled you. At a given total energy the particle populations may be quite different. It's an external parameter to the theory; the theory should work at any energy used in calculations. So the "running constants" is not the physical property of particles or nature but a feature of the human failure do describe the excitation processes properly. Remember the useless self-action we introduce in the theory (for the sake of what?). We are obliged to get rid of its effects in the end. As well, we couple coupled already things, we couple them with a wrong understanding of physics and obtain immediately lots of problems. And you promote all that as "physics". -1 for "... all of the parameters that are inserted by hand must depend on the scale.".

share|improve this answer
    
so you are downvoting answers that are based on the standard physics literature, accepted world-wide in the physics community and are fundamental insights that people (e.g. K. Wilson) earned a Nobel prize for? Well, actually, it makes sense that people like you who don't have a slightest clue about physics downvote correct answers. It just makes me little concerned about the future of this site, if more mystifiers like you appear... –  Marek Jul 25 '11 at 7:29
    
@Vladmir while I found your presentation intriguing and will reread it perhaps several more times- I'm afraid I must accept Marek's answer as it is line with current theory as understood and accepted by most physicists. I'm just not at the appropriate level to validate an answer that goes against what seems to be the mainstream (and even some experimental results adsabs.harvard.edu/abs/1997PhRvL..78..424L) I thank you for giving me something to think about. –  jaskey13 Jul 26 '11 at 21:00
    
@jaskey13: I do not expect you to read and accept my point of view but the toy problem I presented in my essay is exactly solvable, so you can see connection between bare constants, physical constants and cut-off contained in $\eta$. You will see that no $\Lambda$-dependent things is involved in the exact calculations so everything cut-of- or scale-dependent is a big science about nothing. As well, you can consider my opus to be mainstream because there is no error in it. –  Vladimir Kalitvianski Jul 26 '11 at 21:08

The rest masses are certain as they correspond to the ground states of QM systems where the energy is certain. It is valid for any system - "elementary" or compound. High temperatures may create the energy uncertainty and it may influence the "measured" mass.

The mass "measurement" can be indirect via precise transitions expressed with theoretical formulas involving mass.

Many current theories fail to simply use the experimental data on masses and charges and it is obviously not a nature feature but a human being imperfection. If you read Marek's answer, you will learn that, as soon as Marek does not know all the particles the nature can produce, he and his theory cannot be wrong. So masses and charges are running for him, they run his errands. And the main errand is to describe the experimental data with a wrong theory. Difficult task but fortunately feasible with help of running constants.

EDIT: For those who does not know that most of "our theories" are non renormalizable, the usual blah-blah about "scale dependence" is not applicable at all (it does not help).

EDIT 2: I see downvoters do not want to leave the rest masses at rest.

share|improve this answer
    
So would you say then that there is no physical meaning to renormalization? Do charged particles remain cloaked in a cloud of virtual copies of themselves and their own anti-particle in that sense or not? –  jaskey13 Jul 24 '11 at 22:41
1  
@jaskey13: renormalization by its purpose is repairing wrong results. Non yet renormalized theory gives too bad results. Who does renormalizations? It's we, not the nature, not the particles themselves. The physical meaning of renormalizations is to get reasonable results from bad ones. This is a human intervention in bad solutions. There is no physics in it at all. And there is no vacuum polarization. Another thing is that sometimes we get reasonable results in this way. It means that another, "renormalized" theory is good and is somewhere near. We just have not found it non perturbatively. –  Vladimir Kalitvianski Jul 24 '11 at 22:53

@Marek: Wilson never dealt with the true QFT; he used QFT methods in his phase transition problems where QFT methods are not applicable by definition, strictly speaking. That is why his QFT is called an "effective" (smoothed) description. Then QFT people, who could not make up their minds for decades what the value for the cut-off to choose in their RG, decided to call their sorry QFTs "effective" too and choose the cut-off according to the energy. This choice was as good as their previous choices where cut-off tended to infinity but such a choice "explained" renormalizations "a la Wilson". It is a sloppiness and weakness of QFT theorists that you promote here. None received Nobel prise for cut-off dependent constants in QFT, do not exaggerate. I downvote when I see you do not understand physics, purpose of physics, and repeat as a parrot what sinful people use as a pretext to hide their failures.

share|improve this answer
    
Could you please cite something for me to consider why renormalization is so incorrect and any efforts to resolve these problems without it? –  jaskey13 Jul 25 '11 at 11:59
    
@jaskey: Yes, I can. All QED fathers considered renormalizations as non legitimate mathematically prescriptions. P. Dirac insisted in searching better Hamiltonians. R. Feynman invited to guess better equations. L.D. Landau tried hard to arrive at a better theory but failed (Hamiltonian is dead), etc. As to accessible explanation, read my presentation here: docs.google.com/…. –  Vladimir Kalitvianski Jul 25 '11 at 12:10
    
It's a PowerPoint document with comments below slides. It is quite self-consistent. I show how we can advance a wrong coupling and why renormalizations may luckily work. It is renormalization practicers who are happy with renormalizations, not the theory developers. –  Vladimir Kalitvianski Jul 25 '11 at 12:13
    
I don't see the link? Also wasn't Dirac the initial "renormalizer" by filling his sea of electron-holes with electrons? –  jaskey13 Jul 25 '11 at 12:17
    
The link is docs.google.com/… As to P. Dirac, he used renormalizations, as Lorentz, Abraham, and many others before him but he never was satisfied with it. You know, people write equations for physical particles with physical parameters in mind. If the calculations give bad results - it is the theory who is wrong. "Repairing" those bad solutions is repairing that bad theory. So, the logic of QFT fathers was right - we have to find a better (renormalized exactly) theory to obtain the right results directly. –  Vladimir Kalitvianski Jul 25 '11 at 12:27

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.