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According to the Newton mechanics, the force is responsible for changing the body velocity, and the body mass is the body inertia - a property to resist to the applied force. These two things make a clear sense in the Newton equation (Second Law) $ma=F$. Both "parameters" (the body mass and the force strength) are observable and measurable.

Now I am reading an article of G. 't Hooft where he states that "The interactions among particles have the effect of modifying masses and coupling strengths". So not only velocities are modified with forces but the particles masses and the forces themselves.

I see here a contradiction with the Newton's definition of force and mass. As soon as the QFT is more fundamental than Classical Mechanics, does it mean that the 't Hooft's law is stronger than the Newton's ones? What is the impact (effect) of this stronger law on Classical Mechanics? How to define now the masses and forces if they are self-modifiable?

EDIT: OK, is there a classical physics example to demonstrate that a potential interaction can modify mass and charge?

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    $\begingroup$ I agree with the sentiment of @Rasko's comment. However, I don't think it's appropriate to refer to running masses and couplings as "t'Hooft's law". Perhaps I'm being pedandic, but that's how I feel. $\endgroup$
    – qftme
    Jun 14, 2011 at 11:46
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    $\begingroup$ @Vladimir: I don't believe that. You are familiar with quantum theory and running coupling (possibly for decades), based on your other contributions, and yet you pretend to be surprised in this question as if you heard about these well-known and trivial facts for the first time in your life. Sorry, but I don't buy it. $\endgroup$
    – Marek
    Jun 15, 2011 at 12:07
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    $\begingroup$ @Vladimir: I am not going to post an answer because it is already well-known that you dislike correct answers and only want to hear what you decided to hear. The solution here is as follows: Newton didn't bother to think about what it means to interact with the object you measure. Indeed, this is not needed in classical physics because objects are "big". In quantum physics, one runs into all kinds of paradoxes if they forget to contemplate properly on the act of measurement. If you, on the other hand, do think about it, you inevitably arrive at concepts like running couplings. $\endgroup$
    – Marek
    Jun 15, 2011 at 13:28
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    $\begingroup$ (cont.) in particular: how do you measure that mass constant you celebrate so much? If you think about it deeply enough you should realize that it's impossible to measure particle's properties without interacting with it. So the results you get will never be intrinsic to the particle but always depend on the way you interact with the particle. In particular, the results will depend on the center of mass energy of the interaction. So you might as well accept that there's no intrinsic mass constant independent of every measurement... $\endgroup$
    – Marek
    Jun 15, 2011 at 13:34
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    $\begingroup$ @marek come on you lazy genius, post an answer and save us from our ignorance ;) $\endgroup$
    – Larry Harson
    Jun 28, 2011 at 16:32

6 Answers 6

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Hydrodynamics is a whole classical world when you can find most of the effects you ask about; from Archimedes change of weight to renormalisation of turbulent fluid. Were you born in the XIXth century, you would be militant against Hydrodynamics I guess :-)

If you are asking for examples without fields, with only a finite number of degrees of freedom, you are reducing the scope a lot. But generically, the point is that you get some parameter to express departure from Newton's law. Typically, c and h. You need a finite c, besides a non zero h. Then you use many body mechanics to approach to a situation that happens to be near a finite model with small h and very big c, and this we call the classical approximation.

The idea of setting h=c=1 is computationally very practical, but a lot of people gets into problems because of it. For instance, fermionic lagrangians have really a $h$ somewhere multiplying them, so in the classical limit they dissapear; this is a reason to have classical bosonic fields only.

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    $\begingroup$ No, I would not be against Hydrodynamics, of course. To tell the truth, I am doing Hydrodynamics right now. Everything there is calculated, not "fitted". And as soon as it is a calculation based on physical laws, I accept it. I do not accept changing definitions on the go, bla-bla and cheating. To your information, I can get a "renormalization" problem in the regular classical mechanical problem of two interacting bodies, so $\hbar$ and $c$ are not an essential part of the problem. $\endgroup$
    – Vladimir Kalitvianski
    Jun 24, 2011 at 23:42
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Here comes an experimentalist's view:

I accept as premise that four vectors and special relativity exist and are well documented to hold for elementary particles.

I accept that the same holds true for the Heisenberg uncertainty principle.

In special relativity mass is the "scalar length" of the four vector. When one has a specific particle, a photon, one knows the mass is 0. If one has two photons, their combined mass is not additive, it follows four vector algebra and will depend on the angles . The same will be true for two electrons, their combined mass is not additive but forms their invariant mass.

Thus the meaning of mass is different between Newtonian physics and relativistic physics, since in Newtonian masses are additive in relativistic they are not.

Now we have the Heisenberg uncertainty principle and within a specific region delta(p)*delta(x)>h, momentum is not well defined nor position, thus an experiment to get at the force, F=dp/dt will be indeterminate to that extent, and even for a single particle it will measure a different mass depending on the x within the HP. More so when there is an interaction of two particles, and I cannot think of an experiment where a second particle will not be involved in the measurement, where the combination of the indeterminacy of momentum and the vector addition will combine to even higher uncertainty of the combined mass measurement.

This is my simple minded answer without going into QED and virtual photons etc., where still four vector algebra will hold. There is a limit to the range in which Newtonian physics is valid. It is not possible to mix classical and relativistic and quantum mechanical concepts of mass, so there can be no example.

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  • $\begingroup$ @anna v: Thanks, Anna, for your contribution. I do not want to argue about what you said but all you wrote does not cancel the notion of mass. Masses of compound systems are calculated from masses of involved particles and their interactions, so there is no any change of particle masses due to their interactions. There is no one physical phenomenon to make "not surprising" "changes" in masses and charges of involved particles, so I do not trust 't Hooft's words. Look, for example, at the Debye screening in plasma - it is a physical and completely calculable effect. No constant is modified. $\endgroup$
    – Vladimir Kalitvianski
    Jun 22, 2011 at 14:31
  • $\begingroup$ I amnot trying to cancel the concept of mass, I am trying to say that mass is a different animal in classical physics and in quantum relativistic physics. When the HP is ignorable and velocities are low the concepts coincide. When we talk of elementary particles the HP is ever present due to the dimensions of the thingies. $\endgroup$
    – anna v
    Jun 22, 2011 at 15:54
  • $\begingroup$ @anna: The fact that the particle energy and momentum is calculated with different formulas does not touch the meaning of mass but the energy and momentum themselves. When we say an interaction changes the particle velocity, we mean $dV/dt = F/m$. We never write $dm/dt$ or $de/dt$ due to force. $\endgroup$
    – Vladimir Kalitvianski
    Jun 22, 2011 at 16:12
  • $\begingroup$ Maybe I mean Vladimir that the concept of force is not applicable at the level of particle physics except as dp/dt It is momentum that enters the equations that carry any meaning at the micro level.Interactions on elementary particles change momentum and energy , the four vector. $\endgroup$
    – anna v
    Jun 22, 2011 at 18:25
  • $\begingroup$ @anna: I completely agree. The mass is not a dynamical variable. $\endgroup$
    – Vladimir Kalitvianski
    Jun 22, 2011 at 18:37
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When a mass is accelerated unruh radiation is produced. An accelerated particle moves in rindler space so boundary conditions due to the presence of an horizon causes the instability of the vacuum that produces a thermal atmosphere of particles that radiates. Also the accelerating particle produce gravitons. The very nature of inertia, and therefore mass, is veiled in mystery. The Higgs fiels is atributed with the property that it gives mass to particles. The Higgs mechanism is nothing that a well-defined mathematical procedure to give mass to Yang Mills fields while preserving renormalizability. But what is the origin of the Higgs field? Is it gravitational? Is it an artificial construction? If the Unruh radiation scenario is correct when you accelerate a mass you have to subs tract the back reaction due to the emission of gravitons and unruh rad. In response to your answer, I would say that Newtons Law is actually and approximate law of nature that cannot be used rigorously to define mass anymore. The problem is that QFT, String Theory, or M-Theory cannot be used either because there are not universally accepted. The nature of the Higgs boson is unknown, and unruh radiation or gravitons have not been detected. In this sense, the meaning of mass is still an open question.

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  • $\begingroup$ It is not what I wanted to hear. In fact, it is people who changes masses and charges in course of the theory "development" (with coupling again coupled already things). $\endgroup$
    – Vladimir Kalitvianski
    Jun 28, 2011 at 15:05
  • $\begingroup$ I'm afraid that you comment do not make any sense. To provide you with a classical example, let's say what about radiation back reaction. It theoretically lowers the acceleration of matter by a very small amount, effectively making mass a function of acceleration. By the way, physics is a human creation, so is all about peaple. $\endgroup$
    – Ernesto Ulloa
    Jul 1, 2011 at 13:00
  • $\begingroup$ Yes, it is about people, not about interaction. $\endgroup$
    – Vladimir Kalitvianski
    Jul 1, 2011 at 15:57
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The proper force on a particle is clearly defined and independent of its motion in any frame. Its rest mass can be calculated from $F = ma$ in its proper frame. There is no important contradiction to Newton's law of inertia when used this way.

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Consider a binary pulsar with electrostatic charge 1 Coulomb in a constant electric field. As the two stars draw inwards, the potential energy of their interaction decreases their mass. So here is a modification of mass in classical mechanics.

For a modification of force, consider a charged ball falling through air, and going into pure glycerin. The dielectric constant of the glycerin modifies the force.

These examples are ubiquitous, and I don't understand the purpose of this question: Newton's laws are 300 years old, of course 'tHooft's stuff will supersede them, they have been superseded many times already.

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  • $\begingroup$ Ron, but your examples are calculations! I am not against calculations, on the contrary! All resulting masses, forces, etc are calculated from known ones. It is not a renormalization at all. $\endgroup$
    – Vladimir Kalitvianski
    Aug 27, 2011 at 7:39
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    $\begingroup$ You asked for a classical example where interactions modify mass and force. These two examples modify mass and force. These are physical effects, not mathematical calculations. They are in every way analogous to the processes involved in defining a quantum field theory. $\endgroup$
    – Ron Maimon
    Aug 27, 2011 at 9:58
  • $\begingroup$ No modification of masses of constituents happens. It is the total, calculated mass that depends on interactions and it is calculable. Nobody call it a mass renormalization. You mistake regular calculations with renormalizations. They are different by nature and called differently. $\endgroup$
    – Vladimir Kalitvianski
    Aug 27, 2011 at 11:03
  • $\begingroup$ @Ron , Vladimirs purpose with the question is to spread propaganda about his misunderstandings about renormalization. $\endgroup$
    – Heidar
    Aug 27, 2011 at 13:39
  • $\begingroup$ @Vladimir , as I have told you many times before, you are thinking about renormalization in the wrong way. As a FIRST STEP, understand renormalization group calculations in simple models such as the Ising model (scalar field theory at the continuum limit). NOTHING is put in by hand and EVERY modification of masses and coupling constants are CALCULABLE with no problem! $\endgroup$
    – Heidar
    Aug 27, 2011 at 13:39
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I gave a popular explanation of "how our interaction modifies the fundamental constants" here and here.

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  • $\begingroup$ Its amazing to see that you still have no clue about the conceptional and physical aspects of renormalization. I advice you to take a simple condensed matter lattice model and take a continuum limit (will have a natural cut-off). Understand how renormalization group theory works in this model and what it physically means (the constants will run, but the theory is the same and nothing is put in by hand in this case)! Then assume you don't know the continuum model is a low-energy approximation of the lattice model and remove the cut-off, then suddenly infinities appear.Think aboutwhat this means $\endgroup$
    – Heidar
    Aug 21, 2011 at 1:15
  • $\begingroup$ @4tnemele It's amazing to see that you still have no clue about what we do in reality. Well, QFT is hard but I managed to reduce its heaviness and present the essence of the renormalization problem in simple terms. I advise you to follow it first before judging what I know and what I do not understand. $\endgroup$
    – Vladimir Kalitvianski
    Aug 21, 2011 at 8:13
  • $\begingroup$ As long as your criticism displays fundamental misunderstanding about renormalization group theory, I'm not interested in alternatives. What you say was valid in the late 1960's, but since then a deeper understanding of renormalization has emerged. Have you studied RG theory in the study of, say, phase transition in the Ising model? Did you understand the physical meaning of RG in that case? $\endgroup$
    – Heidar
    Aug 21, 2011 at 13:46
  • $\begingroup$ @4tnemele: yes and I know what I say. How can you speak of a "deeper understanding" if you cannot compare mine with a la Wilson's? To facilitate understanding, I designed a realistic but absolutely comprehensible model. You just need to trust in its value and then check everything by yourself. It is not about effective approach being approximate, believe me. Approximate description is approximate by definition, who argues with it? In a la Wilson's there is nothing in common with coupling equations in QED, with conceptual errors and their "fixing". $\endgroup$
    – Vladimir Kalitvianski
    Aug 21, 2011 at 14:16

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