Mass of higgs boson is given by Higgs Field.

a. Mass is a scalar quantity, not a vector quantity
b. Higgs boson is a scalar boson (spin equals zero), not a gauge-vector boson.
c. Higgs Field is a scalar field, not a vector field

We know that scalars, for definition, form a field structure (usually we talk about vector space over rings)
But is true that is possible to restrict scalars to form a ring structure (usually we talk about module over ring)

If mass for Higgs boson is given by Higgs field that is a scalar field, what happens if restrict scalars to form a ring and not a field ?

  • 3
    $\begingroup$ Keep in mind that the term "field" as used in math has nothing to do with the term "field" as used in physics. $\endgroup$ Jul 5 '19 at 4:46
  • $\begingroup$ I'm not sure what you say, for example electric (physics) field is conservative vector (mathematics) field. Or a physics action is a mathematical functional which takes the trajectory, also called path or history, of the system as its argument and has a real number as its result. $\endgroup$
    – Tag
    Jul 5 '19 at 4:52
  • $\begingroup$ This is one of many possible modifications to the mathematics behind the physics. Is there some reason you asked about this particular option? The reason for the question may help us to answer it. $\endgroup$ Jul 5 '19 at 9:34

A mathematical field has a well-defined division (or inverse) operation, a ring does not. So if mass was a ring scalar it would not make sense to divide by mass. While it might be possible to get around that in some formulas, the non-uniqueness would make a formula like $F=ma$ only imply force from mass and acceleration, not imply the mass if you knew the force and acceleration.

  • $\begingroup$ mm...when you say "not make sense to divide by mass" what exactly do you mean, it's like dividing by 0? What do you mean exactly "not imply"? if you knew the force and acceleration what are the physical consequences of excluding the possibility of obtaining the mass from the formula $F=ma$ if you must first have it as a value if you get force from the formula $F=ma$? I am used, when I have a field, to always be able to get 3 quantities, a single triple {F, m, a} but narrowing to a ring, I can have only 2 possible pairs {F, m} or {F, a}. Physical consequences of this thing? $\endgroup$
    – Tag
    Jul 5 '19 at 13:22
  • $\begingroup$ I thought and I think it might not be entirely true when you say if mass was a ring scalar it would not make sense to divide by mass because there is another way to gest solution Formal construction. I believe we must first understand the meaning of having a field $\endgroup$
    – Tag
    Jul 5 '19 at 14:01
  • $\begingroup$ $m^{-1}$ is not defined in a ring since the multiplicative part is just a monoid rather than a group, so for example the formula $a=F/m=Fm^{-1}$ is not defined. Does that have physical effects? Well, it certainly leaves the theory unable to predict what happens when you apply a force to an object. (Rational numbers form a field, so they are not relevant here) $\endgroup$ Jul 5 '19 at 14:11
  • $\begingroup$ If leaves the theory unable to predict what happens when you apply a force to an object is then impossible to make a measure in this sense ? $\endgroup$
    – Tag
    Jul 5 '19 at 16:20
  • $\begingroup$ @Tag - I am not certain what you mean. Mathematical measure and physical measurements are very different things. $\endgroup$ Jul 5 '19 at 20:26

This question is surely somehow a product of not understanding the difference between a field in the sense of physics, and a field in the sense of abstract algebra. If the possible values of a physical field are drawn from some algebraic field (e.g. real or complex numbers), that is just a coincidence of names and has nothing to do with why the physical concept is called a 'field'.

Anyway, although the question focuses on the Higgs boson mass for some reason, I believe a more general question would be better:

"What are the consequences for quantum field theory, if the fields are ring-valued rather than field-valued?"

Though to answer this, perhaps one should first be able to answer an even more straightforward question:

"What properties of quantum field theory depend on the fields being field-valued?"

I don't have answers for any of these questions. They place QFT in a context which is unusually and perhaps even inappropriately generalized. The small generalization to quaternionic QFT is already exotic (but it has been studied); to ask about 'field-valued quantum fields' in general, is to ask about something that very few people would have even considered.

As for 'ring-valued quantum fields', the integers are a ring, and observables with an integer spectrum are commonplace in quantum mechanics; but perhaps the equation of motion would have to be a difference equation rather than a differential equation. In any case, integers are just one example of a ring.

To sum up, such a question appears to require a comparison between two classes of theoretical object that have never been studied. Perhaps some mathematician will take up the challenge. Or perhaps the ring-vs-field distinction can be used to draw some quick clever implication (e.g. with respect to the usual expectation that quantum operators will be hermitian or self-adjoint?).

  • $\begingroup$ thank you for answer, I hope that some mathematician will take up the challenge $\endgroup$
    – Tag
    Jul 8 '19 at 11:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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