Of Particular Significance

1. The Basic Idea (1st version)

© Matt Strassler [September 22, 2012]

This is article 1 in the sequence entitled How the Higgs Field Works: with Math.

If you have read my series of articles on Particles and Fields (with math), you know that all the elementary “particles” are really quanta (i.e. waves whose amplitude and energy are the minimum allowed by quantum mechanics) in relativistic quantum fields. Such fields typically satisfy Class 1 equations of motion (or generalizations thereof, as we’ll see) of the form

  • d2Z/dt2 – c2 d2Z/dx2 = – (2 π νmin)2 (Z-Z0)

where Z(x,t) is the field, Z0 is its equilibrium value, x is space, t is time, d2Z/dt2 represents the change with time of the change with time of Z (and d2Z/dx2 similarly for space), c is the universal speed limit (often called “the speed of light”), and the quantity νmin is the minimum frequency allowed for waves in this field.  (For a review of these equations, see this article.) A few fields satisfy Class 0 equations, which are just Class 1 equations where the quantity νmin is zero. The quanta of such fields have mass

  • m = h νmin / c2

where h is Planck’s quantum mechanics constant. In other words,

  • d2Z/dt2 – c2 d2Z/dx2 = – (2π c2/h)2  m2 (Z-Z0)

Now all of this is only true to a point.  Really, if all that fields did was satisfy Class 0 and Class 1 equations, nothing would ever happen in the universe. Their quanta would just pass by each other and they wouldn’t do anything… no scattering, no smashing, no formation of interesting things like protons or atoms. So let’s put in a modification that is common, interesting, and required by what we know about nature from experiment. We’ll explore it more later.

Let’s think for a moment about two fields, S(x,t) and Z(x,t).  Imagine that the equations of motion for S(x,t) and Z(x,t) are modified versions of the Class 1 equation, where the equilibrium values S0 and  Z0 are zero, but there’s an additional term in the equations involving S(x,t) multiplied by Z(x,t)

  • d2S/dt2 – c2 d2S/dx2 = – (2π c2/h)2 (mS2 S + y2 S Z2)
  • d2Z/dt2 – c2 d2Z/dx2 = – (2π c2/h)2 (mZ2 Z + y2 S2 Z)

There’s a little shorthand here, so read carefully.  S and Z are shorthand for the fields S(x,t) and Z(x,t), which can vary over space and time.  Everything else (c, h, y, mS, mZ) is a constant.  The parameter y is a number, typically between 0 and 1, and called a “Yukawa parameter” or “Yukawa coupling”, for historical reasons. We’ll see how it comes into the story in a few moments.

In almost all circumstances in particle physics, the deviations of fields like S(x,t) and Z(x,t) from their equilibrium values S0 and Z0 are extremely small. Since we’re currently assuming S0=0 and Z0=0, this means S and Z themselves are extremely small: any waves in S and Z typically have small amplitude (they are typically made from a single quantum) and although there are always spontaneous quantum disturbances going on (often referred to as virtual particles, and discussed in the Particles and Fields articles as a sort of quantum jitter) these are also rather small in amplitude (though sometimes  big in importance.) And if S is small and Z is small, then S times Z is really small. [Consider two numbers: if a = 0.01, and b = 0.03, then a times b is 0.0003 — really small, much smaller than either a or b.] Since y isn’t big, the terms y2 S Z2 and  y2 S2 Z are small enough to ignore under many circumstances.

Specifically, we can ignore them in figuring out the mass of the S and Z “particles” (i.e., quanta).  To figure out what an S particle is like, we need to consider a wave in S(x,t), with Z(x,t) assumed to be very small.  To figure out what an Z particle is like, we need to consider a wave in Z(x,t), with S(x,t) assumed very small.    Once we ignore the extra y2 S Z2 and  y2 S2 Z terms, the S and Z fields then both satisfy the simple Class 1 equations of motion we started with, from which we deduce that the S and Z “particles” have mass mS and mZ.

But now imagine a world in which Z0 is zero but S0 is not zero.

  • d2S/dt2 – c2 d2S/dx2 = – (2π c2/h)2 (mS2 [S- S0] + y2 S Z2)
  • d2Z/dt2 – c2 d2Z/dx2 = – (2π c2/h)2 (mZ2 Z + y2 S2 Z)

Again, the S and Z fields are functions of space and time, but everything else, including S0, is a constant.  In this case Z(x,t) itself very small, but S(x,t) is not! Instead it is useful to write

  • S(x,t) = S0 + s(x,t)

where s is the variation of S away from its equilibrium value S0. We can say that s(x,t) is a shifted version of the S(x,t) field. The statement that fields in particle physics stay very near their equilibrium values most of the time is the statement that s(x,t) is very small, and not that S(x,t) is small. Substituting the red equation above into the equations above for S and Z, and remembering that S0 is constant so dS0/dt = 0 and dS0/dx=0, we find the equations become

  • d2s/dt2 – c2 d2s/dx2 = – (2π c2/h)2 (mS2 s + y2 [S0+s] Z2)
  • d2Z/dt2 – c2 d2Z/dx2 = – (2π c2/h)2 (mZ2 Z + y2 [S0+s]2 Z)

As before, if we want to know the mass of the quanta of the S and Z fields, we can drop any term in these equations that involves a product of two or more small fields — terms like Z2 or sZ2 or sZ or s2Z. But we cannot ignore the term y2 [S0+s]2 Z in the second equation entirely, because it contains a term of the form y2 S02 Z, and that contains only one field.  (Note this is not true of the y2 [S0+s] Z2 term in the first equation, every term of which has two factors of Z.) Consequently, although a quantum of the S field still satisfies the equation

  • d2s/dt2 – c2 d2s/dx2 = – (2π c2/h)2 mS2 s

and therefore has mass mS, a quantum of the Z field satisfies a new equation:

  • d2Z/dt2 – c2 d2Z/dx2 = – (2π c2/h)2 [mZ2 + y2 S02] Z

Consequently the quanta of the Z field now have a new mass! not mZ but

  • mZnew = [mZ2 + y2 S02]1/2

(recall that x1/2 means the same as √x.) Because of the simple interaction between the S and Z fields with strength y, when S has a non-zero equilibrium value S0, the mass of the Z quantum shifts by an amount related to y and S0.

And if for some reason the mass mZ of the Z quantum had been zero to start with, then we would say that the S field’s non-zero value has given mass to the Z quantum:

  • mZold = 0    →    mZnew = y S0

Note this mass is proportional both to the equilibrium value S0 and to the Yukawa parameter y; the larger is y, the larger is the resulting mass for the Z quantum.

Well, this is basically how the Higgs field H(x,t) gives mass to particles. It turns out that for each known particle σ [except the Higgs itself], the equation of motion for its corresponding field Σ(x,t) is a Class 0 equation, which naively would imply the σ particle is massless. But for many of these fields there are extra terms in the equation of motion, including a term of the form

  • yσ2 [H(x,t)]2 Σ(x,t) ,

where yσ is a Yukawa parameter, different for each field, that represents the strength of the interaction between the H field and the Σ field. In such a circumstance, a non-zero average value for the Higgs field, H(x,t) = H0, shifts the minimum-frequency of Σ waves, and thus the mass of σ particles, from zero to something non-zero: mσ = yσ H0. Diversity among the Yukawa parameters for the various fields of nature leads to the diversity of masses among the “particles” (more precisely, the “quanta”) of nature.

Notice, by the way, that the Higgs particle has nothing to do with this.  A Higgs particle is a quantum of the Higgs field — a ripple of minimum energy in H(x,t), a wave that depends on space and time.  What gives mass to the other known particles of nature is the non-zero equilibrium constant value for the Higgs field, H(x,t) = H0, all across the universe; this timeless and universally present constant is very different from Higgs particles, which are ripples that vary over space and time, and are both localized and ephemeral.

That’s the basic idea. I’ve left lots of obvious questions unanswered here: why should we expect there to be terms in the equations that involve the product of two or more fields? Why would the known particles be massless if there were no Higgs field? Why is the Higgs field’s equilibrium value non-zero while this is not true of most other fields? How does the Higgs particle enter the story? The ensuing articles will try to address these and other issues.

34 Responses

  1. The Higgs Field average value is ~246 GeV, the W mass is ~80 GeV, so the Yukawa factor for the H-W interation is 80/246 =~ 0.325? Did I get this right?
    So, If the mass contribution comes from the Higgs only, the highest possible mass is 246 GeV?

    1. And why the highest possible value for the Yuwaka factor is 1? What prevents it from being, like, 2?

  2. Just curious about the small terms in the equations that are ignored. How would the classical solution to the coupled field equations behave if the small terms were not ignored?

    If initially I had a vibration in just one field, would the small interaction terms cause the other field to start vibrating? What would the result be over a long period of time.

  3. Vasilis: Your knowledge of physics is way beyond me, I am still trying to understand why physicists observe waves but call waves particles. When you wrote “How can it be big if it has no mass yet (mass is obtained after the interaction with the field)” I thought you were puzzled with the same issue.

  4. Hi Zeynel 🙂 . Thanx a lot for your response. Thanx also for your link, that was great too. I just plugged some numbers in the m = hV/c^2 formula, specifically the mass of the electron, and solved for V (minimum frequency). The result I got for the minimum frequency for an electron is about 12,37×10^10GHertz. This is actually near the soft gamma rays part of the spectrum. I also found that photons at that frequencies turn into electron-positron pairs (through a process called photon to photon physics or gamma – gamma physics). At higher frequencies at the hard gamma rays end they turn into quark – antiquark pairs. Should it be fair to assume that the Higgs field “transforms” very high-energy photons into particles with mass? So lower energy photons (everyday electromagnetic waves) because of their “large” wavelength do not interact with the Higgs field thus they remain massless? Obviously the neutrinos (Mass <0,13eV) get their masses (if they have any) indirectly. I guess my train of thought must be over simplified or probably I’m talking nonsense :- ).

  5. Vasilis, I don’t know what Prof. Strassler’s answer will be, but his first sentence in this post may be of help: “all the elementary “particles” are really quanta (i.e. waves whose amplitude and energy are the minimum allowed by quantum mechanics) in relativistic quantum fields”. So there are no particles only waves. You may also check first few slides in this presentation: .http://www.physics.rutgers.edu/~strassler/QuestHiggs.pdf

  6. Hello Professor. Thanks for the amazing articles. I’m not a physicist so my question might be very naive. I hope I won’t bother you too much. It is said that particles gain their masses through their interaction with the Higgs field. The “bigger” the particle the greater the interaction with the field, the larger the mass obtained by the particle. What I don’t understand is, what it meant by the “bigger” the particle. How can it be big if it has no mass yet (mass is obtained after the interaction with the field). Does it mean that it has larger energy (through E=mc2)? If that is the explanation, then by larger energy does it means higher frequency (through E=hω)?
    Thanks a lot for your time. I hope I didn’t tired you too much.

  7. “Yes. For both the W and Z bosons and the various fermions, there are subtleties that I haven’t yet addressed — these are the ones that make it clear that not only CAN the Higgs particle provide masses for the various particles, it MUST.”

    I’m confused by this statement. I thought the Higgs particle had nothing to do with providing mass; rather it was the Higgs field that provided the mass. Is this a special case you are referring to or am I missing something?
    Don Murphy

  8. This article is really helpful, thank you so much for that!
    I’m just not sure how to link this mathematical explanation to the one you gave in the text “The Known Particles — If The Higgs Field Were Zero”. There your description was that the non-zero Higgs field value is causing mixing of the left-chiral and right-chiral massless fermions and the result is a single massive fermion. I’m looking at the equation, trying to find the 2 massless quanta, however it seems there is just 1 fermion field whose equation of motion is affected by the Higgs field. Could you please shed a bit of light on that?

    1. Good point. The explanation I have given you in the current article is not yet sophisticated enough to match on to the one given in “The Known Particles — If The Higgs Field Were Zero”. I hope to return to that more subtle issue in one of the final articles in this series.

    1. How you define things, both in language and in physics, is always a bit subtle: which things are defined in terms of which others?

      Mass is best defined through the equation E^2 = p^2 c^2 + m^2 c^4; that is a fact about all objects in Einstein’s special relativity. [Here E is energy, p is momentum, c is the speed of light and m is mass.]

      It is a fact that for quanta, m and νmin are related by the equation m = h νmin / c2.

  9. “For example, the Higgs field is not a source of gravity”

    You are correct. What is mistaken for the Higgs field is the aether, which has mass.

    Matter is condensations of ather.

    Particles of matter exist in and displace the aether.

    Displaced aether pushing back and exerting inward pressure toward matter is gravity.

    1. In your previous comment you said “The Higgs Field has mass”; now you contradict yourself. I’m sorry, but even a novice reader can see the pointless and meaningless nature of your remarks. I cannot allow continued cluttering of my site with absurdities, so you’re banned for now.

  10. What is presently referred to as the Higg’s field is the aether.

    Aether has mass and physically occupies three dimensional space.

    [Abridged by host]

    1. This is incorrect on so many fronts I don’t know where to start. For one thing, a uniform Higgs field does not have mass; mass is something that particles can have, but spatially uniform and static fields, on completely basic grounds, do not.

      Please do not make absurdly incorrect assertions about physics on my website. Do it on your own.

      1. ‘What fills space? – Craig Hogan, Director of the Fermilab Center for Particle Astrophysics’

        “For example, the Higgs field is much weirder than the comparisons with molasses or crowds suggest, since it does not actually drag or impede particles, but still somehow shares its mass with them.”

        “the Higgs field … shares its mass with [particles of matter]”

        The Higgs field has mass.

        Aether has mass.

        Matter is condensations of aether.

        1. Thank you for this wonderful comment! It allows me to illustrate so many things at once!

          I know Craig Hogan very well. Before he moved to Fermilab and I moved to Rutgers, he and I were colleagues and friends for five years at the University of Washington. We had lunch about once a week as part of a weekly meeting of astronomers, cosmologists and particle physicists. I cannot possibly count the number of times we had scientific conversations about a wide range of subjects.

          Of course, as at any similar university, the University of Washington had courses on particle physics, including the Higgs boson, the Large Hadron Collider, and the like. Now, who do you think taught those courses? Professor Hogan? or Professor Strassler? Right — I did. I dare say that I, and my other particle physics colleagues, did an ok job, since a few of the students in those courses are now postdocs and professors at major universities. Whereas I would defer to Professor Hogan’s expertise on many subjects — astrophysics and cosmology, in particular — I am confident Professor Hogan would defer to mine in this subject. Lesson: If you want to learn a subject correctly, learn it from the experts.

          Still uncertain about this? Three weeks after the Higgs boson discovery was announced there was a conference on the Higgs held in Orsay, France. Who was asked to give the talk on the theoretical implications of the experimental results? http://indico2.lal.in2p3.fr/indico/conferenceDisplay.py?confId=1747

          In this case, Professor Hogan’s remark that you’ve quoted is an attempt to explain things qualitatively to the public, using words. He is correct that the usual analogies are bad. Unfortunately, he has replaced them with another bad one. What is certain is that if you don’t use words but instead you use math — which in physics is what you must do first — you will see that mass is a precisely defined quantity, and the Higgs field, when it is just a constant non-zero value across space, has none. For example, the Higgs field is not a source of gravity, whereas mass (because it is always associated with energy) always is.

          The reason I have written these articles is that there is no good analogy for the Higgs field, and no substitute for the real equations… simplified a bit, to be sure, to make them comprehensible to a wider public, but the real deal.

      2. Matt Strassler wrote: “What is certain is that if you don’t use words but instead you use math — which in physics is what you must do first — you will see that mass is a precisely defined quantity…”

        In fact, in this case, words are more precise than your “mathematical” expressions.

        This is your mathematical expression: m=(h/c2) v_min.

        You claim that m=(h/c2) v_min is a precise mathematical expression. What does precise mean? Precise means that m=(h/c2) v_min can only be read one way, it has one single and unambiguous meaning. But this is not true for your “mathematical” expression.

        You can read m=(h/c2) v_min at least three different ways: as a definition (that is, the letter m is defined as (h/c2) v_min and now you can use m and (h/c2) v_min interchangeably) or you can read it as “m is proportional to v_min” or you can read it as an identity or as an equation.

        In other words, your expression is not “mathematical” but casuistical. Casuistry simply means that your expression must be interpreted case by case depending on the context.

        But if I wrote, in words, “Prof. Strasssler defines mass as minimum frequency times appropriate unit terms” there is no ambiguity. You and I and all your readers will read and understand the same thing.

        Therefore, a sentence formed with well-defined words is more precise and mathematical than any physics equation.

        If this sounds strange to you, consider that Newton never used an equation in his life, all of his mathematical statements are stated in words and in proportionalities, yet his work is more precisely mathematical than any physics equation. Therefore, your hidden assumption that “mathematics” means “physics equation” is not justified and it is misleading.

        The problem is with your equality sign. Equality sign in physics is loaded like a crooked balance. Galileo introduced geometry to the study of motion to free it from scholastic sophistry of Learned Doctors; but Learned Doctors came back with a vengeance, they eliminated geometry and invented the physics equation and turned mathematics into their Latin and use it to philosophize with “mathmatics.” If you think this is an outrageous statement, consider your sentence: “The real world is quantum mechanical”. Any sentence which starts “The real world is….” is a commentary on the oldest known philosophical subject: reality.

        This is a huge problem in physics, I hope that you take it seriously.

        1. I’m sorry, but I don’t have time for this. There is merit in your points, but it is completely irrelevant to the topic of this article.

          If you don’t find my explanations satisfactory, I encourage you to find better ones elsewhere. Or we can talk about this at another time when I write an article which is more relevant to the issues you raise.

  11. Is there any chance that you could find a latex parser plugin or something equivalent so that reading the equations isn’t quite so painful?

      1. The equations display fine for me. You description of it being painful made me wonder if the browser was not rendering correctly.

  12. Does this description apply equally to both the generation of mass of the W,Z bosons as well as the generation of mass of the fermions e.g electrons, muons, taus and quarks ?

    1. Yes. For both the W and Z bosons and the various fermions, there are subtleties that I haven’t yet addressed — these are the ones that make it clear that not only CAN the Higgs field provide masses for the various particles, it MUST. But the mechanism is the same. Oh, one other interesting thing about the W and Z bosons — the “Yukawa parameters” for the W and Z are directly related (in a relatively simple way) to the strengths of the weak and electromagnetic forces.

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