are massive bosons in fact fermions?

Discussion in 'Physics & Math' started by Arlich Vomalites, Aug 27, 2014.

  1. Arlich Vomalites Registered Member

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    At a naïve theoretical level all gauge bosons are required to be massless, and the forces that they describe are required to be long-ranged.
    According to the Standard Model, the W and Z bosons gain mass via the Higgs mechanism and they can be thought of as 'heavy' photons.


    As opposed to bosons, there are fermions which are not required to be massless.

    How do we distinguish between fermions and massive bosons?

    If there is no way telling them apart, are the massive bosons in fact fermions?
     
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  3. Manifold1 Banned Banned

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    Yes.

    It's called topological knot conjecture: however there have been recent experiments in which we have been able to trap light but on a little larger scale.

    A heavy goldstone boson would have to traverse the lines of path given by geodesics, it would not only have a frequency and a zitter motion, but also in constant free fall in a curved, flat space under the condition there is no boundary system on the geodesic curvature.
     
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  5. Farsight

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    This is a total can of worms. It's difficult to know where to start.

    And it is naïve. Take a look at the Wikipedia gluon article and note "as opposed to virtual ones found in ordinary hadrons". There are no actual real gluons in a proton. Nobody has ever seen a free gluon. In similar vein hydrogen atoms don't twinkle, and magnets don't shine. Because virtual photons aren't real photons. Gauge bosons are not actual bosons flying back and forth. See Matt Strassler's article. He says "a virtual particle is not a particle at all". So what are virtual particles? Field. And this field has its field energy, which is "at rest" and has a mass equivalence. It has mass. So the gauge bosons aren't actual bosons, and they do have mass!

    According to the Standard Model, free neutron decay involves the W boson. Only the neutron mass is 939MeV/c² and the W boson mass is 80GeV/c². Magic!

    Sure. We can create an electron and a positron out of a photons in gamma-gamma pair production. The mass of a body is a measure of its energy content. Then we annihilate the electron with the positron, and a radiating body loses mass. All of it.

    I don't know. The W boson has never actually been observed. Its existence is inferred.

    The distinguishing feature between bosons and fermions is the Pauli exclusion principle. See what Manifold said about knots? Look up TQFT and google on knot and vortex. Then think of it like this: two waves can ride over one another, but two whirlpools cannot overlap.
     
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  7. kurros Registered Senior Member

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    793
    There IS a way to tell them apart. Fermions have spin 1/2, bosons have spin 1 (or 0). This is a measurable difference.

    Also, fermions ARE required to be massless at the fundamental level, much as gauge bosons are. The Higgs mechanism generates masses for both the weak gauge bosons AND all the fermions.

    Finally, W and Z bosons are not really just heavy photons. Well the Z is quite similar to that it is true, but the W's are radically different since they change flavour (i.e. they can change electrons into neutrinos, and change up quarks into down quarks, etc.)

    edit: A more interesting question is whether some boson is really *two* fermions, since two fermions can sometimes bind together into a spin zero or spin 1 particle (such as in the case of two quarks binding into mesons). There is not much support for such a situation in the case of the gauge bosons, however in the case of the Higgs boson there are things called "composite Higgs" models which are quite popular. These indeed postulate that the Higgs boson is actually a bound state of some kind or other, i.e. bound by some new, currently unknown force. To experimental distinguish this possibility from that of a fundamental Higgs boson we "just" need to build bigger colliders and probe the structure of the Higgs boson at higher energies.
     
  8. Arlich Vomalites Registered Member

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    91
    You could be right. A fermion is defined as having spin 1/2, a boson as having spin 1 (or 0). Does this definition also apply to massive bosons?
    The W and Z bosons decay to fermion–antifermion pairs, so that the massive bosons seem to consist of two fermions. Two fermions seem to make a boson.

    I am not sure about this. Are electrons required to be massless?
    Higgs mechanism is based on symmetry breaking which leaves the photon massless whilst giving mass
    to the carriers of the weak force.
    The symmetry breaking explains the distinction between electromagnetism and the weak nuclear force,
    also the distinction between the photon and the W and Z bosons.
    Is it the same as saying that both electron and photon were massless when electromagnetism and weak force
    were unified , and after the symmetry breaking electron was massive whilst the photon was massless?


    Yes, there are composite bosons which suggests that two spin-1/2 fermions may well combine to form a total spin of 1 or 0. In either case, the integer spin implies that the composite object is a boson. Helium, for example, has two isotopes:
    bosonic helium 4He and fermionic helium 3He.

    It could be that massive bosons are in fact composite bosons, their spin is 1. Perhaps the W and Z are not fermions, the
    spin could make a distinction.
     
  9. Arlich Vomalites Registered Member

    Messages:
    91
    Yes, there is a principle called the Pauli exclusion principle. But why? Where does it come from?
    Does it come from the property of mass: two massive objects cannot occupy the same location?
    And what are these massive objects if not fermions? At least at the elementary particle level.

    Yes, it is called beta decay. The W decays into fermion-antifermion pair. The massive W boson consists of two fermions.
    Two fermions make a boson.
    Radioactive decay is thought to be a random process so that, according to quantum theory, it is impossible to predict
    when a particular atom will decay. The random nature of quantum physics means that everything just happen,
    there is no reason for why things happen in quantum physics as opposed to causality in classical physics. It has been written about QED that "electrons, somehow, emit and absorb photons. We do not know how these things happen, but the theory tells us about the probabilities of these things happening."
    It has been said that everything that can happen does happen in quantum physics. But why? Normally everything does
    not happen even if it can.

    The decay process could nevertheless be understood happening because of two fermions occupying the same state in
    the W boson. The decay will happen at the moment when the heavy W boson turns into two fermions and according
    to the Pauli exclusion principle they are not allowed to occupy the same state , so the particle decays in half-life of about 3×10−25 s.
     
  10. Farsight

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    3,492
    Think about the word spinor. A boson is akin to a wave, a fermion is akin to a whirlpool. Two waves can run over one another but two whirlpools can't overlap.

    No, but they are related. A photon is a boson where the path is linear. Again think spinor and the electron is a fermion where the path is closed. See Einstein's E=mc² paper The mass of a body is a measure of its energy-content. And note that he mentions an electron? Think of electron-positron annihilation, and think about this: If a body gives off the energy L in the form of radiation, its mass diminishes by L/c². The electron gives off radiation, its mass diminishes to zero, and it isn't there any more. Ditto for the positron. You know how if you trap a massless photon in a box it increases the mass of that system? Think of the electron as light in a box, only there is no box. Read this: http://www.tardyon.de/mirror/hooft/hooft.htm but note it's not the Nobel 't Hooft.

    At best, you could think of them as waves that are momentarily motionless. Look at the lifetime of the W boson. It's about 10[sup]-25[/sup] seconds. Light doesn't even travel the diameter of a photon in that time. The W bososn just isn't in the same league as the electron or photon.

    It isn't true I'm afraid.

    Things happen for a reason.

    The W boson hardly exists. Don't read too much into it. Instead think about gamma-gamma pair production where we convert two bosons into two fermions, and electron-positron annihilation where the reverse process occurs.
     
  11. arfa brane call me arf Valued Senior Member

    Messages:
    7,832
    A boson can be a fundamental particle or a composite particle.
    -Composite bosons See also: List of...effect on whether it is a boson or a fermion.-
    Bosons have Bose-Einstein statistics, which means every particle (composite or fundamental) can have the same quantum state. Fermions have Fermi-Dirac statistics. "Statistics" is about energy distribution over some collection of particles.

    As for a W boson "hardly existing", this is pretty meaningless; the weak interaction exists, so, you know . . .
     
  12. kurros Registered Senior Member

    Messages:
    793
    Yes the definition applies regardless of mass. As for the decays of W's and Z's, the things they decay into do not indicate that the W's and Z's "consist" of those things. The fermion-antifermion pair are created out of the vacuum and the W or Z is destroyed. The energy simply transfers from one field to another. But yes, angular momentum is conserved, so you will see the spin conserved like this in the decays.

    Yes before symmetry breaking all the fermions must be massless, at least in order to maintain gauge invariance, which is fundamental to how the Standard Model works so we don't want to abandon that. Well actually there are some subtleties with the neutrino masses, but it is true for the other fermions, in the Standard Model anyway. But yes, the symmetry breaking happens in such a way that photons stay massless, but W's and Z's get masses, and all the fermions also get masses, via their Yukawa couplings to the Higgs. This is where most of the free parameters in the Standard Model are by the way; one has to tune all these Yukawa couplings to get the right observed masses for everything. People like to cook up theories about where all these Yukawa couplings come from and how they might be related to each other.
     
  13. Farsight

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    3,492
    Unfortunately it contradicts Einstein and E=mc[sup]2[/sup] wherein "the mass of a body is a measure of its energy-content". Imagine wearing a T-shirt with a crossed-out E=mc[sup]2[/sup] on it.
     
  14. Manifold1 Banned Banned

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    181




    No there isn't, not in the photon knot theory of physics, in fact, the photon can traverse a loop in order to create the observed 1/2 spin of a fermion, like an electron.
     
  15. Arlich Vomalites Registered Member

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    New experimental data shows that neutrinos have mass, although for half a century physicists thought that
    neutrinos, like photons, had no mass. It was the Standard Model of particle physics which assumed that neutrinos are massless.
    It seems that massless neutrinos, also massless fermions, are not something that are observed, although it is again the Standard Model which assumes it.
    There is no proof that massless fermions exist until they are experimentally observed.
     
  16. Farsight

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    3,492
    I think the Standard Model is accidentally correct about this, and that people are misinterpreting that experimental data. Have you ever heard of "photon effective mass"? When you slow down a massless photon in say glass to less than c, it exhibits a little "effective" mass. If you slow it down by catching it in a mirror-box, it adds mass to that system because mass is a measure of energy-content. Then all of the energy is effective as mass. Neutrinos travel so close to c that we can't tell the difference. (When a supernova occurs, the neutrinos actually arrive before the photons because the latter struggle to get out of the star). But if neutrinos slow down for some reason, they would exhibit "effective" mass. The more they slow down, the more effective mass they exhibit. The classic analogue of neutrino oscillation maybe offers some insight to this. If the neutrino lengthens and shortens as propagates, it would be spending some of its time moving at less than c in aggregate, and would therefore exhibit some effective mass.
     
  17. Arlich Vomalites Registered Member

    Messages:
    91
    As soon as it is observed that cows are falling from the sky, I might start considering massless fermions existing. Because it is possible that the cows were originally massless, therefore they were capable of flying, and then somehow the cows were slowed down and acquired "effective" mass and started falling from the sky. We just need to make an observation if this is really happening, theory alone is not enough. Then we need to make measurements to make sure it really is a cow, not a fermion, not a heavy boson which decayed
    into two fermions (or two cows). It could also be a bird mistaken for a cow. Birds could be called heavy bosons.

    I just want to say that even though a theory suggests a particle, it needs also to be observed to prove it exists.
    It is possible to postulate, for example, a boson with a spin-1/2, but apparently it does not exist.
    If a massless fermion is observed, it might turn out to be a spin-1/2 boson because bosons are required to be massless.
    But the impossibility arises because fermion spin 1/2 is not the same as boson spin 1, therefore the observation must be a mistake
    and there should not be such an observation at all.Therefore we don't observe massless neutrinos. The same way we don't observe cows flying and falling from the sky.
    But birds are observed, as well as heavy bosons.
     
  18. Farsight

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    3,492
    There is such a thing as a cowbird, Arlich. And you're looking at it the wrong way. Here's the $64,000 dollar question: on properties alone, what does the neutrino more closely resemble: the electron, or the photon? Well, the neutrino travels at c like the photon. And its and mass and charge are a lot closer to the photon mass and charge than the electron mass and charge. So, it walks like a duck, it swims like a duck, and it quacks like a duck. The neutrino is more like the photon than the electron. It's called a fermion but it's much more like a boson than a fermion.
     
  19. Manifold1 Banned Banned

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    181


    You know farsight the neutrino is best modeled with two photons so that the charge cancels out, unless experiment has let us down and infinitesimally small regions of mass display negligible charges we cannot detect so far. In fact I recall a paper recently mentioning in an idea in his direction. The conclusions of the paper was that the neutrino charge would be very difficult to detect, which means the presence of mass and charge may yet still prove to be the same thing.
     
  20. Arlich Vomalites Registered Member

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    91
    A cowbird? It is probably a hoax. Calling a bird a cow does not make it a cow. The photo you have shown is not evidence that it exists.
    Its real name is something else, a birdcow, a heavy boson. Can you put two massive cows into the same place? What do you get, a cowcow,
    what is its real name?
    View attachment 7321


    Neutrino is a quantum of neutrino field, which is a fermion field. It acts like a field, walks like a field, talks like a field, so is it a field?
    It is matter-field, it is both matter and field. It is a mixture of matter and field. It is not either pure matter or pure field.
    Therefore it is hard to tell what does the neutrino more closely resemble: the electron, or the photon. Because it is both, it is matter
    and it is field, it is their mixture. Does pure matter exist?
     
  21. Farsight

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    3,492
    I'm sure that a stable particle can't have mass if it's got no charge. Somebody might mention the neutron, but that's got no net charge, the magnetic moment says there's charge in there. And I'm confident that the neutrino is more like a photon than an electron, which you can make out a photon in electron-positron pair production. But I've never heard of the neutrino being modelled as two photons. That doesn't sound right. Have you got a reference for that?

    Sorry Arlich, I don't know what you're saying here. You can make and electron and a positron out of photons and convert bosonic field into fermionic field, then you can annihilate the electron and positron to photons. So the fermionic field is just some configuration of the bosonic field. It doesn't have its own quantum.
     
  22. Arlich Vomalites Registered Member

    Messages:
    91
    Matter-field, matter and field, is a mixture of matter and field. Can you purify matter-field so that it is not mixture anymore ?

    You would have pure matter without field. Is it thinkable to have matter but no field?
    Particles of matter are the sources of fields, so in this sense particles are more fundamental than the fields whose sources
    these particles are. Of course if there are fields without matter, these fields are not created by particles of matter.
     
  23. Farsight

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    3,492
    Well, yes. It's quantum field theory, not quantum point-particle theory. The electron is field. It isn't something that's "got" a field.

    No.

    This is the wrong way to think about it, Arlich. In QFT the electron is said to be "an excitation of the electron field". I don't think that says enough as it happens, but nevermind. The point is this: the electron is just field. When you annihilate it with the positron you get two photons, which are just field variations. They're just waves.

    Have you ever looked at topological quantum field theory? See Wikipedia. See where it says it's related to knot theory? I think the best way to think of it is that a particle of matter is a "knot in a field". Have a google on that. And take a look at the Topological Quantum Field Theory Club webpage. See those blue trefoil knots at the top? Pick one, start at the bottom left, and trace around it anticlockwise calling out the crossing-over directions: up down up. Does that ring any bells?
     

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