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Higgs Field and mass?

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  1. Sep 30, 2014 #1
    [Mentor's note: Several related threads have been combined into this thread.]

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    I read that Electron is mainly energy so it can easily travel through the higgs field. Muon has a bit more mass than Electron so it experiences friction when it goes past the higgs field. W particle has more mass than the Muon so it is slowed down more than the muon. At last the Quark has almost all mass and nearly no energy so it is slowed down very much and its kinetic energy converts into mass. So does that mean the more the higgs field slows down a particle the more it adds mass to it?

    The messenger particles that convey the effect of Higgs Field are called Higgs Bosons which are found in the Higgs Field. So where can I find the higgs field? Have the scientists seen the kinetic energy of quarks getting converted to mass? Can matter made up of composite particles acquire mass when moved through higgs field?

    Does this also mean more massive particles will add more mass to themselves when they interact with the higgs boson? Is what we call the rest mass the mass due to interaction of matter with the higgs boson, since higgs bosons create a drag and that drag is the rest mass?
     
    Last edited by a moderator: Oct 1, 2014
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  3. Sep 30, 2014 #2

    phinds

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    Uh ... "slows down" relative to WHAT?

    Do you think that if I shoot a cannonball in a vacuum (away from massive bodies to not complicate the issue) and a ping pond ball at the same velocity, that the cannonball will slow down more because it has more mass or is bigger or is "heavier" or whatever?
     
  4. Sep 30, 2014 #3
    I think that what's at issue here is the misunderstanding that elementary particles must move relative to the Higgs field in order for the particle to have mass conveyed to it. That is simply not the case according to theory and current experimental data.

    It seems to be that the Higgs field is stationary relative to all elementary particles in the universe, regardless of their motion relative to each other - in other words, if electron A is moving towards electron B, the higgs field is still stationary relative to each of the particles, and there is NO relative motion between the electrons and the Higgs field. It's a bit hard to fathom, but that appears to be the way it is.

    Therefore, there is no "friction" that conveys mass to elementary particles. It is simply the existence of the Higgs field in an "on" state (meaning having a positive average value throughout the universe, unlike the electric field which has an average zero value) that conveys mass to the elementary particles. Here's a decent FAQ about it, albeit it could use some updating: http://profmattstrassler.com/articles-and-posts/the-higgs-particle/the-higgs-faq-2-0/
     
  5. Oct 1, 2014 #4
    Now I have read that the speed of electrons in a typical coaxial cable is about 2/3 the speed of light (i.e. 200 million meters per second). So if the Higgs field did not exist, then the particles (Of Matter) would have no rest mass whatsoever and, just as photons (which have no rest mass), would fly off at the speed of light.

    So if an electron was totally made of energy then it would travel at the speed of light. So I am assuming that a electron has some mass which stops it from travelling at the speed of light. Now why does the higgs field not add mass to the electron if it already possesses some mass? Why is mass added to other subatomic particles like quarks more than elctrons if at all the higgs field adds mass to it?

    Also am I right that the speed of electrons in a coaxial cable is 2/3 the speed of light?
     
  6. Oct 1, 2014 #5
    So as far as I know the higgs field converts the kinetic energy of quarks into mass by slowing it down. So what about the things that are not moving? So if an object is stationery then will mass not be added to it by higgs field? But higgs field gives us the rest mass so then an object at rest must have mass which is added to it by higgs field but it is not moving so how can energy be converted to mass according to E= MC2?
     
  7. Oct 1, 2014 #6
    I have mentioned previously that: "If the Higgs field did not exist, then the particles (Of Matter) would have no rest mass whatsoever and, just as photons (which have no rest mass), would fly off at the speed of light".

    So does that imply that electrons are massless and it is only because they interact with Higgs Field that they gain mass? So a photon is also massless then why does the higgs field not add mass to it?

    So then I thought that an electron must have mass before interacting with the Higgs field because if it is massless like the Photon it could easily pass through the higgs field without adding mass to it. So Electron mass is 9.10938291 × 10-31 kilograms. Is this the mass of an electron before interacting with the Higgs Field? If not then whats the mass before the interaction?
     
  8. Oct 1, 2014 #7

    Orodruin

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    The Higgs does not give the photon mass because, unlike electrons, the photons do not interact with the Higgs field. The details of this are in how electroweak symmetry is broken and the photon essentially corresponds to a remaining unbroken symmetry. The broken symmetry instead corresponds to the W and Z bosons which do get a mass through interactions with the Higgs field.
     
  9. Oct 1, 2014 #8
    The Higgs Boson Breaks the symmetry and for those particles whose symmetry is broken mass is added. So why does symmetry braking occur only to particles that have opposite spins? Example two electrons that have opposite spins.
     
  10. Oct 1, 2014 #9

    ChrisVer

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    I have some difficulty in understanding what you are trying to ask....
    First of all, you make some terrible interpretation mistake: the Higgs field doesn't slow down the massive particles... the molasses analogy is not the description of what the Higgs field does. The reason why is because the Higgs field (its vev) is responsible for the particles rest mass (No motion!). If the Higg's field vev happened to be zero, then you wouldn't get a mass term (at least not in the form of a bare mass), but that would be equivalent to having symmetries unbroken... An object at rest has some mass value added by the Higg's vev (and maybe radiative corrections for heavy enough particles).
    Which energy converted to what mass? The [itex]E^2 = p^2 c^2 + m^2 c^4 [/itex] is holding for the particles anytime [for real particles]..the mass terms come from the self-couplings of fields (self-energy diagrams, that's why in the lagrangian the mass comes as the component of a field2 term ).

    If the Higgs field didn't exist, then you would have to break the SU(2)xU(1) symmetry explicitly by adding mass terms by hand. The Higgs field was used because the Higgs mechanism was able to spontaneously break the symmetry down. So without the Higgs your particles' masses (at least in the Standard Model) would need to be put in by hand...

    (sometimes the expression like "if there was no Higgs field" can be misunderstood, because there is a Higgs field )

    Your #8 post makes no sense at all. Which opposite spins? The symmetry breaking also adds mass to the gauge bosons, as well as to all fermions (which had a left and right component- and were in different representations of SU(2) ). In the "classical" Standard Model, where neutrinos are massless for example, you don't get a mass for the neutrinos either...
     
    Last edited: Oct 1, 2014
  11. Oct 1, 2014 #10
    "However, in practice, it is not so puzzling. If you have two particles moving at each other with equal speed, it would seem impossible for either of them to do anything but be symmetrical. However, if each particle has an equal 50-50 chance to be spinning one way or another, it is possible–in theory and in practice–to have this symmetry broken. It begins with symmetry to start with because the particles have an equal and symmetrical 50-50 chance of spinning one way or another."

    This is from wikipedia ChrisVer (Last Para). http://simple.wikipedia.org/wiki/Spontaneous_symmetry_breaking
     
  12. Oct 2, 2014 #11

    ChrisVer

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    Last edited: Oct 2, 2014
  13. Oct 2, 2014 #12

    DrDu

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    I have problems, too. The point is that this article, like most articles on the subject, misses the most important point of spontaneous symmetry breaking (SSB). In fact, SSB is kind of a misnomer. E.g. they write, that SSB leads to the existence of an infinite amount of degenerate ground states for infinite systems. However, this is true for almost all infinite systems. E.g. you can add a particle with mass m to a vacuum. If the particle is homogeneously distributed over all of space, the density will still be 0 and you can't distinguish the new state, which has one more particle, from vacuum. Experimentally, you can create a particle here on earth and this will be independent from whether someone is removing the same kind of particle on the moon. Hence in a normal theory there are also an infinite number of degenerate ground states, but you can ignore it.
    The main point is that this is no longer true in a theory with SSB. There are long range correlations between the creation and anihilation of a Higgs boson on two different places although all the symmetries remain intact. The problem is that this is kind of inconvenient for experimentalists as they would have to care about what their colleagues on Mars are doing (usually they don't even care about what the guys next door are doing). This phenomenon is called ODLRO (off diagonal long range order).
    The only possibility to overcome this dilemma is to take explicitly into account the fact that there are an infinite number of equivalent ground states with slightly different number of Higgs bosons (though the same density) and choose a convenient superposition of them, this at the same time removes the correlations and breaks symmetry, as the new eigenstates are no longer symmetry adapted eigenstates.
    So basically we sacrifice a symmetry, which we can't observe, to not get correlations, which we don't want to observe.

    While mathematically convenient, this is not always the best choice.
    E.g. the formation of atomic Bose-Einstein condensates is often described as a symmetry breaking, but as the systems are finite, it is hard to accept why BE condensation should lead to a state with unsharp particle number.
    In fact, the experimental proof of BE condensation relies often on the detection of long range correlations.
     
  14. Oct 2, 2014 #13
    The quantum field for normal particle species like electrons or quarks is zero everywhere except where there are particles moving around. Particles are wiggles on top of this zero value. The Higgs is different because the value of its quantum field in the vacuum is not zero. We say that it has a vacuum expectation value, or “vev” for short.

    So how does having a non zero vev give mass to particles?
     
  15. Oct 2, 2014 #14
    Explain: "In quantum field theory a particle is an excitation of a quantum field.". So does that mean a non zero quantum field value of the higgs field imparts mass to the elementary particles because the particles are massless and they get excited when near the vev of the higgs field?
     
  16. Oct 2, 2014 #15
    I was thinking on this and I found out that the higgs field gives elementary particles their mass. This happens when the interaction of excitations of the electron field (Electrons are excitations of electron field which are now massless) come in contact with the vacuum state of the Higgs field. Thus mass is created. Am I right or am I just a simpleton or a pseudo- intellectual who is trying to flex his intelligence which cant be flexed?
     
  17. Oct 2, 2014 #16

    phinds

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    Your questions about the Higgs field seem like good ones to me and I wish I could be of some help, but I find it a bit strange myself. About the only thing I feel clear on is that is is not the Higgs Boson that imparts mass but rather the Higgs Field (of which the boson is an excitation) but it sounds like you've already gotten that far.
     
    Last edited by a moderator: Oct 2, 2014
  18. Oct 2, 2014 #17

    phinds

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    You really have a very serious fundamental misunderstanding about motion. You have used the phrase "slowing down" without saying RELATIVE TO WHAT (and you never answered my quesion to you about that) and now you are saying that things can be "not moving" and again not saying RELATIVE TO WHAT?

    Do you think that you are not moving right now? Well you are not moving relative to yourself, you are moving pretty fast relative to the moon, even faster relative to the sun, and from the frame of reference of an accelerated particle at CERN, you are now moving at .999999c, so PLEASE stop talking about motion without saying relative to what.
     
  19. Oct 2, 2014 #18

    mfb

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    This is the speed of signal propagation. It is not the speed of any object. The electron motion from electric currents is surprisingly slow - typically millimeters per second.

    This does not make sense. Is a football field "mainly length" just because it is longer than a bed? To make it worse, different electrons can have different energies.

    It is everywhere. The particle masses are the only visible result apart from the Higgs boson.

    If a football field was totally made out of length...

    This is unrelated to particle spin. A photon has two spin states as well and does not get a mass, while the Higgs field gives a mass to the Higgs boson (with spin 0).

    The best description we have is quantum field theory - formulas. Every attempt to describe those formulas with words will lead to some approximation, some are better some are worse, but none is perfect.
     
  20. Oct 2, 2014 #19

    Drakkith

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    I don't feel like you'll make much headway into this just by asking a few questions about random topics that you've read concerning the higgs field and mass. So far, it seems like members are spending more time correcting your misunderstandings than explaining the subject. You need to learn a lot more of the very basics of quantum mechanics, quantum field theory, and several other topics before you'll be able to grasp exactly what is happening. Most importantly, remember that our theories are primarily math based and any attempt to "translate" it into common language is almost certain to fail. Without knowing how to read the math, it's unlikely you'll get very far. Unfortunately this is not an easy task. People spend years in college learning both the math and the theories.
     
  21. Oct 2, 2014 #20

    Drakkith

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    To be clear, I'm not telling you to stop asking questions, I'm merely suggesting that you start at a more basic level and work your way up. It's going to take a while. Walk before you run and all that.
     
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