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Non-interacting particles?

  1. Oct 2, 2003 #1

    chroot

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    We all know that, according to general relativity, all particles feel the gravitational force. The other forces are not so ubiquitous.

    The proton interacts with the electromagnetic and strong forces, but apparently not with the weak. The electron interacts only with the electromagnetic; the neutrino only with the weak. Let's make a table:
    Code (Text):

                       E&M   S    W
     hadrons            x    x    x
     e.g. protons       x    x
     e.g. muon          x         x
     e.g. electron      x
     ?                       x    x
     gluons?                 x
     neutrinos                    x
     
    I am not sure about the gluon -- I know it is chargeless, but does it interact weakly? I don't think it does.

    I also spent some time thinking about a particle which interacts only strongly and weakly, but not via E&M, and couldn't think of one. My first instinct was the pi-zero, but it decays into photons...

    My real question, however, is this:

    Does the standard model have anything to say about a hypothetical particle which interacts ONLY gravitationally?

    Has such a thing been proposed? I understand that we simply would not be able to detect such a particle at all with current instruments, but can we say anything else about their existence? I'm thinking along the lines of dark matter, of course...

    Thanks for any input.

    - Warren
     
  2. jcsd
  3. Oct 2, 2003 #2

    selfAdjoint

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    Some problems here. You talk about the proton and the gluon, but they are on different levels; the proton is a collection of quarks in a sea of gluons. And the gluons are charged, indeed each one carries two of the six strong force charges, one of thre three charges and one of the three anticharges. Each quark has one of the strong charges and a fractional electric charge. So right there the quarks, and therefore the proton, feel both the strong and the electromagnetic charge, and in the standard model, the electromagnetic interaction is bonded with the weak one into the "electrowaeak" sector of the model.

    But wait, there's more. The quarks, and therefore the proton, do feel the weak force properly so called. The weak interaction can change flavors of quarks - it can cange one kind of quark into another, which none of the other forces can do.

    And finally no, the standard model has nothing about gravity in it.
     
  4. Oct 2, 2003 #3

    Tom Mattson

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    Re: Re: Non-interacting particles?

    I think he knows that, and what he is asking is if there are any particles that do not interact in the Standard Model (that is, they only interactions they could possibly have are gravitational interactions). As for me, I haven't heard of any. The Standard Model is basically a Lagrangian, and the interesting physics occurs in the interaction terms. A particle with no interactions would just appear as a free field, and I have never heard of anyone putting one in.
     
  5. Oct 2, 2003 #4

    chroot

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    Re: Re: Non-interacting particles?

    I should have specified 'electrical charge.' The gluons are not electrically charged. When I mean colour, I say colour.
    Which is exactly what I said in the table.
    However, the proton does not undergo any kind of weak decay. Does this not indicate that it does not interact weakly?

    - Warren
     
  6. Oct 3, 2003 #5

    selfAdjoint

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    The proton, which I can't help but think of as a particular system of quarks and gluons - not primary - does not decay as far as we know. Its natural half life is greater than 10^30 years, according to experiments. But other hadrons do undergo weak interactions. The long lifetime of the proton does not seem to me on the same footing as the fact that the leptons don't feel the strong force. It's more like the proton is the lowest energy baryon, and there's nowhere for a decay to go that preserves the standard model symmetries.
     
  7. Oct 3, 2003 #6

    Tom Mattson

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    Re: Re: Re: Non-interacting particles?

    Decay is not the only kind of weak interaction, though. There is a weak component to n-p and p-p scattering reactions. Because the reaction is dominated by the strong and electromagnetic interactions, the weak component is tough to measure. The one thing that experimentalists have on their side is that the weak interaction violates parity, a discrete symmetry (either parity is violated or it is not; no inbetween). In a reaction such as:

    n+p-->d+g

    we have pn=pp=pd=+1, and pg=-1 for an overall violation of parity, and this reaction does indeed go as written.

    edit: fixed subscript bracket
     
  8. Oct 3, 2003 #7

    chroot

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    Re: Re: Re: Re: Non-interacting particles?

    Aha, this I did not know -- thank you Tom. So the proton is not a non-weakly-interacting particle. Are there any particles which feel ONLY the E&M and strong forces, but not the weak?

    Or how about the strong and weak, but not E&M?

    It certainly seems there's something incestuous afoot in the family of interactions... as selfAdjoint pointed out, the E&M and weak interactions are peers -- yet neutrinos interact only weakly. According to the strong + electroweak unification, it would seem that all coloured particles would also have to interaction with the E&M and weak forces -- but this is not true. Gluons apparently feel only the strong force. (Right?)

    This is perplexing to me. It seems that Nature would have particles that interact with every combination of the forces, or would have some specific combinations that are disallowed by discernable rules. Neither seems to be true. Am I missing something?

    - Warren
     
  9. Oct 6, 2003 #8

    Tom Mattson

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    Re: Re: Re: Re: Re: Non-interacting particles?

    Right. Actually, it occured to me this weekend that neutron decay by itself is sufficient to show that, because in the decay:

    n-->p+e-+ne

    the proton does participate in the weak interaction, as a product.

    None that I know of.

    Right: gluons are both chargeless and flavorless; they have only color.

    Perhaps, because some combinations are disallowed by discernible rules:

    1. If a particle is chargeless, then a photon cannot couple to it
    2. If a particle is flavorless, then a massive gauge boson cannot couple to it.
    3. If a particle is colorless, then a gluon cannot couple to it.

    Once it is determined that a particle can be coupled to a gauge field, then the rules that apply are Electroweak × QCD.
     
  10. Oct 6, 2003 #9

    chroot

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    How about right-handed neutrinos, or left-handed antineutrinos? They should exist, right? They wouldn't interact with any forces at all except gravity.

    - Warren
     
  11. Oct 7, 2003 #10

    Tom Mattson

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    That's right. nR and nL do not couple to charged lepton currents via the weak interaction; and yet those particles are not flavorless. However, that does not mean that there are not "discernable rules" by which weak interactions with right-handed neutrinos are ruled out.

    What now has me thinking is this: A right handed neutrino does not interact weakly. But, since the neutrino is massive there is always a way to Lorentz transform a nR state into a nL state, which does interact weakly. How is it that a force can simply be transformed away like that?

    edit: fixed bold font bracket, subscript brackets
     
  12. Oct 7, 2003 #11

    chroot

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    Wow, you're right -- this *is* very puzzling. I thank you for bringing it to my attention...

    An electron has no intrinsic parity (i.e. it cannot be assign left- or right-handed helicity) because an observer can be moving either slower or faster than it, making its velocity appear in one direction for one observer and in the opposite direction for another.

    The neutrino, if it is massive, also does not have intrinsic parity for the same reason. The neutrino has been concluded to be necessarily massive, due to neutrino oscillation. This means that the weak interaction will be apparent for some observers watching a neutrino, but not for others.

    Can this be explained the same way that the magnetic field is explained? As a purely relativistic effect? The presence of a magnetic field depends upon the observer's frame of reference.

    Magnetism simply steers a particle around, so it seems okay that the field looks different for different observers. But if you can change reference frames and force the neutrino to be the "wrong" helicity, you can magically make the weak force disappear for it.

    This would be a tremendous problem for particles which undergo weak decay -- it would mean that one observer would see a decay, but another wouldn't! I suppose it's...... acceptable.... for neutrinos, since they do not decay, but just get pushed around by the weak force like electrons get pushed around by the magnetic field.

    On the other hand, neutrinos DO oscillate flavors, and this oscillation is a purely weak phenomenon. Does this mean that one observer would see a neutrino oscillate to a new flavor, but another would see it stay the same flavor? How can this be possible?

    Boy Tom... if you can't tell from my meandering here, you have certainly got me thinking too... do you have any ideas?

    - Warren
     
  13. Oct 7, 2003 #12
    If we take for granted that there is an energy where eventually a unified theory of physics holds, wouldn't all particles interact in principle by every force? Furthermore, if they all interact actually at that Planck temperature, 1032K, don't they have a probability (by the Heisenberg uncertainty principle) for such symmetry at lower temperatures?
     
  14. Oct 7, 2003 #13

    Tom Mattson

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    I am going to have to review some of this stuff. My research is centered in the electromagnetic/strong sector of the Standard Model, and the last time I studied neutrinos, they had no mass . In a few days, I'll be posting some stuff both to help me review, and to work this out.

    In case you want to read up, I am going to the following sources:

    Quarks and Leptons, by Halzen and Martin
    Quarks, Leptons and Gauge Fields, by Huang

    If you don't have access to those, then I'll "see" you by the end of the week with something more.
     
  15. Oct 8, 2003 #14

    chroot

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    Hmmm... in the case of electrons emitted in beta decay, the polarisation, P, can be defined as

    P = (NR - NL) / (NR + NL)

    where NR and NL are the number of left- and right-handed electrons, respectively.

    When P = 1, all the electrons are right-handed, and when P = -1, all the electrons are left-handed.

    When the electrons are emitted slowly, they have no net polarisation. As energy increases, however, more and more electrons are left-handed. The formula is simple:

    P = -v/c

    Almost all of the electrons moving relativistically are left-handed.

    I wonder if there is a similar case in play here: due to their extremely small masses, the neutrinos would all normally be moving relativistically, and they'd also be mostly all left-handed -- for most observers. In the limit as they approach the speed of light, they are all left-handed.

    It would seem to answer the question, at least experimentally -- it's a selection effect. Almost all neutrinos are moving relativistically, so almost all neutrinos are left-handed, so almost all neutrinos experience the weak force, so almost all neutrinos oscillate flavor.

    Those that are moving slowly enough to be right-handed to our experimental apparatus would not appear to oscillate -- but that would only very rarely happen, and would surely get swamped out by the overwhelming majority of relativistic left-handed neutrinos.

    What do you think, Tom?

    It still means that the neutrino oscillation is an observer-dependent effect -- for a slow neutrino, some observers see oscillation, others don't -- and that is still a little disturbing to me, considering how greatly dissimilar the reactions between the flavors are.

    - Warren
     
  16. Oct 8, 2003 #15

    Nereid

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    cool neutrinos

    Just to put some numbers on this ... even at the upper observational limit of neutrino rest mass (0.06 eV/c2) neutrinos would have to be very cool to be moving non-relativistically (<~100 K). If nothing, other than the expansion of the universe, could cool neutrinos, there would be very, very few moving at non-relativistic speeds. Relict neutrinos have a temperature of ~1.8 K, and there are a lot of them. Not many folk thrusting designs for a relict neutrino detector at their patent attorneys :wink:
     
  17. Oct 8, 2003 #16

    chroot

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    Nereid,

    You bring up a great point -- the neutrino background temperature of 1.8K. These neutrinos are "very slow," and we should see very many right-handed -- non-interacting ones! Thanks so much for reminding me.

    I have to wonder what sort of contributions they could make to dark matter... Nereid, do you have any tabs on how many there are? I know I've seen figures somewhere... off to arxiv I go....

    - Warren
     
  18. Oct 8, 2003 #17

    Nereid

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    I crunched some numbers a while ago, and even at 1.8K, they're travelling at ~0.1c (that's an OOM).

    I don't have anything immediately to hand on their number etc. Two reasons why they are unlikely to make up more than a small fraction of dark matter:
    1) at 0.1c (or faster) they're too hot to produce the structure we see in the universe (and they would've been a lot faster in early eras, when structure got first established)
    2) WMAP results
     
  19. Oct 8, 2003 #18

    chroot

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    You hush. :D

    - Warren
     
  20. Oct 8, 2003 #19
    a lorentz boost does not change left handed fields into right handed fields. the weak interaction couples to only one chirality.

    thus the fact that massive neutrinos change helicity under a lorentz boost does not mean that the interaction itself is a coordinate dependent phenomenon. all feynman diagrams are lorentz independent, so if there is an interaction in one frame, then there is one in all frames.

    the helicity of a neutrino is only a good eigenvalue in the massless limit, in this limit, chirality and helicity are the same.

    the helicity of a particle does not determine its handedness. handedness is the more fundamental concept here. you say a field is left handed if it lives in the (0,1/2) irrep of the lorentz group. an irrep, by definition, is left invariant by transformations, so it is clear that a lorentz transformation cannot exchange fields from one irrep for another.
     
  21. Oct 9, 2003 #20
    Re: Re: Re: Re: Re: Non-interacting particles?

    remember that the EM force is unified with the weak force. the EM force has a gauge group that is a U(1) subgroup of the weak SU(2) gauge group. so you can t be a trivial irrep of SU(2) without also being a trivial irrep of U(1) as a subgroup of SU(2). therefore any charged particle (which therefore participates in EM) also has isospin, and feels the weak force.

    it is, of course, possible for a particle to be trivial in U(1) but not all of SU(2). the neutrino for example. i don t see any reason why you couldn t talk about a colored neutral weak particle, then you would have strong and weak but not EM, but this is not the standard model. in the standard model, you only have quarks and leptons, none of the leptons carries color, and none of the quarks is neutral.

    the standard model does not unify strong and electroweak forces.
    i don t see where the confusion comes from. are you looking for particles which feel all the forces? quarks fit the bill. they interact weakly, strongly, and electromagnetically.

    don t get mixed up by gauge bosons like the gluon. a gauge boson only participates in the force for which it is the mediator.
     
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