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Higgs particle question anticipating July 4 Cern announcement

  1. Jun 25, 2012 #1
    Soon, I believe on 4 July 2012, CERN is due to make an important announcement regarding the discovery (or not) of the Higgs particle at the LHC. This announcement is likely to be an important milestone in physics. It comes after a long drought of significant fundamental physics discoveries, during which SSC was cancelled in the US and the LHR commissioned in Europe. In this drought there has also been the 40-year rise (and perhaps fall) of much theoretical ratiocination that lacks experimental or observational confirmation. I guess this 4th of July is going to be important for the future of physics and its funding, and I’d very much like to understand the significance of the Higgs, so that I can better appreciate what is going on. I’m not nearly clever enough to do this without help. Hence the question below.

    The best source of information I’ve found so far is a http://minimafisica.biodec.com/Members/k/2011/bcstolhc.pdf [Broken] given by Steven Weinberg to an audience of non-particle physicists (I’m one) at the 50th anniversary of the publication of the successful Bardeen-Cooper-Schrieffer (BCS) theory of superconductivity, in 2007. I’m afraid it’s at the limit of my comprehension and has therefore generated some odd questions. Before I put one of them I hasten to add that I don’t doubt the wisdom of anything Weinberg said.

    He places emphasis on the importance of the spontaneous breaking of an exact symmetry, as recognised by the post-BCS particle-physics community and which, mandated by the Goldstone-Salam-Weinberg theorem, results in the production of an exactly massless particle. Weinberg refers to such a particle as a Goldstone excitation in his mention of cosmological fluctuations; an excitation that because it has no mass, has “zero frequency”.

    If broken symmetry and zero mass are both “exact”, I take him to mean that the field corresponding to the zero-frequency particle is “exactly” constant and not time-varying.

    Now switch to considering a field often taken as a manifestation of broken symmetry — the magnetic field due to a solid lump’s aligned atomic spins. Is this (everyday language) “static magnetic field” associated with a (particle physics language) “massless Goldstone excitation”? Or are these just convenient effective descriptions invoking the mysterious wave-particle duality? I have enough trouble imagining long wavelength EM waves being represented as photons, let alone a static magnetic field masquerading as a particle. Goes against my understanding of harmonic analysis.
    Last edited by a moderator: May 6, 2017
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  3. Jun 25, 2012 #2


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    It does invoke "wave-particle duality". A "massless" particle is a wave that can be excited by an arbitrarily small amount of energy.

    "... although the microscopic Hamiltonian has full rotational symmetry, the low temperature phase does not. ... The original symmetry is still present globally ... a rotation transforms one ordered state into an equivalent one. If a uniform rotation costs no energy ... Such low energy excitations are called Goldstone modes. They are present in any system with a broken continuous symmetry. ... Phonons are an example of Goldstone modes, corresponding to the breaking of translation and rotation symmetries by a crystal structure."
  4. Jun 27, 2012 #3
    Thanks, atty. The link is most helpful. I guess I was puzzled by Weinberg's analogy of zero-frequency "phonons" in the primordial universe, as a static Goldstone excitation persisting unaltered in form right through to the present day. Perhaps a bit of a stretch to so label (for us eternal ) fluctuations, but who am I to quibble at a Weinberg analogy!
  5. Jun 28, 2012 #4


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    The Goldstone bosons corresponding to broken rotational symmetry in a magnet are the magnons.
    As the excitation energy for a magnon goes to zero in the long wavelength limit, they are massless in the language of high energy physics. The reason is that the magnon field is quantized and the energy to create a single magnon becomes arbitrarily small for long enough wavelength just as the energy of a photon becomes smaller and smaller in the limit k->0.
    On the other hand the energy to generate a massive particle like the electron is mc^2 in the limit of vanishing momentum.

    You also have to take in mind that in condensed matter, which is not relativistically invariant, breaking of symmetry does not always imply a massless Goldstone boson.
  6. Jun 28, 2012 #5
    Thanks for that further illumination about "massless particles" being a limit of static excitations (waves?) with zero frequency and infinite wavelength. Like acoustic waves in the early universe, as Weinberg said. Be that as it may, I suspect that "massless particle" is an oxymoron.

    You must here mean the theory of condensed matter, as written, rather than condensed stuff itself ? I'm pretty dense, but hope that I'm still Lorentz invariant!
  7. Jun 28, 2012 #6
    massless particles are excitations which have zero energy in the limit of zero frequency and infinite wave length, not the limiting case of these excitations. The limiting case itself is a massless particle with zero momentum and thus zero energy.
    It is definitley not an oxymoron, as massless particles, such as the photon, exist in nature.

  8. Jun 28, 2012 #7


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    Setting c=1 we have for a particle with mass m, energy E and momentum p that
    E=(m^2+p^2)^(1/2). For a massless particle E=p, or with p=hbar k (de Broglie relation), a linear relation between energy and wavelength.

    As far as Lorentz invariance is concerned you form a distinguished reference frame for the particles moving in your body, or, using different terminology, you break Lorentz invariance. This is quite analogous to the breaking of rotational invariance in a magnet.
    Obviously the magnet or its magnetization axis can still rotate as a whole, but it defines a distinguished direction on a microscopic scale, e.g. for a single spin inside the magnet.
  9. Jun 29, 2012 #8
    Ofir -- you are quite right to correct me. I had meant to refer only to such massless particles as Weinberg was talking of, as being an oxymoron. I think this would have been correct. Very careless of me.

    DrDu: I like the idea of being a distinguished reference frame, but I'm concerned about being a destroyer of Lorentz invariance. When excitement sends the blood rushing through my veins, I still like to think this makes that component of me (slightly) more massive.

    Is Lorentz invariance not thought to be the one unbreakable part of physics?
  10. Jun 30, 2012 #9


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    In fact, strictly speaking, you don't break Lorentz invariance. You would only do so if you filled the whole universe. Nevertheless it is a usefull approximation for the microscopic description of matter to approximate crystals etc. as being of infinite spacial extent.
  11. Jun 30, 2012 #10
    I agree. My ego is such that I often think that the whole universe is just for me, but then I remember the other seven billion. Thanks.
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