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What is a massless particle?

  1. Jul 6, 2012 #1
    Is it possible to explain how a photon can be a particle yet have no mass? I think I need a scientific clarification either of what a particle is or what mass is.
     
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  3. Jul 6, 2012 #2

    PeterDonis

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    Perhaps a clarification of terms will help. A photon is a particle with zero rest mass. That means the invariant length of its 4-momentum vector is zero, unlike a particle with nonzero rest mass, for which the invariant length of its 4-momentum vector is nonzero (and equal to its rest mass).

    But a photon still has energy and momentum; it's just that they're equal (in units where the speed of light is 1). This is what "the length of its 4-momentum is zero" means. In other words, a photon's energy is all kinetic energy--it's all energy due to its momentum. A particle with nonzero rest mass has extra energy (its rest energy), over and above the energy due to its momentum (its kinetic energy). But both are still particles; they both carry energy and momentum.
     
  4. Jul 6, 2012 #3
    To define further... Can a photon be "at rest"? If a photon could be trapped and held I think that means it would no longer have momentum. Without momentum would it be anything measurable?
     
  5. Jul 6, 2012 #4
    No, it cannot.
     
  6. Jul 6, 2012 #5

    Matterwave

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    Peter's description is perfectly valid, but I think it only technically answers a part of your question regarding 0 rest mass.

    The other part of your question, regarding why light is a particle is answered in quantum mechanics.

    We can say a "particle" is anything that a "particle detector" measures. In other words, a particle is anything "corpuscular" which is always measured in basic quantum units (i.e. inseparable). A particle detector ether measures whole number units of this particle or none at all (i.e. you never measure 3/2 of a photon). Even more concretely, given a source of photons, one can always reduce the intensity to such a level that particle detectors detect particles 1 at a time. This is what we usually mean when we say that light is in the form of particles called photons (rather than just stating it is an electro-magnetic wave).
     
  7. Jul 6, 2012 #6

    PeterDonis

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    As Dickfore said, a photon can't be "at rest", but some further clarification may help. First of all, when we say photons move at the speed of light and have zero rest mass, we are, strictly speaking, talking about photons moving in a vacuum. If a photon is moving through a material medium, there are several possible complications:

    (1) One could imagine photons being held inside a container with perfectly reflecting walls (I believe there are some electromagnetic cavity-type experiments that do something like this). The photons would still move at the speed of light and have zero rest mass, but they would interact with the walls, exchanging energy and momentum, so the "average" motion of all the photons could potentially be zero in the rest frame of the container, and the net effect would be to add a small amount to the observed rest mass of the container (due to the kinetic energy of the photons inside it).

    (2) When photons travel through materials which are fully or partially transparent to them (i.e., photons can travel through without being absorbed), they travel slower than the speed of light in a vacuum. However, if we go down to the quantum level, what is really happening is that photons are being absorbed and re-emitted by the atoms in the material; in between each atom, the photons still move at the speed of light and have zero rest mass, but the absorptions/re-emissions add a delay that makes it seem like the photons are moving slower.

    (3) In a superconductor, photons no longer have zero rest mass; they acquire a nonzero rest mass through their interactions with the vibrational modes of the superconductor. So they will move slower than light, and could potentially be seen to be "at rest" (though I don't know if this could actually be observed, since strictly speaking they are virtual photons, not real photons, inside the superconductor), and would still carry energy if they were at rest.
     
  8. Jul 6, 2012 #7
    But, these interacting photons become quasiparticles (cavity modes, plasmon polaritons) of a different kind and acquire (rest) mass. Furthermore, in condensed matter, lorerentz invariance looses its meaning, so there is no necessity of a dispersion relation:
    [tex]
    \omega^2 = c^2 \, k^2 + \left( \frac{m c^2}{\hbar^2} \right)^2
    [/tex]
     
  9. Jul 6, 2012 #8

    PeterDonis

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    If you're talking about my #1, yes, I didn't mean to imply that all photons in cavities or other such setups still have zero rest mass. But I can imagine an idealized container where they would.

    I'm not sure I would say it "loses its meaning", but I agree things get a lot more complicated.
     
  10. Jul 7, 2012 #9
    And #2 as well. To see why (waveguide) modes acquire rest mass, consider the wave equation, and assume that the transversal cordinate dependence separates from the longitudinal coordinate dependence:
    [tex]
    \psi(x, y, z, t) = \phi(x, t) \, \Phi(y, z)
    [/tex]
    [tex]
    \frac{1}{c^2} \, \frac{\partial^2 \phi}{\partial t^2} - \frac{\partial^2 \phi}{\partial x^2} + \lambda^{2}_{n} \phi = 0
    [/tex]
    [tex]
    \frac{\partial^2 \Phi_{n}}{\partial y^2} + \frac{\partial^2 \Phi_{n}}{\partial z^2} + \lambda^{2}_{n} \, \Phi_{n} = 0
    [/tex]
    The discrete eigenvalues for the transversal Laplace operator are determined by the geometry of the waveguide and the boundary conditions.

    Then, performing a Fourier transform of the longitudinal wave equation we get the dispersion relation:
    [tex]
    \omega^2 = c^2 k^2 + c^2 \lambda^2_{n}
    [/tex]
    Then, quantizing these propagating modes, we may assign a mass:
    [tex]
    m_{n} = \frac{\hbar \lambda_{n}}{c}
    [/tex]

    As for the propagation of em waves in a dielectric, the dispersion relation becomes:
    [tex]
    \mathrm{det} \left[ \left(\frac{\omega}{c} \right)^{2} \varepsilon_{i k}(\omega, \mathbf{k}) - k^{2} \delta_{i k} - k_{i} k_{k} \right] = 0
    [/tex]
    where [itex]\varepsilon_{i k}(\omega, \mathbf{k})[/itex] is a frequency and wave vector dependent dielectric tensor that describes the coupling of the system to the external em field.

    It means that the presence of a medium introduces a preferred reference frame (that where the medium is at rest). We no longer require Lorentz invariant expressions as dispersion relations.
     
  11. Jul 7, 2012 #10

    PeterDonis

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    Ah, ok.
     
  12. Jul 7, 2012 #11

    PeterDonis

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    At this level of modeling, though, we aren't really talking about "photons", are we? (Unless you use "photon" to mean "geometric optics approximation", I guess.) At the quantum level, are these (or the Hamiltonians corresponding to them) the equations that get quantized to determine the spectrum of photon modes?
     
  13. Jul 7, 2012 #12
    No, we are. A photon is a quantum of the electromagnetic field. The "classical equation of motion" for the electromagnetic field is simply the wave equation. Then, when you expand the solution in to eigenmodes, you promote the expansion coefficients to creation/annihilation operators and you assume canonical commutation relations between them. This is what is meant by canonical quantization of the electromagnetic field.

    I never assumed geometric optics to be valid. I solved the wave equation with a prizmoidal waveguide, and the boundary conditions for the "transversal part" gave rise to discrete eigenvalues [itex]\lambda_n[/itex].
     
  14. Jul 7, 2012 #13

    PeterDonis

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    So the answer is yes, these *are* the equations that get quantized to determine the spectrum of photon modes. Got it, thanks!
     
  15. Jul 7, 2012 #14
    The "only" approximation is in the treatment of the walls of the waveguide as "classical" objects with no Lagrangian, but only modeled by the boundary conditions that must be satisfied on them.
     
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