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Consider photons and gluons have 0 mass and 0 charge. In many respects they're already understood as the absence of a particle by mathematical models. Couldn't this be interpreted to mean they operate by phenomenon similar to the charge carrying "holes" responsible for transmitting electro-magnetic waves (at the speed of light) in wires?

The implications of this model would seem to be:

  • There is a relativistic aether that is similar to the concept of a theoretical inviscid in fluid dynamics (see en.wikipedia.org/wiki/Inviscid_flow).
  • Differences in "pressure" at this fundamental level may be responsible for gravity, as well as providing possible explanations for a number of other phenomenon currently challenging physics.
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Photons and gluons are bosons. Every "hole" quasiparticle I ever heard of was a fermion. Maybe you could try to make a boson by pairing up two holes... There are a lot of people who want to explain the standard model using ideas from condensed matter physics. There are also very relevant criticisms from particle physicists, e.g. the Weinberg-Witten theorem against the possibility of a composite graviton. –  Mitchell Porter Dec 23 '11 at 2:28
I retitled this, because the implications of an ether are already well considered by physicists, so even a new kind of ether wouldn't be revolutionary, and further, photons can't be holes. –  Ron Maimon Dec 23 '11 at 4:42
Thanks for the very intelligent comments. My comparison with charge carrying holes may have caused confusion. What I had in mind, while analogous to electron-holes is un-related to the charge. Instead if space is made up of a sort of (disconnected) lattice of neutral charge/spin particles, photons would essentially be the gaps in this lattice, which are almost instantly filled by the neighboring particle in the lattice, and so-on down the line. The equivalent for gluons might be if the gaps were perhaps more-or-less stationary or at least running in a tight circle in the lattice. –  user6817 Dec 23 '11 at 8:05
You can have more than one photon in the same state. It isn't clear how you would represent this in your model. –  Harry Johnston Dec 26 '11 at 5:21
Also, while the gluons don't carry electric charge, they do carry color charge, so on one level, it's not accurate to call them 'chargeless' –  Jerry Schirmer Mar 22 '12 at 16:58
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1 Answer

The first part of this idea is impossible. Photons cannot be an absence of a particle because creating a photon and annihilating a photon are not symmetric operations. In a solid, placing an extra electron or removing an electron are symmetric operations because of Pauli exclusion, and the same holds in Dirac's vacuum, when you consider the negative energy states as filled. The hole picture is something that requires Fermi Dirac statistics, as Mitchell Porter commented.

Nevertheless, Bender tries to make sense of the idea of the Dirac sea for Bosons, as part of the program of PT quantum mechanics. The result is a sophisticated mathematical trick, and it would not correspond well in a physical way to photons being the absence of a particle.

The second part of this question is not impossible, but it is known for a long time. There are at least two completely different kinds of "ether" in the vacuum:

  • QCD condensates: these condensates are both gluon condensates due to the short-ranged nature of the strong interaction, and quark condensates which have light oscillations corresponding to sound-like modes, and the phonons of these ether-oscillations are called pions.
  • Higgs condensate: this is the ether that is responsible for taking the SU(2) and U(1) of the standard model to electricity and magnetism at low energies, with a little bit of residual weak interaction.

Both these known ethers are inviscid fluids, superfluids actually, (although a charged superfluid has no low-energy flowing states and so is usually called a gapped superconductor as opposed to a superfluid, a Higgs mechanism rather than a Goldstone boson in the terminology of high energy physics) and further they are relativistic superfluids that do not break Lorentz invariance, that means that they look the same no matter how fast you are going through them.

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protected by Qmechanic Mar 3 '13 at 18:57

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