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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
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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

2 Answers

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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In AWT (dense aether theory) the mass is dual to energy, the gravity is dual to the pressure of radiation (not only photononic, but tachyonic too). It means, the notion of photon antigravity has certain merit in it, but only for tachyonic photons of energy lower than the wavelength of CMBR, which are supposed to exhibit negative radiation pressure in AWT. Above the CMBR energy density scale the photons contain more mass than energy and the energetic neutrinos should be used in this context instead.

Illustratively speaking: the gravity lens appears like sparse blob of more dense vacuum around massive objects, i.e. not bubble. Only the microwaves are repelled from it (Hawking radiation) - the other photons are attracted to it, so they cannot be formed with holes, because they're behaving like any other massive particles in this extent. For neutrinos the same dependence is exactly the opposite: the more massive/energetic neutrinos can escape black hole freely like weak tachyons (compare the OPERA results), these less energetic neutrinos may be swallowed with it.

So in brief: common photons can be considered as bubbles with massive boundary or like the vortex rings of aether: they've some hole inside, but their curvature is still positive as a whole.

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-1: Why do you make nonsense syllables that pretend to have meaning? It is impossible for nonsense to win in the modern media environment. – Ron Maimon Dec 31 '11 at 12:24
What is AWT? Apache Web Toolkit? – Blessed Geek Aug 22 '12 at 3:29

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