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Does photon have size measurement because of its particle nature like electron's 3.86*10^-13m etc..

twistor59
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Manohar Alla
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2 Answers2

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When one reaches energies of GeV and experimentally measures interactions that happen below 1 fermi, the nuclear dimensions, the question of size of an elementary particle becomes separated from the concept of its dimensions.

The elementary particles of the standard model have dimensions 0. This certainty comes because the theoretical standard model fits very well practically all available particle data, and the zero dimension of its constituent particles is one of the basic blocs in the computations.

But, at the level below Fermi, elementary zero dimensional particles have complicated interactions, described by form factors, which create a size for them. In particular for the photon the size is seen as a photon structure function, which changes with the interaction exchange energies. It gives an effective size to the photons, not a fixed one, but one depending on the energy of the probing particle.

High energy photons can transform in quantum mechanics to lepton and quark pairs, the latter fragmented subsequently to jets of hadrons, i.e. protons, pions etc. At high energies E the lifetime t of such quantum fluctuations of mass M becomes nearly macroscopic: t ≈ E/M2; this amounts to flight lengths as large as 1,000 nanometers for electron pairs in a 100 GeV photon beam, and still 10 fermi, i.e. the tenfold radius of a proton, for light hadrons.

Form factors are an experimental measurement of these complicated virtual diagrams that exist whenever one elementary particle interacts with another.

So an elementary particle, including the photon, has a size measurement due to the particular interaction possibilities when scattering or being scattered by other elementary particles. A variable size measurement which depends on the energy transfers of the collisions. It is still of dimension zero.

If you have difficulty visualizing this, think of a spark gap, where an electric field is applied. The higher the field the larger the spark seen, the larger the spatial dimensions it occupies, nevertheless the geometry of the gap is fixed. In the same way the geometrical dimensions of the particle are zero, but depending on the energy of interaction ,it has a size.

anna v
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  • "The elementary particles of the standard model have dimensions 0." How is this conceivable? How to explain what mean a wavelength of 400 nm? For me any electromagnetic wave need space to exist. – dan Jun 26 '15 at 06:35
  • @danielAzuelos Yes, the electromagnetic wave needs space , but it is composed out of zillions of elementary particles/photons see http://motls.blogspot.com/2011/11/how-classical-fields-particles-emerge.html – anna v Jun 26 '15 at 09:32
  • My question was about a photon. Something with sizes (dx, dy, dz, dt) = 0 can't have a wave length. Moreover, as it isn't distinguishable of the "non existing" (which is uniq) it is the "non existing". – dan Jun 26 '15 at 17:50
  • @danielAzuelos Photons are quantum mechanical entities. They are not waves of energy as classical EM waves. They are described by a probability which depends on a wave function squared, i.e. has a sinusoidal form but it is a PROBABILITY to be at (x,y,z,t) and a point can describe it. The link I gave shows how these probability waves build up the classical wave which is an energy wave traveling in space. – anna v Jun 26 '15 at 18:13
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Although the photon appears to exist without physical volume or geometrical size, we can measure the region where the wave's magnitude is non-negligible. This happens at about half a fermi, or roughly $0.5*10^{-15}$ m.

Anubhav Goel
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