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Good question George, You said " particles of light " so you are talking about Photons. I always thought that Photons are exceptions to the rule ( more Velocity = more Mass ) meaning as you know, the faster something goes the more mass it gains. The photon is the gauge boson or electromagnetism (1.) and therefore all other quantum numbers of the photon (such as lepton number: baryon number, and flavour quantum numbers) are zero. (2.) (1.) Role as gauge boson and polarization section 5.1 inAitchison, I.J.R.; Hey, A.J.G. (1993). Gauge Theories in Particle Physics. IOP Publishin: ISBN 0-85274-328-9 (2.) See p.31 inAmsler, C.; et al. (2008). "Review of Particle Physics". Physics Letters B 667: 11340. Bibcode: 2008PhLB..667....1P doi:10.1016/j.physletb.2008.07.018 --------------------------------- The energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ): where k is the wave vector (where the wave number k = |k| = 2π/λ), ω = 2πν is the angular frequency, and ħ = h/2π is the reduced Planck constant. (3.) (3.) Davison E. Soper, Electromagnetic radiation is made of photons, Institute of Theoretical Science, University of Oregon Since p points in the direction of the photon's propagation, the magnitude of the momentum is Also as a gauge field, i.e., as a field that results from requiring that a gauge symmetry holds independently at every position in space time. (4.) ( 4.) Ryder, L.H. (1996). Quantum field theory (2nd ed.). Cambridge University Press. ISBN 0-521-47814-6. For the electromagnetic field, this gauge symmetry is the Abelian U (1) symmetry of a complex number, which reflects the ability to vary the phase of a complex number without affecting observable or real valued functions made from it, such as the energy or the Lagrangian. The quanta of an Abelian gauge field must be massless, uncharged bosons, as long as the symmetry is not broken; hence, the photon is predicted to be mass less, and to have zero electric charge. I guess this is my round a bout way of saying, I am not sure but my guess would be no, I think the there would be no change in energy, but their would be a change in Velocity. So that means I have no idea, LOL. I would like to know one way or another, again good question.

Opinions, comments or exclamation points. Francesco Celani 1, a stalwart LENR (Low Energy Nuclear Reactions) experimentalist has built a sucessful cold fusion generator that has been giving out 5-10 watts for over 6 weeks. It consists of a glass container filled with hydrogen and containing a long thin nickel-copper wire. He says he had to“pump” it up for 2 days before getting it started. It seems that Celani’s idea is to compress as many protons as possible into the nickel-copper lattices. Roughing up the surface of the wire, as Celani does, helps absorb protons into the lattice more easily. The protons may be compressed enough that they could gain enough electrostatic energy to make them as massive as a neutron. The protons would then form a very tight, compressed structure. The lattice the protons are contained in should keep this condensed structure in place. Since the proton now has the mass of the neutron, the following reaction can take place: e-+ u -à d + ve (this turns the proton into a neutron) The reaction is carried out by the weak interaction through the exchange of a virtual W- intermediate vector boson. If a new neutron is formed in this manner, a small vacuum of electric repulsion is formed and the rest of the proton structure around it would cave in on it and one of the protons would join it. The surrounding protons would squeeze this new n-p pair until it responds by shoving back and producing vibrations in the proton structure which are carried to the metal lattice and produce heat. In doing this, the n-p pair loses it excess energy and becomes a deuteron nucleus. There should be no gamma radiation released in this process. Detection of deuterium in the final state of Celani’s generator would help verify this. The power output should be: P = N T (1.2 x 10-12J) where N = # protons, T = transition rate (probability of reaction occurring/s) and 1.2 x 10-12 J is the energy released per event. T is found from: T = 2(pi)[<f/V(0)\i>]2 p(f) where p(f) is the density of final states and the interaction matrix, V(0), is found from the Lagrangian density of the electroweak theory of the Standard Model: V(0) = v+eLd+Lo~ui (ie/(sqrt(2)sinOW))W-u eL uL The left-handed fields are Dirac fermion fields. I get: T = 2(pi)(md/(2Ed))(e/(sqrt(2)sinOW))2(1/(2MW)) (me/(2Ee))(mu/(2Eu))(4(pi)/Ev ~ (6.28 x 10-8)/Ev For Ev~ .511 MeV and N ~ 1020 (reasonable from Celani’s set-up data) then P = 15Watts compared to Celani’s 10Watts A possible method to increase power might be to include a large nucleus (an inert gas most likely) into the hydrogen mix. When the large nucleus is absorbed in the metal lattices along with the protons, the higher electrostatic energy would give the protons their necessary mass more quickly and without being as compressed. 1 pg 97, Steve Featherstone, “Andrea Rossi’s Black Box”, Popular Science Nov2012, Vol281 No5