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I'll answer this question from the perspective of my own non-standard model, so don't put any part of this answer on your homework, or you'll fail the course. In Einstein's general relativity (GR), which is valid only in Minkowski space-time, it is said that light has no mass. That is a result of the definition of straight lines in space-time as being the path of light. Most people who are comfortable using the GR equations will tell you that there is no other kind of space, so the statement "light has no mass" is true in an absolute sense, not just in GR. I say, that's just plain scientific bigotry. Minkowski space-time is highly useful, but it is not the only mathematical analogy for representing the physical universe. In old fashioned Euclidean space (where a straight line is the shortest distance, and where the internal angles of a triangle always add up to 180°) light does not follow straight lines in a gravity field. Instead, light changes direction as well as energy. That happens because light has mass, and like all masses, it is attracted to all other masses. What light doesn't have is rest mass, because light cannot be at rest in any reference frame; it must always move at c. In my model, fundamental particles which can be at rest in their own reference frame consist of orbiting pairs or groups of photons, held in orbit at the speed of light around one another by the Higgs force. The rest mass of the particle is the sum of the masses of the orbiting photons in the reference frame whose origin is the center of the orbital paths. Larger particles form when fundamental particles orbit one another. When photons are grabbed by the Higgs force, they fall into a deap potential well, which increases their energy and mass many fold. (This converts zero-point energy to rest mass.) All other forces are derived from the Higgs force, which results from exchange of momentum between regular energy and dark energy. Free photons are surrounded by a Higgs field; the Higgs field of orbiting photons is spun into a spiral pattern; the spiraling Higgs fields of particles interact resulting in the other forces.In Euclidean space, accelerating a photon in a given reference frame is a matter of changing its direction and energy in that reference frame. (In Minkowski space-time, you can change a photon's energy, but not its direction; so you can't accelerate a photon.) Accelerating a fundamental particle with rest mass is a matter of accelerating the center of the orbital paths of the constituent photons, and thus translating those photons into a different reference frame. The complicate way to calculate the change of the particle's momentum would be to calculate the change of momentum for each orbiting photon, averaged over time. The result would be the same as the much simpler Newton's formula, f = ma. As has already been mentioned in this discussion, f = ma only works for non-relativistic speeds. That is because the mass is not constant. Approaching the speed of light, additional force is needed to change the mass, and f = dp/dt is the formula to use (where dp is the incremental change of momentum, and dt is the increment of time). In Euclidean space, F = dp/dt works for both particles and photons. At relativistic speeds, dp = m•dv + v•dm. At non-relativistic speeds dm is practically zero, so dp = m•dv, and f = ma. I could explain where the Higgs force comes from, but that would mean going more deaply into the nature of regular energy and dark energy and how they exchange momentum with one another. In other words, I'd have to explain my whole model. I have already done that in the New Theories section.

How do you differ 'mass' from 'rest mass' Phractality?

But I disagree with the statement that we can apply a frame of reference to a photon. The photon cannot have any frame of reference. It follows null geodesics.

Well said, especially the part where you mention the zero-point energy being the source of the mass, which is true since the Goldstone boson is a longitudinal photon in it's lowest energy state.

But note that state cannot be ψ|0> because there is no part of space absent of energy, which obviously involves concepts of zero point energy 1/2(h/2π)ω which is the quantum field of residue kinetic energy.

Quote from: Mr. Data on 06/07/2011 17:22:42But I disagree with the statement that we can apply a frame of reference to a photon. The photon cannot have any frame of reference. It follows null geodesics.What I mean is that the same photon has different energies in different reference frames. That's what redshift and blueshift are. The photon has no reference frame of its own because that would compress all space to zero length in the direction of the photon's travel, and it would dilate time infinitely. At least that would be true in the context of nothing being faster than light. However, in the context of my model, I believe dark energy propagates at least 2 x 10^10 times faster than light; so conceivably, one might describe an entirely new kind of reference frame in which a photon may be stationary without the length contraction and time dilation of special relativity. (It hurts my brain to think about this.) This would involve a preferred reference frame, stationary relative to the æther. It might be simplified by assuming dark energy is infinitely fast, but that would introduce a small error, which would have to be corrected, later, when greater precision is needed. (I think that same assumption is tacitly included in GR.) In such a reference frame, centered on a photon, the Higgs field around the photon would have different values in different directions relative to the phase and polarity of the photon. Higgs forces of attraction and repulsion between photons would depend on their phase and polarity relative to one another.

You cannot speak of frame of references for something that does not experience time.

Quote from: yor_on on 06/07/2011 17:36:18How do you differ 'mass' from 'rest mass' Phractality?Interesting question; what is mass, and what is rest mass, because you can get easily concerned with relativistic confusions over relativistic mass and rest energies.

.... Particles are uniquely identical....

It's a difficult thing you do Practicality. Although no different than some respected physicists, when they at one time describe a particle, the next a wave, then putting it all together into a outcome. And somehow very similar to the idea of 'weak measurements', expecting it to be a true description of the universe. How would you explain the photo electric effect, and black body radiation.

Quote from: Mr. Data on 06/07/2011 17:40:06Quote from: yor_on on 06/07/2011 17:36:18How do you differ 'mass' from 'rest mass' Phractality?Interesting question; what is mass, and what is rest mass, because you can get easily concerned with relativistic confusions over relativistic mass and rest energies. Yes; interesting; also difficult. My brain kept going into overload on this one, so I had to take a few coffee breaks before posting my answer. Since E = mc² = hc/λ, the mass of a photon in Euclidean space is m = E/c² = h/λc. The force of gravity between any two masses is f = G•m₁m₂/r². The gravitational attraction between a pair of photons (in Euclidean space) is f = G • h²/(λ₁λ₂c²r²).When a pair of photons are orbiting one another at the speed of light inside a fundamental particle, they are sometimes moving toward an observer and sometimes away; so their wavelengths and corresponding energies (in the observer's reference frame) are constantly changing. Since I am not a mathematician, I can't begin to describe a rigorous solution to show how the individual photon masses add to yield the rest mass of the particle. Suffice it to say that the sum of masses of the photons remains constant, and that sum is the mass of the particle. Accelerating the particle in a given reference frame is a matter of changing the velocity of that reference frame relative to the center of the particle or vice versa. For a small change of relative velocity, dv, you can apply special relativity to calculate the corresponding change of the particle's momentum, dp. The force is f = dp/dt, and the inertial mass is m = f/a = (dp/dt)/(dv/dt) = dp/dv. Theoretically, the same could be done for each of the orbiting photons that constitute the particle. Using a small enough dt, you could derive a function for the mass of each photon at each point in its orbit. Averaging for at least one complete orbit should yield the mass of the particle in the chosen reference frame. Choosing the reference frame whose origin is defined as the center of the particle, you get the rest mass of the particle. Of course it is far, far, far simpler to do the calculation for the whole particle without worrying about what each photon is doing. In all known applications, inertial mass and gravitational mass are equivalent. In my model, the one notable exception is the æther, which has no gravitational mass (since it is the medium of gravity and all the other forces), but its inertial density may be googols of times greater than that of a neutron star. [/quoteSum of the masses of photons?Please, why do you keep referring to photons as though they have a mass?

Sum of the masses of photons?Please, why do you keep referring to photons as though they have a mass?

That's where the mathematicians disgaree with the model you present. There seems no intuitive way to add a mass to a photon by simply looking at it's scalar field equations. You need some kind of symmetry breaking [or by some other method we are yet to understand.] But we do not, and cannot refer to photons as having any mass which can be summed over.

Quote from: Mr. Data on 08/07/2011 05:32:22That's where the mathematicians disgaree with the model you present. There seems no intuitive way to add a mass to a photon by simply looking at it's scalar field equations. You need some kind of symmetry breaking [or by some other method we are yet to understand.] But we do not, and cannot refer to photons as having any mass which can be summed over.I'm not aware that any mathematician has taken a serious look at my model. They probably can't, or won't, go back to thinking in terms of Euclidean space, after having invested so many years learning to think in terms of tensor analysis and warped space-time. Like Einstein said, "The only thing that interferes with [their] learning is [their] education." Once you go back to Euclidean space, it's so very simple. E = mc², so m = E/c². That is as true for a photon as it is for the rest mass of a particle. The scalar field equations don't take the Higgs force into account, and in my model, the Higgs force is asymmetrical, so there is your broken symmetry. When photons fall into a Higgs potential well, they move closer together under the attractive influence of the Higgs force, so that increases their energy and their mass many fold. They become blueshifted so that their wavelength is shortened enough to fit inside the radius of the particle, which might millions of times smaller than the original wavelength of the free photons, before they were captured by the Higgs force. This blueshifting is similar to the way photons gain energy, in GR, as they enter a gravity well, but the Higgs force is not accounted for in GR. To express the same scenario in GR, you would have to make some radical changes to GR. You would have to mathematically describe the asymmetrical warp of space-time due to the Higgs force, which is perhaps a googol times stronger than the warp caused by gravity. I say the Higgs force between two photons is asymmetrical because it has different values in different directions, relative to the phase and polarity relationship between the photons. It can be either attractive or repulsive. The vast majority of the time, photons may pass without being affected by the Higgs force; it only affects them when their distance, phase, polarity and wavelength are a good match to a strange attractor. Each species of particle is a strange attractor. I don't know if there is a Higgs boson to act as a catalyst for the formation of particles. If so what catalyst is needed to form a Higgs boson? I think it is more likely that the catalyst is just a highly improbable coincidence of perhaps four or more of the right photons meeting at the correct angles and distances simultaneously. (Kind of like four old friends simultaneously bumping into each other by accident in a revolving door in a country none of them have ever visited before, and they all just happen to be wearing the same exact clothes which none of them ever wore before, and they're all taking a bite out of the same kind of sandwitch that none of them ever ate before.) That being the case, perhaps new particles are popping into existence all over the universe, but probably more so where light intensity is high, like inside a star. Or perhaps not. I'm not sure what you mean by "summed over". The mass of a free photon, is simply m = E/c², and the mass of two free photons is simple m₁+ m₂. (As I said before, mass is relative to the observer's reference frame; that goes for both particles and for free photons.) Perhaps I have been a bit sloppy when talking about summing the masses of a pair or group of orbiting photons. Each photon has a constantly changing direction, so it energy-mass is constant (assuming circular orbits) only in reference frames which are stationary relative to the center of the particle. (At such small scales, gravity is irrelevant; only the Higgs force matters.) If the orbits are elliptical, the energy-mass might fluctuate with an extremely short period, but the kinetic energy plus potential energy should remain constant (unless the particle is unstable).

But if there is no Higgs boson?