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Offline Phractality

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Split: what is Mass?
« on: 06/07/2011 16:48:28 »
Split from here: http://www.thenakedscientists.com/forum/index.php?topic=7762.0

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.
« Last Edit: 06/07/2011 17:28:21 by JP »


 

Offline Mr. Data

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Re: Split: what is Mass?
« Reply #1 on: 06/07/2011 17:19:45 »
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.

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.
 

Offline Mr. Data

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Re: Split: what is Mass?
« Reply #2 on: 06/07/2011 17:22:42 »
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.
 

Offline yor_on

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Split: what is Mass?
« Reply #3 on: 06/07/2011 17:36:18 »
How do you differ 'mass' from 'rest mass' Phractality?
 

Offline Mr. Data

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Split: what is Mass?
« Reply #4 on: 06/07/2011 17:40:06 »
How 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.
 

Offline Phractality

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Re: Split: what is Mass?
« Reply #5 on: 06/07/2011 19:40:23 »
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.

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.
 

Offline Phractality

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Re: Split: what is Mass?
« Reply #6 on: 06/07/2011 20:45:57 »
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.
I had to Google "Goldstone boson". After reading the Wikipedia article on it, I still have only the vaguest idea what it is. They mention "spontaneously broken internal symmetry generators", which might be related to my concept of the Higgs field having different strengths in different directions relative to a photon's phase and polarity. In other words, the Higgs field is asymmetric.

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.
In college, 43 years ago, my brain rebelled against learning what a Psi function is. The nomenclature is Geek to me. So I can't tell if we're talking about the same concept or not.

The energy of a free photon is E = hc/λ until it meets another photon of just the right energy, at just the right distance, phase and polarity, to generate a Higgs force of attraction strong enough to cause the two photons to orbit one another. This Higgs force then draws the photons closer together. When photons move toward an attractive force, their photonic energy increases, and their wavelength decreases. This is similar to the length contraction and time dilation of photons entering a deep gravity well; the difference is that the well is related to the Higgs force instead of gravity. Without introducing the concept of zero point enery, the mass of a particle would seem to violate conservation of mass-energy.

Zero point energy is the energy required to separate the orbiting photons which constitute a fundamental particle, freeing them to follow a straight path (in Euclidean space). Also it is the increase in energy-mass when free photons become a particle.

A fundamental particle is like a mini black hole. It doesn't fit the formulas for gravitational black holes because the force to mass ratio is so great. The weak force is perhaps 44 orders of magnitude greater than gravity; the strong force is many more orders of magnitude stronger than that, and the Higgs force many orders of magnitude stronger still. So it doesn't take much mass to make this kind of black hole.

A photon with the energy equivalent of an electron's mass has a wavelength perhaps a million times greater than the diameter of an electron. This suggests to me that a pair of photons becoming an electron would have to be blueshifted by a factor of a million or so. The trouble is that the ration of mass to wavelength should remain constant. I sprain my brain when I try to figure out how to squeeze that much photonic energy into such a small space. Perhaps the formula, E = hc/λ, breaks down near the bottom of a Higgs potential well.
 

Offline Mr. Data

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Split: what is Mass?
« Reply #7 on: 07/07/2011 00:03:58 »
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.

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.

Impossible. Particles are uniquely identical. Photons do not change any frame of reference when there is no relativistic frame of reference for them.
 

Offline Mr. Data

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Split: what is Mass?
« Reply #8 on: 07/07/2011 00:05:41 »
You cannot speak of frame of references for something that does not experience time.
 

Offline Phractality

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« Reply #9 on: 07/07/2011 02:14:47 »
You cannot speak of frame of references for something that does not experience time.

The photon does not experience time, but the person observing it does. Of course we don't actually observe a photon except when it interacts with a particle, but we may infer its continuous existance between being emitted and absorbed. 

Are you saying that a photon exists only at the instant when it is emitted and again at the instant when it is absorbed? That is a philosophical choice, like whether there is sound when a tree falls in a forest and no one hears it. To loosely quote Bill Clinton, it depends on what the definition of IS is.

I prefer to believe the photon exists continuously in the form of a wave; the particles that comprise the æther wiggle from side to side as a photon passes, even though that wiggle may or may not affect of any particle of our universe. If you accept the politically correct view that there is no æther, then it follows that photons don't exist either.

If a photon exists continuously as a wave in the æther, then its energy and wavelength are different in different frames of reference. Suppose identical photons, A and B, are emitted by a source. Space ship Alpha is moving toward the light source and absorbs blueshifted photon A; space ship Beta is moving away from the light source and absorbs redshifted photon B. Just prior to being absorbed, both photons were blueshifted in the reference frame of Alpha and redshifted in the reference frame of Beta. If the two ships had changed course at the last second, Alpha might have absorbed blueshifted B and Beta might have absorbed redshifted A. The energy that is absorbed when the photon hits the ship depends only on which ship the observer is in. 
 

Offline Phractality

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« Reply #10 on: 07/07/2011 04:20:01 »
How 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.
 

Offline Phractality

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« Reply #11 on: 07/07/2011 04:41:24 »

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

In my model, each species of particle is a strange attractor in the chaotic mix of regular energy (æthereal shear waves) and dark energy (æthereal pressure waves). This allows for extremely small variability among individuals of a given species of particle (something like Heisenberg's uncertainty principle). Only photon pairs (or groups) of just the right energies can form stable orbits around one another. Each species of particle has its own particular half life, as well as a particular very narrow range of rest mass. The rest mass of a particular species of particle may be known to 12 digits accuracy; but that is an average. We can't measure one individual particle that accurately. It is the tiny variability among individual particles which makes the universe interesting and unpredictable.
 

Offline yor_on

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« Reply #12 on: 07/07/2011 16:57:21 »
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.
 

Offline Phractality

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« Reply #13 on: 07/07/2011 20:09:44 »
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.
I don't expect I will ever develop my model sufficiently to demonstrate precisely how such higher order effects arise from more fundamental principles. I can only provide qualitative answers; perhaps others can take the baton from me and quantify my concepts. 

As I have stated many times, the universe is chaotic, and each species of fundamental particle is a strange attractor. A sea of fundamental particles becomes the medium for a higher order of chaos, in which other strange attractors group the fundamental particles into quarks, leptons, mesons, etc.

Photoelectric effect and black body radiation were described very well by Einstein in his miracle year. He treated the electron as a fundamental particle. It may turn out that an electron exists at the third, fourth or fifth tier of complexity above fundamental particles. Then Einstein's analysis might be refined, one layer of complexity at a time, perhaps requiring a century of research for each layer. Explaining photoelectric effect in terms of fundamental particles might be like explaining a football game in terms of individual blades of grass on the field.

Occams razor is a good guide for practical theories, but it can get in the way of discovery.
 

Offline Mr. Data

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« Reply #14 on: 08/07/2011 03:05:26 »
How 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.
[/quote

Sum of the masses of photons?

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

Offline Phractality

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« Reply #15 on: 08/07/2011 04:16:23 »

Sum of the masses of photons?

Please, why do you keep referring to photons as though they have a mass?
I explained that, yesterday, in my first post to this thread. Without repeating the whole argument, I'll say this:

Minkowski invented a new kind of mathematical analogy to represent the physical universe in four dimensions. In his space-time, the path of light is the definition of a straight line. Gravity bends the path of light in old-fashioned Euclidean space (where the internal angles of a triangle add up to 180°), but not in Minkowski space-time. Consequently, the space-time continuum is warped by gravity. It is a common misconception that the warp of space-time is the cause of gravity; it's the other way around.

Explicitly redefining "straight line" tacitly redefines all the old familiar parameters, including mass. Mass in Minkowski space-time ain't the same thing as mass in Euclidean space. Since gravity doesn't alter the direction of a photon in space-time, it doesn't exert a force, and the formula, f = ma, becomes meaningless. At least that seems to be the consensus among people who are comfortable solving complex 4D problems in GR (as I am not). Not every GR expert agrees, though.

Personally, I don't understand why they don't just apply the correct formula, f = dp/dt. By that formula, gravity does exert a force on a photon; that force is parallel to the photon's path, regardless of which dirction the gravity potential gradient vector points. But I never took tensor analysis in kalludge, so duh, who am I to question the GR experts on their own game?

I will, however, insist that a photon obviously has mass in Euclidean space. The gravity of a star changes both the direction and the energy of a photon. It takes force to change momentum; for every force, there is an equal and opposite reaction force; the photon has to exert an equal and opposite force on the star. The star's momentum is changed by an amount equal and opposite to the change in the photon's momentum. Otherwise, momentum would not be conserved. So a photon definitely has mass in Euclidean space.

Minkowski space-time is extremely useful, expecially when solving trajectories of large objects over long distances and times. Small spherical objects remain spherical in Minkowski space-time, even where the space-time is severely warped by gravity. That is not true of Euclidean space, which may be the reason that many GR experts insist that Euclidean space doesn't exist where gravity is that strong. However, my model of particles, consisting of orbiting pairs or groups of photons, looks very strange if the path of a photon is a straight line. How can anyting go around in circles and follow a straight line at the same time? That is why I prefer to describe my model in Euclidean space.

Also, if the path of light is the definition of a straight line, then anything that bends the path of light in Euclidean space can warp space-time. I believe the Higgs force, bends the path of light to a far greater extent than gravity does (perhaps a googol times more strongly). So GR would need to be modified to account for these sources of space-time warp in order to reconcile GR with the standard model of particle physics, as well as to reconcile it with my model. Doing so might be necessary to make my model mathematically manageable, but it is not necessary to a conceptual understanding of my model. I'll handle the concepts; I leave the math to the mathematicians.
 

Offline Mr. Data

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« Reply #16 on: 08/07/2011 05:32:22 »
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.
 

Offline Phractality

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« Reply #17 on: 08/07/2011 07:13:20 »
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.

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).
 

Offline CPT ArkAngel

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« Reply #18 on: 08/07/2011 12:05:36 »
But if there is no Higgs boson?

How do you see a black hole?

The only explanation for the cause of Relativity in GR is the properties of spacetime. But how can you explain entanglement in such a spacetime where nothing can propagate faster than light? The only conclusion is that Euclidean space or a very strange modified spacetime is more fundamental. The use of spacetime by Einstein was to explain why nothing propagates faster than light...

In my model, there is no need for a strange attractor, the wave (fields) and entanglement between particles regulate the laws of Relativity and Entanglement. With it, the charge produces electromagnetism and inertia. In the same way as you say, charge polarity, polarization, but with adding symmetry and specific mass density (curvature of light and apparent spacetime), are necessary to produce a specific particle. I agree with you about gravity causing the light to bend. But by including the black hole to the theory, we find something strange, the timerate is zero at the event horizon. If the timerate is zero, it is zero for any observer with a non zero timerate. GR tensors cannot explain otherwise because you need continuity, this is where Susskind explanation is flawed. You can theorize anything you want inside Susskind Black hole without a possibility to deny it. At zero timerate, either you have infinite gravity or zero. Which one you choose? If you have no gravity, it is because gravity is generated there. What possibly could have a zero timerate and generate gravity? Your particles like mine are made of photons and they have a mass. Black hole exhibits a particle like behaviour when they are coupled with another one. I don't even bother about thinking of infinite gravity.

I must say, i like the way you think and you have very good ideas, but try to have a larger perspective that would ground macroscopic and microscopic scales together. And by the way, as QM is actually interpreted, it simply fails to generate the macroscopic reality we live in. I know you don`t like to think in terms of energy momenta but it is my explanation for the bending of light by gravity.

One way to simply describe rest mass of an elementary particle is: Inertia produced by the particle measured in its own frame. This way, it implies inertia=mass.
« Last Edit: 08/07/2011 12:44:19 by CPT ArkAngel »
 

Offline Mr. Data

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« Reply #19 on: 08/07/2011 19:57:24 »
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.

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).

Photons = rest mass = ???????

How is it possible for a photon to have a rest mass. It only shows you cannot know what a rest mass is as you are giving it to a photon, which is impossible. A photon never rests. Nor does it have a mass.
 

Offline yor_on

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« Reply #20 on: 08/07/2011 20:06:47 »
I enjoy the way you think Phractality, as far as I know I've never seen anyone except you trying to explain 'relativity' in a Euclidean space :) It's highly original and quite refreshing reading. You seem to have a good grasp on what differs the geometries, and the way you describe them is intriguing to me. I can see your argument for mass, and if you by geometries define something possible to 'exist' on its own, then a Euclidean space should exist too.

I see space as a 3d expression myself, with gravity being the color that flows in three dimensions under the expression of our arrow. But differing from the Euclidean I see space as 'being the arrow' in some fundamental way. If we accept that mass changes 'time' relative some other ones observation (frame of reference) and if we also accept that relative motion has this ability, then it seems to me as if we have localized a 'arrow' both in mass and in 'motion'.

The problem I've had with that a long time is a uniform motion relative a acceleration. It would have been very nice if I could define a 'time dilation' only relative accelerations as that would have given us a simple definition of it as being 'gravity', but it's not that simple. Uniform motion will have a 'time dilation' intrinsic to it too, at least as I understands it for the moment. I've been wobbling there, trying to find some way to define it to only 'accelerations' aka gravity, but I haven't succeeded.

In what way do you think mass exist, as a result of a Higg field solely? It's reasonable as you define us as 'bosons', but where comes the dichotomy in play? Where does 'bosons' becomes defined 'particles', and where do the transition come from the probabilistic definition of a outcome, to what we see macroscopically with a football getting kicked between players, macroscopically?
« Last Edit: 08/07/2011 20:22:57 by yor_on »
 

Offline yor_on

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« Reply #21 on: 08/07/2011 20:09:37 »
Good question Mr Data :) As I understands it, he sees the 'mass' as a direct result of translating a Minkowski 4D representation geodesic to a Euclidean 3D.
 

Offline Mr. Data

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« Reply #22 on: 08/07/2011 22:26:15 »
As I said to bored chemist once, mass is a behaviour of the system. If a photon does not exhibit this behaviour, then what good is there in saying it has the essential ingredients to make the behaviour work? It seems paradoxical, and backward.
 

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« Reply #23 on: 09/07/2011 00:58:26 »
In this case it is a question of explaining the change of geometry, that is if I got the idea right. And as we define 'rest mass' from a concept of where it is 'at rest', as defined by a geodesic, not caring of how many time you changed 'speed', relative some arbitrarily fixed point, to get into those various geodesics, I can't help but finding it interesting. You might see it as a alternative way of trying to define a relative universe, in where you then must get a 'mass' as the photon 'bends' as defined from a Euclidean space. But this is how I understand the idea, I'm not sure if this is the way Phractality sees it?
 

Offline Phractality

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« Reply #24 on: 09/07/2011 06:29:14 »
I'm not ignoring you guys. Those are tough questions, and I don't want to give you flippant answers which I will later regret. Besides, my attention span is limited, these days. Please be patient with me, and feel free to talk behind my back.
 

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« Reply #24 on: 09/07/2011 06:29:14 »

 

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