c, the speed of light, and the BPhoton
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Re: c, the speed of light, and the BPhoton
Very good points all around, Nevyn. I hadn't visualized it that way, that axial spins won't double the radius, until now. But it seems apparent. So beyond the first axial, are there no further axials? Wouldn't another stack necessitate another endover spin?
I completely agree that this radius is just a volume, especially given our recent animations. Would the particle be moving so fast that its potential hits could be treated as this radius, as a matter of heuristics perhaps?
But given that it is still just The Particle (radius of 1) moving through a volume, it's easy to see how even spunup photons could dodge most other photons in the field (assuming a low enough density). And thus, retain linear velocity c.
I completely agree that this radius is just a volume, especially given our recent animations. Would the particle be moving so fast that its potential hits could be treated as this radius, as a matter of heuristics perhaps?
But given that it is still just The Particle (radius of 1) moving through a volume, it's easy to see how even spunup photons could dodge most other photons in the field (assuming a low enough density). And thus, retain linear velocity c.
Jared Magneson Posts : 444
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Re: c, the speed of light, and the BPhoton
Nevyn, I recall our previous discussion concerning your observation that the bphoton’s first axial spin should be its only axial spin. Miles seems certain, and his radius doubling calculations require higher axial spins. I suggested that endoverend spinning of the complete X,Y and Z orthogonal spin set might collapse, or simplify into the next higher axial spin. I still believe that is a possibility. I’m surprised you haven’t asked or received any additional thoughts from Miles about it.
It seems the electron is the smallest radius bphoton stacked spin charged particle that is too large to achieve a lightspeed linear velocity due to increased collisions from charge fields emitting more numerous and smaller photons. I question that assertion. We may find that the ultraviolet photon Jared cites above, 1821 times smaller than the electron, can easily reach lightspeed but can also be found at lower velocities due to increased collisions when crossing higher photon density charge fields. I see no reason not to accept that large photons can interact at subc linear velocities; that is certainly the case with larger charged particles, protons and neutrons (atomic matter). Your particleantiparticle musing assume both have zero linear velocities.
You asked, “How do photons travel?” Photons travel when they acquire a linear velocity from collisions. They continue to gain or lose energy, or spins, with each collision, but larger photons probably do not reach lightspeed when they are subject to an increased number interfering collisions.
After all our previous discussions, the idea that photon collisions are comparable to billiard ball collisions seems ludicrous. True, the bphoton is in a larger volume when it is part of a larger spin, but photons don't simply cross paths, they weave paths that become impossible to avoid for larger photons. The slower the linear velocity, the greater the likelihood that the larger photons are involved in many simultaneous extended collisions which we call charge recycling.
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It seems the electron is the smallest radius bphoton stacked spin charged particle that is too large to achieve a lightspeed linear velocity due to increased collisions from charge fields emitting more numerous and smaller photons. I question that assertion. We may find that the ultraviolet photon Jared cites above, 1821 times smaller than the electron, can easily reach lightspeed but can also be found at lower velocities due to increased collisions when crossing higher photon density charge fields. I see no reason not to accept that large photons can interact at subc linear velocities; that is certainly the case with larger charged particles, protons and neutrons (atomic matter). Your particleantiparticle musing assume both have zero linear velocities.
You asked, “How do photons travel?” Photons travel when they acquire a linear velocity from collisions. They continue to gain or lose energy, or spins, with each collision, but larger photons probably do not reach lightspeed when they are subject to an increased number interfering collisions.
After all our previous discussions, the idea that photon collisions are comparable to billiard ball collisions seems ludicrous. True, the bphoton is in a larger volume when it is part of a larger spin, but photons don't simply cross paths, they weave paths that become impossible to avoid for larger photons. The slower the linear velocity, the greater the likelihood that the larger photons are involved in many simultaneous extended collisions which we call charge recycling.
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LongtimeAirman Admin
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Re: c, the speed of light, and the BPhoton
Jared Magneson wrote:So beyond the first axial, are there no further axials? Wouldn't another stack necessitate another endover spin?
Ever since I started building my first stacked spin simulator, I realised that there could not be any axial spins above the first. When you are trying to model these things you have to dig deep into what they are and how they work, as you know, and I quickly hit a problem when dealing with higher axial spins. Either the higher axial spin doubles the radius or it does not. There is no middle ground and I don't see how it can be called axial if the spin axis doesn't go through the center of what is spinning. To me, that is what axial rotation means. The object must spin about its own center which in the case of a higher axial spin would mean through the center of the last Z spin.
If we assume that it does double the radius in some way, what dimension is it spinning about? It can't be X, Y or Z because they come next. There are only 3 dimensions so what is the direction of the higher axial spin axis? It just doesn't make any sense.
I just ignored them. I programmed them in but never used them. I don't believe removing higher axial spins causes any problems with Miles theory except changing some of the math occasionally. It certainly hasn't halted my own progress.
Jared Magneson wrote:Would the particle be moving so fast that its potential hits could be treated as this radius, as a matter of heuristics perhaps?
I don't think so. This is the realisation I hit recently. I had this vague idea that the spins happened so fast that they would appear as a boundary to charge photons. But once I started to work with both spin and linear velocity, I realised that they can't move that fast because both the linear velocity and the tangential velocities are the same, c. Add in the radius of each spin level and you realise that they rotate slower the higher they are. When you reach an electron, with about 14 spin levels, the top spin level is rotating quite slowly with respect to c.
That seems like such a weird statement: something moving at c can be considered slow with respect to c. It is true nonetheless because it is the rotation, not the tangential velocity, that is slow.
Even if we just consider the collisions we don't find that radius because collisions can happen at any point in the cycle, whether the BPhoton is on the outside edge or in the middle. The only way to find that radius through collisions is to look at the maximum radius that a collision occurs at, assuming you could know what the center is.
The problem of the higher axial spins is actually affecting me at the moment. I am working on calculating the position and orientation of a BPhoton given any number of spin levels. This removes the need for all of those translation and rotation groups (or pivot points) we have used in our stacked spin models. Instead, I just calculate a single matrix that puts the BPhoton in the correct place (which is just what the 3D system does with those groups anyway). I am doing this on the GPU for efficiency reasons so, in theory, I could have (2.1 / max spin levels) billion photons within the same data structure, each with its own spin levels (but a common maximum number of spin levels, but they can be individually turned off so you can still have 1 photon with 4 spins and another with 6 spins, for example).
I have to deal with those higher axial spins knowing that I won't personally use them. They usually don't get in the way, but they do take up space in the data and time in the calculations so I would prefer to remove them completely. Maybe it's time to confront Miles about them. I need a definitive answer as to whether they exist or not and if so, how they operate. Personally, I just can't see how they can exist.
Nevyn Admin
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Re: c, the speed of light, and the BPhoton
LongtimeAirman wrote:Nevyn, I recall our previous discussion concerning your observation that the bphoton’s first axial spin should be its only axial spin. Miles seems certain, and his radius doubling calculations require higher axial spins.
Miles current math uses those higher axial spins, but it doesn't require them. We could easily rewrite the equations without them.
LongtimeAirman wrote:I suggested that endoverend spinning of the complete X,Y and Z orthogonal spin set might collapse, or simplify into the next higher axial spin. I still believe that is a possibility.
I don't understand this. The X and Y spins don't collapse or simplify into the Z spin, so why would the Z spin do so? Miles explicitly calls the higher axial spin an endoverend spin so it is just like any other spin level. There is nothing special about it apart from the name. But a spin level has both a radius and a rotation axis and that axis must be outside the gyroscopic influence of the inner spins. That's what gives it the endoverend motion. So how can it possibly be axial? Endoverend spin means rotating about an edge whereas axial spin means rotating about a center. The two just don't equate.
LongtimeAirman wrote:I’m surprised you haven’t asked or received any additional thoughts from Miles about it.
That's because I haven't asked. I try to figure things out on my own in the first instance. I ask you guys for help in the second instance (with respect to Miles work). As I said in my response to Jared, I think it is time I did so though.
LongtimeAirman wrote:It seems the electron is the smallest radius bphoton stacked spin charged particle that is too large to achieve a lightspeed linear velocity due to increased collisions from charge fields emitting more numerous and smaller photons. I question that assertion. We may find that the ultraviolet photon Jared cites above, 1821 times smaller than the electron, can easily reach lightspeed but can also be found at lower velocities due to increased collisions when crossing higher photon density charge fields. I see no reason not to accept that large photons can interact at subc linear velocities; that is certainly the case with larger charged particles, protons and neutrons (atomic matter).
This is a tricky question to answer definitively. When we measure light, we don't measure an individual photon, we measure a lot of them. So the speed of light is just an average. There could certainly be specific photons moving slower than c.
Maybe a better question to ask is what would a slow photon look like?
The answer is, it depends on what you are measuring. If we are measuring the frequency or wavelength, then it would just look like a smaller photon because its internal frequency/wavelength would not be stretched out as much as it would be if it was travelling at c so the frequency would be measured as faster than a UV photon, for example, and the wavelength would be shorter.
LongtimeAirman wrote:Your particleantiparticle musing assume both have zero linear velocities.
It assumes they have a common linear velocity, not necessarily zero. But yes, I was thinking of them spinning without any linear velocity.
LongtimeAirman wrote:You asked, “How do photons travel?” Photons travel when they acquire a linear velocity from collisions. They continue to gain or lose energy, or spins, with each collision, but larger photons probably do not reach lightspeed when they are subject to an increased number interfering collisions.
That was just a link to Miles paper containing the quotes I used. I wasn't actually asking the question.
LongtimeAirman wrote:After all our previous discussions, the idea that photon collisions are comparable to billiard ball collisions seems ludicrous. True, the bphoton is in a larger volume when it is part of a larger spin, but photons don't simply cross paths, they weave paths that become impossible to avoid for larger photons. The slower the linear velocity, the greater the likelihood that the larger photons are involved in many simultaneous extended collisions which we call charge recycling.
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Yes, billiard ball mechanics is true to a certain extent, but our particles enjoy more freedom than a billiard ball so it is a bit more complicated.
The weaving paths don't make it impossible to avoid a collision, just less likely.
Nevyn Admin
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Re: c, the speed of light, and the BPhoton
I'm very interested in your progress programming this phenomenon, Nevyn, since you're one of the few in any position to help me diagram it as well. Your model/programming is by no means less... important or critical than mine, mine just makes it easier to use in "movie" animations is all. If I could just import your setup directly into Maya we'd be so much further along, and ready to tackle electricity and interatomic motions too, but it's just so tedious on my end. I really appreciate your work on this and your patience with my misunderstandings and errors.
Keep calling it like you see it, sir.
Keep calling it like you see it, sir.
Jared Magneson Posts : 444
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Re: c, the speed of light, and the BPhoton
My hope is to implement stacked spin collisions using this new model. That's why I need so many photons (well, I don't need billions of them). I have the first level of math worked out, in theory as I haven't tested it yet. It's all based on a single equation for determining the rotation angle of a spin level. I call it the Pivoter Equation because of its form: Pivt/4r (Pi=3.14 but it uses both version of Pi since the 4r is referring to kinematic Pi) and it is used to pivot between tangential velocities and angles. Where v is the tangential velocity, t is the time since last update (ie delta t) and r is the radius of the spin level. I work out the rotation angle for each spin level and then create a matrix containing the translation and rotation per level. Then I just multiply all of those matrices together to get the final matrix for that photon. Fairly simple, really. The implementation is proving a bit troublesome though. Working on the GPU is a bit of a pain when things don't work as you expect. Very hard to debug the code when it runs somewhere that you can't reach and can't do simple things like print out a debug message. It's fun though.
I'm sure we could get this working in Maya once you have some experience with scripting. It would be even better if you can write to the GPU in Maya but we don't need that to implement a single photon or even just a few photons. I've had to move back to Java for this work as I couldn't get direct access to the GPU in a browser. In theory I can through an API called WebCL but no browsers have implemented it and the third party versions are a little bit of a pain. I tried to use the WebGL API (which ThreeJS is based on) which gives me shaders to work with, but they are a bit too cumbersome for this type of work. I'll see where it leads and look at working back towards a browser at some point in the future. Hopefully the WebCL API has been accepted by then.
I'm sure we could get this working in Maya once you have some experience with scripting. It would be even better if you can write to the GPU in Maya but we don't need that to implement a single photon or even just a few photons. I've had to move back to Java for this work as I couldn't get direct access to the GPU in a browser. In theory I can through an API called WebCL but no browsers have implemented it and the third party versions are a little bit of a pain. I tried to use the WebGL API (which ThreeJS is based on) which gives me shaders to work with, but they are a bit too cumbersome for this type of work. I'll see where it leads and look at working back towards a browser at some point in the future. Hopefully the WebCL API has been accepted by then.
Nevyn Admin
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Re: c, the speed of light, and the BPhoton
We should have no problem writing to the GPU in Maya  all my Nucleus™ dynamics (nDynamics) already use the GPU readily, as well as the Viewport using it exclusively. All the vids I've done have been GPUbased so far, in this regard.
But how do we test this at light speed? I've never made a scene anywhere near the length (size) that light travels in one second, so that may be another barrier for me down the road. Your math seems pretty clean so far. I'll keep plugging away at PyMEL until I can get somewhere useful with it.
What I'd like to do ideally would be attach the camera to one, main photon and then "film" its travels as it gets bombarded by other photons, or misses them, as the case may be. In one second alone there could be many hits/misses, so if we stretch that second out into 60 (timelapse it, basically), we may get a good visual representation to work with. Just some ideas.
But how do we test this at light speed? I've never made a scene anywhere near the length (size) that light travels in one second, so that may be another barrier for me down the road. Your math seems pretty clean so far. I'll keep plugging away at PyMEL until I can get somewhere useful with it.
What I'd like to do ideally would be attach the camera to one, main photon and then "film" its travels as it gets bombarded by other photons, or misses them, as the case may be. In one second alone there could be many hits/misses, so if we stretch that second out into 60 (timelapse it, basically), we may get a good visual representation to work with. Just some ideas.
Jared Magneson Posts : 444
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Re: c, the speed of light, and the BPhoton
Jared Magneson wrote:We should have no problem writing to the GPU in Maya  all my Nucleus™ dynamics (nDynamics) already use the GPU readily, as well as the Viewport using it exclusively. All the vids I've done have been GPUbased so far, in this regard.
From my perspective, there is a big difference between using the GPU and having access to it. I'm sure Maya uses the GPU as much as it can, and since GPUs are designed for such applications, that is a lot. But it is another thing to give Maya users access to the GPU through code. If it can accept OpenCL code then it does (or it might use CUDA, but I hope not) and we can do some truly magical things with it.
Jared Magneson wrote:But how do we test this at light speed? I've never made a scene anywhere near the length (size) that light travels in one second, so that may be another barrier for me down the road.
You don't need to work at light speed, you just need to make sure that everything is moving at the correct speed relative to each other. We want to see what is going on at this small scale so we don't want them rushing around at extremely fast speeds. We want to see them in slow motion so we can study what they do. You set time in the model by specifying the value of t in my equation above. You might tell the model that 1ns has gone by but in reality, 10ms have actually passed. On the next frame, 12ms might have passed but we still tell the model that 1ns has passed. At least, if you were recording the frames for a video that is what you would do. For viewing by a human, you let the model time vary based on the real time that has varied so that it doesn't slow down and speed up (but can suddenly jump).
Jared Magneson wrote:Your math seems pretty clean so far. I'll keep plugging away at PyMEL until I can get somewhere useful with it.
Yeah, there isn't much to it at the moment. I want to make the spin calculations as efficient as possible to make room (time) for the collision calculations which will be much more expensive. That is the real work in this project. I have a reasonable idea of how it works but am yet to attempt any math and logic.
If you want to play with my equation, then it might help if I break it down a bit so you can see what it is doing and why.
A = Pi * vt / 4r
There are 3 parts to the equation: a circle and two distances. The full equation is 2Pivt/8r which shows us a bit more than the reduced form. You can see that 2Pi represents a circle. 8r represents the circumference of the spin level (2*KPi*r). That is the kinematic Pi being used. The tangential velocity and the time are multiplied together to give us the distance traveled at that velocity in that time since v = d/t => d = vt.
We take the distance traveled (if we were moving in a straight line) and divide it by the circumference of the level. This gives us the distance traveled as a percentage of the circumference. We then multiply that by 2Pi to give us an angle. That angle can be larger than 2Pi as it represents the complete motion which could include multiple revolutions of the spin level. I keep 2 copies of that value around: one asis so I can always tell the full rotation if I want it and another that is between 0 and 2Pi so that I can use that in any graphics math since I just want to know where it is to render it.
The angle represents the amount of rotation in the time given. You can give the equation an absolute time (more of an elapsed time) and it will give you an absolute angle or you can give it delta time and it will give you a delta angle. It is the angular motion equivalent of d=vt and can be used in the same way. It basically converts between tangential velocity and angular velocity as defined by the mainstream, not Miles, but it is doing it in a Mathis way. Strange, I know.
Nevyn Admin
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Re: c, the speed of light, and the BPhoton
At a cursory glance that seems to follow just fine, quite elegant really. Thank you for taking the time to explain your thinking and math, and it certainly seems useful to simplify the math so we can feed more instances into the calculation later. I do need to study your equation more as it's hard to visualize for me, but your logic seems sound for sure.
(Tech Note) Inside Maya, Bullet and a few other PhysXbased sims use CUDA but OpenCL should be far more accessible. Either way, there's quite a bit of CPU firepower around here to be used if need be to get things started, 34 cores in play if I pile them all up. Still, a hundred times that in GPU cores, so of course that would be ideal.
Pretty genius "Mathism" at the end there too, by the way.
(Tech Note) Inside Maya, Bullet and a few other PhysXbased sims use CUDA but OpenCL should be far more accessible. Either way, there's quite a bit of CPU firepower around here to be used if need be to get things started, 34 cores in play if I pile them all up. Still, a hundred times that in GPU cores, so of course that would be ideal.
Pretty genius "Mathism" at the end there too, by the way.
Jared Magneson Posts : 444
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Re: c, the speed of light, and the BPhoton
I still need to read the last 3 posts but I have to write this:
(I'm using rough numbers for brevity)
A Bphoton with 0 spins moves 300,000 km in 1 second.
If it has 1 spin, in that second it also completes 1300 cycles. It will have travelled 300,000 km  10^25 m. There's so tiny a difference that it can be ignored, and also it can't be measured.
Stack spin after spin and you will double that tiny difference every time. You'll surely reach a number of stacked spins where that difference can't be ignored any more, and also it will be measured in experiments.
This is why infraredphoton and ultravioletphoton are treated like light and electronphoton is treated like a particle, even if they all are Bphotons with a stable number of stacks. (IR has 3 spins, UV has 6 and electron has 9)
I'm not sure about the number of spins, but that is the principle I'm trying to explain. And that is taken directly from the photon2 paper, from the part where Mathis explains why he is taking c as linear speed.
(I'm using rough numbers for brevity)
A Bphoton with 0 spins moves 300,000 km in 1 second.
If it has 1 spin, in that second it also completes 1300 cycles. It will have travelled 300,000 km  10^25 m. There's so tiny a difference that it can be ignored, and also it can't be measured.
Stack spin after spin and you will double that tiny difference every time. You'll surely reach a number of stacked spins where that difference can't be ignored any more, and also it will be measured in experiments.
This is why infraredphoton and ultravioletphoton are treated like light and electronphoton is treated like a particle, even if they all are Bphotons with a stable number of stacks. (IR has 3 spins, UV has 6 and electron has 9)
I'm not sure about the number of spins, but that is the principle I'm trying to explain. And that is taken directly from the photon2 paper, from the part where Mathis explains why he is taking c as linear speed.
Ciaolo Posts : 133
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Re: c, the speed of light, and the BPhoton
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Airman wrote:I suggested that endoverend spinning of the complete X,Y and Z orthogonal spin set might collapse, or simplify into the next higher axial spin. I still believe that is a possibility.
Nevyn wrote. I don't understand this. The X and Y spins don't collapse or simplify into the Z spin, so why would the Z spin do so? Miles explicitly calls the higher axial spin an endoverend spin so it is just like any other spin level. There is nothing special about it apart from the name. But a spin level has both a radius and a rotation axis and that axis must be outside the gyroscopic influence of the inner spins. That's what gives it the endoverend motion. So how can it possibly be axial? Endoverend spin means rotating about an edge whereas axial spin means rotating about a center. The two just don't equate.
Airman. Assume we have a complete orthogonal spin set, ie. X,Y and Z. The next endoverend spin radius doubling must be another X or Y spin – reorient that to the top horizontal for reference. I believe that a vertically oriented tangential collision will convert that horizontal toroidal volume into a spherical one.
You had shared your observation that you could not find a physical cause for higher axial spins in http://milesmathis.forumotion.com/t118p25stackedspinmotionsimulator#1237.
Airman. I'm sorry it's taken me so long to reply. I agree. The resulting spherical axis will form along a horizontal line passing through the toroid center and also through the toroid’s body, but it does not pass through the bphoton which is at the “tangential” collision point near the outer edge of the torus as the new spherical axis is formed.
Any such vertical collision can give any torus a spherical spin. The problem is that a stable axial spin cannot occur without a complete orthogonal spin set.
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Airman wrote:I suggested that endoverend spinning of the complete X,Y and Z orthogonal spin set might collapse, or simplify into the next higher axial spin. I still believe that is a possibility.
Nevyn wrote. I don't understand this. The X and Y spins don't collapse or simplify into the Z spin, so why would the Z spin do so? Miles explicitly calls the higher axial spin an endoverend spin so it is just like any other spin level. There is nothing special about it apart from the name. But a spin level has both a radius and a rotation axis and that axis must be outside the gyroscopic influence of the inner spins. That's what gives it the endoverend motion. So how can it possibly be axial? Endoverend spin means rotating about an edge whereas axial spin means rotating about a center. The two just don't equate.
Airman. Assume we have a complete orthogonal spin set, ie. X,Y and Z. The next endoverend spin radius doubling must be another X or Y spin – reorient that to the top horizontal for reference. I believe that a vertically oriented tangential collision will convert that horizontal toroidal volume into a spherical one.
You had shared your observation that you could not find a physical cause for higher axial spins in http://milesmathis.forumotion.com/t118p25stackedspinmotionsimulator#1237.
Nevyn. Let's try this from a different perspective. I want you to see that the spin axis required to turn a torus into a sphere does not go directly through the central hole but must pass through the body of the torus.
Airman. I'm sorry it's taken me so long to reply. I agree. The resulting spherical axis will form along a horizontal line passing through the toroid center and also through the toroid’s body, but it does not pass through the bphoton which is at the “tangential” collision point near the outer edge of the torus as the new spherical axis is formed.
Any such vertical collision can give any torus a spherical spin. The problem is that a stable axial spin cannot occur without a complete orthogonal spin set.
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LongtimeAirman Admin
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Re: c, the speed of light, and the BPhoton
Ciaolo wrote:
A Bphoton with 0 spins moves 300,000 km in 1 second.
If it has 1 spin, in that second it also completes 1300 cycles. It will have travelled 300,000 km  10^25 m. There's so tiny a difference that it can be ignored, and also it can't be measured.
Stack spin after spin and you will double that tiny difference every time. You'll surely reach a number of stacked spins where that difference can't be ignored any more, and also it will be measured in experiments.
I'm totally with you here. The spin offset (we can call it) of the first stacked spin is then 0.0000000000000000000000002 meters, and as you can see we have to double it many times to get to any appreciable size at the macro level, but relative to the singleaxialspin photon the electron is still very large.
Ciaolo wrote:This is why infraredphoton and ultravioletphoton are treated like light and electronphoton is treated like a particle, even if they all are Bphotons with a stable number of stacks.
I think this may have been true traditionally, but in "our" physics they're all the same particle. Light is just a word. The electron's volume of influence (travel volume?) is much bigger, but it's also still dodging most of the charge around it simply because that fundamental particle is still the same size it always was; it's just traveling in a larger volume and happens to be more recursive (doubling back on its previous motions) than smallerspin particles. So I think what's happening is that it collides with a smaller photon and then collides with it again, and again, and is doing this with multiple photons over a given dt.
We know from Mathis that the electron is recycling/emitting ~35,000 times its own mass in photons per second (19*1821) which could indicate that, given my hypothesis above, it's possibly colliding with a given photon multiple times before emitting it. But does it mean that the electron has collided with EVERY photon that is being recycled?
A great many photons may pass through this volume untouched and unscathed, but those wouldn't be part of its emission proper. And what would this look like? This is what we're hellbent on diagramming, here, to understand how photons traveling at c interact with larger particles traveling subliminal.
Jared Magneson Posts : 444
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Re: c, the speed of light, and the BPhoton
@Airman:
Keep in mind that the spin volume only "forms" an actual torus on the x1 spin, or the first stacked spin beyond the axial. All higher stacked spins form much more complex shapes and the torus is not accurate or useful as a descriptor after that.
Refer to my video again if it helps, although we do need to update it and (hopefully) take it all the way through to the electron level, at some point.
https://vimeo.com/188447627
Keep in mind that the spin volume only "forms" an actual torus on the x1 spin, or the first stacked spin beyond the axial. All higher stacked spins form much more complex shapes and the torus is not accurate or useful as a descriptor after that.
Refer to my video again if it helps, although we do need to update it and (hopefully) take it all the way through to the electron level, at some point.
https://vimeo.com/188447627
Jared Magneson Posts : 444
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Re: c, the speed of light, and the BPhoton
Here is a photon with A, X, Y and Z spins.
(Click on the image for a higher resolution version)
We want to rotate that around the Y axis to add an axial spin level.
Which gives us this:
The higher axial spin does not produce a sphere from a full spin set and it does not increase the radius. This is because the motions are integrated, not merely added together.
(Click on the image for a higher resolution version)
We want to rotate that around the Y axis to add an axial spin level.
Which gives us this:
The higher axial spin does not produce a sphere from a full spin set and it does not increase the radius. This is because the motions are integrated, not merely added together.
Nevyn Admin
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Re: c, the speed of light, and the BPhoton
It appears that the path is changed, but not the radius, which makes sense. So does that mean we keep the higher axial spins in the physics, but exclude them from radius calculations?
Jared Magneson Posts : 444
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Re: c, the speed of light, and the BPhoton
My assumption stated that we start with a full X,Y,Z set. To produce a new axial spin requires two actions: 1) The next spin added will be an endoverend X or Y spin that doubles the radius from the previous Z; 2) Convert that second X (or Y) into a new A by a tangential collision. I understand that we are talking about a bphoton moving through its integrated spin states while limited to the toroidal volume.Nevyn wrote: Here is a photon with A, X, Y and Z spins. …
We want to rotate that around the Y axis to add an axial spin level. …
The higher axial spin does not produce a sphere from a full spin set and it does not increase the radius. This is because the motions are integrated, not merely added together.
I've given you my best guess. I'm not convinced that there aren’t any higher A spins. I believe they are required in order to properly understand the 1851 energy multiplier for each complete spin set (A,X,Y,Z). Before developing your competing theory, please ask Miles for a comment.
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LongtimeAirman Admin
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Join date : 20140810
Re: c, the speed of light, and the BPhoton
Would an additional axial spin add any more energy to the equation, though? Does it add mass somehow, or velocity? To add energy, we either need to increase mass or velocity (e=mc², of course).
I can see how the first axial spin would increase energy in a collision, since we have both a linear velocity and a tangential velocity in play (c²). A nonspinning, moving particle would not have that additional velocity to add in. But would a higherspin set/level add any more tangential velocity to our particle?
I can see how the first axial spin would increase energy in a collision, since we have both a linear velocity and a tangential velocity in play (c²). A nonspinning, moving particle would not have that additional velocity to add in. But would a higherspin set/level add any more tangential velocity to our particle?
Jared Magneson Posts : 444
Join date : 20161011
Re: c, the speed of light, and the BPhoton
The higher axial spin I added is actually illegal because the spin axis (the arrow) goes through the previous path. I just wanted to show that an axial spin does not make a sphere from a torus, in this case. It is an interesting shape though.
If I had added another endoverend spin instead of an axial spin then it would just end up with the same shaped path but on a different axis. That is, the images I presented spin around the Z axis but the next endoverend spin would spin about the X axis. Same shape, longer path, different axis.
If I had added another endoverend spin instead of an axial spin then it would just end up with the same shaped path but on a different axis. That is, the images I presented spin around the Z axis but the next endoverend spin would spin about the X axis. Same shape, longer path, different axis.
Nevyn Admin
 Posts : 1401
Join date : 20140911
Re: c, the speed of light, and the BPhoton
Now I want to show that the same general path is presented for all spin levels, no matter how many spin sets are used. Here is a series of images showing complete spin sets (X, Y, Z, only the first spin level is axial). This ranges from the smallest photon up to an electron or somewhere close to it.
Spin Set 1
Spin Set 2
Spin Set 3
Spin Set 4
Spin Set 5
As we add more spin sets, the path becomes more erratic, in a way. It is like the artist keeps having a few drinks in between drawings and can't keep the pencil straight.
This just shows that we don't really need to animate all the way up to an electron, since the same general path will be found at every level. Showing only 2 or 3 spin sets is enough to get the idea of how they form.
Spin Set 1
Spin Set 2
Spin Set 3
Spin Set 4
Spin Set 5
As we add more spin sets, the path becomes more erratic, in a way. It is like the artist keeps having a few drinks in between drawings and can't keep the pencil straight.
This just shows that we don't really need to animate all the way up to an electron, since the same general path will be found at every level. Showing only 2 or 3 spin sets is enough to get the idea of how they form.
Nevyn Admin
 Posts : 1401
Join date : 20140911
Re: c, the speed of light, and the BPhoton
That sure makes things easier, Nevyn!
But wouldn't that only be true along one spinset... Uhm, direction? What if we have a x spin instead of a positive in there, or combinations of these? It seems like there are four possible nextspinstack motions, as there's no reason a zspin couldn't come before an xspin (that I can think of).
So we have an axial spin along y as the first spin, then:
+x
x
+z
z
as possible next spins. We've only done it in x,y,z order for ease of reference, correct?
Or am I way off here? Is there any reason an axialspin particle must go +x next, as opposed to x or either z?
And more importantly, if it did go either direction depending on the collider's vector, would this change your volumeshape among the various stacksets?
But wouldn't that only be true along one spinset... Uhm, direction? What if we have a x spin instead of a positive in there, or combinations of these? It seems like there are four possible nextspinstack motions, as there's no reason a zspin couldn't come before an xspin (that I can think of).
So we have an axial spin along y as the first spin, then:
+x
x
+z
z
as possible next spins. We've only done it in x,y,z order for ease of reference, correct?
Or am I way off here? Is there any reason an axialspin particle must go +x next, as opposed to x or either z?
And more importantly, if it did go either direction depending on the collider's vector, would this change your volumeshape among the various stacksets?
Jared Magneson Posts : 444
Join date : 20161011
Re: c, the speed of light, and the BPhoton
LongtimeAirman wrote:I'm not convinced that there aren’t any higher A spins. I believe they are required in order to properly understand the 1851 energy multiplier for each complete spin set (A,X,Y,Z).
.
Having higher axial spins does not affect the 1821 energy multiplier because that is caused by an endoverend spin, which we still have. The only thing it does is affect the general equation it comes from by removing the axial spins so we have sets of 3 instead of 4. Since the 1821 number comes from the energy difference between 2 spin levels, differing by an endoverend spin and a radius doubling, the axial spins do not actually have that energy difference. I can't see how a higher axial spin can increase the energy at all. It changes the motion but does not increase the energy and that sounds like a paradox to me.
To be clear, if you take an axial spin and add an endoverend spin to that, then you will increase the energy, double the radius, etc. However, if you take an existing stacked spin and give it an axial spin on top, then I don't see how the energy is added to the particle.
Nevyn Admin
 Posts : 1401
Join date : 20140911
Re: c, the speed of light, and the BPhoton
Using negative spins (ie. they rotate in the opposite direction to a positive one) only changes the path, not the shape. It basically mirrors the path. Looking at the Z axis (lower left on images showing 4 views) you can see how the path is along the diagonal from lower left to upper right. If we add in some ve spins, it can change that diagonal from lower right to upper left. The same shape is found though.
The way to change the actual path and find a different shape is to change the initial rotation offset of a spin level. All of those images have the spin level start at 0 (which direction that is depends on the spin axis: X > Y, Y > Z, Z > X). That is, for an X spin, the particle is translated in the Y dimension and then rotates around the X. But we could translate it in the Z dimension or Y or Z or anywhere on that circle. Changing these rotation offsets gives a very different shaped path. I'll whip one up for you in a sec.
The way to change the actual path and find a different shape is to change the initial rotation offset of a spin level. All of those images have the spin level start at 0 (which direction that is depends on the spin axis: X > Y, Y > Z, Z > X). That is, for an X spin, the particle is translated in the Y dimension and then rotates around the X. But we could translate it in the Z dimension or Y or Z or anywhere on that circle. Changing these rotation offsets gives a very different shaped path. I'll whip one up for you in a sec.
Nevyn Admin
 Posts : 1401
Join date : 20140911
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