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Quote from: Malamute Lover on 10/07/2020 21:17:23In short, do you really know what you are talking about as opposed to just throwing fancy words around? I know what I'm talking about: I got it in pictures at an attachment with the post: "Quantum Gravity Follows?" in this forum. You sound like you would be interested in it.

In short, do you really know what you are talking about as opposed to just throwing fancy words around?

Quote from: Malamute Lover on 10/07/2020 21:17:23It is improper terminology to say ‘a set with additional structures’So it's a set endowed with features. The statement is standard in Mathematics.

It is improper terminology to say ‘a set with additional structures’

Quote from: Malamute Lover on 10/07/2020 21:17:23A circle drawn on representation of a Riemann sphere will generally not look like a circle but will be ‘squashed’ in the direction of the ∞ pole.No. A circle through the point at infinity will look like a circle, it will just be denser closer to infinity but a denser line is still a line. This is in the standard representation of a sphere in a coordinate system with the z-axis pointing 45 degrees downwards. Maybe in some other projection, it will appear squashed.

A circle drawn on representation of a Riemann sphere will generally not look like a circle but will be ‘squashed’ in the direction of the ∞ pole.

Quote from: Malamute Lover on 10/07/2020 21:17:23Please explain what you intended to be understood by ‘a circle in the Riemann Sphere and how it fits into your explanation. Presumably, you meant ‘on’ and not ‘in’. There is no ‘in’ relative to a Riemann Sphere. It is all surface.I can talk about a line "in" a plane if the plane is infinitesimally thick (the thickness is non-zero).

Please explain what you intended to be understood by ‘a circle in the Riemann Sphere and how it fits into your explanation. Presumably, you meant ‘on’ and not ‘in’. There is no ‘in’ relative to a Riemann Sphere. It is all surface.

Quote from: Malamute Lover on 10/07/2020 21:17:23By describing a ‘left out spacetime node’ as a hole in the Riemann Sphere. It would appear that you mean some region of the Riemann Sphere cannot be the result of any calculation, that the coordinates interior to this region are forbidden in some fashion. Can you shed any light on what these coordinates might be and how you came to determine them? And how does the ‘circle’ come into play?The circle is like a line of longitude. The holes are holes on a line through the center of the sphere and perpendicular to the momentum vector of the particle, where the holes are located where this line intersects the sphere. I determined them by reasoning about the production of an accelerating force on a photon (protophoton: a photon not yet traveling at the speed of light).

By describing a ‘left out spacetime node’ as a hole in the Riemann Sphere. It would appear that you mean some region of the Riemann Sphere cannot be the result of any calculation, that the coordinates interior to this region are forbidden in some fashion. Can you shed any light on what these coordinates might be and how you came to determine them? And how does the ‘circle’ come into play?

Quote from: Malamute Lover on 10/07/2020 21:17:23Also why are you representing spacetime as a Riemann Sphere? The time dimension requires the use of imaginary numbers. But you need three real dimensions to represent spatial coordinates. How do you map these three dimensions onto a Riemann Sphere?I propose copying one dimension of space and one of time into a Riemann Sphere, resulting in a particle when endowed with charges.

Also why are you representing spacetime as a Riemann Sphere? The time dimension requires the use of imaginary numbers. But you need three real dimensions to represent spatial coordinates. How do you map these three dimensions onto a Riemann Sphere?

Quote from: Malamute Lover on 10/07/2020 21:17:23And what features are you endowing to the complex (involving imaginary numbers) structure (set with features) of the Riemann Sphere that results in a ‘hole’?Points of space on the circle versus left out points (infinitesimal breaks in the circle).

And what features are you endowing to the complex (involving imaginary numbers) structure (set with features) of the Riemann Sphere that results in a ‘hole’?

You would just be able to reproduce the proof if you are telepathic, knows Earth's voice, and if your mind knows how to copy spacetime into a Riemann Sphere.

Can you provide any actual evidence for those claims?

A circle through the ∞ pole is not possible because a center point from which it is equidistant cannot be defined.

There is no such thing as an axis ‘pointing 45 degrees downwards’

Also, a hole though a Riemann sphere (which does not have an inside anyway) is not a valid concept since the surface does not have a linear metric. A straight line through a Riemann sphere would only be definable at the equator. Otherwise there are no straight lines.

A particle has three translational degrees of freedom in space. You cannot represent the state of a particle in one dimension. Good luck on defining momentum vector. And how do ‘charges’ (in your expanded definition) fit into the picture? Yes, I read your paper. No, it does not answer the question.

Also, you need to talk more about what structure(s) you want to endow to the set of points on the surface of the Riemann sphere to build a coherent picture. In particular in what fashion do the required features get endowed?

In addition, infinitesimally small holes do not work well with quantum theory.

Just my experience, so no objective proof. You may ask a Psychic person.

Quote from: Malamute Lover on 11/07/2020 17:06:24A circle through the ∞ pole is not possible because a center point from which it is equidistant cannot be defined.The projection allows a circle through the pole: the point at ∞ is projected at a unit distance from the center of the sphere. I've seen a picture of a Riemann sphere. The distance along a longitude line from the equator to the north pole is finite in the projection (not to scale).

Quote from: Malamute Lover on 11/07/2020 17:06:24There is no such thing as an axis ‘pointing 45 degrees downwards’The projection of the z-axis onto 2 dimensions is such.

Quote from: Malamute Lover on 11/07/2020 17:06:24Also, a hole through a Riemann sphere (which does not have an inside anyway) is not a valid concept since the surface does not have a linear metric. A straight line through a Riemann sphere would only be definable at the equator. Otherwise there are no straight lines.By the projection, one could imagine the sphere to be an ordinary sphere with a linear metric by reassigning distances. There is a one-to-one mapping between a Riemann sphere and an ordinary sphere. The 3-dimensional space it is in allows for straight lines as chords intersecting the Riemann sphere.

Also, a hole through a Riemann sphere (which does not have an inside anyway) is not a valid concept since the surface does not have a linear metric. A straight line through a Riemann sphere would only be definable at the equator. Otherwise there are no straight lines.

Quote from: Malamute Lover on 11/07/2020 17:06:24A particle has three translational degrees of freedom in space. You cannot represent the state of a particle in one dimension. Good luck on defining momentum vector. And how do ‘charges’ (in your expanded definition) fit into the picture? Yes, I read your paper. No, it does not answer the question.You have it wrong: my model defines a particle as equivalent to a Riemann sphere with holes. Actually more than one superimposed Riemann sphere for particles other than the photon. I just need 3 dimensions to put the particle in. I simply encode the charges as added or left out points. That way spacetime does not need a table of particle names and properties at each event of spacetime. If you have ever considered what it would take to implement calculated laws of physics this would seem logically expedient.

Quote from: Malamute Lover on 11/07/2020 17:06:24Also, you need to talk more about what structure(s) you want to endow to the set of points on the surface of the Riemann sphere to build a coherent picture. In particular in what fashion do the required features get endowed?The structure of a 2-sphere.

Quote from: Malamute Lover on 11/07/2020 17:06:24In addition, infinitesimally small holes do not work well with quantum theory.Call them little Planck-length openings then.

Just my experience, so no objective proof.

The distance from zero to ∞ of the sphere as you measure it on the projection with a yardstick that doesn't get denser is finite. It is just conceptually equal to ∞. You are caught up in the conceptual. One can make a finite model of the Riemann Sphere (RS One could conceptually say that one can line up the RS with events of spacetime such that there is a plane that cuts the RS at a circle, on which (plane) one can draw a line. Spacetime just needs to be locally flat. ).

The holes must be smaller than a photon. The size of a quantum of spacetime. Yes, a photon has a size in my model.

It seems to me that somehow it is not satisfying to say a particle is a vibration in a field.

I thought about implementing the laws of physics and came to the conclusion that particle properties must be communicated between fields. Just think about an electron scattering off another electron.

What physical experiment could be performed to demonstrate that particles are Riemann spheres?

So what does “communicated between fields (plural)” mean?

The Riemann sphere IS conceptual.

Quote from: Kryptid on 12/07/2020 17:29:11What physical experiment could be performed to demonstrate that particles are Riemann spheres?Orientate an Electron such that it's axis (pointing from the point 0 to the point ∞) points in the "up" direction. Then it will not emit a photon in the "down" direction.Also: particles seen from their northern hemispheres will look different as seen from their southern hemispheres.

Quote from: Malamute Lover on 12/07/2020 21:17:50So what does “communicated between fields (plural)” mean?Think of the Feynman diagram of an electron scattering off an electron. What happens is: one electron must tell the electromagnetic field that it wants to scatter of another electron following a path λ, and it must specify its momentum. This must be read off the electron field interacting with spacetime. Then the electron must specify it's expected momentum change. Then the electromagnetic field must compute the virtual photon direction and wavelength and the positions for invoking the creation and annihilation operators, then the virtual photon must communicate the change of direction of momentum to the other electron. Only then can the virtual photon be emitted.

Quote from: Malamute Lover on 12/07/2020 21:17:50The Riemann sphere IS conceptual.Wrap your conceptual head around this: the RS is conceptually a distortion of the Complex plane such that the point at ∞ sits at a sphere's north pole. As such it sits at a finite distance as measured with an undistorted measuring rod.

The Riemann sphere is NOT a distortion of the complex plane.

Now explain what differences there may be between the two hemispheres of an electron.

Also, since the question Kryptid asked is about a physical experiment, how would you determine which photon came from which electron and how to determine the axis orientation of the individual electrons?

Your insistence that the metrics of a Riemann sphere are those of an ordinary sphere and not the varying metrics of a complex plane mapping argues against any difference.

Quote from: Malamute Lover on 13/07/2020 17:46:35The Riemann sphere is NOT a distortion of the complex plane.I'll ask a Mathematician on the other forum and refer you.

Quote from: Malamute Lover on 13/07/2020 17:04:58Now explain what differences there may be between the two hemispheres of an electron.The northern hemisphere would have a higher density of spacetime events than the southern hemisphere.

Quote from: Malamute Lover on 13/07/2020 17:04:58Also, since the question Kryptid asked is about a physical experiment, how would you determine which photon came from which electron and how to determine the axis orientation of the individual electrons?Isolate a single electron in a magnetic trap with a known magnetic field imposed. They can do the double-slit experiment with one electron at a time.

Quote from: Malamute Lover on 13/07/2020 17:04:58Your insistence that the metrics of a Riemann sphere are those of an ordinary sphere and not the varying metrics of a complex plane mapping argues against any difference.No, I was wrong. It must be the RS metric. The particle lives in locally flat space so there is a line intersecting the RS perpendicular to the momentum, where the holes are located.

But then you do not understand how a Riemann sphere works. It was just an impressive sounding phrase that you did not think anyone could challenge you on.

Since you previously said that the surface of a Riemann sphere is really linear

What does it mean to have spacetime events on the surface of an electron? And how do you reconcile representing an electron being a Riemann sphere with your definition of a particle as the holes in a Riemann sphere?

I have done some experiments with my mind

Did you think that actually meant something?

I can tell if any information is not from my mind.