Naked Science Forum
On the Lighter Side => New Theories => Topic started by: Yahya A.Sharif on 04/05/2020 18:41:02

What is a photon ?
A photon whether it is mass or energy it shouldn’t move at c.

A photon whether it is mass or energy it shouldn’t move at c.
Yet it does.
If you say that your view does not agree with reality, it is not because reality has made a mistake,

A photon is a mathematical construct that helps us predict the behaviour of electromagnetic radiation.

A photon is a mathematical construct that helps us predict the behaviour of electromagnetic radiation.
It contains energy.
Einstein says no mass nor energy should move at the speed c.

A photon is a mathematical construct that helps us predict the behaviour of electromagnetic radiation.
It contains energy.
Einstein says no mass nor energy should move at the speed c.
That's silly.
Einstein knew perfectly well that light
(1) has energy and
(2) travels at the speed of light
So he can't have said
Einstein says no mass nor energy should move at the speed c.
You need to read more carefully or something

Let me get some clarification on this: are you simply looking to understand how light can move at the speed of light? If so, then the current location of this topic is acceptable. If you are, instead, trying to assert that light doesn't move at the speed of light and you are trying to convince us that it doesn't, this is belongs in the New Theory section.

If you are, instead, trying to assert that light doesn't move at the speed of light and you are trying to convince us that it doesn't, this is belongs in the New Theory section.
I'm speaking about a contradiction in special relativity.

Einstein knew perfectly well that light
(1) has energy and
(2) travels at the speed of light
He knew light is energy and it travels with c but he says mass or energy "light" shouldn't move at c.
Can you see any contradiction here ?

I'm speaking about a contradiction in special relativity.
Time to move this to New Theories then...
he says mass or energy "light" shouldn't move at c.
Please provide a reputable source that supports this statement. I am not aware of anywhere in special relativity that states that energy cannot move at c.

He knew light is energy and it travels with c but he says mass or energy "light" shouldn't move at c.
Can you see any contradiction here ?
Yes, my understanding of relativity contradicts what you say.

he says mass or energy "light" shouldn't move at c.
I am not aware of anywhere in special relativity that states that energy cannot move at c.
Einstein considers a moving object with its rest mass and other energy as a whole mass in his equations , then mass is just energy and energy is just mass if a photon is energy then it is just a mass , it is a mass moves at c which it shouldn't. He does't distinguish between mass and energy in his theory and equations or mass/energy equivalence

he says mass or energy "light" shouldn't move at c.
I am not aware of anywhere in special relativity that states that energy cannot move at c.
Einstein considers a moving object with its rest mass and other energy as a whole mass in his equations , then mass is just energy and energy is just mass if a photon is energy then it is just a mass , it is a mass moves at c which it shouldn't. Or mass/energy equivalence
There are , in effect, two sorts of mass.
There is the mass you understand, and the mass you do not understand.
https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_vs._rest_mass

it is a mass moves at c which it shouldn't.
Again, show us from a reputable source that no form of mass can move at c.

The speed of propagation of electromagnetic energy is calculated from first principles by Maxwell's equations. The only solution in vacuo is a constant, c.
The quantisation of electromagnetic energy is demonstrated in various experiments and consistent with Planck's calculation of the black body spectrum.
Relativity begins with the assumption that information cannot travel faster than c, and derives a number of experimentally testable results from gravitational red shift to the hydrogen bomb and satellite navigation, that have been shown correct to exceptional accuracy.
Plenty of people don't understand Maxwell, Planck or Einstein, but that's no reason to think they were wrong, as experimental data shows they were correct.

What is a photon ?
A photon whether it is mass or energy it shouldn’t move at c.
A photon is a quantum of EM radiation. It has mass via its momentum since mass is defined as m = p/v. Since fore a photon v = c we have m = p/c or p = mc. For a photon E = pc we now have E = mc^2
Einstein did in fact show that radiation in another (1907) relativity paper. that light has mass. What Einstein meant was that no body can accelerate from v < c to v > c in is 1905 paper on relativity.
Whenever people don't clarify what they mean by the term "mass" confusion arises and people make mistakes.
There are two kinds of inertial mass in relativity: proper mass (aka rest mass) and relativistic mass. I wrote a paper on this subject. It's online at
On the concept of relativistic mass
https://arxiv.org/abs/0709.0687
There is a great deal of scientific literature out there who use mass to mean relativistic mass.

Einstein knew perfectly well that light
(1) has energy and
(2) travels at the speed of light
He knew light is energy and it travels with c but he says mass or energy "light" shouldn't move at c.
Can you see any contradiction here ?
The problem is that you're wrong. Einstein never said that.

A photon is a Riemann Sphere with four charges encoded into it on a circle in the Riemann Sphere. The charges are encoded as "left out events of spacetime". I can picture it.

I can picture it.
I can find a picture of a unicorn. That does not make it real.A photon is a Riemann Sphere with four charges encoded into it on a circle in the Riemann Sphere. The charges are encoded as "left out events of spacetime".
None of that actually makes sense.

That does not make it real.
It can explain why and how a photon is emitted at the speed of light.

That does not make it real.
It can explain why and how a photon is emitted at the speed of light.
Since nobody understands it, it doesn't explain anything.
On the other hand we already knew how photons were formed.
https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)#Timedependent_perturbation_theory

On the other hand we already knew how photons were formed.
I don't see a mechanism at that reference.

I don't see a mechanism at that reference.
Nobody can see anything in your so called explanation.
Just in case anyone missed the question.
What part of it does not make sense?
Is there anyone reading this who knows what any part of tantalum1's diagram means?

It just takes a little getting used to.
Anyway my model explains refraction and predicts refraction in the right direction. Doesn't total internal reflection suggest that a photon has structure?

Doesn't total internal reflection suggest that a photon has structure?
No.
You can (and they did) model total internal reflection as a classical phenomenon.
It just takes a little getting used to.
Thus far, precisely zero people have got used to it and precisely one person thinks it's worth the effort.

Mass/ energy is just confusion and have no meaning .The K.E due to motion shouldn't be treated as mass of " compounds and elements " and affect the motion . Relativistic mass is just an invention.
The goal from below is if rest mass can't move at c then a photon shouldn't and if a photon moves at c then a rest mass should
Let me start with:
Let me refer to energy due to motion as " pure energy which is different from rest mass compounds and elements or matter"
Relativistic mass is rest mass " compounds and elements " added to it energy due to motion. Both can be treated as mass this mass is " relativistic mass " with the same effect " preventing to reach c "S.R
If the above is true.Energy due to motion and rest mass both can be treated as matter.
In this case then why this specific energy can be treated as matter ? why not a photon ? if this is for the photon then photon can be treated as matter and shouldn't move at c" A photon does move at c.
If the above is wrong and photon can not be treated as mass then energy due to motion also can not be treated as mass " because a photon and energy due to motion are both energy" .Energy due to motion is energy and photon is energy and if photon doesn't prevent itself from reaching c then this energy due to motion can not prevent rest mass to reach c .photon is not a special case it is energy and the energy preventing rest mass to reach speed of light is also energy.It is not scientific to describe some phenomenon that contradict physics as a special case only if we have an explanation if we do not have an explanation then we should a apply physical laws on that phenomenon since a photon and energy due to motion are both energy then any thing applied to energy must be applied to a photon , we can't say F=ma for any mass but iron there must an explanation for that and a new knowledge that explains this .Not because we want to accept Einstein' theory of special relativity we refer to something contradicts it as a special case
If they are the same, rest mass " compounds and elements " is just pure energy like a photon and energy due to motion is just pure energy then energy plus energy can move at c just like a photon Energy plus energy is just a photon that can move at c, in this case a rest mass can move at c
If mass and energy are one,what is energy is mass and what is mass is energy then rest mass is a photon and can move at c
Einstein just made energy mass " moving object" and energy energy " a photon " he didn't know where energy is mass and where energy is energy and why.

Relativistic mass is just an invention.
Then why, at places like CERN, do they have to bolt the magnets down to strongly?
The magnets produce a field which bends the path of the protons from a straight line into a circle.
And, of course there is a reaction force which pushes back on the magnets.
But protons are really light they have little mass so, there shouldn't be much reaction force.
But there is that's why they need to use big bolts.
So, we know that relativistic mass is real.
So we know you are wrong.

Mass/ energy is just confusion and have no meaning
The fact that nuclear bombs work is proof that you are wrong.
If the above is true.Energy due to motion and rest mass both can be treated as matter.
Matter is not interchangeable with mass. Objects can have mass and not be matter (case in point, the Z boson that carries the weak nuclear force).
why not a photon ?
The term "photon" is very specific. It refers to the quantum of the electromagnetic force. It is a boson particle with a spin of one, is electricallyneutral and does not have a rest mass. It something does not have those characteristics, then it isn't a photon.
In any case, every test of special relativity to date has passed. So if you are positing that special relativity has a flaw, then it is significantly more likely that you are mistaken than relativity being mistaken. Unless, of course, you can point out the flaws in the experiments. Can you?

Mass/ energy is just confusion and have no meaning
The fact that nuclear bombs work is proof that you are wrong.
They work
E=mc² .but the energy due to motion and rest mass shouldn't be added together. They are different .One is energy and the other is mass "compounds and elements" .So the Energy/Mass equivalence here is meaningless.
In any case, every test of special relativity to date has passed. So if you are positing that special relativity has a flaw, then it is significantly more likely that you are mistaken than relativity being mistaken. Unless, of course, you can point out the flaws in the experiments. Can you?
There is not a proof for relativistic mass .That is there is not a proof that mass increases and decelerate and prevent itself from reaching c.
The term "photon" is very specific. It refers to the quantum of the electromagnetic force. It is a boson particle with a spin of one, is electricallyneutral and does not have a rest mass. It something does not have those characteristics, then it isn't a photon.
Energy due to motion and energy of photon are both energy .I'm comparing between energy due to motion and the energy we know that a photon contains

There is not a proof for relativistic mass .That is there is not a proof that mass increases and decelerate and prevent itself from reaching c.
So, you are going to ignore the evidence from accelerator experiments?
If Einstein had not come up with his theory, evidence from acceleration of nuclear particles would have made relatively and the mass/energy equivalence obvious and the equation has been tested to the degree that noone working in the field has any reason to doubt it.

Every photon is a wizard.
The universe is filled with an infinite number of photons. and each of these photons moves away from its source at c speed. They even drags their source after them. They have the feature of shortening their own paths to keep their speeds in c value.
* It is the mentality of special relativity theory.

Mass/ energy is just confusion and have no meaning
The fact that nuclear bombs work is proof that you are wrong.
They work
E=mc² .but the energy due to motion and rest mass shouldn't be added together. They are different .One is energy and the other is mass "compounds and elements" .So the Energy/Mass equivalence here is meaningless.
In any case, every test of special relativity to date has passed. So if you are positing that special relativity has a flaw, then it is significantly more likely that you are mistaken than relativity being mistaken. Unless, of course, you can point out the flaws in the experiments. Can you?
There is not a proof for relativistic mass .That is there is not a proof that mass increases and decelerate and prevent itself from reaching c.
The term "photon" is very specific. It refers to the quantum of the electromagnetic force. It is a boson particle with a spin of one, is electricallyneutral and does not have a rest mass. It something does not have those characteristics, then it isn't a photon.
Energy due to motion and energy of photon are both energy .I'm comparing between energy due to motion and the energy we know that a photon contains
The mass is real that's why there's about 26 tonnes of force acting on the magnets without the relativistic mass it would only be about a six thousandth of that.
You can't just ignore evidence because you don't want to accept that it shows you are wrong.

Every photon is a wizard.
The universe is filled with an infinite number of photons. and each of these photons moves away from its source at c speed. They even drags their source after them. They have the feature of shortening their own paths to keep their speeds in c value.
* It is the mentality of special relativity theory.
The fact that you don't understand it doesn't mean it's wrong.

One is energy and the other is mass "compounds and elements"
What is "mass compounds and elements"? Is it something you invented?
There is not a proof for relativistic mass .That is there is not a proof that mass increases and decelerate and prevent itself from reaching c.
If it wasn't for relativistic mass, gold wouldn't be yellow: https://www.fourmilab.ch/documents/golden_glow/
If it wasn't for relativistic mass, mercury wouldn't be a liquid: https://www.physics.rutgers.edu/grad/601/CM2019/ed068p110.pdf
The universe is filled with an infinite number of photons.
Citation needed.
and each of these photons moves away from its source at c speed.
Only in the reference frame of the source.
They even drags their source after them.
No, they don't.
They have the feature of shortening their own paths to keep their speeds in c value.
No, they don't.
* It is the mentality of special relativity theory.
No, it isn't. Your continual strawmanning of relativity is tiresome.

Quote from: xersanozgen on Yesterday at 12:37:00
They have the feature of shortening their own paths to keep their speeds in c value.
No, they don't.
If a star has the speed 0.60 c , the way of a photon becomes L' = L (1  v^{2}/c^{2})^{1/2} = 0.80 L.
The distance (that the photon does not travel yet) decreases.
As if MAGIC TRİCK

Quote from: xersanozgen on Yesterday at 12:37:00
The universe is filled with an infinite number of photons.
Citation needed.
Citation: LOGIC

Quote from: xersanozgen on Yesterday at 12:37:00
The universe is filled with an infinite number of photons.
Citation needed.
Citation: LOGIC
OK, we know that the universe is finite.
So we know that there is only space (and energy) in it for a finite number of photons.
so LOGIC, says you are wrong.
Yet you think logic says you ar right.
So, we can deduce that you do not understand logic.
That explains a lot.
Please refrain from further posting until you have learned to be rational.

The equation E=mc2 is connected to a conservation of energy equivalency between mass and energy. If you wanted to know the energy output within a nuclear bomb design, you need to know the difference in mass, before and after, the boom. The mass burn will cause some of the mass to convert to energy. Energy conservation applies and can be calculated with E=mc2. Energy cannot be destroyed, but it can convert between different states/phases.
E=mc2 does not mean mass and energy are the same phenomena. They are two different phases with different properties. In Special relativity, mass cannot move at the speed of light, since that would require infinite relativistic mass.
Relativistic mass is a different phase from mass and energy. It is closer to the concept of dark matter, than it is to rest mass. Infinite dark matter or infinite relativistic mass would create a moving target universe, that would accelerate into a big crunch, and not allow a rest mass phase to be sustainable. The loose analogy is ice and water remain at 0C until all the ice melts or all the water solidifies, then only one phase will remain instead of two.
Many physicists do not like the concept of relativistic mass, since it implies a connection to energy conservation. Energy conservation messes up the concept of relative reference, since energy conservation is an absolute and cannot be relative. For example, If we have moving train and a stationary person, one can see the relative motion in time and space.
However, the energy would not be the same for both reference scenarios. A 50,000 ton train train needs more energy than a stationary man to move 10 mph. An energy balance of this two component system, will tell us who is moving in absolute terms. This energy balances messes up the relative reference magic trick, that can fool the eyes and instruments.
Relative reference, only works if you use only two out of the three equations of Special Relativity; t and d, but you cannot use m. The reason is the energy phases is connected to d and t, but energy is not the m phase, The direct evidence of our universe is based visual evidence; energy, which is not the same as mass. We infer mass from energy, but cannot measure mass directly, due to time lag and distances. Energy, when it reaches us, has separated from the mass since mass moves much slower.
If we could directly measure the mass and the relativistic mass of the universe, since these are different different phases from photons, the universal energy balance be different from only measuring the energy. If mass and energy were the same thing, then two or three equation SR would work the same, and physics would not need to exclude relativistic mass to make relative reference work. The illusion of relative reference is an artifact of mass and energy not being the same.
The photon phase is interesting in that it moves at the speed of light c, while also having finite expressions in space and time; wavelength and frequency. If mass moved at c, space, time and mass for any and all quanta of mass, would always be uniform and maximized; infinite. There would be no variety, since infinite relativistic time, distance and mass would be the same for any mass quanta.
Energy is not limited this way, since it is a different type of phase from mass. It has both inertial and c characteristics, that are connected, by can act independently of each other. Photons are the bridge between c and the inertial states of mass/matter, which are all restricted to below c.
Another source of confusion in physics is the c reference; speed of light, is really the ground state of the universe and not a ceiling state, even the speed of light is fast. Matter and antimatter only form at the upper limits of energy. We cannot form matter with radio waves, but rather we need to increase the energy level to beyond gamma. Matter appears is at the ceiling, while radio waves are at the floor. The speed of light reference exists at the floor, and this floor partially anchors all photons; balloons, on a string, with matter clouds at the ceiling.
Photons are not entirely in the ground state of the creference. They also have potential within their finite inertial attributes; wavelength and frequency. The creference ground state have no inertial potential, similar to the photons. This can be approximated with only infinite wavelength photons. Infinite wavelength is a far as you can red shift photons and remain the photon phase. After that one is in the creference ground state, where photons are not a valid phase.
The ground state at c, is like sealevel relative to water. Just as all the liquid water on the earth, from mountains, to rain clouds, to rivers, try to flow toward sealevel to minimize potential, matter, energy, and force move the ceiling toward the floor at clevel.
Gravity cause spacetime to contract toward the c reference; singularity of the black hole. The other three forces convert matter to energy/photons which moves the inertial phases into a phase with one leg at c. The universal red shift is then moving all the photons in the direction of infinite wavelength. Can you see the pattern from celling to the floor at clevel?
The hard part is learning to stand upright; matter clouds and clevel floor, after being told to stand on your head for so long. The blood will flow from your brain, and make you dizzy at first.

They are two different phases with different properties.
The word "phase" has a specific meaning in science.
And mass and energy are not "phases"
Calling them both "things" may look less sciencey, but at least it isn't wrong.
Relativistic mass is a different phase from mass and energy.
Tosh.
It's mass. since energy conservation is an absolute and cannot be relative.
Energy is relative.
It depends on your point of view.
The middle of a tree has low gravitational energy compared to the top, but high energy compared to the bottom.
Another source of confusion in physics is the c reference; speed of light, is really the ground state of the universe and not a ceiling state
No.

@Colin2B
Yes a proton in a particle accelerator will approach c with no reach to c. And the proton decelerate in its motion . That doesn't mean proton has a relativistic mass.Deceleration does not necessarily is due to mass increment.It is due to inertia increment. Mass/Energy equivalence is just a mess as I mentioned Reply#24
Inertia is proportional to mass .Bigger mass is bigger inertia and bigger deceleration.
As inertia is proportional to mass .We can replace mass with inertia in the equation of S.R
I=I0/√(1v²/c²)
The I0 is the inertia of a mass at rest. For each rest mass M0 kg there is a rest inertia I0, in which I=cM , c is a constant.
The factor 1/√(1v²/c²) in Einstein equation above doesn't change because of the idea that velocity never reach c .But there will not be change in mass.The change is in mass inertia.
Think of a mass m at motion pushed by force F=ma , all these are constant , F, m and a. The velocity will multiply and will not reach c. And now think of this mass reached v velocity.This mass can collide a spring with some compression force and if v is bigger the force will be bigger.So the force by the spring to stop motion or to overcome inertia will be bigger for a bigger v. Inertia increases by the increment in v the equation above. . There is not increment in mass here.There is increment in inertia for the mass at motion .
Inertia is the resistance to move a mass and the resistance to stop this mass .Force in the spring or inertia increases by the increment in velocity and that is according to Einstein's equation above.
There is not something called kinetic energy .Kinetic energy doesn't exist. Instead there is a work done .The work done by a force F to push a mass m is the work done by this force to a distance d this distance is at which v stops in a maximum value " no acceleration" .
We can consider work as force times distance d at this distance velocity reached a maximum constant v" no acceleration"
If I push a mass m with force F the work done is force times distance d.
The work done is :
W=ma*d
W= 1/2 (mv)/(t*d)
v=s/t
W=1/2mv²
1) W=1/2mv² v is constant when acceleration stops and force stops
2)W=ma*d d is constant . its value is the distance at which v stops increasing.
Equation 1 is work done on a spring.Work done by mass at constant velocity when it collides a spring.
Equation 2 is work done or force times d , d is the distance at which v reaches maximum" the same v as in the first equation "
The energy 1/2mv² only appears when the mass collides something .Then the 1/2mv² energy is just another way to calculate W=Fd .Because there is a relation between distance and constant velocity we can write W=Fd in another way with respect to v we can write it W=1/2mv²
The mass resists motion .Force overcame this inertia .Making a mass to move at constant speed . Then a mass possess this inertia this inertia a mass has possessed appears in colliding spring.The inertia it possess is not a thing it is how this mass resists motion to reach constant v and how it will resist to reach 0 velocity.
.
A photon is energy and with no rest mass" massless" . A photon then does not have an inertia .In this case it can move at c .Nothing will decelerate it because no inertia and no resistance to its motion.Mass is mass energy is energy they are different things . E=mc² is true but that doesn't mean energy behaves as mass .A photon is energy moves at c rest mass is mass doesn't move at c.This explains well a photon issue and a void the Energy/Mass wrong equivalence .

If a star has the speed 0.60 c , the way of a photon becomes L' = L (1  v2/c2)1/2 = 0.80 L.
Length contraction is reference frame dependent. In the reference frame where the star is actually measured to have a velocity of 0.6 c, there will be no length contraction. In the star's reference frame, there will be length contraction.
As if MAGIC TRİCK
Just because something seems magical doesn't mean that it's wrong.
Citation: LOGIC
What is your logical reasoning?
A photon then does not have an inertia .
Actually, it does. A photon falls towards a source of gravity at the same rate as an object with mass does. A flashlight held horizontally will have its beam deflected towards the Earth (although this will be far too slight to notice with the human eye). It would be far more noticeable near a black hole.

deceleration does not necessarily is due to mass increment ,it is due to inertia increment. Mass/Energy equivalence is just a mess as I mentioned Reply#24
Inertia is proportional to mass .Bigger mass is bigger inertia and bigger deceleration.
As inertia is proportional to mass .We can replace mass with inertia in the equation of S.R
I=I0/√(1v²/c²)
Did you read that before you posted it?
You start by saying that mass and inertia are different, then you say they are proportional, then you say they are the same.
While you are at it you say
"Bigger mass is bigger inertia and bigger deceleration."
which is plainly the wrong way round.
So, lets get back to the point I made.
There's an enormous reaction force on the magnets used in accelerators.
And that's because they are accelerating rather large masses.
But the thing they are accelerating is just a bunch of protons.
If the mass of the proton didn't increase with energy, the forces needed to bend their path into a circle would be much smaller.
But the forces are real, and so we know that the mass increase is real.
This remains true whether you are confused about it or not.
A photon then does not have an inertia
And yet there is a transfer of momentum when it is emitted, reflected or absorbed.
Your view doesn't agree with reality.
This is not because reality messed up.
E=mc² is true but that doesn't mean energy behaves as mass .
Yes it does.
If you have enough of it, and don't take account of it, it will tear the magnets out of their mountings.

A photon then does not have an inertia .
Actually, it does. A photon falls towards a source of gravity at the same rate as an object with mass does.
A photon will not have any acceleration in any circumstances in the universe . Its speed can change but it jumps from a velocity value to another velocity value .Inertia is the resistance to change in velocity or to gain acceleration .A Photon also has gravity because a photon is energy , both energy and rest mass have gravity .
A photon moves .As it has no rest mass it will move at the maximum speed which is c this is without any obstacles. that prevents it from reaching c .This obstacle is rest mass which if a photon has it will prevent a photon from reaching c.. The rest mass is an obstacle because a rest mass never reaches c.

A photon will not have any acceleration in any circumstances in the universe .
Ah, so you are denying the existence of gravitational lensing. That's good to know.

If it wasn't for relativistic mass, gold wouldn't be yellow: https://www.fourmilab.ch/documents/golden_glow/
With an atomic number of 79, gold is in the last row of the periodic table containing stable elements, and only four stable elements (mercury, thallium, lead, and bismuth) have greater atomic number.
Why those heavier elements look grey? Why copper look orange?

The term "photon" is very specific. It refers to the quantum of the electromagnetic force. It is a boson particle with a spin of one, is electricallyneutral and does not have a rest mass. It something does not have those characteristics, then it isn't a photon.
Do transformers, induction heaters, wireless chargers, radio antennae release photons?
If a single positive pulse is fed to an antenna, will it release electricallyneutral photons?

A photon will not have any acceleration in any circumstances in the universe .
Ah, so you are denying the existence of gravitational lensing. That's good to know.
Acceleration the rate in increment in velocity .A photon doesn't accelerate .The photon velocity can not increase more than c .Deceleration is different .Deceleration is the rate of reduction in velocity a photon can have a reduction in its speed and this true ..
Pushing a flashlight forward will not increase light speed . So photon has no acceleration or inertia .
pushing a flashlight forward is a good example pushing will not accelerate a photon because a photon has no rest mass.
Mass and inertia are proportional .And both can prevent mass to reach c. Einstein used mass instead of inertia and made the mess in reply #24

OK, we know that the universe is finite.
So we know that there is only space (and energy) in it for a finite number of photons.
so LOGIC, says you are wrong.
I would say almost infinite ; so opportunism is not significant.
Yet you think logic says you ar right.
So, we can deduce that you do not understand logic.
That explains a lot.
Please refrain from further posting until you have learned to be rational.
You could not comprehend requirement of a comparison material/object for the the lifetime experiment of muons. Therefore...your refeering is problematic for any subject.

You could not comprehend requirement of a comparison material/object for the the lifetime experiment of muons. Therefore...your refeering is problematic for any subject.
Do any of the professional scientists here understand "requirement of a comparison material/object for the the lifetime experiment of muons."?
I can't even parse it.
It's true that, if stuff is written badly enough, I can't review it.

If a single positive pulse is fed to an antenna,
What do you think you mean by that?

Quote from: Kryptid on Today at 07:50:34
Quote from: Yahya A.Sharif on Today at 07:43:52
A photon will not have any acceleration in any circumstances in the universe .
Ah, so you are denying the existence of gravitational lensing. That's good to know.
Acceleration the rate in increment in velocity .A photon doesn't accelerate .The photon velocity can not increase more than c .Deceleration is different .Deceleration is the rate of reduction in velocity a photon can have a reduction in its speed and this true ..
Pushing a flashlight forward will not increase light speed . So photon has no acceleration or inertia .
pushing a flashlight forward is a good example pushing will not accelerate a photon because a photon has no rest mass.
Mass and inertia are proportional .And both can prevent mass to reach c. Einstein used mass instead of inertia and made the mess in reply #24
OK, so you don't understand what acceleration is.
It the rate of change of velocity.
Velocity is a combination of speed and direction.
The speed of light is constant but, in a gravitational field the direction changes, and thus the velocity changes.
Light traveling on a "curved" path is accelerated towards the inside of the curve.

Do transformers, induction heaters, wireless chargers, radio antennae release photons?
Yes.

If a single positive pulse is fed to an antenna,
What do you think you mean by that?
For example, a battery is grounded on its negative terminal. The positive terminal is momentarily connected to the antenna, and then disconnected again after some time.

So thats a bidirectional pulse where the voltage flows into the antenna, hits the end and bounces back and to for a while, followed by a similar pulse when you disconnect it.
Yes that will radiate EM radiation.

Do any of the professional scientists here understand "requirement of a comparison material/object for the the lifetime experiment of muons."?
Yes absolutely, any one asks to compare what is the normal value of muons' lifetime.
Is this a discussion subject in a science forum?
You are great.

Acceleration the rate in increment in velocity .A photon doesn't accelerate .The photon velocity can not increase more than c .Deceleration is different .Deceleration is the rate of reduction in velocity a photon can have a reduction in its speed and this true ..
Pushing a flashlight forward will not increase light speed . So photon has no acceleration or inertia .
pushing a flashlight forward is a good example pushing will not accelerate a photon because a photon has no rest mass.
Mass and inertia are proportional .And both can prevent mass to reach c. Einstein used mass instead of inertia and made the mess in reply #24
Counterintuitively, an object can travel at constant speed and be in a state of acceleration at the same time. Take the teeth of a spinning gear as an example. The gear can be spinning at a constant speed but the teeth with be experiencing an angular acceleration. This is the same thing as a ray of light being trapped in orbit around a black hole. The ray of light is still traveling at c, but it is also experiencing angular acceleration. That, in turn, it just an extreme case of light being deflected by a gravitational field (which is what causes gravitational lensing). So light can indeed experience an acceleration due to gravity. That acceleration is the same acceleration experienced by matter.
Inertia is defined as the resistance of an object to a change in velocity. An object's mass (and therefore inertia) will cause it to resist being accelerated by a gravitational field. However, this resistance is identical to the resistance to acceleration that a light ray will experience in a gravitational field. Take a look at the equivalence principle. If we are in an elevator accelerating upwards, then its effects are indistinguishable from that of a uniform gravitational field.
Let's say that we are in that elevator and we have a pebble in one hand and a flashlight in the other. We orient the flashlight horizontally and flip it on at the exact same time that we release the pebble. Because the elevator is accelerating upwards, the floor of the elevator will accelerate towards both the pebble and the flashlight beam. In our reference frame, it will look like the pebble is accelerating towards the floor. In turn, it will look like the light beam is being bent more and more towards the floor of the elevator as well (assuming a very, very wide elevator to keep it from hitting the wall, of course). The pebble will then hit the floor at the same moment that the flashlight beam will hit the floor.
Since this is equivalent to a gravitational field, a beam of light will be bent towards a source of gravity at the same rate as the falling pebble. If the pebble is defined as having mass and inertia, then the light beam can also be defined as having mass and inertia because it behaves the same way in that same gravitational field.

Do any of the professional scientists here understand "requirement of a comparison material/object for the the lifetime experiment of muons."?
Yes absolutely, any one asks to compare what is the normal value of muons' lifetime.
Is this a discussion subject in a science forum?
You are great.
Why didn't you say that the first time.
OK, we know what the lifetime of muons is because we can make them.
The important issue is that the half life varies with speed.
And it's calculable
https://en.wikipedia.org/wiki/Muon#Theoretical_decay_rate
So, what was that about this
Is this a discussion subject in a science forum?
?
Did you think you were doing science or someting?

Counterintuitively, an object can travel at constant velocity and be in a state of acceleration at the same time. Take the teeth of a spinning gear as an example. The gear can be spinning at a constant velocity
That's not a constant velocity, it's a constant speed.
That's the difference.

That's not a constant velocity, it's a constant speed.
That's the difference.
Ah, you are right. I'll go back and correct it.

For example, a battery is grounded on its negative terminal.
That is merely a matter of convention. On American cars the negative terminal is "grounded" by attaching the negative terminal of the battery to the cars frame. In some European cars its the opposite, i.e. positive terminal is set to ground. There is nothing sacred about either convention.

For example, a battery is grounded on its negative terminal.
That is merely a matter of convention. On American cars the negative terminal is "grounded" by attaching the negative terminal of the battery to the cars frame. In some European cars its the opposite, i.e. positive terminal is set to ground. There is nothing sacred about either convention.
Nobody said there was.
But, since I asked him for a specific example, and he provided one, he was doing what was actually needed.

So thats a bidirectional pulse where the voltage flows into the antenna, hits the end and bounces back and to for a while, followed by a similar pulse when you disconnect it.
Yes that will radiate EM radiation.
Let's say that the antenna is connected to the positive terminal of battery for 1 minute. During that time, the antenna has higher electric potential than the ground. Does it release photons in this period? Or photons are only released during the change of electric potential, i.e. when connecting and disconnecting it?

The antenna has inductance capacitance and resistance.
In principle it takes "forever" for the voltage and currents to settle.
In practice it probably takes a few times the time required for light to travel the length of the antenna.
After that, it stops emitting EM radiation except of course, that arising from the fact that it is (very slightly) warm.

If you look at the three equations of Special Relativity; SR, photons are modeled with two equations; distance and time for wavelength and frequency. Red shift only needs these two equations. Mass or rest mass uses all three equations; time, distance and relativistic mass. Photons and mass are two different phases, since they do not use the same number of SR parameters. Like ice and water two phases can exist side by side and interconvert into each other depending on the conditions; E=mc2.
When we look out the universe, our instruments and eyes receive EM radiations from the distant parts of the universe. We see the photons from quasars but we do not directly see the mass of the same quasars. We infer the mass from theory. The net affect is we use two equation Special Relativity to infer mass, which actually requires three equations to do properly.
This creates the illusion called relative reference, which is real if we limit ourselves to two equation SR. It does not work with three equation SR. There is a push in some area of physics to get rid of relativistic mass, since it makes it harder to do the relative magic trick.
As an example, say you had a train that was moving at velocity=V. There is a person who is stationary at the train station. Both references will see a relative velocity V. The operative word is "see" which implies photons and two equation SR. This relative velocity assumption will be valid as long as we only time and distance to define what we "see".
It does not work if we use mass, which is connected to what we "feel". Momentum will cause us to feel motion. No train os so smooth as to damp out all feelings. If the man at the station does not feel anything, he may infer he is stationary. However, his eyes will see what appears to be relative motion; magic trick. Relative reference confuses many people because many will visualize both the sense of sight and the sense of feeling and get confused by the possible conflict. You need to be conditioned to only see, but not feel anything, before the magic works and you become an expert.
The seeing part is well established, Let us look at the feeling part connected to mass and momentum. If the train ran into you at velocity=V, versus the stationary person running into you at velocity=V, the pain level will be different, due to the impact of different masses and momentum. This is due to conservation of energy. The difference is connected to the energy needed for each reference to reach V.
If we know the experiment used 5000 liters of fuel to account for all the motion; energy balance in the two reference system, we know the train has to be in motion, even if the two parameter SR appears to show us moving relative to each other. If the man had been propelled with 5000 liters of fuel, he would be going much faster than V. It would violate energy conservation if he only ended up going V. You would know what you appear to see is an two parameter illusion.
In terms of stars in the universe, we cannot touch and feel the stars. We are stuck at the relative reference of sight and two parameter SR, without anyway to do an energy balance with feeling, mass, momentum, and three parameter SR. This could change if we knew how to measure relativistic mass directly. This would allow us to directly infer a universal energy balance. This measurement has been very elusive, so many in physicists would prefer to make it go away and use only two parameter SR. However, this gives us the relative illusion affect and no way to settle the energy balance. It also confuses mass and photons since both are attributed to two parameter SR; one phase.

Photons and mass are two different phases,
That's still wrong.
Why say it again?

The antenna has inductance capacitance and resistance.
In principle it takes "forever" for the voltage and currents to settle.
In practice it probably takes a few times the time required for light to travel the length of the antenna.
After that, it stops emitting EM radiation except of course, that arising from the fact that it is (very slightly) warm.
Is there a minimum limit for photon's frequency? Can it be 0?

Is there a minimum limit for photon's frequency? Can it be 0?
The em wave/photon is caused by the acceleration of the electrons, if there is no frequency there is no acceleration, hence no energy (E=hf).
The leading edge of a pulse (+ve going or ve going) contains a wide range of frequencies and as @Bored chemist says, will take finite time to build and settle and includes reflections from the far end of the antenna.

Is there a minimum limit for photon's frequency? Can it be 0?
The em wave/photon is caused by the acceleration of the electrons, if there is no frequency there is no acceleration, hence no energy (E=hf).
The leading edge of a pulse (+ve going or ve going) contains a wide range of frequencies and as @Bored chemist says, will take finite time to build and settle and includes reflections from the far end of the antenna.
Is it possible to produce 1 microHertz photon? What about 1 nanoHertz? Is there any lower limit?

Creating EM radiation with wavelengths bigger than the Earth is tricky.
Proving that you have created it probably even trickier.

So, there is only practical limit then, while the fundamental limit is 0.

Is there a minimum limit for photon's frequency? Can it be 0?
No. It can't be zero.
The em wave/photon is caused by the acceleration of the electrons, if there is no frequency there is no acceleration, hence no energy (E=hf).
An atom can emit photons and in QM there is no such thing as an electron acceleration
Is it possible to produce 1 microHertz photon? What about 1 nanoHertz? Is there any lower limit?
Sure its possible. What is a gigahertz in one frame can be a microhertz in another frame.

An atom can emit photons and in QM there is no such thing as an electron acceleration
Can a free electron emit photons?

An atom can emit photons and in QM there is no such thing as an electron acceleration
Can a free electron emit photons?
define "free" but...
https://en.wikipedia.org/wiki/Freeelectron_laser
https://en.wikipedia.org/wiki/Synchrotron_radiation

define "free" but...
Not attached to an atom/ion.

An interesting way to look at the photon is connected to its waveparticle duality in the light of Special Relativity.
A photon will move at the speed of light. Say we applied Special Relativity; SR, to a photon and use wavelength for distance, and frequency for 1/time. We then plug in the speed of light into the SR equations for time and distance Based on the solution, photons should all show uniformity, since the solutions at c will the same for all wavelengths and frequencies.
However, what we observe is diversity at all wavelengths even though we know the photon is moving at the speed of light. This tells me the diversity nature of the photon, is not connected to the speed of light. Rather is has to be an inertial affect, not connected to the speed of light. The red and blue shift are inertial affects. They do not need the speed of light to be induced. Motional less that c will induce them.
One way to explain the waveparticle duality in light of the Special Relativity; SR paradox, is the wave aspect of the photon is inertial, while the particle aspect, is at the speed of light. For this to be possible, the wave would need to be generated in a cyclic inertial way, as the particle moves at c. A sine wave crosses the yaxis three times each wave. Where the amplitude is zero is where the wave energy is uniform and consistent with the speed of light.
In other topics, I have contended that the speed of light is the ground state of the universe. The inertial wave is at higher potential than the speed of light particle, except where the wave crosses the yaxis. There the inertial potential become zero; uniformity.

A photon is a Riemann Sphere with momentum charges on a circle in the sphere.

A photon is a Riemann Sphere with momentum charges on a circle in the sphere.
No it isn't.

Have you got a better idea that has momentum charges encoded in the particle?

momentum charges encoded in the particle?
That doesn't mean anything.

momentum charges encoded in the particle?
That doesn't mean anything.
Very convenient for you, that you can't interpret it. Any idiot can interpret it.

Very convenient for you, that you can't interpret it. Any idiot can interpret it.
Okay then. Please provide a link explaining what a "momentum charge" is.

"Momentum charge" is the charge necessary for a particle to be able to carry momentum. It's like an electric charge. A photon has to have momentum charge to be able to follow a curved path in curved spacetime.

"Momentum charge" is the charge necessary for a particle to be able to carry momentum
No
"Momentum charge" is something you made up.
The thing that allows particles to carry momentum is called "mass".

The thing that allows particles to carry momentum is called "mass".
Yes, but a photon is massless, yet it must have something like mass. I chose "momentum charge" as a reasonable name.
So, I take it you don't have a better idea.

yet it must have something like mass.
Yes... it has... mass.
Confusingly, it is also massless.
No need to make up names.

"Momentum charge" is the charge necessary for a particle to be able to carry momentum.
Please provide evidence that a "charge" is necessary in order for an object to carry momentum.
yet it must have something like mass.
That "something" is called energy. E=mc^{2}.

That "something" is called energy. E=mc2.
Energy is just a number. It must have something tangible: something a force can push on.
Please provide evidence that a "charge" is necessary in order for an object to carry momentum.
Momentum or mass is a basic property of particles. If it has no momentum charge or mass charge a force cannot act on it: F = dp/dt. So one can't prove it by finding a particle without momentum or mass charge.

Energy is just a number.
No, it's a quantity.
It has units Joules.momentum charge or mass charge
Inventing another phrase doesn't make things better.

No, it's a quantity.
You can't apply a force to a quantity. "Quantity" is abstract about something.
So it doesn't make sense to say a particle with quantity follows a curved path because it has quantity. It must have something tangible corresponding to the quantity.
Inventing another phrase doesn't make things better.
What else are you going to call it: you can't call it "mass charge".

You can't apply a force to a quantity. "Quantity" is abstract about something.
Charge is a quantity as well.
So it doesn't make sense to say a particle with quantity follows a curved path because it has quantity. It must have something tangible corresponding to the quantity.
How are you defining "tangible"?
What else are you going to call it: you can't call it "mass charge".
How about simply "mass"? Photons have relativistic mass.

How are you defining "tangible"?
A left out spacetime event on a circle in a Riemann Sphere. Thus equivalent to a set with additional structure.
How about simply "mass"? Photons have relativistic mass.
I guess one can call it "relativistic mass". I just think it is paradoxical since one also calls a photon "massless".

You can't apply a force to a quantity.
You also can't apply force to energy.
So that fits just fine with energy being a quantity.
What else are you going to call it: you can't call it "mass charge".
First, before you invent a name for it, you have to demonstrate that it's actually real.
A left out spacetime event on a circle in a Riemann Sphere. Thus related to a set with additional structure.
How are you defining
"A left out spacetime event on a circle in a Riemann Sphere. Thus related to a set with additional structure."
?
I just think it is paradoxical since one also calls a photon "massless".
Photons are uncharged, yet you seem happy to say they have a charge.

So it doesn't make sense to say a particle with quantity follows a curved path because it has quantity. It must have something tangible corresponding to the quantity.
That's so general as to be hard to interpret but, for example, charge is a quantity.
And you can say
" a particle with quantity charge follows a curved path because it has quantity charge "
Mass is also a quantity.
"a particle with quantity mass follows a curved path because it has quantity mass"

First, before you invent a name for it, you have to demonstrate that it's actually real.
They showed it is real with the experiment about bent starlight.
How are you defining
"A left out spacetime event on a circle in a Riemann Sphere. Thus related to a set with additional structure."
A circle and Riemann Sphere are defined in mathematics books. For "left out spacetime node" I will say it is a hole in the circle in the Riemann Sphere. This is equivalent to a set with additional structures.
Photons are uncharged, yet you seem happy to say they have a charge.
I call all properties (not spin, position or velocity) "charges" since I code them using identical methods on a circle in a Riemann Sphere.
charge is a quantity.
I define them as quantities with additional structures.

They showed it is real with the experiment about bent starlight.
Yes, they showed that it has mass.
A circle and Riemann Sphere are defined in mathematics books.
And they are abstract concepts with no physical reality.
I didn't think you liked abstract things...
"Quantity" is abstract about something.
For "left out spacetime node" I will say it is a hole in the circle in the Riemann Sphere.
So, you have defined it in terms of a hole in something which does not exist.
That's not really progress...
This is equivalent to a set with additional structures.
How do you define "a set with additional structures."?
I call all properties (not spin, position or velocity) "charges" since I code them using identical methods on a circle in a Riemann Sphere.
OK, so you deliberately choose to misuse a word, so you can apply it to something that does not exist.
Have you mistaken this for a creative writing forum?
I define them as quantities with additional structures.
That's not a definition; it is word salad.

A left out spacetime event on a circle in a Riemann Sphere.
Can you rephrase that in a way that the average person can understand?
I just think it is paradoxical since one also calls a photon "massless".
The kind of mass that a photon doesn't have is "invariant" or "rest" mass.

Yes, they showed that it has mass.
It can be interpreted that they showed it has momentum. E = pc.
And they are abstract concepts with no physical reality.
I have proof they are real. Not objective proof.
How do you define "a set with additional structures."?
Just as the words say, it comes from a mathematics book. The structures are a metric plus a map that maps physical space to a Riemann Sphere, presumably the edges and nodes of a graph too.
OK, so you deliberately choose to misuse a word,
I've seen the term "mass charge" used, I don't remember where.
That's not a definition; it is word salad.
Yes, it is not a definition, just a description of a definition.
Can you rephrase that in a way that the average person can understand?
A hole in a circle in a Riemann Sphere. A Riemann Sphere is the Complex plane rolled into a sphere such that the plane maps to the sphere by stereographic projection. A sphere is a 2dimensional surface of a ball.

I have proof they are real. Not objective proof.
Please show this proof.
A hole in a circle in a Riemann Sphere. A Riemann Sphere is the Complex plane rolled into a sphere such that the plane maps to the sphere by stereographic projection.
That didn't help. At all.

It can be interpreted that they showed it has momentum. E = pc.
If you like.
But it certainly doesn't say anything about "charge", does it?
I have proof they are real. Not objective proof.
So, not actually proof then.
Why did you think that would help?
The structures are a metric plus a map that maps physical space to a Riemann Sphere, presumably the edges and nodes of a graph too.
Yes, but how do you define it in this context in a way that actually helps to say what the thing actually is?
I've seen the term "mass charge" used
It's probably in the police manual
So what , that's no excuse to misuse it.
A hole in a circle in a Riemann Sphere.
So, it's (assuming it's small) a circle on a plane where one of the axes is imaginary.
Do you understand how that rules out the idea that it is "real" in either the colloquial or the mathematical senses?
just a description of a definition.
No.
"word salad" is a description of your definition.
Please try harder.

A circle and Riemann Sphere are defined in mathematics books. For "left out spacetime node" I will say it is a hole in the circle in the Riemann Sphere. This is equivalent to a set with additional structures.
A structure is a set endowed with feature(s). It is improper terminology to say ‘a set with additional structures’, A very simple example of a structure would be the alphabet, arranged in alphabetical order. The 26 letters are the set independent of any order. The endowed feature is the ordering.
A circle is the set of points equidistant from a given point. A circle can be defined on the surface of a Riemann Sphere by specifying the central point and the radius. Note that the surface of a Riemann Sphere does not have a constant metric but compresses asymptotically as the ∞ pole is approached. 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.
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.
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?
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?
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’?
In short, do you really know what your are talking about as opposed to just throwing fancy words around?

In short, do you really know what your 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.
It is improper terminology to say ‘a set with additional structures’
So it's a set endowed with features. The statement is standard in Mathematics.
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.
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 zaxis pointing 45 degrees downwards. Maybe in some other projection, it will appear squashed.
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.
I can talk about a line "in" a plane if the plane is infinitesimally thick (the thickness is nonzero).
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?
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).
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?
I propose copying one dimension of space and one of time into a Riemann Sphere, resulting in a particle when endowed with charges.
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’?
Points of space on the circle versus left out points (infinitesimal breaks in the circle).
Why did you think that would help?
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.
Yes, but how do you define it in this context in a way that actually helps to say what the thing actually is?
It is actually a Riemann Sphere with left out points.
Do you understand how that rules out the idea that it is "real" in either the colloquial or the mathematical senses?
No, there are many physical objects that are modeled with imaginary numbers.
Please try harder.
I got it defined in a picture at the post listed above.
Please show this proof.
See 10 lines of text upwards.
That didn't help. At all.
The Complex plane is two real lines at right angles to one another, with one line multiplied by . Stereographic projection is: take a point in the plane and draw a line from the point to the north pole of the sphere. Then the point where the line intersects the sphere is where the point maps to.

In 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.
The article cited makes no sense. It is a collection of buzzwords from the field of Physics that is full of problems. The journal in which it was published was originally mostly dedicated to condensed matter physics where it was fairly well respected. But it was too limited a field with too much competition and to stay above water they gradually started publishing anything and everything that even looked like physics and/or mathematics regardless of actual value. In the same issue that the article you linked appears, there was another article that claimed that prime numbers were predictable and one of the sources it cited claimed that Pi was really a rational number.
If you think your work is so groundbreaking why not send it to Physical Review Letters? They publish oddball speculative things. Of course the first thing they do is check your academic background in the field of physics. :
You have yet to demonstrate that you know what you are talking about.
It is improper terminology to say ‘a set with additional structures’
So it's a set endowed with features. The statement is standard in Mathematics.
You did not know that until I told you. Yet it is an important part of your argument.
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.
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 zaxis pointing 45 degrees downwards. Maybe in some other projection, it will appear squashed.
A circle through the ∞ pole is not possible because a center point from which it is equidistant cannot be defined.
A geometric line has zero thickness. It cannot be thicker or thinner.
The Riemann sphere is a mapping of the extended complex plane onto a sphere. It consists entirely of surface. There is no zaxis.
A coordinate system based on axes is orthogonal by definition. There is no such thing as an axis ‘pointing 45 degrees downwards’.
In the Riemann sphere, the Origin (where all values are zero) is located at the ‘south pole’. All complex values contained on the complex plane on or within a circle of radius 1 lie on or below the ‘equator’ of the Riemann sphere. All other values are in the ‘northern hemisphere’ with ∞ at the ‘north pole’. There are no finite complex values on this pole and in fact the distinction between real and imaginary vanishes, since the infinitely distant edges of the complex plane are rolled up into a point.
Since all complex numbers with x and/or y values greater than 1 and i respectively are above the ‘equator’ and the ‘north pole’ is ∞, the geometric distance between successive values of a given positive increment becomes smaller and smaller as these values approach the pole. Using the definition of a circle as the set of points equidistant from a given point, a circle drawn on the surface of a Riemann sphere appears more and more squashed as the center point moves ‘northward’ because the geometric distance between value differences on the surface of the sphere becomes smaller and smaller.
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.
I can talk about a line "in" a plane if the plane is infinitesimally thick (the thickness is nonzero).
The Riemann sphere is a projection of a plane onto a sphere. A plane has zero thickness and so does the surface of a sphere. An infinitesimally small but nonzero thickness makes no sense unless you are trying to introduce nonstandard analysis. Which I do understand, so if you are going to import more buzzwords, I am ready for you. Hint: A Large Cardinal is not a Big Bird.
.
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?
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).
On a Riemann sphere, a line of longitude intersects both 0 and ∞. A circle is the set of points equidistant from a given point. On an ordinary sphere, a line of longitude is a pole intersecting Great Circle. There are two possible center points, both on the equator lying on another pole intersecting Great Circle 90° around from the given longitude. However, on a Riemann sphere the length from pole to pole is infinite, from zero to ∞. There is no unique point or pair of points that can be defined as the center point of the circle. Any point you choose has a varying distance to the different parts of the circle. The equidistant requirement cannot be satisfied. It is not possible to construct a pole intersecting circle on a Riemann sphere. It may look like a circle on a globe but it is not one on a Riemann sphere.
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.
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?
I propose copying one dimension of space and one of time into a Riemann Sphere, resulting in a particle when endowed with charges.
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.
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’?
Points of space on the circle versus left out points (infinitesimal breaks in the circle).
We have already seen that a line of longitude on a Riemann sphere is not a circle. So either change your terminology or change your model.
In addition, infinitesimally small holes do not work well with quantum theory.
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?

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?

Can you provide any actual evidence for those claims?
Just my experience, so no objective proof. You may ask a Psychic person.
A 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).
There is no such thing as an axis ‘pointing 45 degrees downwards’
The projection of the zaxis onto 2 dimensions is such.
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.
By the projection, one could imagine the sphere to be an ordinary sphere with a linear metric by reassigning distances. There is a onetoone mapping between a Riemann sphere and an ordinary sphere. The 3dimensional space it is in allows for straight lines as chords intersecting the Riemann sphere.
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.
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.
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?
The structure of a 2sphere.
In addition, infinitesimally small holes do not work well with quantum theory.
Call them little Plancklength openings then.

Just my experience, so no objective proof. You may ask a Psychic person.
No evidence = no science.

A 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).
There is no center of a Riemann sphere. It consists only of the surface. It is a projection of a plane and such has only 2 dimensions, just like the plane. The length of a line from the 0 point to the ∞ point is infinite. If you think it is finite, then there is some maximum value for x and y, which would mean that the entire complex plane is not projected onto the sphere. The whole point of the Riemann sphere is that ∞ is defined and a bunch of otherwise nasty functions become well behaved without needing to put arbitrary constraints on them.
You do not understand what a Riemann sphere is. Trying to make it into an ordinary sphere fails big time. You are just using impressive sounding words in order to sound impressive.
There is no such thing as an axis ‘pointing 45 degrees downwards’
The projection of the zaxis onto 2 dimensions is such.
You are talking about a picture. Just like with the Riemann sphere, you are looking at a picture and thinking it somehow is the real thing. The placement of the zaxis on the xy plane in such pictures is entirely arbitrary. It is simply a way of showing three dimensions on a twodimensional page. In fact, there is no such thing as mathematical projecting a third orthogonal axis onto a plane.
Projection of points having nonzero z coordinate values onto the xy plane is a simple matrix operation. For vector (x,y,z) the projection vector is (x,y,0).
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.
By the projection, one could imagine the sphere to be an ordinary sphere with a linear metric by reassigning distances. There is a onetoone mapping between a Riemann sphere and an ordinary sphere. The 3dimensional space it is in allows for straight lines as chords intersecting the Riemann sphere.
The picture is simply for visualization purposes. Imagining a Riemann sphere to be an ordinary sphere removes all its unique properties. A Riemann sphere is surface only and is not embedded in three dimensions. Even an ordinary sphere which is a ‘rolled up’ projection of a finite section of a plane still only has a surface and is not embedded in three dimensions.
Stop looking at pictures and learn some math.
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.
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.
If you need three dimensions, you cannot use a Riemann sphere. It only has two dimensions just like the plane it maps to. It consists only of the surface.
Spacetime does not contains names of particles at each point, only the properties as they interact with whatever fields may apply. A field theory predicts what forces will be in effect at each point. It is the field that controls what happens at the points. No need for each point to figure out what should happen. And properties fall nicely in the symmetries of the Standard Model. These properties are in effect regardless of where the particle is. The interaction of these properties with the field(s) determines what happens, including whatever transformations may take place. Spacetime does not need a table of particle names and properties for each point. The imagined interior has no coordinates assigned to it.
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?
The structure of a 2sphere.
A 2sphere is a circle. As we have seen, your longitudinal circles do not work on a Riemann sphere. There is no definable center point. A conventional sphere can have longitudinal circles.
In addition, infinitesimally small holes do not work well with quantum theory.
Call them little Plancklength openings then.
I would make them larger than that. Space at the Planck length is unstable because its Heisenberg Uncertainty selfenergy is enough to make a black hole.

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.

Just my experience, so no objective proof.
There aren't any.
Did you somehow think that was the way adults discuss things?

What physical experiment could be performed to demonstrate that particles are Riemann spheres? Keep in mind that if it can't be experimentally tested, then it isn't a scientific hypothesis.

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 Riemann sphere IS conceptual. One cannot make a finite model of it. You think that because it is pictured as a globe that it is a globe. The surface is a mapping of the complex plane. There is no inside and the surface is infinite. The image of a globe appears to be the sum total of your knowledge of the Riemann sphere. You are just playing games with impressive sounding words to try to make yourself sound impressive. It is not working.
You have no real knowledge of serious mathematics. Give it up.
The holes must be smaller than a photon. The size of a quantum of spacetime. Yes, a photon has a size in my model.
A quantum of spacetime is not a meaningful concept, at least not the way you want to use it. The concept of a quantum generally involves a fundamental unit, for example, the electric charge of an electron. All electric charges are multiples of this fixed quantum.
At the scale of a Planck volume, the selfenergy due to the Uncertainty Principle combined with the extremely small dimensions results in a black hole. One might want to speak of the size of the black hole in Euclidean terms, e.g., the diameter. However, a black hole does not have a diameter. It is all surface. There is no mapping of that region of space into a Euclidean space because, as with macro black holes, division by zero happens. Moreover, near that scale spacetime fluctuations, also due to the Uncertainty Principle, do not allow a fixed length to be defined.
‘The size of a quantum of spacetime’ is not a meaningful term.
You have no real knowledge of serious quantum physics. Give it up.
It seems to me that somehow it is not satisfying to say a particle is a vibration in a field.
Quantum field theorists speak of particles as disturbances or excitations in a field. The term ‘vibration’, although often used in popularized science explanations, can be misleading since it could imply that a particle can be described simply by its wavelength. All particles with mass have a Compton wavelength but they also have other characteristics.
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.
The idea of interactions of excitations in a field is the basis for field type physical theories, the first serious example of which was Maxwell’s equations formulated over a century and a half ago. Quantum Field Theory (QFT) is a newer and much more complicated example.
I am confused by your phrase “communicated between fields”. The field is the underlying background in which the excitations (particles) interact. In QFT the field incorporates classical mechanics and electromagnetism with quantum mechanics (including QED and QCD) and Special Relativity. No one has yet figured out how to get General Relativity to play nice with the other children.
So what does “communicated between fields (plural)” mean?
Or is it that you have no real knowledge of what a field theory is, either?

What 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.
So 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.
This is the story that must be considered when trying to simulate the scattering on a computer.
The 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.

What 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.
Wrong. The direction of photon emission when an electron drops from excited state to ground state is random with respect to the electron's spin state, which is the only way of talking about an electron’s axis. The direction of photon emission is determined in a probabilistic manner by multiple factors including the prevailing magnetic field, of which the nearby nucleus is a major component, and preserving energy momentum conservation over event accumulation.
In addition, electron spin is a matter of angular momentum, in effect spinning clockwise or counterclockwise. It is not a matter of which way a pole is pointing. Electrons with opposing spin values are not ‘upside down’ with respect to each other. They are spinning in opposite directions. To drive that home, the only formulation for spin states that matches measurements is the formalism of angular momentum. For that to be the case, the clockwise or counterclockwise spin must be with respect to a fixed axis. That is, your hypothetical 0 and ∞ axis must always point in the same direction, negating your argument.
Now explain what differences there may be between the two hemispheres of an electron. Oceans, continents, ice caps? 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.
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?
So 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.
Wow, where do I start?
To begin with, you are using two different meanings of the term ‘field’. Originally you used the term with respect to a particle being a ‘vibration’ in a field, which would imply field theory, a field being a mathematical apparatus for determining the properties at any given point. Now you are using the term to refer to the electromagnetic fields of electrons. Same word, different meanings.
The term λ (lambda) does not appear in Feynman diagrams. That term is used in General Relativity for the Cosmological Constant. What does appear in Feynman diagrams is the term γ (gamma), which is the symbol for the photon. In Feynman diagrams, it is the label applied to the wavy line that indicates the exchange of virtual photons. This diagram show electronelectron scattering.
(https://protonsforbreakfast.files.wordpress.com/2014/04/feynmandiagram1.jpeg)
In Quantum ElectroDynamics (QED) for which Feynman diagrams were first invented, the path is not first chosen. In theory, any path at all might be used, each having its own probability amplitude. This is a wavelike entity. As such, probability amplitudes constructively and destructively interfere with each other. The effective path is the one with the greatest surviving probability amplitude, and usually the only survivor with a resulting probability (square modulus of amplitude) of unity. Paths happen. They are not chosen. There are experiments demonstrating multiple paths being used such as the famous doubleslit example.
There are people here who are familiar with deep Physics and deep Mathematics. You cannot make it up as you go along and not get called out.

The 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.
Wrap your head around this. The Riemann sphere IS the complex plane with ∞ defined. ∞ does not exist on the complex plane. Having ∞ defined allows the use of functions that might for some values involve division by zero as long as the end result of the evolution of the function does not result in division by zero or otherwise carry forward any infinite term, since this is an unusable result.
The Riemann sphere is NOT a distortion of the complex plane. If it were it would yield different answers than the complex plane and that would be pointless. The distance from (x=1, y=0i) to (x=2, y=0i) is the same as the distance from (x=1000001, y=0i) to (x=1000002, y=0i) regardless of whether we are talking about the Riemann sphere or the complex plane. There is no distortion. Thinking the Riemann sphere acts like an ordinary geometric sphere does not work. If you want to look at the Riemann sphere as an ordinary sphere, then the metric is not constant. If you want to insist there is a constant metric, then you cannot have ∞ at the pole. And if you do not have ∞ at the pole then you do not have a Riemann sphere.
Your problem is that you need to see things as pictures rather than trying to grasp the concepts. This is how you ended up with the idea that the zaxis is at 45° because you looked at a picture and misunderstood it. You want your particles to be spheres with circles drawn on them and holes in the circles. You saw a picture representing the Riemann sphere and thought that was an impressive sounding thing that you could get away with because nobody would understand it. Surprise! :)

The Riemann sphere is NOT a distortion of the complex plane.
I'll ask a Mathematician on the other forum and refer you.
Now 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.
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?
Isolate a single electron in a magnetic trap with a known magnetic field imposed. They can do the doubleslit experiment with one electron at a time.
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.
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.

The Riemann sphere is NOT a distortion of the complex plane.
I'll ask a Mathematician on the other forum and refer you.
A Riemann sphere is not a distortion. It is an identity map plus a ∞ point. f(z) = 1/z. A distortion would imply that it gives different answers than the complex plane and in some irregular manner.
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.
I am waiting for that referral.
Now 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.
Since you previously said that the surface of a Riemann sphere is really linear, as you have said earlier and as is clear from your use of the term ‘higher density’, then please tell me at what latitude the size of a spacetime event goes below the Planck area. And how you manage to have a realized infinity at the ∞ pole with an infinite number of spacetime events at one point. If you leave out ∞ then you do not have a Riemann sphere, that being the real power of a Riemann sphere, that 1/0 is defined. BTW that is complex ∞ at the pole. What meaning do you assign to that?
Earlier you referred to an electron as a Riemann sphere. Are there more spacetime events in the northern hemisphere than in the southern hemisphere. 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?
Please give a concise explanation of how your model works. Never mind justifications or ramifications. Just describe the model itself clearly and concisely.
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?
Isolate a single electron in a magnetic trap with a known magnetic field imposed. They can do the doubleslit experiment with one electron at a time.
That has nothing to do with the direction a photon would go from an electron in an excited state. The magnetic field would take part in determining the direction, not some hypothetical ∞ pole on the electron. How do you know which way the pole is pointing? If you do not know that, you have no way to verify your claim.
Were you talking about the photon or the electron going through the slits? Presumably the photon, which of course would require that you know which direction the photon is going. However, unless you know which way your hypothetical ∞ pole is pointing – something you have not yet addressed – it is unrelated to your claim. Regardless, as I said earlier, the magnetic field is going to influence the direction.
On top of all this, whether the accumulation of data in the double slit experiment will yield a shotgun (no selfinterference) pattern or a concentric circles (selfinterference) pattern on the receiver depends on whether or not there are detectors in the slits.
What I see here is that you have grabbed another impressive sounding term to add to your collection, ‘double slit experiment’, with no comprehension of what it means.
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.
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.
When you clearly and concisely explain your model as asked above, be sure to include how the above terms fit in. And once again, explain how with two dimensions you manage to depict three spatial dimensions, with six degrees of freedom, and time. And mass, frequency etc.
And does this mean that things you said earlier in this post and in earlier posts are wrong?

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.
I got it drawn see below.
Since you previously said that the surface of a Riemann sphere is really linear
I was wrong it should have the metric of the Riemann sphere, but live in Euclidean spacetime even if events of spacetime then don't align with the events on the particle.
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?
You got it wrong: the RS is the particle, the holes encode its charges. Actually an electron is two superimposed RSs superimposed with two antiRS, with another particle (a gravitron) superimposed on that.
The whole model is posted at "Quantun Gravity Follows?" in this forum (see the newest reply attachment: "Physics from Axioms clean,pdf" and "Physics from Axioms Advanced.pdf").

You seem to suggest space can compute at the same time as outputting a particle and that the computation takes an infinitesimally long time to execute. I don't think this is the case.

I have done some experiments with my mind and now it seems an electron is a single RS, not a superposition of RSs.

I have done some experiments with my mind
Did you think that actually meant something?

Did you think that actually meant something?
I believe so, I can tell if any information is not from my mind.

I can tell if any information is not from my mind.
No, you can not.
That was sorted out a few hundred years ago.

No, you can not.
That was sorted out a few hundred years ago.
I assure you, I can.
How was it sorted out?
It couldn't have been shorted out for me since I wasn't born yet.

I assure you, I can.
I assure you that you are mistaken.
How was it sorted out?
https://en.wikipedia.org/wiki/Dream_argument
It couldn't have been shorted out for me since I wasn't born yet.
That's just silly.
Do you think that Pythagoras' theorem doesn't apply to you because it's very old?

I carefully examined it, and it can be rigorously tested by anyone.

it can be rigorously tested by anyone.
So far, you have not even described it.

What is a photon ?
A photon whether it is mass or energy it shouldn’t move at c.
A photon is a type of bridge state between the speed of light reference and inertial references. The photon has characteristics of both references at the same time.
Photons travel at the speed of light, yet all photons do not appear homogeneous in appearance. In other words, according to SR as velocity increases toward the speed of light finite distance and time expressions will appear to contract, At the speed of light, distance and time will appear to contract to a pointinstant in our reference,
Yet, even though all photons travel at the speed of light, they do not all show the homogeneity of the point instant as predicted by SR. Instead they will display different expressions of finite distance and time; wavelength and frequency even at the speed of light. Photons have one leg in the speed of light reference; v=c, and one leg in inertial reference; finite wavelength and frequency not characteristic of the c reference.
Photons are a bridge between inertial reference and the speed of light reference. When photons form from the interaction of matter and force, they are born and appear at the speed of light with finite wavelengths and frequencies to reflect their dual nature.
Photons appear when matter lowers potential. In this case we go from only inertial reference to inertial plus the speed of light reference; two legged photon. If matter was to increase potential photons will be absorbed. In this caee we go from inertial plus speed of light to just inertial when matter increases potential.
If you of the math, the speed of light aspect or leg reflects lowered inertial potential since it only appears if matter lowers potential. If matter gains potential the speed of light leg disappears. The speed of light leg is connected to the ground state of the universe. Matter has to lower potential for this leg to appear via the photon. To fully reach the ground state at the speed of light, the inertial leg of the photon also has to lower potential to zero; red shift toward infinite wavelength.

In today's installment of Puppy power gets things wrong he says
Yet, even though all photons travel at the speed of light, they do not all show the homogeneity of the point instant as predicted by SR.
In reality, SR makes no such prediction.
On the other hand
If you of the math, the speed of light aspect or leg reflects lowered inertial potential since it only appears if matter lowers potential.
doesn't make enough sense to even be wrong.

In today's installment of Puppy power gets things wrong he says
Yet, even though all photons travel at the speed of light, they do not all show the homogeneity of the point instant as predicted by SR.
In reality, SR makes no such prediction.
On the other hand
If you of the math, the speed of light aspect or leg reflects lowered inertial potential since it only appears if matter lowers potential.
doesn't make enough sense to even be wrong.
Finite expressions of distance and time, such as the wavelength and frequency of visible light, make no sense in a speed of light reference. At the speed of light, the entire universe would appear to be pointinstant and all wavelength would need to be contained in that point reference; homogeneous. Photons are not entirely in the speed of light reference, since they also show finite expressions of space and time that can only be seen and make sense in inertial reference. The inertia of photons is connect to the inertial reference leg affect and is not a speed of light reference leg affect.
Photons are a type of bridge phenomena between the speed of light reference and all inertial reference. Matter and mass, which are inertial affects, cannot travel at the speed of light according to SR. When matter lowers potential, photons are released and a bridge to the speed of light reference forms. Photons lose their inertial leg through the expansion of the universe; red shift. In the limit, photon wavelength will expand to infinite wavelength and the photon will approach the universal ground state at the speed of light; only one leg and a speed of light reference.
The question one may ask is how can higher potential matter form from the lower potential ground state at the speed of light? It has to do with entropy. Entropy contains energy that is not entirely useable in inertial reference, The second law states that entropy has to increase and therefore entropy is forming a dead pool of unusable energy. This dead pool of energy is the potential within the speed of light reference; infinite entropy. Since this dead energy is not entirely useable by inertial reference it is not part of the traditional energy balance. Based on the traditional inertialcentric energy balance the speed of light reference has least live pool energy, but the most dead pool energy, since live pool energy is an inertial leg affect and not a speed of light reference affect.
At the speed of light spacetime breaks down such that time and space can act independent of each other. This precludes photons since they need space and time to be connected. One can move in time without space limitations and move in space without time limitations; omniscience and omnipresent, respectively. This allows for infinite variety, complexity and infinite entropy and dead pool energy. There is an infinite dead pool of potential, relative to what we call inertial reference, but there are no two legged photons at the speed of light, therefore zero live energy; ground state of the universe relative to all inertial centric references.

in a speed of light reference.
That's the bit that makes no sense.
You keep saying it as if it means something.
It does not.

So far, you have not even described it.
I described it at: New Theories, post: Quantum Gravity Follows? at the attachment: "Physics from Axioms clean.pdf" at reply #6 of the post.

When photons form from the interaction of matter and force, they are born and appear at the speed of light with finite wavelengths and frequencies to reflect their dual nature.
By what mechanism?
Just like I need a method and equipment to pick up an object, the Universe needs a mechanism to emit photons.

in a speed of light reference.
That's the bit that makes no sense.
You keep saying it as if it means something.
It does not.
You do not seem to know how to visualize from the POV of the speed of light reference. I will give you a hint. We know matter cannot go the speed of light, therefore visualizing the speed of light reference using moving matter; rocket ship approaching c, is the wrong approach. This will fall short. One has to go the other way; toward the ground state.
If we would see the universe as a pointinstant at the speed of light, only infinite wavelength could be seen, since only this wavelength will not be seen as less than a point. While infinite wavelength, in internal reference, has almost zero inertial energy; approaches the c reference of the ground state.
When photons form from the interaction of matter and force, they are born and appear at the speed of light with finite wavelengths and frequencies to reflect their dual nature.
By what mechanism?
Just like I need a method and equipment to pick up an object, the Universe needs a mechanism to emit photons.
One possible way to explain the mechanism is with an experimental analogy. Say we had a wave tank with two wave generators, one on each side. The wave generators each generate the same size waves, but they are 180 degrees out of phase. We are adding energy to the tank, from both sides, but the water in the tank appears to be still due to the wave cancelation. We have hidden energy. It is there, but hidden by the wave addition.
To make the hidden energy appear, that is hidden due to the wave cancelation we need to add a partition to the center of the water tank. If I place a board, of any length, in the stillness of the water wave tank, a wave front will build on one side of the partition, and a different wave front will descend on the other side, since the wave generators are 180 degree out phase. The partition releases the hidden energy.
In the case of matter, waves and the release of photons of energy, the partition is the speed of light reference interacting with the hidden inertial waves. When this happens the two legged photon is at the speed of light, rides the partition with an inertial energy leg.
In the case of the election and proton; + and  charge, to use an example, they are equal and opposition in terms of EM waves. There is stillness due to wave cancelation; quantum energy level, at least until the C reference partition appears, causing the photon to emerge at the partition at c.
In our wave tank, if I added a board as a partition, the side of the board with the building wave front will push the board in the direction of the sinking wave front. Photons go away from the positive nucleus toward the election cloud and beyond. There is difference between the two charges in terms of the timing and partition placement.

Today's collection of meaningless/ undefined compound nouns from Puppy power:
"speed of light reference"
" ground state."
"pointinstant"
" c reference "
"hidden energy"
" speed of light reference"
"internal reference"
"zero inertial energy"
" inertial energy leg."
" C reference partition"
And possibly a few I missed.
Please provide a clear explanation of what those phrases mean using real physics.
Then we can examine the question of whether or not you are saying anything useful.

Say we had a wave tank with two wave generators, one on each side.
The analogy does not hold. For water waves we have waves in matter made by matter. In the photonemmision case, we have vibrations in the electron field that must translate into vibrations of the electromagnetic field.