19 hours ago, Markus Hanke said:

I don’t know what this is supposed to show, to be honest. One wire is infinite, the other one is not (it returns to the same starting point). In both cases they look locally flat. But globally these situations differ.

What am trying to say is that at any point in a manifold it can be just be a point of another manifold...it depends on how they are connected and to the extend it's possible....as you have said 'In both cases they look locally flat. But globally these situations differ.' locally the point becomes almost indistinguishable from a point on the flat manifold and a point on circular manifold.

Back to my point of discussion...we are deriving Riemann tensor from parallel transporting a vector in this case a parallelogram loop....it's a loop or in the shape of parallelogram loop coz every other place in the loop is a point arranged, joined,fused( I don't know more better term to use) to form the parallelogram, therefore, the basis vectors has to close for lie bracket to be zero and since to move from one bases vector of parallelogram to there should be no gap for flow curves to close...that point of closure where [X,Y] =0 is what am interested in and what am saying is that, that point can be a point in another manifold... alternatively when we subtract the covariant derivates along the basis vector we remain with [X,Y] lie bracket...this lie bracket can be a point in another manifold.

20 hours ago, Markus Hanke said:

This is where the connection (and thus the Riemann tensor) come in, because it tells is how these local patches are globally related to one another.

Am trying to introduce another local manifold...where principle of uncertainty exist....into local patches that are globally related as you have stated,where Riemann tensor matters.

Minkowski is tangent to this 'another local manifold '.... To extract information from this ' another local manifold' Minkowski becomes a limiting factor.

12 hours ago, KJW said:

That depends on the connection. For a Riemannian connection, normal coordinates exist which simplify the proofs of identities. But for a general connection, for example if the torsion tensor is non-zero, then normal coordinates do not exist in general. Also, if one is using normal coordinates to simplify the proofs of Riemann tensor identities, then one also needs to include the proof of the existence of the normal coordinates themselves. Thus, in the end it may actually be simpler to prove the Riemann tensor identities directly without invoking normal coordinates. And while one is at it, one may as well prove the Riemann tensor identities for the general connection.

True.

Am also trying carefully not to enter into the rabbit hole of mathematics where I keeping digging and can't get out.

3 hours ago, MJ kihara said:

locally the point becomes almost indistinguishable from a point on the flat manifold and a point on circular manifold.

A point is just a point; taken in isolation it carries no further information, so the above is trivially true. But in GR we don’t work with isolated points, we work with semi-Riemannian manifolds endowed with a connection and a metric. There’s a lot of additional structure here, which carries the geometric information that eventually makes up gravity.

3 hours ago, MJ kihara said:

Back to my point of discussion...we are deriving Riemann tensor from parallel transporting a vector in this case a parallelogram loop....it's a loop or in the shape of parallelogram loop coz every other place in the loop is a point arranged, joined,fused( I don't know more better term to use) to form the parallelogram, therefore, the basis vectors has to close for lie bracket to be zero and since to move from one bases vector of parallelogram to there should be no gap for flow curves to close...that point of closure where [X,Y] =0 is what am interested in and what am saying is that, that point can be a point in another manifold... alternatively when we subtract the covariant derivates along the basis vector we remain with [X,Y] lie bracket...this lie bracket can be a point in another manifold.

I can’t make any heads or tails of this, I’m afraid.

3 hours ago, MJ kihara said:

Am trying to introduce another local manifold...where principle of uncertainty exist

If by “uncertainty” you mean non-commutation of certain geometric quantities, then this has already been attempted:

https://en.m.wikipedia.org/wiki/Noncommutative_geometry

I’m not up to date on where this is at right now, but I don’t think it got us any closer to quantum gravity.

3 hours ago, MJ kihara said:

Am also trying carefully not to enter into the rabbit hole of mathematics where I keeping digging and can't get out.

You won’t get very far then, since this whole area is inherently mathematical in nature.

7 hours ago, Markus Hanke said:

A point is just a point; taken in isolation it carries no further information

It depends on the information we have about it...to the necked eyes far away stars are just bright points in the night sky...use a powerful telescope you gain more information and they cease being just bright points.

8 hours ago, Markus Hanke said:

But in GR we don’t work with isolated points, we work with semi-Riemannian manifolds endowed with a connection and a metric.

It's true.

Am not trying to change GR,am scrutinizing it's framework...a manifold is a topological space that locally resembles Euclidean space near each point.according to Wikipedia.

7 hours ago, Markus Hanke said:

I can’t make any heads or tails of this, I’m afraid.

Unfortunately,this is fundamental to what am saying...hope misspelling is not the cause(out of auto suggestion)

7 hours ago, Markus Hanke said:

If by “uncertainty” you mean non-commutation of certain geometric quantities, then this has already been attempted:

https://en.m.wikipedia.org/wiki/Noncommutative_geometry

I’m not up to date on where this is at right now, but I don’t think it got us any closer to quantum gravity.

Uncertainty from the principal of uncertainty;quantum mechanic....from what you have said, it's then uncertainty in commutation and non commutation.

8 hours ago, Markus Hanke said:

You won’t get very far then, since this whole area is inherently mathematical in nature.

The idea is already developed what am doing is that am in the process of formalizing it... gaining established terms ..so far, as long as am concerned it has helped me unimaginably to get understanding within a short period and have deep intuition about concepts of physics...am scrutinizing the basic framework of physics...I will keep following math but not to the point of losing interest to what am doing .... The details are so huge, to the point of not getting to what I want to achieve if I dip down to the ocean of mathematics.....the best thing is,what is correct is correct irregardless of the mathematics formalism used..I will strive to use simple mathematics concepts where possible(Occam's razor be my guiding principle)

12 hours ago, MJ kihara said:

Am also trying carefully not to enter into the rabbit hole of mathematics where I keeping digging and can't get out.

Really? You've been mentioning the Lie derivative, and the Riemann tensor, and manifolds, curvature, etc, for quite a while now.

What are those about if not mathematics? Thai cuisine?

Seems to me you're going in circles around the rabbit hole of mathematics without quite venturing to stick your head in it.

36 minutes ago, MJ kihara said:

Uncertainty from the principal of uncertainty;quantum mechanic....from what you have said, it's then uncertainty in commutation and non commutation.

Ok. So what's up with that?

There's no limit to how small one can be for any pairing of variables, while the other leads to a combined uncertainty (for the non-commuting variables) of approx. h-bar.

Where do we go from there and how does it relate to a theory of all free parameters of the standard model + gravity, which is what TOE means?

58 minutes ago, joigus said:

What are those about if not mathematics? Thai cuisine?

Seems to me you're going in circles around the rabbit hole of mathematics without quite venturing to stick your head in it.

Ha ha ha😂...to get that it was somehow a headache...thanks to repeated online lectures.

given that experience...am afraid to get stuck... however,I have do where I can before Age start interfering with my mental strength and memory.

1 hour ago, joigus said:

Ok. So what's up with that?

Let's take the torsion free formula for example using parallelogram loop...we subtract covariant derivatives along the vector fields after parallel transporting a vector a long the basis vector of infinitesimal loop in this case parallelogram...if the parallel transported vector using a given connection closes we get lie bracket equals zero....now the point is here,to satisfy special theory of relativity the transported vector transfer has to be limited to the speed of light...now the universe is expanding at every other point, however it's effects is negligible at such a point...by the time you are parallel transporting the vector to the other basis vector the length will be different hence as you are doing that uncertainty develops in measurement and therefore,there will be uncertainty in covariant derivatives being commutating or non commutating...all this plays out at the surface topology of a graviton according to what I have...

1 hour ago, joigus said:

There's no limit to how small one can be for any pairing of variables

Yes somehow ,maybe to some limit,It will need further clarification,am not there yet.

1 hour ago, joigus said:

Where do we go from there and how does it relate to a theory of all free parameters of the standard model + gravity, which is what TOE means?

The free parameters of standard model will have to be shown how its related to the graviton and how those parameters contribute to the uncertainty happening on surface topology of a graviton then how graviton is related to being quantum of gravity..i.e related to gravity...for gravity GR is a master.... according to what am doing the link between the two worlds might be or is through lie bracket..

11 minutes ago, MJ kihara said:

hence as you are doing that uncertainty develops in measurement and therefore,there will be uncertainty in covariant derivatives being commutating or non commutating

However since we are working with infinitesimal loop parallelogram...approximating lie bracket to zero will give the correct answer for a flat manifold locally.

1 hour ago, MJ kihara said:

However since we are working with infinitesimal loop parallelogram...approximating lie bracket to zero will give the correct answer for a flat manifold locally.

For a flat manifold "locally" and "globally" carry no significant difference. You may have non-vanishing Christoffel symbols as a consequence of your choice of coordinates (like picking polar coordinates to describe the plane R²), but nothing's going on as concerns curvature and topology, as calculating the Riemann tensor will tell you.

I, among others, fail to see what you're getting at. Every smooth manifold is locally flat. I'm even more at a loss as to what any of this has to do with a theory of all the parametrics of elementary particles, AKA TOE.

44 minutes ago, joigus said:

For a flat manifold "locally" and "globally" carry no significant difference. You may have non-vanishing Christoffel symbols as a consequence of your choice of coordinates (like picking polar coordinates to describe the plane R²), but nothing's going on as concerns curvature and topology, as calculating the Riemann tensor will tell you.

I, among others, fail to see what you're getting at. Every smooth manifold is locally flat. I'm even more at a loss as to what any of this has to do with a theory of all the parametrics of elementary particles, AKA TOE.

Actually not true. Flat manifolds can have non-trivial topology. But the rest of my comment is quite alright.

I can feel @KJW breathing down my neck... 😅

6 hours ago, joigus said:

Every smooth manifold is locally flat.

I think it depends on how local it is...remember to derive Riemann tensor we take a loop where we are parallel transporting a vector on, in this case a parallelogram,all this information is condensed/ compressed as we take the limit of basis vector of the parallelogram to zero..therefore,it's like we are constructing an overlay that becomes our manifold to give us a tensor field, that is,at every point of this manifold a value is assigned depending on the value of Riemann tensor...My argument is that, whatever is on the parallelogram loop, including uncertainties, all is condensed to a point, therefore,if it's effects are negligible on the loop,it's effects will be useless when we condense it to a point by taking the limit to zero...hence..our so called Riemannian manifold or pseudo Riemannian manifold the bed rock of GR becomes insensitive to the most local things(quantum issues)....we should remember by the time GR was developed by Einstein, quantum theory was not fully developed,therefore, I doubt all this is by design..it depended on the information that was there at that particular time, otherwise,they could have stated this manifold is an overlay.

6 hours ago, joigus said:

I'm even more at a loss as to what any of this has to do with a theory of all the parametrics of elementary particles, AKA TOE.

It hasn't gotten complicated yet...we have to know how spinors form,the source of electromagnetism, superconductivity, how entanglement is related to gravity...e.t.c...e.t.c..I know you think am mad...it's a TOE..you know,I was once advised to take baby steps in this forum,I was furious at that time,however,it has somehow rewarded me...I never thought I will fully understand GR mathematics...but now I can argue using it ...it has been a reward.

53 minutes ago, MJ kihara said:

remember to derive Riemann tensor we take a loop where we are parallel transporting a vector on, in this case a parallelogram

I said that one can define the Riemann tensor this way. I didn't say that one has to define the Riemann tensor this way. I prefer to view GR analytically rather than geometrically, and in this view, the Riemann tensor is defined in terms of the integrability conditions of the coordinate transformation equation of the connection.

21 hours ago, MJ kihara said:

I think it depends on how local it is...

How local what is?

Properties of a manifold can be local or global. An evolution law or differential equation can be local or non-local. There's not such concept as being 'a little local'... Locality

21 hours ago, MJ kihara said:

it's like we are constructing an overlay that becomes our manifold to give us a tensor field,

Say what?

On 5/19/2025 at 4:45 AM, MJ kihara said:

...remember to derive Riemann tensor we take a loop where we are parallel transporting a vector on, in this case a parallelogram,all this information is condensed/ compressed as we take the limit of basis vector of the parallelogram to zero..

The arguments am presenting hereafter I think will tend to be controversial,anyway,we are still dealing with TOE... welcome back.

Let's take a photon to be the vector to be parallel transported hence it's polarization will be taken to be the indicator of this vector direction.we are still using parallelogram loop in this argument. Sides of parallelogram is X,Y and the vector parallel transported (in this case a photon) is Z. As we take the limit of X and Y side to zero to obtain Riemann tensor...the lie bracket [X,Y] significance increases.. therefore,when the manifold is flat [X,Y]=0, however,there is still zero point energy in the background.

Let's make a correspondence between the lie bracket used in derivation of Riemann tensor and the canonical commutation lie bracket used in quantum mechanics,that's according to my thinking.The photon being parallel transported it's initial measurements of polarization indicates it's position and as it moves a long the parallelogram loop it's momentum will be affected by presence of any curvature, therefore,it's final polarization measurements difference to the initial measurements,back to the origin point will be not be zero.However,if the manifold is flat the difference will be zero. In canonical commutation the lie bracket is non zero i.e [x^i,p^j]=iℏδij

By establishing this correspondence i will equate to X and j will equate to Y.... earlier in my arguments I introduced issue of uniformity of information by that I meant i=j=1 . However perturbation on surface of graviton decouple this relationship i=j=1 leading to i=1=j=-1=0 and therefore,δij=0.

This leads to [x^i,p^j]=iℏ*0.hence,[x^i,p^j]=0 therefore,x^i=p^j. This condition reduces ℏ→0. From quantum world of uncertainty to the classical world where observables commute.

" THE BIG DEAL is,this ℏ don't just reduce to zero it decomposes/decays to virtual particles i.e 1.054571817...×10³⁴ virtual particles or 6.62607015×10^34 virtual particles for for h "...mmm..!

Is it strange?...yes...it's strange somehow. Continuous operation/measurements on every point of the manifold the same thing happens.

THE IMPLICATIONS of this;virtual particles becomes elements of the metric tensor... metric is emergent!

I intend to include all this on next edition of my book.

[x^,p^x]=iℏI

I=δij. Where l is the unity operator. In this reasoning these equates to graviton i.e l≡ δij≡graviton.

2 hours ago, MJ kihara said:

The arguments am presenting hereafter I think will tend to be controversial,anyway,we are still dealing with TOE... welcome back.

Let's take a photon to be the vector to be parallel transported hence it's polarization will be taken to be the indicator of this vector direction.we are still using parallelogram loop in this argument. Sides of parallelogram is X,Y and the vector parallel transported (in this case a photon) is Z. As we take the limit of X and Y side to zero to obtain Riemann tensor...the lie bracket [X,Y] significance increases.. therefore,when the manifold is flat [X,Y]=0, however,there is still zero point energy in the background.

Let's make a correspondence between the lie bracket used in derivation of Riemann tensor and the canonical commutation lie bracket used in quantum mechanics,that's according to my thinking.The photon being parallel transported it's initial measurements of polarization indicates it's position and as it moves a long the parallelogram loop it's momentum will be affected by presence of any curvature, therefore,it's final polarization measurements difference to the initial measurements,back to the origin point will be not be zero.However,if the manifold is flat the difference will be zero. In canonical commutation the lie bracket is non zero i.e [x^i,p^j]=iℏδij

By establishing this correspondence i will equate to X and j will equate to Y.... earlier in my arguments I introduced issue of uniformity of information by that I meant i=j=1 . However perturbation on surface of graviton decouple this relationship i=j=1 leading to i=1=j=-1=0 and therefore,δij=0.

This leads to [x^i,p^j]=iℏ*0.hence,[x^i,p^j]=0 therefore,x^i=p^j. This condition reduces ℏ→0. From quantum world of uncertainty to the classical world where observables commute.

" THE BIG DEAL is,this ℏ don't just reduce to zero it decomposes/decays to virtual particles i.e 1.054571817...×10³⁴ virtual particles or 6.62607015×10^34 virtual particles for for h "...mmm..!

Is it strange?...yes...it's strange somehow. Continuous operation/measurements on every point of the manifold the same thing happens.

THE IMPLICATIONS of this;virtual particles becomes elements of the metric tensor... metric is emergent!

I intend to include all this on next edition of my book.

[x^,p^x]=iℏI

I=δij. Where l is the unity operator. In this reasoning these equates to graviton i.e l≡ δij≡graviton.

There are so many things wrong here. Where to start?

First, even though nothing you say here makes much sense, I gather that you have a fundamental confusion between non-commutativity in (pseudo)Riemannian geometry (non-commutativity of similar quantities at different but infinitesimally close points) and non-commutativity in QM (non-commutativity of essentially different quantities at the same point).

Also, how in the name of blazing hell is a collection of ones and zeroes a good representation for gravitons?

Plus you cannot define a sensible position operator for photons. Let alone gravitons. In a relativistic gauge theory x and p must be replaced for something else, more akin to E and B (electric and magnetic field).

Jeez!

5 hours ago, joigus said:

There are so many things wrong here. Where to start?

First, even though nothing you say here makes much sense, I gather that you have a fundamental confusion between non-commutativity in (pseudo)Riemannian geometry (non-commutativity of similar quantities at different but infinitesimally close points) and non-commutativity in QM (non-commutativity of essentially different quantities at the same point).

Don't conclude it's wrong because it doesn't make sense...the math is there for you to see,the correspondence argument is there for you to follow...the confusion you are talking about ain't there...am doing what has not been done, it's not that am repeating what has been established...pliz it could be more helpful if you follow my arguments keenly from that point I introduced the issue of Riemann tensor,the issue of overlay as the manifold is being constructed i.e the tensor field.

Remember Ricci tensor is a contraction of Riemann tensor, therefore,by dealing with Riemann tensor am dealing with GR..the correspondence joins GR and QM...how else do you think should be done?

5 hours ago, joigus said:

Plus you cannot define a sensible position operator for photons. Let alone gravitons. In a relativistic gauge theory x and p must be replaced for something else, more akin to E and B (electric and magnetic field).

As I have stated prior the result of all this argument is metric is emergent leading to what you are saying 'you cannot define a sensible position operator for photons'...metric is emerging from uncertainty developing on surface of graviton inform of perturbations...just look at my prior argument..it's not just photon any other massless particle can be used instead, provided it has something measurable as polarization is for a photon.Initial measurements provide for x and of course,photon has momentum that is being affected by presence of any curvature,that will a have eventually an effect on initial photon polarization measurements.i.e your are taking initial measurements let the photon go through the parallelogram loop back to the origin then you make the final measurements...this is done at every point of the manifold...mmm...doing this physically can be challenging therefore a thought experiment is appropriate.

6 hours ago, joigus said:

Jeez!

That's how strange it is...how else do you explain expansion of the universe...the metric itself is changing..as stated in the prior conclusion metric is emergent....the manifold is expanding...the universe is expanding.

19 minutes ago, MJ kihara said:

Don't conclude it's wrong because it doesn't make sense

This kind of says it all, doesn't it.

I'm too tired now. Maybe tomorrow.

The reason I keep pushing is because the idea am developing keep leading me to things that have already been done and things I haven't learned, for instance I had said;

On 7/25/2025 at 6:01 AM, MJ kihara said:

...am doing what has not been done, it's not that am repeating what has been established...

to clarify this Paul Dirac developed correspondence between poisson brackets and commutator.... wikipedia canonical quantization;

Classical and quantum brackets

Dirac's book^[2] details his popular rule of supplanting Poisson brackets by commutators:

{A,B}⟼1iℏ[A^,B^] .

One might interpret this proposal as saying that we should seek a "quantization map" Q mapping a function f on the classical phase space to an operator Qf on the quantum Hilbert space such that Q{f,g}=1iℏ[Qf,Qg]

It is now known that there is no reasonable such quantization map satisfying the above identity exactly for all functions f and g.

For me,am making correspondence to the lie bracket arguments used in the derivation of Riemann tensor with the reasoning that since GR is already established once quantum mechanics couples to GR the rest fall in line including newton's laws.

Also am learning what am doing is somehow tantamount to geometric quantization.

A good theory should lead you to what you know, don't know(has been established or not yet established), what is to be known and what will be known.

I will say it again (apparently I need to say things three or four times before they get through):

You have a fundamental confusion between non-commutativity in (pseudo)Riemannian geometry (non-commutativity of similar quantities at different but infinitesimally close points) and non-commutativity in QM (non-commutativity of essentially different quantities at the same point).

Geometric quantisation is about commutation of dynamical functions (= phase-space functions = functions of x and p) evaluated at the same point (x,p) of phase space.

Lie brackets, connections, curvature, etc, are about commutation of local functions in infinitesimally close points of space.

I'm not saying there couldn't be a bridge between both notions. I'm saying both things are different and you haven't built that bridge AFAICS.

Then there's the small business of connecting all that to the free parameters of the standard model.

11 hours ago, joigus said:

I will say it again (apparently I need to say things three or four times before they get through):

You have a fundamental confusion between non-commutativity in (pseudo)Riemannian geometry (non-commutativity of similar quantities at different but infinitesimally close points) and non-commutativity in QM (non-commutativity of essentially different quantities at the same point).

My friend...you are wrong on you conclusion....from my arguments of GR being an overlay while constructing it's tensor field i.e the Riemann/ pseudo Riemann manifold what did you get out of it?

For me what am getting from you is that your not getting what am doing...I while tell you once again am not repeating the the the route that has always been used.

1 minute ago, MJ kihara said:

(non-commutativity of similar quantities at different but infinitesimally close points) and non-commutativity in QM

Similar quantities are the same quantities unless you're introducing other quantities that am not aware of.

How many definitions are there for a position and momentum?

The issue of infinitesimal close points are your idea not mine.

I am talking about the same point in a manifold not close points or infinitesimal close points.

11 hours ago, joigus said:

Then there's the small business of connecting all that to the free parameters of the standard model.

Am yet there,you are jumping to the gun.

Corrections to any misspelling brought by auto suggestions...I will tell you...not I while tell you.....and lots of repeating the the the.

11 hours ago, joigus said:

I'm saying both things are different and you haven't built that bridge AFAICS.

The bridge has already been constructed...

On 7/24/2025 at 8:55 PM, MJ kihara said:

This leads to [x^i,p^j]=iℏ*0.hence,[x^i,p^j]=0 therefore,x^i=p^j. This condition reduces ℏ→0. From quantum world of uncertainty to the classical world where observables commute.

" THE BIG DEAL is,this ℏ don't just reduce to zero it decomposes/decays to virtual particles i.e 1.054571817...×10³⁴ virtual particles or 6.62607015×10^34 virtual particles for for h "...mmm..!

ℏ→0...this process is not a permanent process i.e once off event..it's a continuous cyclic process at Planck's time scale..ℏ→0.....ℏ→1.....ℏ→0.....ℏ→1....

Don't for get the issue of zero point energy...it's always there in the background.

42 minutes ago, MJ kihara said:

I am talking about the same point in a manifold not close points or infinitesimal close points.

The same lie bracket used in derivation of Riemann tensor (...from classical world....) is the same that am using in quantum mechanics (... quantum world canonical commutator..) following the correspondence arguments that I made earlier on the thread.

The arguments and conclusion that I have presented,in one way or the other begs the question:-what are this virtual particles?

Okay by using canonical commutator essentially I have been using the Heisenberg picture that has led me to the Big deal that when ℏ→0...ℏ doesn't just disappear but to the conclusion that; ℏ don't just reduce to zero it decomposes/decays to virtual particles,that become elements of the metric that according to the idea am putting across.

Therefore,let us look from Schrodinger's picture perspective to see if the identify of the virtual particles am talking about will be revealed...I will use Schrodinger's equation.

10 hours ago, MJ kihara said:

Corrections to any misspelling brought by auto suggestions...I will tell you...not I while tell you.....and lots of repeating the the the.

Then edit your post and correct it, as I do. Members shouldn't be striving to understand what you're trying to say. Also, don't put my words in your mouth, and use the quote function correctly. It wasn't you who made that comment about similar quantities at infinitesimally close point and different quantities at the same point. I was.

Your physics doesn't make sense either. You don't understand the difference between the configuration space and the phase space.

It's obvious you don't even understand the meaning of Planck's constant. Besides, it's not 1.054571817...×10³⁴, but 1.054571817...×10^-34. And it has dimensions of (energy)x(time), or equivalently, (momentum)x(distance), or equivalently, (angular momentum). Therefore, it wouldn't, shouldn't, and can't represent a cardinal (number of objects).

I'll say it again:

Jeez!

2 hours ago, joigus said:

Then edit your post and correct it, as I do. Members shouldn't be striving to understand what you're trying to say. Also, don't put my words in your mouth, and use the quote function correctly. It wasn't you who made that comment about similar quantities at infinitesimally close point and different quantities at the same point. I was.

Your physics doesn't make sense either. You don't understand the difference between the configuration space and the phase space.

It's obvious you don't even understand the meaning of Planck's constant. Besides, it's not 1.054571817...×10³⁴, but 1.054571817...×10^-34. And it has dimensions of (energy)x(time), or equivalently, (momentum)x(distance), or equivalently, (angular momentum). Therefore, it wouldn't, shouldn't, and can't represent a cardinal (number of objects).

I'll say it again:

Jeez!

IN CAPITAL LETTERS YOU ARE TOTALLY WRONG.

Don't bring trivial issues of misspelling here,I can't have reached at this point of discussion without knowing what is a Planck's constant...am not owning, I haven't owned any words of yours esp given that it's clear to me you haven't grasped what I have been explaining.

I feel sorry 😔 for you given the fact that am talking about metric emerging...and you still holding on connections and curvature,when the discussion is past that...mmm....GR is already developed, Einstein did the job.

My ADVICE TO YOU GO THROUGH AGAIN THIS THREAD to atleast comprehend what am saying.

Are you seeing any negative in this 1.054571817...×10³⁴

3 hours ago, joigus said:

I'll say it again:

Jeez!

The irony....you should say that to yourself, coz so far you haven't comprehended what have been doing after several pages of the thread.

You are at the meaning of Planck's constant...I'm at the source of Planck's constant.

Just now, MJ kihara said:

Are you seeing any negative in this 1.054571817...×10³⁴

No.

So go back to your drawing board and think again.

3 hours ago, joigus said:

Therefore, it wouldn't, shouldn't, and can't represent a cardinal (number of objects).

Just do a simple math given the dimensions of energy times time for ℏ...and given the number of virtual particles am talking about...calculate the energy of a single virtual particles and you will start to see the essence of me calling them VIRTUAL particles.

Then multiply them given the same dimension with ℏ to see the essence of ℏ→1.

I haven't just walked out of the bush then magically started to formulate all these...I advised you not to panic without a cause and recklessly jump the gun.

20 minutes ago, studiot said:

No.

So go back to your drawing board and think again.

There is No going back to the drawing board...the number is as it is specifically,their is no negative there! Not by mistake...it's by design ( not me however...mmm.....) to suite the arguments,the idea,the theory e.t.c..what am working on, it works like that.

On 7/24/2025 at 1:55 PM, MJ kihara said:

This leads to [x^i,p^j]=iℏ*0.hence,[x^i,p^j]=0 therefore,x^i=p^j. This condition reduces ℏ→0. From quantum world of uncertainty to the classical world where observables commute.

That’s not how commutation relations behave. [a,b] = ab-ba

If it’s zero that in no way mean a=b it just means the order you do the operation doesn’t matter

On 7/24/2025 at 1:55 PM, MJ kihara said:

" THE BIG DEAL is,this ℏ don't just reduce to zero it decomposes/decays to virtual particles i.e 1.054571817...×10³⁴ virtual particles or 6.62607015×10^34 virtual particles for for h "...mmm..!

How does a constant decompose into anything? And why does the exponent get inverted but not the argument?

1 hour ago, swansont said:

That’s not how commutation relations behave. [a,b] = ab-ba

If it’s zero that in no way mean a=b it just means the order you do the operation doesn’t matter

Nonetheless either way it leads to the same outcome when the result is zero.

In the thought experiment we were using a photon for parallel transporting, initial photon polarization times(×) it's momentum measurements is used to indicate a in this case, therefore,we can either measure first measure polarization followed with momentum or first measure momentum followed with polarization.then we parallel transport the photon through parallelogram loop back to the origin make final measurements of polarization then momentum or final measurements of momentum then polarization,multiply the two this will give us b...then apply commutation relationship [a,b] = ab-ba we expect if the manifold is flat we should get zero.

1 hour ago, swansont said:

How does a constant decompose into anything?

We get Planck's constant from measurements....it doesn't tell us why exactly that number..the phenomenon that leads to that is the one am trying to clarify...Note that it's a theory being developed,at this juncture consistency matters a lot.

1 hour ago, swansont said:

And why does the exponent get inverted but not the argument?

Lets use h.

6.62607015×10³⁴ Virtual particles=6.62607015×10⁻³⁴ J⋅Hz−1

^{Therefore, one virtual particle will have} 1×10⁻⁶⁴ J⋅Hz⁻¹

Michelson–Morley experiment could not detect such a particle.

2 hours ago, swansont said:

And why does the exponent get inverted but not the argument?

To ensure it meets the condition; ℏ→1

If one virtual particles has 1×10⁻⁶⁴ J⋅Hz⁻¹

^Then

6.62607015×10³⁴ Virtual particles will have 6.62607015×10⁻³⁴ J⋅Hz−1 which is equivalent to Planck's constant.

7 hours ago, MJ kihara said:

Just do a simple math given the dimensions of energy times time for ℏ...and given the number of virtual particles am talking about...calculate the energy of a single virtual particles and you will start to see the essence of me calling them VIRTUAL particles.

Then multiply them given the same dimension with ℏ to see the essence of ℏ→1.

I haven't just walked out of the bush then magically started to formulate all these...I advised you not to panic without a cause and recklessly jump the gun.

Your ignorance of physics is Mariana-Trench deep. Perhaps only comparable to your hubris.

"Virtual particles" is just a fancy name for fluctuations of quantum relativistic fields that can go off-shell for fleeting times. The number operator in QFT for virtual particles gives zero for the expectation value, as it should. Suggesting there are something in the ballpark of 10³⁴ virtual particles (per what?, per unit volume?, in the whole universe?) just because the value of Planck's constant suggests you so strikes me as a feverish hallucination.

Oh, and Planck's constant can be said to to go to 0 in a certain sense (although that's not exactly what we mean when we say that)*. But it makes absolutely no sense to say that it tends to 1. You see, it's not a variable:

IT'S A CONSTANT!!!

Never mind.

You might not read a lot from me on any of your threads from now on.

And please do correct your syntax and spelling, it seriously interferes with your clarity.

Good luck.

---

*What we mean when we say that is that every characteristic quantity with the dimensions of action in the problem under consideration is huge in comparison to \( \hbar \).

4 hours ago, MJ kihara said:

Nonetheless either way it leads to the same outcome when the result is zero.

In the thought experiment we were using a photon for parallel transporting, initial photon polarization times(×) it's momentum measurements is used to indicate a in this case, therefore,we can either measure first measure polarization followed with momentum or first measure momentum followed with polarization.then we parallel transport the photon through parallelogram loop back to the origin make final measurements of polarization then momentum or final measurements of momentum then polarization,multiply the two this will give us b...then apply commutation relationship [a,b] = ab-ba we expect if the manifold is flat we should get zero.

But your claim is that polarization and momentum are equal is clearly bogus, and if that claim doesn’t matter, anything built on it is dubious.

4 hours ago, MJ kihara said:

We get Planck's constant from measurements....it doesn't tell us why exactly that number..the phenomenon that leads to that is the one am trying to clarify...Note that it's a theory being developed,at this juncture consistency matters a lot.

A foolish consistency is the hobgoblin of little minds

— Emerson

Being consistent doesn’t matter if you’re wrong. But it’s not even clear what you’re being consistent with.

4 hours ago, MJ kihara said:

Lets use h.

6.62607015×10³⁴ Virtual particles=6.62607015×10⁻³⁴ J⋅Hz−1

Well, no. That’s the point. There is no way to get from one to the other. They are not equal.

4 hours ago, MJ kihara said:

^{Therefore, one virtual particle will have} 1×10⁻⁶⁴ J⋅Hz⁻¹

Michelson–Morley experiment could not detect such a particle.

Why not? You don’t have to detect angular momentum. We’ve detected particles with zero angular momentum — the Higgs boson

4 hours ago, MJ kihara said:

To ensure it meets the condition; ℏ→1

If one virtual particles has 1×10⁻⁶⁴ J⋅Hz⁻¹

^Then

6.62607015×10³⁴ Virtual particles will have 6.62607015×10⁻³⁴ J⋅Hz−1 which is equivalent to Planck's constant.

h=1? Is there any evidence of h having a value other than what we currently understand it to be? Doesn’t that require breaking rotational symmetry?

But also: no. That’s not true. I can have two spin 1/2 particles and not have their total angular momentum be spin 1; it can also be spin 0. We see this in e.g. hydrogen. In alphas, we have 4 particles, but total spin is zero.