A solution to cosmological constant problem?

[Mordred]

Why does the particle keep changing ?

 Sounds to me a bunch of guesswork. How many pages spent on trying to determine what is meant by SU(3) atom. If it's just gluons why didn't the author simply state a gluon field to begin with...

Though you would still need quarks for the gluons to mediate. So actually doesn't solve the issue.

Formula already provided for the field strength between two or more quarks without quarks field strength is zero.

Edited Sunday at 12:09 AM by Mordred

[JosephDavid]

1 minute ago, Mordred said:

Why does the particle keep changing ?

 Sounds to me a bunch of guesswork. How many pages spent on trying to determine what is meant by SU(3) atom. If it's just gluons why didn't the author simply state a gluon field to begin with...

Though you would still need quarks for the gluons to mediate. So actually doesn't solve the issue.

In fact the author indicated that in the paper, when he introduced the  lagrangian of su(3) field divided by these number of units.  

[Mordred]

no 

16 minutes ago, JosephDavid said:

In fact the author indicated that in the paper, when he introduced the  lagrangian of su(3) field divided by these number of units.  

Not by any valid method I wont waste time repeating myself on the valid method via the Langrangian equations of motion, the relevant creation annihilation operators and Maxwell Boltzmann.

I wasted enough time repeating myself and being ignored on those points.

meson formula between two quarks page 5

example formula for quark quark interaction ground state of a bound system.

\[E(r)=2m-\frac{\alpha_s}{r^2_o}+br+\frac{p^2}{m}\]

where m is the mass p the momentum the radius of the ground state is

\[\frac{2}{mr^3_o}=\frac{\alpha_s}{r^2_o}+b\]

here is a table for you

http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/quark.html

apply any quark combination in that table then do the conversion with the eV

Then think back to those calculations on total mass.

1 ev times 10^{123} still gives too much mass you wont get an interaction field of gluons less than 1 ev

 

Edited Sunday at 12:28 AM by Mordred

[JosephDavid]

29 minutes ago, Mordred said:

no 

Not by any valid method I wont waste time repeating myself on the valid method via the Langrangian equations of motion, the relevant creation annihilation operators and Maxwell Boltzmann.

I wasted enough time repeating myself and being ignored on those points.

meson formula between two quarks page 5

example formula for quark quark interaction ground state of a bound system.

 

E(r)=2m−αsr2o+br+p2m

 

where m is the mass p the momentum the radius of the ground state is

 

2mr3o=αsr2o+b

 

here is a table for you

http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/quark.html

apply any quark combination in that table then do the conversion with the eV

Then think back to those calculations on total mass.

1 ev times 10^{123} still gives too much mass you wont get an interaction field of gluons less than 1 ev

 

Gluons are massless. 

[Mordred]

24 minutes ago, JosephDavid said:

Gluons are massless. 

no kidding guess I didn't know that (hear the sarcasm in my response ?

did you forget 

\[E^2=\sqrt{(pc^2)+(m_0 c^2)}\]

qluons will be whatever number density is required to mediate the color interaction between any number of quark combinations the energy equivalence to mass can be applied to above formula.

That is precisely why the Langrangian includes the four momentum Momentum includes mass and velocity.

After all the times I mentioned Bose-Einstein statistic's have you ever bothered to look at how it gets applied?

How many times now ? 

for the record I tried doing the lowest order of quark combination to get a total energy for volume using the 10^{123} the lowest value I could find using the lightest meson was 10^{81} kg equivalence which is still too high.

(that was using my access to feycalc at the university I do assistance instruction at.)

care to tell me which meson I used ?

before you mention SUSY here are the epions for SUSY for MeV energy levels

https://citeseerx.ist.psu.edu/document?repid=rep1&type=pdf&doi=b86e7520a33a331471d8b4ba6e76a3d106f5c70a

Edited Sunday at 01:09 AM by Mordred

[JosephDavid]

51 minutes ago, Mordred said:

no kidding guess I didn't know that (hear the sarcasm in my response ?

did you forget 

 

E2=(pc2)+(m0c2)−−−−−−−−−−−−√

 

qluons will be whatever number density is required to mediate the color interaction between any number of quark combinations the energy equivalence to mass can be applied to above formula.

That is precisely why the Langrangian includes the four momentum Momentum includes mass and velocity.

After all the times I mentioned Bose-Einstein statistic's have you ever bothered to look at how it gets applied?

How many times now ? 

for the record I tried doing the lowest order of quark combination to get a total energy for volume using the 10^{123} the lowest value I could find using the lightest meson was 10^{81} kg equivalence which is still too high.

(that was using my access to feycalc at the university I do assistance instruction at.)

care to tell me which meson I used ?

 

let’s first understand  gluons. These guys are massless. So here’s the deal. Normally, gluons keep quarks bound in particles like protons and neutrons, bouncing back and forth like little springs between quarks. But unlike other force carriers, like photons, gluons are social. They don’t just interact with quarks; they’re also interested in each other! It’s like you’re at a party, and these gluons are so friendly, they start dancing together. This dance gives us something wild called a **gluon condensate**—a whole sea of these massless gluons hanging out in empty space.  And here’s another trick: if you get a bunch of gluons together, they can clump up into something we call a **glueball**. Yeah, that’s right—no quarks, just a wad of pure glue. Imagine a baseball made entirely out of sticky tape with nothing in the middle. That’s glueball, just pure gluon stickiness bundled up. The author here cites some work on glueballs,

Now here’s where it gets interesting: these gluons, despite doing all this work, don’t add any mass to the universe. It’s like having an invisible scaffold stretching across the cosmos. This unbreakable SU(3) symmetry is always there, holding everything in place without actually weighing anything down. The author’s point here is that SU(3) doesn’t just hold quarks together; it stretches across the universe, massless but unbreakable, like an invisible safety net. So, next time you’re worried about putting on a few pounds, just remember the gluons. They’re out there, holding the universe together, doing all this hard work, without gaining a gram. Makes you appreciate the whole massless, invisible glue keeping the cosmos in shape, doesn’t it?

[Mordred]

One day you will actually look at the math itself instead of trying to lecture those that have I'm fully aware of glueballs

Have you ever bothered to look at the lightest theoretical value for a glueball ?

https://en.wikipedia.org/wiki/Glueball

try that in the method I described earlier.

Multiply that by the number of SU(3) atoms then use the energy momentum formula to calculate its mass equivalence.

The sheer problem is the power of 123.

There is no getting around that even using photons and number of quanta a unit of quanta is in joules per hertz 

How many joules in one eV ? Even ignoring wavelength and just using joules if you multiply by 10 ^123 you get 10^{73} that's without the corresponding wavelength. Which we know would only get worse if you apply the wavelength.

Edited Sunday at 02:12 AM by Mordred

[JosephDavid]

23 minutes ago, Mordred said:

One day you will actually look at the math itself instead of trying to lecture those that have I'm fully aware of glueballs

Have you ever bothered to look at the lightest theoretical value for a glueball ?

https://en.wikipedia.org/wiki/Glueball

try that in the method I described earlier.

Multiply that by the number of SU(3) atoms then use the energy momentum formula to calculate its mass equivalence.

The sheer problem is the power of 123.

There is no getting around that even using photons and number of quanta a unit of quanta is in joules per hertz 

How many joules in one eV ? Even ignoring wavelength and just using joules if you multiply by 10 ^123 you get 10^{73} that's without the corresponding wavelength. Which we know would only get worse if you apply the wavelength.

 

If you’re after a massless, stable backdrop for the universe, gluon condensation is the way to go. It’s like an invisible field, filling space without adding weight, simple, clean, and seamless. Glueballs, by contrast, are little balls of pure glue, but they carry mass due to their binding energy. So, if you’re aiming for a truly massless vacuum structure, gluon condensation is the better choice. I looked into the author’s another paper on the cosmological constant and noticed he describes dark energy as a kind of Bose-Einstein distribution.

 

https://papers.ssrn.com/sol3/papers.cfm?abstract_id=4598396

[Mordred]

great What do you thing was used originally to calculate 10^90 photons of the standard model ? That number was calculated back in the 70's and is still the theoretical bounds for the universes energy budget to this very day

not 10^123

using Bose-Einstein both qluons and photons have the same spin so the result will be identical

 

Edited Sunday at 02:23 AM by Mordred

[JosephDavid]

10 minutes ago, Mordred said:

great What do you thing was used originally to calculate 10^90 photons of the standard model ? That number was calculated back in the 70's and is still the theoretical bounds for the universes energy budget to this very day

not 10^123

using Bose-Einstein both qluons and photons have the same spin so the result will be identical

 

Photons are completely decoupled from gluons and, fundamentally, have no interaction with dark energy either.

[Mordred]

Gluons are not DE either you cannot have an expanding universe where any particle density stays constant.

Lambda is constant. Stop trying to apply the authors crap to me please. He is wrong on so many levels from what you are describing.

 

 

Edited Sunday at 02:43 AM by Mordred

[JosephDavid]

6 minutes ago, Mordred said:

Gluons are not DE either you cannot have an expanding universe where any particle density stays constant.

Lambda is constant. Stop trying to apply the authors crap to me please. He is wrong on so many levels from what you are describing.

 

 

The number of these SU(3) vacuum atoms remains constant, which explains the cosmological constant. 

Edited Sunday at 02:47 AM by JosephDavid

[Mordred]

that number does not work get that through your head even if you use the lowest possible energy/ mass equivalence its wrong. PERIOD

[JosephDavid]

Just now, Mordred said:

that number does not work get that through your head even if you use the lowest possible energy/ mass equivalence its wrong. PERIOD

It is correct, as it explains the lowest energy density we have ever measured, the vacuum energy density, based on observations of the universe's expansion.

[Mordred]

BS it is

the cosmological constant has a value of roughly \[6.0*10^{-10}\] joules per cubic meter. Multiply that and see if 10^123 is equivalent.

we're done its obvious you don't know what your talking about and do not wish to learn and is just trolling GL have fun

 

Edited Sunday at 02:56 AM by Mordred

[MigL]

2 hours ago, JosephDavid said:

he author used the proton volume to define the unit of space that the massless gluon field may occupy at each point in the spacetime fabric of the universe.

For several pages you've been making the argument that the proton is the fundamental SU(3) 'particle, because it doesn't decay, and associated 3rd law considerations.
Then you make the jump to using the volume of the proton as the fundamental unit in the calculations.
What thermodynamic considerations allow for a volume to be a fundamental unit ?
Certainly not the same ones that allow fir a proton to be a fundamental unit.
After all, other particles share a similar volume, and they are not fundamental as they decay.

If you don't quit going round in circles attempting to explain yourself, you're going to get dizzy.

[JosephDavid]

5 minutes ago, MigL said:

For several pages you've been making the argument that the proton is the fundamental SU(3) 'particle, because it doesn't decay, and associated 3rd law considerations.
Then you make the jump to using the volume of the proton as the fundamental unit in the calculations.
What thermodynamic considerations allow for a volume to be a fundamental unit ?
Certainly not the same ones that allow fir a proton to be a fundamental unit.
After all, other particles share a similar volume, and they are not fundamental as they decay.

If you don't quit going round in circles attempting to explain yourself, you're going to get dizzy.

The volume of the proton is the volume that confines gluons, the massless particles that represent the unbreakable su(3) symmetry. 

[MigL]

But the argument was specific to the proton, and its associated stability.
Are ubstable neutrons not held together by those same gluons that 'represent the unbreakable SU(3) symmetry' ???

[JosephDavid]

34 minutes ago, Mordred said:

BS it is

the cosmological constant has a value of roughly

6.0∗10−10

joules per cubic meter. Multiply that and see if 10^123 is equivalent.

 

we're done its obvious you don't know what your talking about and do not wish to learn and is just trolling GL have fun

 

Are you not getting it yet? Look, when we’re talking about **vacuum energy density**, we’re not just dealing with energy per cubic meter. Nope, it’s **energy per cubic meter per SU(3) vacuum atom.  Think of it this way: imagine you’ve got a giant pizza meant to feed an entire town. Now, instead of handing the whole thing to one person (which would be ridiculous), you slice it up and spread it out to everyone in town, giving each person a manageable slice. That’s what’s going on here with vacuum energy. In this analogy, the enormous vacuum energy from QFT is like that pizza. But instead of loading it all into one place, it’s **spread across \(10^{123}\) proton-sized SU(3) vacuum atoms** scattered throughout the universe. Each atom holds just a fraction of the energy—just like each person gets a slice of pizza. So, what we’re actually observing is the **energy per cubic meter per atom**. This breakdown per atom is what makes the observed vacuum energy density match what we see in the real universe. It’s not just energy per cubic meter, it's distributed per atom, and that’s what keeps everything nice and balanced.

11 minutes ago, MigL said:

But the argument was specific to the proton, and its associated stability.
Are ubstable neutrons not held together by those same gluons that 'represent the unbreakable SU(3) symmetry' ???

Yes, because we need something stable and measurable to define the volume that confines gluons. The neutron could work, but it decays into a proton. So, the proton's volume is more reliable as the one that truly confines massless gluons, since the proton doesn’t decay through any channels in the Standard Model.

[Mordred]

50 minutes ago, JosephDavid said:

 **vacuum energy density**, 

Define energy density I want to hear what you think it is.

I also want your definition of vacuum in regards to that energy density definition.

You might recall I provided the equations for QCD vacuum.

Show me the formula to determine the cosmological constant value for equation of state w=-1 and showing how it is derived using that formula.

For all your claims you have yet to show any relevant formulas or calculations and I have done so this entire thread.

So don't try to preach to me or tell me I don't know what I am talking about.

1 hour ago, Mordred said:

BS it is

the cosmological constant has a value of roughly

6.0∗10−10

joules per cubic meter. Multiply that and see if 10^123 is equivalent.

 

we're done its obvious you don't know what your talking about and do not wish to learn and is just trolling GL have fun

 

What formula did I use to calculate that value ?

Hint I mentioned it in this thread but never latexed it in.

Lol though that's the value today what's the value at z=1100.

Your relatively new here so don't really know my skills however Cosmology and particle physics I have credentials in both fields. I know the equations I mentioned they are of fundamental importance to everything I mentioned. SuSY is part of my studies. Some of those Langrangians are included by me in this thread (easily missed though if you don't know them).

(Pati-Salam)

 

 

 

Edited Sunday at 04:19 AM by Mordred

[JosephDavid]

45 minutes ago, Mordred said:

Define energy density I want to hear what you think it is.

I also want your definition of vacuum in regards to that energy density definition.

You might recall I provided the equations for QCD vacuum.

Show me the formula to determine the cosmological constant value for equation of state w=-1 and showing how it is derived using that formula.

For all your claims you have yet to show any relevant formulas or calculations and I have done so this entire thread.

So don't try to preach to me or tell me I don't know what I am talking about.

What formula did I use to calculate that value ?

Hint I mentioned it in this thread but never latexed it in.

Lol though that's the value today what's the value at z=1100.

Your relatively new here so don't really know my skills however Cosmology and particle physics I have credentials in both fields. I know the equations I mentioned they are of fundamental importance to everything I mentioned. SuSY is part of my studies. Some of those Langrangians are included by me in this thread (easily missed though if you don't know them).

(Pati-Salam)

 

 

 

QFT hands us this massive number for vacuum energy density, huge, way too big. But the author’s got a neat idea: maybe that big number isn’t all piled up in one place. Instead, based on **solid physical reasons**, he’s saying it’s **spread out over \(10^{123}\) tiny SU(3) atoms** filling the whole universe.

Now, when you spread that QFT energy over all these little SU(3) “atoms,” suddenly you end up with a vacuum energy density that actually fits what we observe. So, the **vacuum energy density we’re seeing isn’t just energy per cubic meter**; it’s **energy per cubic meter per SU(3) atom**.

And that, my friend, is what quantum spacetime is all about here. We’re not looking at one big, smooth field of energy. We’re looking at energy carefully spread out, divided across countless little units. It’s this distribution that makes everything line up with what we measure, simple, logical, and right there in front of us.

[Mordred]

What energy level is contained in each SU(3) atom.

no matter what value you choose you will violate energy mass conservation when compared to the total energy of the observable universe. When you have 10^{123} SU(3) atoms. its as simple as that it doesn't matter what SU(3) is meant to describe the power 123 is the very issue here.

Do you  not understand rudimentary math I gave a precise means to see I am correct on that.

lets try your terminology the number of microstates SU(3) atoms is too high for the observable  universe that no matter what energy is contained in each microstate you surpass the total energy budget of the Observable universe error margin taking 10^93 in kg first case your proton example (yes you could do the same with the energy equivalent.) and dividing by 10^53 for total mass of the universe. WoW 10^40 times the total mass of the universe. Basic math there quite obvious.

just as obvious to apply the energy momentum equation for conversions between mass and energy

see that error margin yet ? try it at 1 ev per SU(3) atom Massive massless doesn't matter its the energy contained yet you argued about massless 

lets see

\(10^{123}\) ev =\( 16.02*10^{104}\) joules  apply the complete most famous equation 

\[E^2=\sqrt{(pc^2)+(m_oc^2)}\] that equation handles both massive and massless cases.

gives \(1.783*10^{87} kg\) still an error margin of 10^34.

simple math 

 

Edited Sunday at 05:03 AM by Mordred

[MJ kihara]

It seems people have different perspective on author's article...maybe he needs to defend it more actively....for me and my concepts I see  a smoking gun ...it's just a bit in a big puzzle...I wish he could defend it thoroughly so that it can be used for reference.My issue was what made him to use the term SU(3) atom?...then the global vacuum (universal) how is related to SU(3) unit vacuum,because I doubt if it could be 100% stable,since in such a case, it can't have any connections to the global vacuum...(bringing the issue of multiverse)...that's why I was bringing the issue of holography, vacuum leakages and quantum noise.

Finding solutions to fundamental problem is a struggle.....let the struggle continue.

[Mordred]

1 hour ago, MJ kihara said:

It seems people have different perspective on author's article...maybe he needs to defend it more actively....for me and my concepts I see  a smoking gun ...it's just a bit in a big puzzle...I wish he could defend it thoroughly so that it can be used for reference.My issue was what made him to use the term SU(3) atom?...then the global vacuum (universal) how is related to SU(3) unit vacuum,because I doubt if it could be 100% stable,since in such a case, it can't have any connections to the global vacuum...(bringing the issue of multiverse)...that's why I was bringing the issue of holography, vacuum leakages and quantum noise.

Finding solutions to fundamental problem is a struggle.....let the struggle continue.

That's a huge part of the problem but I clearly show it doesn't matter how he defines the SU(3) atom if it contains any energy even of the lowest orders he still exceeds the energy budget.  Anyone is welcome to experiment with different values with he method I just described to see that and confirm for themselves. You literally cannot place a single massless or massive particle in each SU(3) atom and not do so. By whopping factors.

So using SU(3) atoms whatever it contains how could it possibly solve the problem with similar error margins to the the catastrophe value ?

That error margin due to the 123 power has been obvious to several of our members right from the start. 

lets try another route lets compare Current cosmological constant value to SU(3) atoms using one ev per 10^{-15} cubic meters which is the volume described in article shall we. Lets get to equivalent units.

 

converting

\[6.0*10^{-10} joules/m^3=3.75*10^9 ev/m^3\] 

now take 1 ev per SU(3) atom in volume \(10^{-15}\) meters supplied for for volume for each SU(3) atoms

gives \(1.0*10^{138} ev/m^3=1.602 *10^{119} joules/m^3\)

looks strikingly like the old Cosmological constant problem it was suppose to solve

That 1 ev value is equivalent to a photon or any massless particle the ev value will depend on its wavelength. For 1 ev it's wavelength is 1239 nm.

All I'm doing is basic mathematics and conversions here nothing fancy

Edited Sunday at 06:59 AM by Mordred

[MJ kihara]

10 hours ago, Mordred said:

per 10^{-15} cubic meters

 

10 hours ago, Mordred said:

now take 1 ev per SU(3) atom in volume 10−15 meters supplied for for volume for each SU(3) atoms

gives 1.0∗10138ev/m3=1.602∗10119joules/m3

That's not the actual volume if you have used it. 10^-15*10^-15*10^-15 cubic meter.