A solution to cosmological constant problem?

[Mordred]

2 hours ago, joigus said:

So these atoms are there. They're not a vacuum. They're a background. Why is it high-precission tests of the standard model haven't found them yet? Hadrons of all kinds, from high-energy beams and jets would certainly scatter off SU(3) bound states, with a very sizable cross section. On top of that, they are 10⁴³ times more abundant than ordinary neutrons, so the luminosities would be over the roof. Why haven't we seen the littlest inkling that they're there?

Don't think for a moment I haven't noticed you aren't answering any of this. This thread is starting to smell with a really foul stink.

Not too mention all particles regardless of type contribute to CMB blackbody temperature including weakly interactive.  So why aren't these SU(3) atoms detected is a very relevant question.

I also noticed no one paid any attention to the SU(3) gauge I posted this involves protons and neutrons as well as mesons.

Guess they don't want to think about what energy levels would be involved  with those composite particles.

Edited October 31 by Mordred

[MigL]

5 hours ago, JosephDavid said:

So, when we talk about the proton as representing an SU(3) vacuum atom, it’s not just about its size; it’s about its role as the ultimate low-entropy, stable unit in the structure of the vacuum.

Then why are you dividing the volume of the observable universe by the volume of a proton to get the vacuum energy adjustment ?
Why not divide by the number of protons in the observable universe ?

What justifies the volume consideration ????

[joigus]

On 10/30/2024 at 4:43 PM, TheVat said:

This strikes me as having a Popperian "black swan" issue, in terms of a definitive observation establishing that protons don't decay.  

Quote

Currently, the most precise results come from the Super-Kamiokande water Cherenkov radiation detector in Japan: a lower bound on the proton's half-life of 2.4×10³⁴ years via positron decay, and similarly, 1.6×10³⁴ years via antimuon decay, close to a supersymmetry (SUSY) prediction of 10³⁴–10³⁶ years.

They certainly are the Methuselahs of the strongly-interacting matter. Everything else decays in a tiny fraction of a blink or in the time it takes to sip out a coffee.

But I agree. It's probably a green[?]-swan kind of issue.

With every year that Super Kamiokande keeps running, it becomes increasingly unlikely that we will see any single proton decay.

Looks like stars will die long before protons do.

4 hours ago, MigL said:

Then why are you dividing the volume of the observable universe by the volume of a proton to get the vacuum energy adjustment ?
Why not divide by the number of protons in the observable universe ?

What justifies the volume consideration ????

Yes, that's funny. One is related with the speed of light and the age of the universe. The other with the strong coupling. I wouldn't say it's impossible for them to be related, but it's not very compelling, to say the least.

This idea of dividing the size of the universe by the size of a proton is not new, btw. Dirac thought of it many years ago. This hypothesis has fallen out of favour for a series of reasons. 10⁴⁰ seemed to be the central scaling factor.

Some of these ratios had to be fixed with a root, like (10⁴⁰)^9/4=10⁹⁰ which is the number of photons, or the entropy of the observable universe. The whole 'theory' reeked of numerical mysticism, aka numerology.

 

[Mordred]

I'm still wondering how long it's going to take some people , to recognize there is a HUGE distinction between a ground state and an excited particle state.....

I've given up trying to get that across to certain people. The clue that all quantum fields has a ground state should have indicated those people might just be missing a detail...

It literally doesn't matter what quantum  field is used.  They all have a ZPE. The ground state isn't the particles themselves.

Lol providing the field strength formula for QCD obviously wasn't a strong enough clue.

The conversation is still discussing particles and not the interaction between particles which is where the ground state is applied in terms of superconductivity.

here is the simplest mathematical statement describing the above. Maybe just maybe this will work

\[\hat{a}(\vec{k})|0\rangle=0\]

the \(\vec{k}\) is the wavefunction the \(|0\rangle\) is the ground state the \(\hat{a}\) is the creation annihilation operators that give the creation and annihilation of particles (ladder operators) From the ground state.

the creation/annihilation operators are determined via the quantum harmonic oscillator. So is the Hamilton

\[\hat{H}=\omega(\hat{a}^\dagger a+\frac{1}{2})\]

the number Operator is

\[\hat{N}=\hat{a}^\dagger \hat{a}\] 

gives the Hamilton a nice simple form

\[\hat{H}=\omega(\hat{N}+\frac{1}{2})\]

 

I will leave it at that.

 

Edited November 1 by Mordred

[JosephDavid]

5 hours ago, MigL said:

Then why are you dividing the volume of the observable universe by the volume of a proton to get the vacuum energy adjustment ?
Why not divide by the number of protons in the observable universe ?

What justifies the volume consideration ????

 

 It all comes down to stability, entropy, and what we’re really trying to get at. The author is talking about vacuum energy, not just the stuff we can see, like stars, gas, and galaxies that light up the night sky. We’re talking about everything: dark matter, dark energy, and the whole underlying fabric that makes the universe tick. It’s about understanding what makes the whole space.

According to the third law of thermodynamics, as things get colder and closer to absolute zero, they need to become more ordered, less chaotic. Think about a cup of hot coffee: the atoms are buzzing around, jittery and restless. Now, imagine cooling it down to near absolute zero. Everything quiets down, the chaos fades, and you’re left with order, no extra movement, no noise. That’s what the author is trying to describe for the vacuum energy, finding a unit that’s super stable with low entropy, and that’s where the free proton comes in. The free proton is like a brick that just doesn’t crumble. It doesn’t have any natural decay channels that conserve both charge and energy.  It’s solid, unchanging, the kind of fundamental building block you want when you’re describing something stable.

Now, let’s talk about the numbers. You’ve got 10^80, which is how many protons are in all the ordinary matter we can see, stars, planets, and everything out there that shines or glows. But that’s just 5% of the universe. It’s like standing on a beach and only counting the waves you can see on the surface—that’s 10^80. But 10^123 SU(3) vacuum atoms? That’s the entire ocean: the waves, the deep currents, and everything beneath the surface, including the dark matter and  dark energy we can’t directly see. It’s the whole cosmic picture.

The author uses the volume of a proton because it’s the most stable, low-entropy unit there is. If you want to get a handle on the vacuum structure, what fills the space in the universe, you need something that doesn’t break down, something fundamental that can act as a real building block.

Here’s the really neat part: the number 10^123 isn’t just some shot in the dark. It actually explains precisely  what we see—the expansion of the universe or the cosmological constant. The match is too perfect to ignore. It’s like seeing trees sway in the wind; you may not see the wind, but the way the leaves move tells you something real is making it happen. The author’s number for the SU(3) vacuum atoms lines up with the actual expansion rate, so it’s telling us we’re onto something fundamental. That’s why the author’s approach is compelling. It’s not just a thought experiment, it’s an insight backed by observation and fundamental laws. 

 

[Mordred]

The harmonic oscillator  has absolutely  nothing to do with stability read post above. Neither does the cosmological constant vacuum catastrophe even though its obvious you've chosen to ignore everything I stated.

Thankfully there are other readers even those that haven't gotten involved.

 

Edited November 1 by Mordred

[joigus]

8 hours ago, Mordred said:

Not too mention all particles regardless of type contribute to CMB blackbody temperature including weakly interactive. 

Absolutely. Once they posit those things are there. How can we prove they're not there? By as many ways as we can. If you think something is right, you try to prove it's wrong!

"Why do we want to attack the argument?", our interlocutors may think. Because that is the heartbeat of science: Attack the argument from as many directions as you can. If it survives the attacks, you got yourself a theory or a principle.

[Mordred]

46 minutes ago, joigus said:

Absolutely. Once they posit those things are there. How can we prove they're not there? By as many ways as we can. If you think something is right, you try to prove it's wrong!

"Why do we want to attack the argument?", our interlocutors may think. Because that is the heartbeat of science: Attack the argument from as many directions as you can. If it survives the attacks, you got yourself a theory or a principle.

This matches a policy any good physicist follows. Spend more time trying to prove your theory wrong otherwise it will never become robust enough under cross examination.

For those interested this is about the best article I have been able to locate on QCD dual superconductivity.

https://arxiv.org/abs/hep-lat/0510112

This link shows how to incorporate to string theory 

https://arxiv.org/abs/hep-ph/0301032

As I mentioned numerous times type 1 vs type 2 superconductors the distinction has to do with vortex penetration depth

http://lampx.tugraz.at/~hadley/ss2/problems/super/s.pdf

Edited November 1 by Mordred

[joigus]

19 minutes ago, Mordred said:

This matches a policy any good physicist follows. Spend more time trying to prove your theory wrong otherwise it will never become robust enough under cross examination.

For those interested this is about the best article I have been able to locate on QCD dual superconductivity.

https://arxiv.org/abs/hep-lat/0510112

The sorriest part of this is they don't even realise how big a number 10¹²³ actually is. There's nothing that we can see or grasp, or intuit, the counting of which is 10¹²³ in the observable universe. You must get into combinatorics of countable things in the universe to get to (and surpass) a number like this. Even the number of photons is ridiculously small in comparison. So coldly stating that there are 10¹²³ Helium-like things among us and nobody has ever noticed takes some gall.

[Mordred]

16 minutes ago, joigus said:

The sorriest part of this is they don't even realise how big a number 10¹²³ actually is. There's nothing that we can see or grasp, or intuit, the counting of which is 10¹²³ in the observable universe. You must get into combinatorics of countable things in the universe to get to (and surpass) a number like this. Even the number of photons is ridiculously small in comparison. So coldly stating that there are 10¹²³ Helium-like things among us and nobody has ever noticed takes some gall.

Agreed it would have been more intelligent to use

\[\rho_R=\frac{\pi^2}{30}{g_{*S}=\sum_{i=bosons}gi(\frac{T_i}{T})^3+\frac{7}{8}\sum_{i=fermions}gi(\frac{T_i}{T})}^3\]

To determine how many up and down quarks would be available to form protons and neutrons in the first place.

Maxwell Boltzmann above takes into account the laws of thermodynamics.

https://en.m.wikipedia.org/wiki/Maxwell–Boltzmann_distribution

Edited November 1 by Mordred

[MigL]

2 hours ago, JosephDavid said:

According to the third law of thermodynamics, as things get colder and closer to absolute zero, they need to become more ordered

We understand the underlying Thermodynamics, we just don't understand the 'connections' made to other phenomena.

2 hours ago, JosephDavid said:

The author uses the volume of a proton because it’s the most stable, low-entropy unit there is

The neutron is similar, and the same as the SU(3) minimum volume; should it not be just as stable according to SU(3) stability considerations???

2 hours ago, JosephDavid said:

The author uses the volume of a proton because it’s the most stable, low-entropy unit there is. If you want to get a handle on the vacuum structure, what fills the space in the universe, you need something that doesn’t break down,

So if the neutron is NOT stable, IOW, it isn't the volume that matters, but the proton itself, "Why are you not using the number of protons in the calculation, instead of the volume ?"

[Mordred]

I was curious as to a fundamental question, we all know the CMB temperature today is 2.73 Kelvin with the radius of the Observable universe being 46.3 Gly.

So I asked what would the radius need to be to reach 1 Kelvin. It turns out the universe would need to have a radius of 140 Gly. The universe would be roughly 28.8 Gyrs old far far into the future assuming nothing changes with the cosmological parameters 

Edited November 1 by Mordred

[Genady]

10 minutes ago, Mordred said:

4.63

Typo?

46.3

[Mordred]

5 minutes ago, Genady said:

Typo?

46.3

yeah thanks for the catch corrected above.

Edited November 1 by Mordred

[JosephDavid]

28 minutes ago, MigL said:

The neutron is similar, and the same as the SU(3) minimum volume; should it not be just as stable according to SU(3) stability considerations???

here's the key difference: both neutrons and protons are bound by SU(3) interactions, the strong force. However, the neutron is inherently unstable because of the SU(2) gauge field, which is the weak force. The **neutron undergoes beta decay via the SU(2) weak interaction, breaking down into a proton**, an electron,  and an antineutrino. This process adds entropy, increasing disorder which makes the neutron unsuitable for what the third law of thermodynamics requires. The third law says that as systems approach absolute zero, they need to be in a state of minimal entropy, as ordered and stable as possible. The neutron with its tendency to decay and create disorder, doesn’t meet that standard.

The proton on the other hand, doesn’t have natural decay channels under natural conditions, it’s inherently stable This stability makes the proton a much better fit for representing a low-entropy building block of the universe’s vacuum structure. The proton stays put and keeps everything in order, fitting the minimum entropy condition that the third law demands.

 

35 minutes ago, MigL said:

So if the neutron is NOT stable, IOW, it isn't the volume that matters, but the proton itself, "Why are you not using the number of protons in the calculation, instead of the volume ?"

 

The number of protons we can see in the universe, that’s what’s in  ordinary matter like stars, planets, and galaxies. But that’s just 5% of the entire universe. The other 95% is made up of dark matter and dark energy, things we can’t see directly but know are there because they affect how galaxies move and how the universe expands. It’s like looking at an iceberg. The part above water—the visible 5%, is like the ordinary matter we can see. But the much bigger part below the surface, that 95%, that’s the **dark matter and dark energy.  If we just counted what’s above water, we’d be missing the real size and structure of the whole iceberg. That’s why the author used the volume of a proton instead of just counting the visible protons. It’s not about counting what we can see; it’s about using a unit that represents the entire cosmic structure. The proton volume allows the author to calculate the number of SU(3) vacuum atoms, which account for both the visible and invisible parts of the universe, and which explained precisely the cosmological constant.

 

[Mordred]

23 minutes ago, JosephDavid said:

here's the key difference: both neutrons and protons are bound by SU(3) interactions, the strong force. However, the neutron is inherently unstable because of the SU(2) gauge field, which is the weak force. The **neutron undergoes beta decay via the SU(2) weak interaction, breaking down into a proton**, an electron,  and an antineutrino. This process adds entropy, increasing disorder which makes the neutron unsuitable for what the third law of thermodynamics requires. The third law says that as systems approach absolute zero, they need to be in a state of minimal entropy, as ordered and stable as possible. The neutron with its tendency to decay and create disorder, doesn’t meet that standard.

The proton on the other hand, doesn’t have natural decay channels under natural conditions, it’s inherently stable This stability makes the proton a much better fit for representing a low-entropy building block of the universe’s vacuum structure. The proton stays put and keeps everything in order, fitting the minimum entropy condition that the third law demands.

 

I would like to see a mathematical proof of the above as I know you do not know the math show a peer review article that the neutron would not satisfy the third law of thermodynamics. Both protons and neutrons are degeneracy systems.

That peer review should show the relevant Fermi- energy for each

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

Edited November 1 by Mordred

[MJ kihara]

4 hours ago, joigus said:

This hypothesis has fallen out of favour for a series of reasons. 10⁴⁰ seemed to be the central scaling factor.

Some of these ratios had to be fixed with a root, like (10⁴⁰)^9/4=10⁹⁰ which is the number of photons

This a perfect example of numerology....it's not related at all to what the author is doing.

[JosephDavid]

21 minutes ago, Mordred said:

I would like to see a mathematical proof of the above as I know you do not know the math show a peer review article that the neutron would not satisfy the third law of thermodynamics. Both protons and neutrons are degeneracy systems.

That peer review should show the relevant Fermi- energy for each

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

It seems like there is a fundamental point being missed here. The number of microstates for the neutron is significantly greater than that of the proton, and this directly impacts the entropy. According to the third law of thermodynamics, as systems approach absolute zero, they must reach a state of minimal entropy, meaning the system should have as few microstates as possible, leading to maximum stability and order.

The expression for entropy \( S \) in terms of the number of microstates \( \Omega \) is given by Boltzmann's entropy formula:

\[
S = k_B \ln(\Omega)
\]

 

In this equation, \( \Omega \) represents the number of different ways the system’s energy can be arranged. The more microstates available, the greater the entropy.

Since the neutron can undergo beta decay, it has many potential microstates, which leads to increased entropy. On the other hand, the proton is inherently stable and has fewer available microstates, resulting in lower entropy. This lower entropy aligns perfectly with the requirement for minimum entropy as dictated by the third law of thermodynamics.

Thus, in terms of the third law, the proton, with its lower number of microstates and resulting lower entropy, is a better fit for the condition of minimal entropy compared to the neutron, which inherently has more microstates and higher entropy.
 

Edited November 1 by JosephDavid

[MJ kihara]

On 10/30/2024 at 3:56 PM, MJ kihara said:

as discussed by Steven Weinberg [19], the vacuum energy
density arises from summing the zero-point energies of quantum fields, including massless
bosons like photons and gravitons.

From the author article.

I don't see why some people are against the article,am not convinced otherwise...what we need is more additions such as holographic principle and issue concerning vacuum leakages.

[Mordred]

16 minutes ago, JosephDavid said:

It seems like there is a fundamental point being missed here. The number of microstates for the \textbf{neutron} is significantly greater than that of the \textbf{proton}, and this directly impacts the entropy. According to the \textbf{third law of thermodynamics}, as systems approach \textbf{absolute zero}, they must reach a state of \textbf{minimal entropy}, meaning the system should have as few microstates as possible, leading to maximum stability and order.

The expression for entropy S in terms of the number of microstates Ω is given by \textbf{Boltzmann's entropy formula}:

 

S=kBln(Ω)

 

 

In this equation, Ω represents the number of different ways the system’s energy can be arranged. The more microstates available, the greater the entropy.

Since the \textbf{neutron} can undergo \textbf{beta decay}, it has many potential microstates, which leads to increased entropy. On the other hand, the \textbf{proton} is inherently stable and has fewer available microstates, resulting in \textbf{lower entropy}. This lower entropy aligns perfectly with the requirement for \textbf{minimum entropy} as dictated by the third law of thermodynamics.

Thus, in terms of the third law, the proton, with its lower number of microstates and resulting lower entropy, is a better fit for the condition of minimal entropy compared to the neutron, which inherently has more microstates and higher entropy.
 

No there isn't a misunderstanding on my part Are you familiar with  the S matrix for protons and neutrons ? The number of up down quarks for each is merely the mean average color charge relations

If you examine the number of microstates contained within either the proton and neutron via the S matrix your earlier statement makes little sense based on stability Protons and neutrons are composite particles plain and simple so they have internal microstates. The number of microstates within the Proton does not change in the interaction you described. An electron is not a microstate contained within the proton to begin with

19 minutes ago, MJ kihara said:

From the author article.

I don't see why some people are against the article,am not convinced otherwise...what we need is more additions such as holographic principle and issue concerning vacuum leakages.

holographic principle won't help what many fail to understand is is that the holographic principle is conformal The laws of physics would be the same if you use the holographic Principle or QM/QFT or even classical physics. That is the basis of its premise.

Edited November 1 by Mordred

[JosephDavid]

17 minutes ago, Mordred said:

No there isn't a misunderstanding on my part Are you familiar with  the S matrix for protons and neutrons ? The number of up down quarks for each is merely the mean average color charge relations

If you examine the number of microstates contained within either the proton and neutron via the S matrix your earlier statement makes little sense based on stability Protons and neutrons are composite particles plain and simple so they have internal microstates. The number of microstates within the Proton does not change in the interaction you described. An electron is not a microstate contained within the proton to begin with

 You're right that both protons and neutrons are made of quarks, and both have complex internal structures. They have all these quarks, color charges, and strong interactions that keep everything together, and that means they both have internal microstates. In that sense, they are pretty similar. But here's the key difference. The neutron is unstable, it has a habit of falling apart, and that happens because of the weak interaction. The neutron undergoes beta decay, turning into a proton, an electron, and an antineutrino. Every time it does that, you have more possible outcomes, more microstates to deal with. It is like having a machine with a lot of different ways it can break down, and that means more disorder, more entropy. So when you look at the neutron's S-matrix, you are not just seeing internal quark stuff, you are also seeing all these possible decay channels, and that adds up to a lot more accessible states.

Now let’s look at the proton. The proton, on the other hand, is stable. It does not just fall apart on its own. It stays put. No natural decay paths under normal conditions. That stability means the proton has fewer ways to evolve, fewer accessible final states. So if you look at the proton's S-matrix, it is much simpler—fewer transitions, fewer accessible microstates, and therefore, lower entropy compared to neutron. 

So, even though both particles have similar internal structures, what really matters here is how they behave over time. The neutron has this extra layer of complexity because it can decay, and that gives it a higher number of accessible microstates. The proton, by staying stable, has fewer accessible states and lower entropy. When you need minimal entropy, like what the third law of thermodynamics is talking about, the proton is just the ideal choice. It stays steady, does not add chaos, and keeps everything orderly. It is not just about what is inside, it is about what they do over time.

[Mordred]

 The decay does not affect the number of microstates any neutron examined will have identical number of microstates otherwise it wouldn't be a neutron. What alters is the transition amplitudes via CKMS mass mixing matrix a simpler method though being Beit Wigner where under that treatment you treat the proton or neutron as a single particle.

 

Edited November 1 by Mordred

[JosephDavid]

17 minutes ago, Mordred said:

 The decay does not affect the number of microstates any neutron examined will have identical number of microstates otherwise it wouldn't be a neutron. What alters is the transition amplitudes via CKMS mass mixing matrix a simpler method though being Beit Wigner where under that treatment you treat the proton or neutron as a single particle.

 

You are correct that any neutron, when examined in isolation, retains its inherent internal microstates—it remains a neutron, composed of up and down quarks and held together by the strong interaction.

However, my point pertains to the concept of **accessible microstates over time**, rather than just the internal microstates at a particular instant. The S-matrix is used to describe the possible transitions a particle can undergo. For the neutron, due to the weak interaction and the CKM mixing, it has a decay path where it transforms into a proton, electron, and antineutrino. These decay pathways represent additional accessible configurations, effectively increasing the number of possible microstates that the neutron can occupy over time.

The proton, by contrast, is stable under normal conditions and does not undergo spontaneous decay. Consequently, its S-matrix is simpler, with fewer pathways to alternative states, resulting in fewer accessible microstates overall compared to the neutron.

The distinction I am making is not about the inherent microstates that define the neutron as a neutron, but rather about the broader set of accessible states that arise due to the decay process. The neutron's ability to decay into other particles introduces additional possible states, which increases its entropy. The proton, lacking such decay channels, remains in a lower entropy state with fewer accessible outcomes. This distinction is key when considering entropy and stability, particularly in relation to the third law of thermodynamics, which favors minimal entropy in stable systems.

[Mordred]

What your missing is an essential aspect that I have repeated  numerous times.

Take a bottle and fill it with neutrons treat each neutron as a microstate. Lets completely ignore internal microstates. With decay over time the number of neutrons will be less over time. This is precisely why I repeatedly mention number density calculations. The number density is also affected by temperature

So in terms of entropy you really can't look at stability alone as the stability of neutrons is gained through the formation of deuterium. So if that same bottle is filled with deuterium there is no change in number density. Regardless of any instability of free neutrons.

This is something the author never took into consideration he used volume when he should have used number density via Maxwell-Boltzmann which would have given him all the needed requirements to apply for calculating the number of protons or neutrons at any given temperature to apply to entropy.

The entropy formulas are a direct result of Maxwell Boltzmann in regards to particle physics and cosmology applications.

So why wasn't that method applied to begin with ?

 

 

Edited November 1 by Mordred

[JosephDavid]

17 minutes ago, Mordred said:

What your missing is an essential aspect that I have repeated  numerous times.

Take a bottle and fill it with neutrons treat each neutron as a microstate. Lets completely ignore internal microstates. With decay over time the number of neutrons will be less over time. This is precisely why I repeatedly mention number density calculations. The number density is also affected by temperature

So in terms of entropy you really can't look at stability alone as the stability of neutrons is gained through the formation of deuterium. So if that same bottle is filled with deuterium there is no change in number density. Regardless of any instability of free neutrons.

This is something the author never took into consideration he used volume when he should have used number density via Maxwell-Boltzmann which would have given him all the needed requirements to apply for calculating the number of protons or neutrons at any given temperature to apply to entropy.

The entropy formulas are a direct result of Maxwell Boltzmann in regards to particle physics and cosmology applications.

So why wasn't that method applied to begin with ?

 

 

Imagine you have a bottle full of neutrons. You’re treating each neutron as a microstate, and that makes sense if you’re just counting them in that bottle. But here’s the thing: neutrons don’t just stay the same over time. They decay into protons, electrons, and antineutrinos. This means the number of neutrons is going to decrease over time because they are unstable. So, if we’re talking about entropy, we cannot ignore that decay. The neutron’s decay gives it more ways to change and increases the number of possible states, which means higher entropy.

Now, let’s compare that to deuterium. Deuterium is a combination of a proton and a neutron bound together. When they are bound, they stabilize each other. So, if you have a bottle full of deuterium, it’s more stable. The number density of deuterium will stay more or less the same because they are not decaying like free neutrons. But deuterium still has more possible microstates than a proton by itself because you have two particles interacting, and that adds complexity.

A proton, on its own, is very stable. It doesn’t have natural decay pathways under normal conditions. It just stays a proton, with fewer ways to change compared to a neutron or even deuterium. That’s why the author chose the proton for this analysis. He wanted the most stable, simplest unit possible, with the lowest entropy, to help understand the vacuum energy of the universe.

 

Edited November 1 by JosephDavid