JosephDavid

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.

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.

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.

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.

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. 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.

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.

Your questions made me understand the author arguments more clearly. The proton and neutron might seem similar since they’re both built from quarks held together by the strong force, but they’re not the same when it comes to stability and entropy. Neutrons? They decay, breaking down into a proton, an electron, and a neutrino. That breakdown matters, it means more ways for things to fall apart, more “messiness” or entropy in the system. The proton, though, is different. It doesn’t decay (at least, we’ve never seen it happen), and it doesn’t have any available decay channels in the standard model that wouldn’t violate fundamental conservation laws, like charge conservation. Simply put, it’s stable because the rules of physics don’t give it a way out. So, the proton’s stability is directly tied to these conservation laws. This stability means the proton has minimal entropy, it doesn’t have all those extra states it could move into. And here’s where the third law of thermodynamics kicks in: as a system gets down to close to absolute zero, it demands a setup with as little entropy as possible. This is why the proton fits the bill perfectly. When we’re looking to break down the vacuum energy into stable, low-entropy “units” in the context of SU(3), it’s not just about taking any volume, we need something as stable as they come, with minimal entropy. The proton’s locked-down, conservation-law-protected structure makes it the ideal candidate for this job. 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.

Good question indeed. As you get closer to absolute zero, the third law of thermodynamics is saying, "I want a system that’s as orderly and low-entropy as possible." Now, look at liquid helium. Sure, it’s fascinating stuff—flows without friction, does all sorts of quantum tricks. But even down there near zero, it’s still got a lot of atomic hustle and too many possible states, which means higher entropy. Enter the proton. This thing is rock-solid, It’s stable, and has very few microstates to mess with, exactly what the third law ordered for a low-entropy, high-order setup at the coldest extremes. So, if we are looking for the perfect fit for a minimal-entropy remnant, the proton’s our guy. This is an interesting note. Can you give more details? It’s not just about volume, what really matters is remnant volume with minimal entropy. The third law of thermodynamics is looking for something stable and with the least possible disorder. Now, the neutron might look similar in size to the proton, but here’s the catch: free neutrons don’t stick around, they decay, and that means extra entropy. The proton, on the other hand, is rock solid. It’s stable, no decay, and has minimal entropy because it doesn’t have extra states to fall into. So if you’re looking for a low-entropy remnant that satisfies the third law, it’s the proton that wins.

Well, the choice of the proton isn’t about convenience, it’s about its physical significance. The proton is stable; we’ve never seen it decay, and it stands strong under SU(3) symmetry, even when U(1) breaks down. That’s what gives it physical meaning beyond just being a handy reference. It’s not a random choice, it’s a backed by what we know from experiment and theory.

Good questions. Why not the hydrogen atom? Well, hydrogen is a little more complicated. It’s made of a proton and an electron and that electron’s got U(1) symmetry written all over it, which breaks at low temperatures. So, hydrogen isn't the best candidate because it's tied to electromagnetic interactions, not exactly stable if you're looking at what happens when U(1) breaks. The proton, though? That’s a different story. Now, about quarks. Sure, up and down quarks are part of the SU(3) game, but they’ve got a problem: you never find them alone. They’re always confined, stuck inside protons, neutrons, whatever. Quarks are like that friend who never leaves the party alone, they’re always bound together, thanks to quark confinement. The proton, on the other hand, is the smallest, most stable package of SU(3) stuff we can actually observe. It’s been tested again and again, and it’s never been seen to decay. So why pick the proton as the remnant volume? Because it’s the simplest stable bound state that’s pure SU(3). It’s the one structure that stays unbriken under the third law of thermodynamics, when you get to those really low temperatures, it’s still standing. That’s what makes it perfect for this context. We need something stable, something minimal, and something that stays intact with SU(3) symmetry. That’s the proton. It’s not just a random pick, it’s the only choice that makes sense if you want a fundamental remnant that’s consistent with what we observe in the vacuum.

If you can solve something with a simple relation, why complicate it by adding more language ? As Newton once said, "Truth is ever to be found in simplicity, and not in the multiplicity and confusion of things." It’s like in language: why say "a round object used in games" when you can just say "ball"? The author found a simple, logical relationship that ties together the relevant measurements to address the issue.

Physics is fundamentally an empirical science, its foundation is built on measurements, observations, and experiments. It’s not about math for math’s sake; instead, math is just a language we use to describe physical phenomena. The real core of physics exists in connecting these descriptions to what we can actually measure. In this paper, the author uses the math of spontaneous symmetry breaking and Snyder’s quantum spacetime, not as ends in themselves, but as tools to explain real, observable phenomena. The achievement here is in how the author finds logical connections between these physical measurements, uniting them in a simple, coherent relationship. This isn’t about whether there’s enough math and adding several complications, it’s about whether the math helps us understand and explain the empirical data we observe. The author managed to do exactly that, building a bridge between very solid physical concepts that leads to a clearer understanding of the universe, grounded in what we can actually measure.

The author’s proposal is tied directly to some very solid experimental facts. We all know that the proton has never been observed to decay. That’s crucial. It means that, practically speaking, the proton is a stable structure, which strongly suggests that there’s something fundamental about its existence. Now, pair that with the third law of thermodynamics, which implies that as we approach absolute zero, there should always be some kind of remnant volume that doesn’t just vanish or break down. This tells us that there’s an unbroken, stable structure remaining even at the lowest energy states. Now, what the author is doing is connecting these dots: if there’s a remnant volume, and we know that protons don’t decay and are associated with SU(3) symmetry, then it makes perfect sense to use the volume of the proton as the benchmark for this remnant state. The SU(3) symmetry is unbreakable, it remains stable even when other symmetries like U(1) are broken near zero kelvin. That’s what makes protons so fundamental. The author matches this remnant volume, implied by the third law with the proton volume, defining SU(3) units or vacuum atoms. These SU(3) units can be used to explain why the vacuum energy doesn’t explode to some ridiculous value like QFT tells us it should. Instead, it’s spread out across these stable SU(3) units, bringing the predicted value right down to what we actually observe as dark energy. So, what’s being overlooked here is that this whole approach is anchored in solid principles: the experimental stability of the proton, the third law of thermodynamics, and the fundamental unbreakability of SU(3) symmetry. The author isn’t skipping the foundations—he’s grounding the whole argument in them. It’s about finding a stable, unbroken remnant that matches what we see in nature, and that’s why using the proton volume as the basis makes so much sense. This is what lets the author solve the cosmological constant problem with high precision.

You know, this thread really shows two kinds of folks. On one side, you’ve got people who are actually trying to understand things by sticking to simple, solid physical principles. And then, you’ve got others who just can’t resist making it all way more complicated than it needs to be, adding confusion on top of confusion. It’s funny how so many people get excited about these complicated theories, like the 10^{500} possible multiverses, when the answer might just be staring us in the face, rooted in something as fundamental as the SU(3) confinement scale.

Actually, vacuum dilute as well according to recent study by DESY https://www.quantamagazine.org/dark-energy-may-be-weakening-major-astrophysics-study-finds-20240404/

The author cited this paper. They are very related.

That is interesting. You are touching a topic that the same author investigated recently and published in https://www.worldscientific.com/doi/10.1142/S0218271824500366

Alright, here’s the thing. The misunderstanding here is thinking of the universe and the proton as three-dimensional objects, they’re not. They’re both four-dimensional, with three spatial dimensions plus one time dimension. Now, if you take the boundary (surface area) of the universe and divide it by the boundary of a proton, you get a huge number, the author obtained, 10^(123). That number is supposed to tell us, by holography, about how many su(3) atoms could filling up the universe. Here’s where it gets interesting: the “holographic principle” says that everything inside a volume can be described by information on its boundary. Now, in quantum field theory (QFT), if you integrate over all the vacuum energy in four dimensions, you get this massive energy density, way bigger than what we actually observe as dark energy. But by applying the holographic idea, that huge number from our boundary comparison lets us spread out this QFT vacuum energy across all those SU(3) vacuum atoms. The result? The vacuum energy density drops down to the observed dark energy precisely. This gives a holographic explanation for dark energy confined in our four-dimensional universe and effectively solving the cosmological constant problem. I read in the paper that the author used in Eq 27 a distribution implied by Snyder quantum spacetime which may be more fundamental than the statistical distributions such as Bose-Einstein and Fermi-Dirac.

That is very interesting hint, can you elaborate more ?

Do you think this paper relevant ? https://link.springer.com/article/10.1140/epjc/s10052-024-12481-7

Ah, that's an interesting point you've brought up! The author citing papers connected to the holographic principle and mentioning in the abstract that this solution might shed light on the origin of gauge/gravity dualities adds a whole new dimension to the discussion. You see, the holographic principle is this fascinating idea that suggests all the information contained in a volume of space can be represented as a theory on the boundary of that space, kind of like how a hologram works. It's a deep concept that bridges quantum mechanics and gravity. The author might be aiming to provide insights into how gauge theories (SU(3)) relate to gravitational theories through cosmological constant lens. This ties into the gauge/gravity duality, which is a powerful concept in theoretical physics suggesting that certain gauge theories are equivalent to gravity theories in higher dimensions. If the author's approach can highlight the origin of these dualities, it could be a significant step toward understanding how different forces in the universe are connected at a fundamental level. It might even offer clues about how spacetime and gravity emerge from more basic quantum processes. So, not only is the paper addressing the cosmological constant problem by proposing that dark energy behaves like a superconductor state of matter, but it's also potentially shedding light on deep connections between quantum field theories and gravity. That's pretty exciting stuff! It's always amazing to see how ideas from different areas of physics can converge, offering new perspectives and solutions to longstanding problems. As Einstein said, "We can't solve problems by using the same kind of thinking we used when we created them." Exploring these new connections might be just what's needed to push our understanding forward.

Exactly, that’s the point! He is doing the former, comparing volumes, not dividing the number of SU(3) atoms by the universe volume.

The equation you derived is (energy density)*(SU(3) effective volume)/(universe volume). So, you're saying SU(3) effective volume by universe volume is the same as SU(3) atoms by universe volume? The first is about comparing volumes, but the second is like dividing the number of these atoms over the universe volume. There's a clear difference in meaning there, right?

Exactly! You got the division twisted. The author ain’t dividing SU(3) atoms by the universe volume. he’s calculating it by dividing the universe volume by the SU(3) effective volume. Big difference. What the author’s really pushing is that dark energy is like a superconductor state of matter. All it needs is two composite electrons acting like a scalar field to break U(1) symmetry and leave SU(3) intact. That’s the real deal he’s talkin’ about.

Gotcha, and I respect that. My way of talking just helps me express more clearly, but I get that it might not work for everyone here. Apologies if it came off the wrong way.