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March 4

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Accelerating (massless) gluons by the strong force

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Here are two facts, AFAIK:

Fact #1. When it comes to photons, the impact of gravity on them - which actually must change their momentum (according to the very definition of "force" in mechanics), is reflected, either by accelerating them - i.e. by changing their direction (if their motion has just been perpendicular to the direction of the gravitational field), or by their redshifting or blueshifting (if their motion is parallel to the direction of the gravitational field).

Fact #2. When it comes to gluons, they have a color charge; while this color charge - with respect to the strong force, plays the same role, as the electric charge does - with respect to the electric force. Hence, the strong force has an impact on gluons.

My question is: What's the impact of the strong force on gluons, bearing in mind that gluons have no mass, so apparently the strong force can't accelerate them by changing - their speed - i.e. the absulote value of their velocity.

In other words: Is the way gluons are influenced by the strong force, identical to the way photons are influneced by gravity, even though the explanation of the impact of gravity is usually ascribed to the curved spacetime - which is not the case when it comes to the strong force?

But maybe I'm wrong, so although the gluon is massless, its speed (i.e the absolute value of the gluon's velocity) is not as "constant" as the photon's speed is, so also the gluon's speed can be changed by the strong force. Can it? HOTmag (talk) 05:07, 4 March 2024 (UTC)[reply]

Basically you are asking: Since Gluons are massless they always travel at the speed of light, so how can they accelerate/change energy? But so far as I know Gluons are virtual only, there are no free Gluons, and virtual particals do not need to conserve momentum or energy in isolation. Ariel. (talk) 06:06, 4 March 2024 (UTC)[reply]
Gluons were first detected in 1978, and I guess they were detected as free. Quark–gluon plasma, in which gluons are free, was first detected in the year 2000. Additionally, glueballs are assumed to have their gluons free.
By the way, I won't be surprised if a given (massless) gluon's speed turns out to be non-constant (hence different from the speed of light). See my thread here. HOTmag (talk) 06:36, 4 March 2024 (UTC)[reply]
The observations that could best be explained as involving gluons, made starting in 1978, did not "detect" gluons. One can say they were "discovered". The researchers studied the hadronic decay of the Y boson ϒ meson and identified that this decay was probably mediated by gluons. Only in 1979 had enough evidence been collected to assign a sufficiently high statistical significance to the gluon explanation to confirm the physical existence of this thus far theoretical particle.  --Lambiam 12:49, 4 March 2024 (UTC)[reply]
It was the ϒ meson (Greek upsilon), not the (hypothetical) Y boson. --Wrongfilter (talk) 13:02, 4 March 2024 (UTC)[reply]
Fact 1 doesn't work. A photon has no mass, so the gravitational force acting on it is zero. The resulting acceleration is zero divided by zero, which is undefined. In other words, classical mechanics doesn't apply to photons. Newton's laws can be bent in such a way that they predict deflection of photons in a gravitational field, but the predicted deflection is only half that predicted by General Relativity. Observations are consistent with GR. PiusImpavidus (talk) 11:18, 4 March 2024 (UTC)[reply]
Due to your comment, I realized I had to strike out the content of parentheses containing the word "force" in my original post, and that's what I've just done. However, I also realized that your comment had no impact on my question, being about gluons, rather than about photons. HOTmag (talk) 12:11, 4 March 2024 (UTC)[reply]
On a scale less than about 0.8 fm gluons are the force carriers of the strong force; see Strong interaction § Behavior of the strong interaction. It is not particularly meaningful to assert that the strong force has an impact on gluons; in some sense they are the strong force.  --Lambiam 13:01, 4 March 2024 (UTC)[reply]

Before asking my question, let me begin with three facts regarding it:

1. By the classical definitions in Mechnaics, a mass of a given particle, reflects the particle's resistance to accelaration: The bigger mass the particle carries, the bigger its resistance to acceleration is.

2. By the classical definitions in Mechnaics, the angle of deflecting the light by gravitational lensing, can be regarded as a kind of acceleration, actually a normal/radial one: Indeed, it does not change the speed (the absolute value of the veloicity), yet it does change the velocity's direction.

3. By General relativity, the angle of deflecting the light by gravitational lensing, only depends on properties of the gravitational lens (actually its mass), and on the distance between the lens and the light, yet not on any propery of the light itself (e.g. its momentum/color).

Question: Am I allowed to infer - from these three facts alone (regardless of special relativity), that light has a constant mass (e.g. zero mass), independent of the light's momentum/color? For, if the light's mass varied - e.g. depended on the light's momentum/color - so that blue photons were "heavier" than red ones, then by #1 - a red photon's resistance to this photon's normal/radial accelaration - would be smaller than a blue photon's resistance to this photon's normal/radial accelaration, and so by #2 - the angle of deflecting red photons by a given gravitational lens - would be wider than the angle of deflecting blue photons by that very gravitational lens, as opposed to #3. Hence (apparently), light has a constant mass, independent of the light's momentum/color. Is my deduction correct, from a logical point of view (regardless of special relativity)? HOTmag (talk) 10:19, 4 March 2024 (UTC)[reply]

Light has no mass. But any masses going in the same direction and speed as each other would be deflected the same. The orbit of a satellite around the earth does not depend on its mass - well not unless we had one that had a mass comparable to the earth! NadVolum (talk) 10:54, 4 March 2024 (UTC)[reply]
Oops, I forgot that a given body's gravitational acceleration does not depend on the body's mass. Sorry for this huge (contemporary) mistake... HOTmag (talk) 11:05, 4 March 2024 (UTC)[reply]

Do you have in mind examples, of an exceedingly weak gravitational field (if not the weakest one), ever measured?

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HOTmag (talk) 12:48, 4 March 2024 (UTC)[reply]

Well there's [1]. Or maybe Modified Newtonian dynamics is the sort of thing that you're interested in. NadVolum (talk) 13:30, 4 March 2024 (UTC)[reply]
Thx. HOTmag (talk) 15:28, 4 March 2024 (UTC)[reply]

Regardless of the following four formulas, is there any other way (whether an empirical one or a theoretical one) to prove that light has no mass?

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Denoting: the speed of light by the velocity by the gamma factor by the mass by the momentum by and the energy by the four formulas I'm currently ignoring are:

1.

2.

3. (bearing in mind that

4. (while denotes the so-called "relativistic mass").

I'm asking the question in the title, because the formulas mentioned above were originally developed/deduced for bodies slower than light, so these formulas won't be reliable (in my eyes) for determining whether light has any mass - e.g. a mass for creating a gravitational field.

Additionally, for defining force, I only use the formula (which actually defines the force as the change in momentum with respect to time), rather than the formula (which actually defines the force as the product of mass and acceleration).

I would also like to ignore the theory of elementary particles (including gauge theory), according to which photons have no mass. HOTmag (talk) 15:33, 4 March 2024 (UTC)[reply]

It cannot be proved on theoretical grounds, but in general: Rest mass is invariant (it's measured to be the same quantity in all inertial reference frames) and massless energy is not. A cloud of photons has both gravity and rest mass. Taken individually a photon's frequency is not invariant, it's arbitrary; its blueness/redness depends on the observer. There are empirical constraints on its mass, see Photon#Experimental_checks_on_photon_mass. Modocc (talk) 17:52, 4 March 2024 (UTC)[reply]
Has it ever been empirically proved that a cloud of photons has gravity/mass? I emphasize I'm asking from an empirical point of view, because I guess it would be impossible to theoretically prove this - without relying on any of the formulas mentioned above which I currently ignore. HOTmag (talk) 18:22, 4 March 2024 (UTC) HOTmag (talk) 18:22, 4 March 2024 (UTC)[reply]
Does conservation of mass-energy and Electron-positron annihilation suffice? Modocc (talk) 18:32, 4 March 2024 (UTC)[reply]
Are you referring now to my new question about the cloud of photons? If you are, then - using the conservation law you've mentioned - along with the annihilation you've mentioned, how do you prove that a cloud of photons has gravity - without relying on any of the formulas I've mentioned in my first post? Note that Relativity theory allows mass to disappear and to re-appear, as long as the total energy is conserved. HOTmag (talk) 18:38, 4 March 2024 (UTC)[reply]
I didn't notice your question(s) earlier. Certainly, the photons' gravity can and should be empirically demonstrated: for example, by measuring the deflection of ultra-thin beams of energetic light which cross paths in extremely close proximity to each other. The gravitational deflection(s) will be small and these can be measured by redirecting the beams into an interferometer. In the confines of a small Earth-bound setting a ring of mirrors can be added such there are a multitude of repeated crossings of the beams (in a vacuum) that would accumulate their deflections if necessary. One would need to calibrate the beams independently and then turn them on together to see if there is an observable effect. Ideally one would repeat the experiment and vary its parameters to get a more complete picture of their interactions if any. Modocc (talk) 22:31, 4 March 2024 (UTC)[reply]
Do you think such an experiment (or any similar one) has ever been carried out? HOTmag (talk) 22:48, 4 March 2024 (UTC)[reply]
Not to my knowledge. Perhaps we simply need to instigorate it. :-) Modocc (talk) 23:09, 4 March 2024 (UTC)[reply]
The Eddington experiment measured the gravitational deflection of starlight passing near the Sun. Philvoids (talk) 01:34, 5 March 2024 (UTC)[reply]
That showed that the sun gravitates light. The unanswered question is whether or not photons reciprocate and gravitate. The difficulty with that is photons are not known to interact with each other via gravity, and if they do their angular deflections would be very very small anyway, perhaps too small to detect. Modocc (talk) 02:45, 5 March 2024 (UTC)[reply]
The two words "reciprocate and" in your recent response, are needless, in my opinion. Even without them, the "unanswered question" (as you name it) basically remains the same question. HOTmag (talk) 08:55, 5 March 2024 (UTC)[reply]
You cannot use relativity to derive the mass of the photon or anything else; relativity per se doesn't know what a friggin photon is. You tell relativity: "Here's a particle with mass and energy " (info comes from other theories, e.g. de Broglie for the energy), and then you ask "How does it behave?", and relativity will tell you that. For instance, it tells you that the momentum of the photon is , using equation (3), which is the only relevant equation here. --Wrongfilter (talk) 12:40, 5 March 2024 (UTC)[reply]
1. Do you think that a given frigging photon has no mass?
If you do, then my original question was:
2. How do you know that? Do you only know that by (some of) the equations I've indicated in my first post?
HOTmag (talk) 13:19, 5 March 2024 (UTC)[reply]
I can only repeat myself. All these equations are part of the theory of relativity. Relativity doesn't know what a photon is, therefore these equations cannot tell you what the mass of a photon is. You need to look elsewhere for that. --Wrongfilter (talk) 13:25, 5 March 2024 (UTC)[reply]
Let's leave Relativity theory. Personally, do you think (or estimate) that light has no mass? HOTmag (talk) 13:32, 5 March 2024 (UTC)[reply]
This is a reference desk and not a place for personal opinions. --Wrongfilter (talk) 13:36, 5 March 2024 (UTC)[reply]
Let's leave personal opinions. Can science determine, that a given body - whose speed does not depend on any observer - has no mass? HOTmag (talk) 13:40, 5 March 2024 (UTC)[reply]
See Photon#Experimental_checks_on_photon_mass. Modocc (talk) 15:49, 5 March 2024 (UTC)[reply]
I've read this chapter many times over the recent years. Yet, I don't see how it answers my question, about "no mass", rather than about a mass smaller than a given tiny limit. HOTmag (talk) 15:58, 5 March 2024 (UTC)[reply]
See Invariant mass for the details on how physicists calculate it. Modocc (talk) 17:46, 5 March 2024 (UTC)[reply]
The article only states:
1. Systems whose four-momentum is a null vector (for example, a single photon or many photons moving in exactly the same direction) have zero invariant mass and are referred to as massless. But unfortunately, the article doesn't explain why "a single photon" mentioned in the parentheses is an "example" of that. In other words, the article doesn't explain why a photon's four momentum is a nul vector. It wouldn't have been a null vector, had we assumed that a photon does have a (positive) mass.
2. Thus, the mass of a system of several photons moving in different directions is positive, which means that an invariant mass exists for this system even though it does not exist for each photon...For example, rest mass and invariant mass are zero for individual photons. But unfortunately, the article doesn't explain why this is true.
HOTmag (talk) 18:40, 5 March 2024 (UTC)[reply]
Consider two red photons a and b from stationary sources approaching you with velocities +|v| and -|v| from opposite directions. Now switch to another reference frame such that the source of a is moving towards you and the source of b is receding. a is bluer and b is redder. Doppler shift changes what energy is measured for each photon, but it doesn't change their total energy. That way the invariant mass of the proton and my newspapers didn't change if they were on my bike, I was walking with them, or being read by the numerous customers I delivered them to. Physicists are simply tallying (empirically) an invariant mass-energy for each particle. Modocc (talk) 19:18, 5 March 2024 (UTC)[reply]
I understand your thought experiment (a well known one), but I don't understand why you think it proves that the mass is zero. Why zero? HOTmag (talk) 20:50, 5 March 2024 (UTC)[reply]
I don't think that at all. Your 1st quote about zero mass is simply empirical, and I explained the 2nd quote to give it some context. Modocc (talk) 21:53, 5 March 2024 (UTC)[reply]
I thought you'd meant to answer my question about "no mass". HOTmag (talk) 10:10, 13 March 2024 (UTC)[reply]

When an electron and a positron collide - annihilating each other, does the gravitational field - having been created by them - disappear?

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Assuming that light has no gravity. HOTmag (talk) 18:33, 4 March 2024 (UTC)[reply]

The photons carry away the mass-energy from the collision and redistributes it to other parts of the system. So the field itself has to change (in accordance with its mass-energy distribution). Modocc (talk) 18:53, 4 March 2024 (UTC)[reply]
Their combined energy and momentum do not change but are now carried by photons. In general relativity, the gravitational field is not ascribed to individual bodies but a universal field entirely determined by the stress–energy tensor, which only depends on the distribution of energy and momentum.  --Lambiam 18:56, 4 March 2024 (UTC)[reply]
Is it a response to me or to Modocc?
Anyway, bearing in mind that the two original massive particles no longer exist after the light was emitted, is it carrying now the original gravitational field? If this light is not, then which entity is, in this case? Is this gravitational field carried now by no entity? In other words: does spacetime get curved, yet by no entity? I'm quite surprised... HOTmag (talk) 19:01, 4 March 2024 (UTC)[reply]
It was meant to be a response to the original question; I had not noticed Modocc's response. Your question above ignores the nature of the gravitational field iin general relativity, which is not ascribed to individual bodies. So the photons do not carry a gravitational field, but they influence the gravitational field.  --Lambiam 19:44, 4 March 2024 (UTC)[reply]
I thought your first response had been meant to be a response to Modocc, because of its indent. Anyway, back to your main response: Assuming theoretically - that the only gravitational field that existed in the whole universe before the collision of the two massive particles - was the gravitational field created by them, you actually claim it still exists after the collision - although the whole universe (in our theoretical case) currently contains light only - actually being the light emitted after the collision, so you actually claim we've got a curved spacetime - without any mass - we assuming that light has no mass. That's why I was so surprised, having read your first response. I'd always thought, that spacetime could only be curved by a presence of mass. Had I been wrong? HOTmag (talk) 19:56, 4 March 2024 (UTC)[reply]
Yes, if you only consider only matter to have mass and do not equate other forms of energy with mass. The concept of mass in general relativity is notoriously difficult to define; see the section Mass in general relativity § Defining mass in general relativity: concepts and obstacles. Spacetime is curved by the presence of energy and momentum; see the second paragraph of Einstein field equations in combination with Stress–energy tensor. Photons have energy and momentum.  --Lambiam 07:08, 5 March 2024 (UTC)[reply]
I'm surprised again. If (as you actually claim) light curves spacetime, i.e. creates a gravitational field, i.e. has (by defintion) an active gravitational mass, so what's the physical meaning of a given photon having "no mass" - as commonly accepted to claim? On the second hand, if - by saying that light has "no mass" - one actually means light has no passive gravitational mass, then one actually says nothing, because the change in a given body's momentum due to the gravitatoinal force (while defining any force as rather than as ) - as well as the body's gravitational normal/radial acceleration, are not influenced by the body's passive gravitational mass - nor even by whether it's a positive one or a zero one. On the third hand, if - by saying that light has "no mass" - one actually means light has no inertial mass, then what does a given photon's inertial mass mean, bearing in mind that light is not influenced by any force other than the gravitational one (e.g. when a given photon's path is deflected when it approaches the sun)...
This surprise, as described in the previous paragraph, takes me back to the title of my previous thread.
By the way: regarding your first ten words: "Yes, if you only consider only matter to have mass", please see the first paragraph in our article Einstein field equations: "In the general theory of relativity, the Einstein field equations (EFE; also known as Einstein's equations) relate the geometry of spacetime to the distribution of matter within it". This first claim in the lede, is sourced by a reliable source. So, would you suggest that the lede in this article be fixed somehow, e.g. by replacing "matter" by "mass" (or by "mass-energy"), along with replacing the source (ibid.) by another one? HOTmag (talk) 08:49, 5 March 2024 (UTC)[reply]
A photon has a rest mass equal to zero.
Have you considered getting a proper textbook on relativity? I just pulled my Rindler from the shelf. PiusImpavidus (talk) 11:08, 5 March 2024 (UTC)[reply]
I wonder how your remark has anything to do with my question in the current thread. Have I ever claimed the photon has a positive mass?
As for your question: Have you read my previous thread? HOTmag (talk) 11:16, 5 March 2024 (UTC)[reply]
In your post above my last post, you asked about the physical meaning of "a photon has no mass", followed by some discussion on active or passive gravitational mass and inertial mass. I answered that it's the rest mass that's zero. In other words, I gave a direct answer to the question you asked just above. If that's irrelevant, you asked the wrong question.
Yes, I read your previous thread. It sounds like you don't care too much about actually understanding physics, so I assume the answer to my question is no. PiusImpavidus (talk) 18:43, 5 March 2024 (UTC)[reply]
Yes, I know that the photon's rest mass must be zero (because the photon can't be at rest). That said, my question you've quoted, which should have been quoted in its full version: "what's the physical meaning of a given photon having 'no mass' - as commonly accepted to claim?", was actually meant to be addressed to all those physicists (not me but for example Okun) who claim that there's only one kind of mass, which I and you call "rest mass", and which they call "the mass" - i.e. "the only one mass possible". They don't make any distinction between, what I and you call "rest mass", and any other kind of mass. If you think (like me) that there is more than one kind of mass, then we can become good friends. However, as far as they are concerned, they actually claim that "a photon has no mass", without making any distinction between a photon's rest mass and a photon's relativistic mass. In their opinion, if a photon's mass is zero (e.g. when considering the "rest mass"), then any mass of the photon must be zero, simply because there's only one mass possible.
As for your second remark ("It sounds like you don't care too much about actually understanding physics"). Unfortunately, you didn't explain why you thought so (about a person who thought she could become your friend), so I can't respond to this remark with the same elaboration appearing in my previous thread. HOTmag (talk) 19:08, 5 March 2024 (UTC)[reply]
Before general relativity, mass was thought to be a well-defined and reasonably understood physical concept, unambiguous with respect to a preferred inertial frame of reference. That is no longer the case. I referred you to the section Mass in general relativity § Defining mass in general relativity: concepts and obstacles. Did you read this? Do you expect us, instead of referring you to sources, to come up with solutions to these obstacles?  --Lambiam 21:21, 5 March 2024 (UTC)[reply]
One week before you referred me to this article, User:Graeme Bartlett did the same in a previous thread, so I thanked him for the article and also stated I had read it. That was long before you referred me to this article for your first time (now it's your second time). Anyway, after having read it, I asked him some further questions (which you can find in that previous thread). HOTmag (talk) 10:05, 13 March 2024 (UTC)[reply]
What do you mean by "the system"? The only system has been the original pair of electron-positron, which turned out to be light. HOTmag (talk) 19:00, 4 March 2024 (UTC)[reply]
Although you didn't constrain them to a closed system or to a hypothetical one they still gravitate and affect each other (at least in theory).Modocc (talk) 19:06, 4 March 2024 (UTC)[reply]
After the collision, they no longer exist, do they? HOTmag (talk) 19:13, 4 March 2024 (UTC)[reply]
The photons which are created contribute to the stress–energy tensor of the gravitational field of spacetime. They gravitate. Modocc (talk) 19:38, 4 March 2024 (UTC)[reply]
Thank you for this link. I was not aware of this theoretical effect. I still wonder, though, if it has ever been empirically detected. I also wonder if it's mentioned in Wikipedia. HOTmag (talk) 19:43, 4 March 2024 (UTC)[reply]
And the destructive collision will make some tiny gravitational waves. So not all energy will end up as photons. Graeme Bartlett (talk) 07:16, 5 March 2024 (UTC)[reply]
Gravitational waves can only be created, if the curvature in spacetime is already given - the waves characterizing the manner of how this given curvature propogates in spacetime.
That said, I'm asking if the very curvature really still exists after the collision, assuming that the whole universe had only contained the electron and the positron before they collided and annihilated each other while emitting light only, about which I'm actually asking whether it can curve spacetime (along with creating gravitational waves with respect to this curvature), assuming that the whole universe no longer contains matter after the collision. HOTmag (talk) 09:50, 5 March 2024 (UTC)[reply]
In theory, all massless particles gravitate. Consider protons: Their gluons are massless but these have enough energy to account for the proton's mass which gravitates. Modocc (talk) 11:45, 5 March 2024 (UTC)[reply]
A given proton is composed of quarks and gluons. Every quark "inside" the proton (i.e. contrary to a "free" quark) carries, not only what would be its mass if the quark were free, but also another component of mass which reflects the potential energy associated with the strong force carried by the gluons. So why do you think the proton gravitates due to the gluons, rather than due to those two components of mass of the quarks contained in the proton? HOTmag (talk) 12:40, 5 March 2024 (UTC)[reply]
Oops. It was in my thoughts but I didn't type "most of" when I said they have "...enough energy to account for the proton's mass...". Note, I didn't specify why or how. Modocc (talk) 15:24, 5 March 2024 (UTC)[reply]
Also after you add "most of", the question I asked you in my previous response remains the same. HOTmag (talk) 15:29, 5 March 2024 (UTC)[reply]
See Proton: "Using lattice QCD calculations, the contributions to the mass of the proton are the quark condensate (~9%, comprising the up and down quarks and a sea of virtual strange quarks), the quark kinetic energy (~32%), the gluon kinetic energy (~37%), and the anomalous gluonic contribution (~23%, comprising contributions from condensates of all quark flavors)". Modocc (talk) 16:35, 5 March 2024 (UTC)[reply]
Note, that the quarks this experiment refers to, have become "free" quarks, after the stationary proton split into several moving particles carrying "kinetic energy", like: "up and down quarks", a sea of "virtual strange quarks", and moving gluons. On the other hand, my question to you was about the quarks "inside" the stationary proton, rather than about "free" quarks outside it. The question also explains why there's a difference between, the mass of a quark "inside" the proton, and the mass of a "free" quark. HOTmag (talk) 17:53, 5 March 2024 (UTC)[reply]
The mass (invariant) of a proton is what is measured when it is stationary. They calculated the gluons contributions per the invariant mass article. Modocc (talk) 18:22, 5 March 2024 (UTC)[reply]
Yes, the mass of the proton is the mass of the stationary proton, but the masses of the quarks and gluons are masses of those components after they became free, i.e. after the proton split into them while they (including "sea of virtual strange quarks") became particles carrying "kinetic energy". HOTmag (talk) 18:49, 5 March 2024 (UTC)[reply]
My sandwich contains numerous bound and free particles. That fact doesn't change their contributions to its mass. Modocc (talk) 19:38, 5 March 2024 (UTC)[reply]
The difference between a quark inside a proton (i.e. a bound quark) and a quark outside a proton (i.e. a free quark) is as follows: All agree that a given proton is composed of quarks and gluons. However, contrary to a "free" quark, every quark "inside" the proton carries, not only what would be its mass if the quark were free, but also another component of mass which reflects the potential energy associated with the strong force carried by the gluons.
Note that the article equates the mass of the stationary proton (i.e. before it split into plenty of particles) with the sum of the masses/energies of those many particles (including a "see of strange" ones) into which the proton split. My question to you was: Why do you rule out the following option: If the calculation had only referred to the quarks "inside" the proton before it split, the mass of the proton would have turned out to be exactly the sum of all the bound quarks inside it alone, so that its gravity would have only depended on this sum alone, without relying on any gluon. HOTmag (talk) 20:00, 5 March 2024 (UTC)[reply]
See Color confinement. Quarks are always in a bound state with each other. Modocc (talk) 20:30, 5 March 2024 (UTC)[reply]
Yes, but after a proton splits into moving quarks (including "a sea of strange" ones), they become free - even for a millionth of second - until they create new hadrons. Without their being free - even for a millionth of second, they couldn't have been detected, but they have! Further, the calculation of the mass of a proton - as the sum of the masses/energies of those many particles (including a "see of strange" ones) into which the proton split, actually relies on this millionth of second - during which those particles were free (untill they created new hadrons because of the color confinement). HOTmag (talk) 20:37, 5 March 2024 (UTC)[reply]
Pumped with enough energy we have observed Quark matter that decays. That doesn't change their contributions to the proton mass. BTW, I've other things to tend to so I will be disappearing soon. Modocc (talk) 21:16, 5 March 2024 (UTC)[reply]
When you disappear, what will happen to your gravitational field?  --Lambiam 21:24, 5 March 2024 (UTC)[reply]
A quark inside a proton can't decay. It's only the free one which can. On the other hand, my question to you was about bound quarks, i.e. ones inside a proton, when I asked you why you thought the proton's gravitational mass relied also on any bound gluon's gravitational mass rather than on the mass of bound quarks alone. The experiment you referred me to, was not about bound quarks but rather about quarks detected when they were free, even for less than a milionth of second (before they created new hadrons because of the color confinement). HOTmag (talk) 10:34, 13 March 2024 (UTC)[reply]
Gluons contribute to the proton mass per [2], a reliable source. Modocc (talk) 16:00, 13 March 2024 (UTC)[reply]
Now you're ignoring my response to your response in which you presented the source. HOTmag (talk) 16:36, 13 March 2024 (UTC)[reply]
Certainly "...the contributions to the mass of the proton are..." refers to the proton's bound constituent particles. Your responses simply keep ignoring that. Modocc (talk) 16:50, 13 March 2024 (UTC)[reply]
So let's remain in dispute about whether the values of those contributions refer to the values of bound constituent particles (as you claim) or to the values of those particles had they been free (as I claim). HOTmag (talk) 17:35, 13 March 2024 (UTC)[reply]