Why Gravity is Significant in Proton & Neutron Nuclei

[ZapperZ]

If the proton was a perfect sphere with positive charge distributed uniformly over the surface, the 1/r^2 force would apply only down to the surface (EM force would be 0 inside). But the proton is not a perfect sphere. It is a wave function that has rather fuzzy boundaries. It seems that the existence of the strong nuclear force is based on the assumption that EM repulsion continues to follow the 1/r^2 relationship to a point that appears to be within that fuzzy boundary.

But I can ask you the same thing: what evidence do you have that the valid EM laws simply break down at a certain scale? While EM laws that we have are known to work, your guess work hasn't. So the "burden of proof" (something YOU should know about) is in your court. It is up to you to show that there ARE evidence to suggest that your idea might be valid.

I am just asking questions. All questions are based on ignorance. Otherwise, why ask the question?

But you already stated that you ARE questioning the validity of QCD. Maybe I am simply a foolish person, but if I were to question the validity of something, I would want to make sure I have understood as much as I can about that thing. So these simply aren't just "questions" out of curiousity.

My question was: What evidence is there that protons repel protons with enormous EM force (\propto 1/r^2) that continues down to the 'surface of the proton'.

Andrew Mason

The validity of EM and QED. On the other hand, what evidence do YOU have that Maxwell equation and/or QED do not work down to the "surface of the proton"?

Zz.

 

[Andrew Mason]

But I can ask you the same thing: what evidence do you have that the valid EM laws simply break down at a certain scale? While EM laws that we have are known to work, your guess work hasn't.

But EM laws do break down at the atomic level, which is why we have quantum theory. So the question is not whether they break down. The question is: at what point are they no longer valid?

So the "burden of proof" (something YOU should know about) is in your court. It is up to you to show that there ARE evidence to suggest that your idea might be valid.

At this stage I am cross-examining the evidence. I don't have to come up with a valid theory at this stage to test the prosecution's case.

But you already stated that you ARE questioning the validity of QCD. Maybe I am simply a foolish person, but if I were to question the validity of something, I would want to make sure I have understood as much as I can about that thing. So these simply aren't just "questions" out of curiousity.

In an ideal world I would have read and understood textbooks on quantum theory, tensor analysis, and general relativity and know how to develop Schrodinger's equation from first principles. I don't live in an ideal world.

The validity of EM and QED. On the other hand, what evidence do YOU have that Maxwell equation and/or QED do not work down to the "surface of the proton"?

But IF <A=the existence of the strong nuclear force> is in some part based on the assumption that <B=the EM field equations apply down to the surface of the proton>, and IF <C=EM field equations are known to break down at some point at the atomic level> THEN the onus would be on the proponent of A to show B is true. I don't have to lead evidence for a verdict of 'not proven'.

Andrew Mason

 

[ZapperZ]

But EM laws do break down at the atomic level, which is why we have quantum theory. So the question is not whether they break down. The question is: at what point are they no longer valid?

Hello? I asked you earlier what that "V" is in the Schrodinger equation that is USED to find all those solutions to an atom. Where do you think this came from?! This is exactly the coulombic potential from E&M!

At this stage I am cross-examining the evidence. I don't have to come up with a valid theory at this stage to test the prosecution's case.

This assumes that you have the ability to understand the evidence. Presumably, if you are not an expert in the field, you bring in experts that can evaluate the evidence. The experts in the fields have spoken, in VOLUMES of work in peer-reviewed journals.

So now what?

But IF <A=the existence of the strong nuclear force> is in some part based on the assumption that <B=the EM field equations apply down to the surface of the proton>, and IF <C=EM field equations are known to break down at some point at the atomic level> THEN the onus would be on the proponent of A to show B is true. I don't have to lead evidence for a verdict of 'not proven'.

Andrew Mason

Read above on why your so-called evidence that EM fields break down at the atomic level isn't valid. Thus, there are no such instances in this case.

Zz.

 

[humanino]

But IF <A=the existence of the strong nuclear force> is in some part based on the assumption that <B=the EM field equations apply down to the surface of the proton>, and IF <C=EM field equations are known to break down at some point at the atomic level> THEN the onus would be on the proponent of A to show B is true. I don't have to lead evidence for a verdict of 'not proven'.

Andrew, seriously, let us be honest :
C) holds for sure. There is no doubt. I told you everybody uses everyday the standard model down to 10^{-16}m and even less ! We do not see any problem with it. We wish we could find a problem, some would kill to find a small clue for a problem !

B) Sure holds. Here at the lab we throw electrons on protons. No problem. We too apply the standard model more than one order of magnitude smaller than the size of the proton. We probe it down to a scale about one hundredth its size. Its working. I told you we will soon deliver a snapshot of the constituents, and you keep telling us "but you don't know what you're doing"

A) holds too. It explains so many things. We work nuclear plants with it, and satisfactory make calculations about the power of the Sun. We have a simple, symmetry-based model called QCD, from which we can actually prove the most successful model of the nuclear forces : the Skyrme force. You need not understand all the details. This model has been discovered after a long work, mainly based on both satisfying the quantum rules and using semi-classical approximations, and the model was proven to work great ! Only recently did we understand how the model was justified by the fundamental QCD. So now we understand how the forces (Skyrme) between protons and neutrons in the nucleus emerge as residual forces of the interaction between the constituents (QCD) which are quarks and glue. Those are facts. I am telling you, this is worth studying it, there is no reason to doubt about it. Every day in other lab they collide heavy ions, and check the laws in the regime where perturbative technics apply : it works. The instanton models are based on the fundamental QCD, they are able to predict the mass of maybe more than 80% of hadrons within less than 10% accuracy, and we know from the beginning that they are approximations. Really, I swear, it does not make sens to say the strong interaction is spurious ! We can't display in front of your eyes more than half a century of planetary efforts, involving hundreds of thousands of passionated people, all of them eager to find the least flaw anywhere.

Andrew, please, ask serious questions. There are so many informations everywhere. We told you why your first calculation is wrong : you do not even take into account EM. And you answer "but all of you guys are wrong with this" you throw this at our face. This is not fair. We are ready to help you understand what we are doing, but you can't tell us "the entire middle-age history is fake. I did not read it, but I know those little green men erased all they've been doing during thousand years, and replaced it with a fairy tale" What are we supposed to do ? Tell you the all middle-ages history and you point every little detail until you understood we were right ? This is insane ! I am unable to do it anyway ! You have to trust your peers at some point, and as long as nothing comes wrong with what has been said, you cannot redo all the experiments at home !

 

[Andrew Mason]

Andrew, seriously, let us be honest :
C) holds for sure. There is no doubt. I told you everybody uses everyday the standard model down to 10^{-16}m and even less ! We do not see any problem with it. We wish we could find a problem, some would kill to find a small clue for a problem !

I never said there was a problem with the experimental evidence. I am wondering whether there might be more than one explanation for some of it ie. something other than a force. I am trying to determine, without doing a post-graduate program in nuclear physics, what the evidence is for the existence of these strong nuclear forces.

B) Sure holds. Here at the lab we throw electrons on protons. No problem. We too apply the standard model more than one order of magnitude smaller than the size of the proton. We probe it down to a scale about one hundredth its size. Its working. I told you we will soon deliver a snapshot of the constituents, and you keep telling us "but you don't know what you're doing"

I never said that. I simply asked the questions: how do we know that there is a strong nuclear force? and how do we know that protons exert EM repulsive force within the nucleus?

I just thought there might be a simple answer to those questions.

So now we understand how the forces (Skyrme) between protons and neutrons in the nucleus emerge as residual forces of the interaction between the constituents (QCD) which are quarks and glue. Those are facts. I am telling you, this is worth studying it, there is no reason to doubt about it. ...

... Really, I swear, it does not make sense to say the strong interaction is spurious ! We can't display in front of your eyes more than half a century of planetary efforts, involving hundreds of thousands of passionated people, all of them eager to find the least flaw anywhere.

Andrew, please, ask serious questions.

I'll admit my study of physics is rather out of date. I took a couple of courses on quantum physics but most of my understanding of the subject came from Richard Feynman. I will give you a quote from Feynman's Lectures on Physics (Feynman, of course, won the Nobel Prize for his work in quantum electro-dynamics, so I think that he understood QED fairly well):
Vol 1, page 12-12,

"12-6 Nuclear Forces:...

These forces are within the nuclei of atoms, and although they are much discussed, no one has ever calculated the force between two nuclei and indeed at present there is no known law for nuclear forces. These forces have a very tiny range which is just about the same as the size of the nucleus, perhaps 10^{-13} centimeter. With particles so small and at such a tiny distance, only the quantum-mechanical laws are valid, not the Newtonian laws. In nuclear analysis we no longer think in terms of forces, and in fact we can replace the force concept with a concept of the energy of interaction of two particles a subject that will be discussed later. Any formula that can be written for nuclear forces is a rather crude approximation which omits many complications; one might be somewhat as follows: forces within a nucleus do not vary inversely as the square of the distance but die off exponentially over a certain distance r, as expressed by F = (l/r^2) exp(-r/r_0) where the distance r_0 is of the order of 10^{-13} centimeter. In other words, the forces disappear as soon as the particles are any great distance apart, although they are very strong within the 10^{-13} centimeter range. So far as they are understood today the laws of nuclear force are very complex: we do not understand them in any simple way and the whole problem of analysing the fundamental machinery behind nuclear forces is unsolved. Attempts at a solution have led to the discovery of numerous strange particles, the \pi mesons, for example, but the origin of these forces remains obscure."


If I can start by updating this paragraph, I might be able to stop annoying everyone with my naive questions. For example, have we been able to measure the force between a proton and a neutron in, say, the deuterium nucleus? Have we been able to measure the nuclear force between two protons in, say, the He nucleus?

Andrew Mason

 

[ZapperZ]

I'll admit my study of physics is rather out of date. I took a couple of courses on quantum physics but most of my understanding of the subject came from Richard Feynman. I will give you a quote from Feynman's Lectures on Physics (Feynman, of course, won the Nobel Prize for his work in quantum electro-dynamics, so I think that he understood QED fairly well):
Vol 1, page 12-12,

"12-6 Nuclear Forces:...

These forces are within the nuclei of atoms, and although they are much discussed, no one has ever calculated the force between two nuclei and indeed at present there is no known law for nuclear forces. These forces have a very tiny range which is just about the same as the size of the nucleus, perhaps 10^{-13} centimeter. With particles so small and at such a tiny distance, only the quantum-mechanical laws are valid, not the Newtonian laws. In nuclear analysis we no longer think in terms of forces, and in fact we can replace the force concept with a concept of the energy of interaction of two particles a subject that will be discussed later. Any formula that can be written for nuclear forces is a rather crude approximation which omits many complications; one might be somewhat as follows: forces within a nucleus do not vary inversely as the square of the distance but die off exponentially over a certain distance r, as expressed by F = (l/r^2) exp(-r/r_0) where the distance r_0 is of the order of 10^{-13} centimeter. In other words, the forces disappear as soon as the particles are any great distance apart, although they are very strong within the 10^{-13} centimeter range. So far as they are understood today the laws of nuclear force are very complex: we do not understand them in any simple way and the whole problem of analysing the fundamental machinery behind nuclear forces is unsolved. Attempts at a solution have led to the discovery of numerous strange particles, the \pi mesons, for example, but the origin of these forces remains obscure."


If I can start by updating this paragraph, I might be able to stop annoying everyone with my naive questions. For example, have we been able to measure the force between a proton and a neutron in, say, the deuterium nucleus? Have we been able to measure the nuclear force between two protons in, say, the He nucleus?

Andrew Mason

If you are any more out of date than this, you would have to read works chiselled in stone tablets. It might be helpful to read the publication dates on this.

At the time Feynman wrote this, the pion (or pi meson) was thought to be the mediator of the strong force based on the Yukawa theory. It isn't. We know MORE (lots more) about it now since the present-day development has clearly point out the gluons as this mediator.

You also must keep in mind that you are trying to comprehend something that is highly complex. Even as someone who is a trained physicist but with expertise in condensed matter, I would never be silly enough to come up to another nuclear physicist/high energy physicist and start spewing out my own theory about nuclear forces, even though my knowledge of nuclear physics and high energy physics are more than what a typical quack has. Are you able to comprehend the idea that protons and neutrons do not maintain their rigid indentity when they are part of a nucleus? Can you understand the parton structure and how the so-called "color forces" interact between them? What about the tantalizing hint at the observation of the quark-gluon plasma at RHIC?

These may appear to you to be a disjointed set of information that have nothing to do with what you're asking, but they are EXACTLY the "evidence" that you are asking for. While quantum mechanics isn't JUST the Schrodinger equation or the uncertainty principle, the same way QCD isn't JUST about quarks and the standard model. It is about a whole body of knowledge in dealing with the strong interaction. You can't just pick one thing and ignore the rest because there is a whole zoo of consequences via the prediction of the existence of the strong interaction. These consequences are the ones we observe and verify via experiments that confirm the validity of QCD. And we continue to do that with better accuracy and high degree of certainty.

Zz.

 

[Andrew Mason]

If you are any more out of date than this, you would have to read works chiselled in stone tablets. It might be helpful to read the publication dates on this.

I am not assuming that his lectures (ca. 1961) are a source of up-to-date information. But his explanations of the concepts are very useful.

You also must keep in mind that you are trying to comprehend something that is highly complex. Even as someone who is a trained physicist but with expertise in condensed matter, I would never be silly enough to come up to another nuclear physicist/high energy physicist and start spewing out my own theory about nuclear forces, even though my knowledge of nuclear physics and high energy physics are more than what a typical quack has.

I don't think I am giving you my own theory about nuclear forces. I am a long way from understanding the elementary stuff, let alone developing a theory. I am just asking questions.

Are you able to comprehend the idea that protons and neutrons do not maintain their rigid indentity when they are part of a nucleus?

This is actually the purpose of some of my questions. We can certainly talk about coulomb forces between separate protons. How do we really know that these same coulomb forces exist between protons in the nucleus?

Can you understand the parton structure and how the so-called "color forces" interact between them? What about the tantalizing hint at the observation of the quark-gluon plasma at RHIC?

I am sure I could... in another lifetime. I confess I do not really understand how an exchange of elementary particles between protons / neutrons creates an attractive force.

These may appear to you to be a disjointed set of information that have nothing to do with what you're asking, but they are EXACTLY the "evidence" that you are asking for. While quantum mechanics isn't JUST the Schrodinger equation or the uncertainty principle, the same way QCD isn't JUST about quarks and the standard model. It is about a whole body of knowledge in dealing with the strong interaction. You can't just pick one thing and ignore the rest because there is a whole zoo of consequences via the prediction of the existence of the strong interaction. These consequences are the ones we observe and verify via experiments that confirm the validity of QCD. And we continue to do that with better accuracy and high degree of certainty.

I do appreciate your help.

Andrew Mason

 

[humanino]

Please read [thread=41110]this thread[/thread] about nuclear interactions.

Andrew, I think you did not take time to read [thread=41110]this thread[/thread] which is short, and which might help you asking questions.

 

[quantumdude]

A simple calculation should reveal the reason that this gravity idea cannot be right. Take a 2-nucleon system, and a typical nucleon radius of about 1 fermi (that's 10^-15m), and calculate the gravitational potential energy. I got something on the order of magnitude of 10^-30eV, which is far, far below the scale of nuclear binding energies (which are on the order of 1 to 10² MeV).

What more could you want?

 

[Andrew Mason]

Andrew, I think you did not take time to read [thread=41110]this thread[/thread] which is short, and which might help you asking questions.

I am trying to understand the nature of the forces within the nucleus. It seems to me that we have several consistent mathematical models which, together, allow us to predict the behaviour of fundamental constituents of matter and energy very accurately. It is not necessarily the case, however, that a consistent model describes reality. It may be just a helpful device to work with. The concept of a 'particle' for example, when speaking of fundamental particles, is just a mental device that allows us to think about mathematical wave functions as discrete entities. We know that these are not particles in the classical sense.

The 'particle' is still a very useful model to use, just as lines of force is a useful concept for mathematically modelling the magnetic and electric field. We don't pretend that they really exist. The concept of the exchange of virtual particles creating a force between quarks, and the residual effect of this exchange creating strong forces between protons and neutrons (perhaps because protons and neutrons in close proximity share , seems to me to be a similar device. If it works to explain behaviour, it is a very useful model to use. But that does not mean it is the only possible way of looking at it.

Gravity can be conceived as a attractive force that one particle exerts on another particle by virtue of their respective masses. Newton provided a very consistent mathematical model for gravity, which is still very useful. Einstein provided another very different mathematical model which has also been very useful. Einstein's model, however, involves a reexamination of our concepts of distance and time and the essence of matter itself. Einstein's theory of general relativity is, in that sense, the more fundamental model.

Now it may be that gravity has very little to do with the structure of the nucleus at the level that we are probing now. But I am not so sure that will be the case as we probe more deeply.

Now, dealing with the concept of a force created by the exchange of virtual particles, I have a conceptual diffculty here. How is it that such an exchange is attractive? Doesn't an exchange of particles carry momentum that would tend to move the sender and recipient farther apart?

Andrew Mason

 

[Orion1]

Classical Consistency...



In Classical Quantum Electrodynamics, the Fine Structure Constant is a classical tool used to measure the relative strength between any force and that of the classical Strong Nuclear Force.

Given this classical model which approximates for any two particles on a nuclear scale, the following equations results. (note that these obey the 'Andrew Model')

Classical Quantum Electrodynamics gravitational Fine Structure Constant:
\propto_g = \frac{Gm_p^2}{\hbar c}
m_p - proton mass

Classical Quantum Electrodynamics Electromagnetism Fine Structure Constant:
\propto_e = \frac{K_e q^2}{\hbar c}
K_e - Coulomb's proportionality constant
q - proton charge

Andrew, based upon these equations, what are the SI (International Standard) units for \propto_e and \propto_g?

Andrew, based upon these equations, what is the relative difference in magnitude between \propto_e and \propto_g?

Andrew, how do these magnitudes compare with that of the classical Strong Nuclear Fine Structure Constant?

These equations obey the original 'Andrew Model'. (post #1)



"The fine structure constant measures the strength of the electromagnetic force that controls how charged elementary particles (such as electrons and photons) interact."...

Reference:
http://whatis.techtarget.com/definition/0,,sid9_gci866284,00.html

 

[selfAdjoint]

In Classical Quantum Mechanics, the Fine Structure Constant is a classical tool used to measure the relative strength between any force and that of the classical Strong Nuclear Force.

I think you'd better look up the definition of the fine structure constant again.

 

[quantumdude]

Now it may be that gravity has very little to do with the structure of the nucleus at the level that we are probing now. But I am not so sure that will be the case as we probe more deeply.

It is the case that gravity has very little to do with it. As I said in my last post, the gravitational potential energy between 2 nucleons that are separated by any acceptable nuclear radius (as determined by scattering expeirments) cannot possibly account for the binding energy between them (as determined by fission experiments). This idea can be rejected on the basis of a "back of the envelope" calculation.

Now, dealing with the concept of a force created by the exchange of virtual particles, I have a conceptual diffculty here. How is it that such an exchange is attractive? Doesn't an exchange of particles carry momentum that would tend to move the sender and recipient farther apart?

Imagine that 2 ice skaters are the particles, and the force carrier is a basketball. If they exchange it such that one skater throws it to the other, then indeed the force is repulsive. But if they exchange it in such a way that one grabs it from the other, then the two are drawn together, and the force is attractive.

 

[Ian]

When making gravitational calculations on atomic and nuclear scales, why use the magnitude of G as quoted in the original post?

There is no proof that it is correct to use the 'solar scale' value of G at the smaller scale, neither has any physicist ever given a theoretical definition to Newton's constant that adequately predicts G.

However, G has been defined as the 'torsion' of spacetime according to quantum theory, therefore matter of a higher density should produce a greater compression of the vacuum. Since nuclear matter has a far greater density than the sun or earth, the distortion of the vacuum in the vicinity of nuclear matter must be far greater than we observe at the solar scale and G at that tiny scale must be much greater than we assume.

If you increase the observed value of G by a factor equal to the perceived difference in strength between gravitational and electric forces in atoms (and use this in gravitational calculations) then the electric and gravitational mathematical expressions all become unified.

Therefore the easiest and only way to unify Gravitation and electromagnetism is to show that G changes with matter density, or to produce a competent theoretical model that predicts or explains how G changes with increasing density.
(a simple re-working of Einstein's flat-sheet + ball analogy will do it, explain G that is)

 

[humanino]

When making gravitational calculations on atomic and nuclear scales, why use the magnitude of G as quoted in the original post?

This is obviously a very relevant objection, with which I fully agree. We have no clue of gravity at small scales, and we wish we had !
Yet :

However, G has been defined as the 'torsion' of spacetime according to quantum theory

Could you be more precise, or elaborate ?
Torsion is a well defined quantity already in differential geometry, it is the non-symmetrical part of the connection coefficients (Christoffel symbols) (torsion in MathWorld)
T(u,v)=\nabla_uv-\nabla_vu-[u,v]
and it is assumed zero in Einstein's theory :
[URL='https://www.physicsforums.com/insights/author/john-baez/']John Baez on zero torsion[/url]

 

[Nereid]

On the one hand, we have a theory (or set of mutually consistent theories) which account for ALL the experimental and observational results, some with breath-taking degrees of accuracy.

On the other hand, we have some speculation and hand-waving, a few curious and possibly interesting 'what if?'s.

On the third hand (us .pas omoH sentient sgnieb from htraE are cursed/blessed with more sdnah than Homo sap. ), our two best (classes of) theories - GR and QM - cannot both be, in domains far, far beyond what we've been able to test to date.

Your mission, should you choose to accept it, is to convert the hand-waving into the outline of a sketch of a hypothesis, and show - to even 5 OOM (I'm feeling generous today) - that it is consistent with relevant observational results.

 

[marlon]

At the time Feynman wrote this, the pion (or pi meson) was thought to be the mediator of the strong force based on the Yukawa theory. It isn't. We know MORE (lots more) about it now since the present-day development has clearly point out the gluons as this mediator.

Just to be accurate...

The pions are indeed not the strong force mediators as originally assumed by Yukawa. The strong force mediators are elementary particles themselves : the gluons. These gluons make sure the quarks constituting a meson or a baryon are all bound together so the meson (quarkdoublet) and the baryon (quarktriplet) does not "fall" apart. The lightest meson is the pion (quark anti-quark combination) which is NOT an elementary particle yet it DOES mediate a "part" of the strong force. I am referring to the socalled residual strong force that holds atomic nuclei together...

Also keep in mind that QCD describes at best the behavior of colour-confinement, although this problem is not yet "solved". Many propositions exist among theoretical physicists like the dual abelian higgs-model using the nice magnetic monopoles introduced by Dirac. Also keep in mind that gluons can interact with each other via the colour-confinement (although not every gluon exhibits a colour; meaning they can be colour-neutral), which is a big difference with the mediators of the EM-force ie the photons...

regards
marlon

 

[ZapperZ]

When making gravitational calculations on atomic and nuclear scales, why use the magnitude of G as quoted in the original post?

There is no proof that it is correct to use the 'solar scale' value of G at the smaller scale, neither has any physicist ever given a theoretical definition to Newton's constant that adequately predicts G.

However, G has been defined as the 'torsion' of spacetime according to quantum theory, therefore matter of a higher density should produce a greater compression of the vacuum. Since nuclear matter has a far greater density than the sun or earth, the distortion of the vacuum in the vicinity of nuclear matter must be far greater than we observe at the solar scale and G at that tiny scale must be much greater than we assume.

On the other hand, what could be the possible impetus to consider that G might be different at those scales? I haven't seen any. In fact, going by recent trends in the latest set of experimental measurement of G to test the Arkani-Hamed hypothesis of millimeter scale compaction, no deviation has been found up to 10 micrometer scale![1] Couple this with an earlier measurements [2], and the recent report from the 2004 APS April meeting from the U. of Mainz on "neutron bouncing" experiment, there is clearly every indication that gravity that we know of at the macroscopic scale still works the same way at the microscopic scale. Granted that these scales are still not within the nuclear length scales, but if we simply go by experimental observations and trends, there are ZERO indications that G would deviate from the known value.

Zz.

[1] J. Chiaverini et al., PRL v.90, p.151101 (2003).
[2] PRL v.86 , 1418 (2001); J.C. Long et al., Nature v.421, p.922 (2003).

 

[Andrew Mason]

It is the case that gravity has very little to do with it. As I said in my last post, the gravitational potential energy between 2 nucleons that are separated by any acceptable nuclear radius (as determined by scattering expeirments) cannot possibly account for the binding energy between them (as determined by fission experiments). This idea can be rejected on the basis of a "back of the envelope" calculation.

Could the binding energy not simply be stored in the form of matter rather than a force x distance concept of potential energy? The pairing of a proton and a neutron is a very stable configuration from an energy perspective. But when enough energy is added so that the particles are lifted out of that deep energy well, the result is a conversion of some of that matter back into energy.

Do we really have to be wedded to a concept of a 'force' within the nucleus? I say this knowing full well that there is probably something quite basic that I am missing so please be gentle on me.

Imagine that 2 ice skaters are the particles, and the force carrier is a basketball. If they exchange it such that one skater throws it to the other, then indeed the force is repulsive. But if they exchange it in such a way that one grabs it from the other, then the two are drawn together, and the force is attractive.

Or perhaps hockey players grabbing a puck ... but I appreciate your analogy. As I understand your analogy, the skaters would only get closer together if the other was still holding onto the ball. Otherwise, if it is just the momentum of the ball that moves the 'grabber' toward the 'grabbee', the two don't get very close - not unless the ball is much more massive than the player. So that begs the question: what is the skater/quark holding onto the ball/gluon with? Would you not have to assume some additional 'sticky' force there between the ball/gluon and the skater/quark?

I have tried to wrap my mind around virtual particles with negative momentum traveling faster than c in a cloud of uncertainty to explain the nuclear force. I am getting the impression that trying to give a physical meaning for what is essentially a mathematical solution to a wave function is probably hopeless. Forgive my naiveté in thinking that there must be other, as yet undiscovered, model or fundamental principle that we are missing in all of this. But I find it all very fascinating.

Andrew Mason

 

[ZapperZ]

Could the binding energy not simply be stored in the form of matter rather than a force x distance concept of potential energy? The pairing of a proton and a neutron is a very stable configuration from an energy perspective.

This is not correct. A hydrogen atom (which has NO neutron) is a lot more stable than any of its isotopes. Thus, the pairing of a proton and a neutron does NOT produce a "very stable" configuration. The stability of a nuclear configuration is a lot more complex than that.

Or perhaps hockey players grabbing a puck ... but I appreciate your analogy. As I understand your analogy, the skaters would only get closer together if the other was still holding onto the ball. Otherwise, if it is just the momentum of the ball that moves the 'grabber' toward the 'grabbee', the two don't get very close - not unless the ball is much more massive than the player. So that begs the question: what is the skater/quark holding onto the ball/gluon with? Would you not have to assume some additional 'sticky' force there between the ball/gluon and the skater/quark?

The problem here is that there is a lack of understanding of quantum field theory. Keep in mind that QFT has been successfully used to obtain an accurate description of the band structure of the semiconductors that you are using in your modern electronics. So the validity of its methodology is well-known, even when you do not realize you are using it. The materials that you are using are littered with descriptions involving virtual phonons, magnons, spinons, polarons, etc.

So the fundamental issue now is to clearly understand QFT. Unfortunately, I think it is impossible to teach QFT online! :)

Zz.

 

[Andrew Mason]

This is not correct. A hydrogen atom (which has NO neutron) is a lot more stable than any of its isotopes. Thus, the pairing of a proton and a neutron does NOT produce a "very stable" configuration. The stability of a nuclear configuration is a lot more complex than that.

So is this incorrect?: http://hyperphysics.phy-astr.gsu.edu/hbase/particles/deuteron.html

The problem here is that there is a lack of understanding of quantum field theory. Keep in mind that QFT has been successfully used to obtain an accurate description of the band structure of the semiconductors that you are using in your modern electronics. So the validity of its methodology is well-known, even when you do not realize you are using it. The materials that you are using are littered with descriptions involving virtual phonons, magnons, spinons, polarons, etc.

So the fundamental issue now is to clearly understand QFT. Unfortunately, I think it is impossible to teach QFT online! :)

Next on my reading list is "An Introduction to Quantum Field Theory"

Andrew Mason

 

[quantumdude]

Could the binding energy not simply be stored in the form of matter rather than a force x distance concept of potential energy? The pairing of a proton and a neutron is a very stable configuration from an energy perspective. But when enough energy is added so that the particles are lifted out of that deep energy well, the result is a conversion of some of that matter back into energy.

Did you not see that the binding energy due to gravitation differs from the observed binding energy by about 38 orders of magnitude? Can you not perform a calculation to determine how much mass you would need to make up the difference? Just try it, and you'll see that it doesn't work.

Do we really have to be wedded to a concept of a 'force' within the nucleus? I say this knowing full well that there is probably something quite basic that I am missing so please be gentle on me.

I didn't say anything about forces. I spoke of binding energy.

Or perhaps hockey players grabbing a puck ... but I appreciate your analogy. As I understand your analogy, the skaters would only get closer together if the other was still holding onto the ball.

Nope. That's why I put them on ice skates. Even if the ball was dropped and rolled away, the skaters would move together, once the exchange had taken place.

Otherwise, if it is just the momentum of the ball that moves the 'grabber' toward the 'grabbee', the two don't get very close - not unless the ball is much more massive than the player.

Nope again. Look at the law of conservation of momentum. If the exchange is violent enough, the momentum transfer can be as high as you like.

So that begs the question: what is the skater/quark holding onto the ball/gluon with? Would you not have to assume some additional 'sticky' force there between the ball/gluon and the skater/quark?

The particles aren't holding onto the quanta with anything. Charged objects are sources of quanta. But as a side note, it is true in the case of the strong force that the quanta are also charged. But that is not necessary for the quanta to mediate the interaction, as demonstrated in the EM case.

 

[ZapperZ]

So is this incorrect?: http://hyperphysics.phy-astr.gsu.edu/hbase/particles/deuteron.html

You need to read what I said more carefully. I said

"This is not correct. A hydrogen atom (which has NO neutron) is a lot more stable than any of its isotopes."

"A lot more stable" does not mean one is stable and the other isn't. The fact that you can find more "H" or "H2" rather than "D" or "D2" imlies that H is more favorable to be formed than D.

So you are learning QFT? Have you also studied Second Quantization, which is practically the language of QFT?

Zz.

 

[Andrew Mason]

You need to read what I said more carefully. I said
"This is not correct. A hydrogen atom (which has NO neutron) is a lot more stable than any of its isotopes."

And I had said that a proton neutron pair is very stable. That is what I understood you to say was incorrect. I never said it was more stable than a proton. Probably nothing in the universe is as stable as a proton.

"A lot more stable" does not mean one is stable and the other isn't. The fact that you can find more "H" or "H2" rather than "D" or "D2" imlies that H is more favorable to be formed than D.

Or it simply implies that the D nucleus decays more rapidly than ^1_1H. Perhaps we started with only neutrons in the universe and that resulted in more ^2_1H than ^1_1H.

So you are learning QFT? Have you also studied Second Quantization, which is practically the language of QFT?

Not yet. I am brushing up on my undergraduate quantum physics text first and then we'll see how far I get. Thanks for the encouragement.

Andrew Mason

 

[Orion1]

Grainy Gravitation...

Did you not see that the binding energy due to gravitation differs from the observed binding energy by about 38 orders of magnitude? Can you not perform a calculation to determine how much mass you would need to make up the difference? Just try it, and you'll see that it doesn't work.

\propto_g = \frac{Gm_p^2}{\hbar c}

m_p - proton mass

\propto_g = 5.904*10^{-39}

Andrew, how much mass is required to increase the gravitational fine structure constant to 1 (\propto_g = 1)?

Probably nothing in the universe is as stable as a proton.

Some GUT theories predict the proton as 'unstable'.

Actually, the electron is the most 'stable' particle known, however this is tangential because electrons cannot carry colour.

Andrew, I recommend studying Quantum Electrodynamics prior to Quantum Field Theory.

[/color]

 

[ZapperZ]

Or it simply implies that the D nucleus decays more rapidly than ^1_1H. Perhaps we started with only neutrons in the universe and that resulted in more ^2_1H than ^1_1H.

And if you know this already, then this ring a bell that CLEARLY should have told you that a proton-neutron set up isn't as stable as a proton. Please note that originally, you are indicated that this set up is stable, as if it is the preferred ground state of a nuclei. I merely point out that it isn't when compared to a just a single proton as in the H atom. This means that the rest of your guesswork on the nucleon model falls apart. The nucleon structure isn't as naive as this. THAT is what I am trying to convey.

Zz.

 

[Andrew Mason]

Did you not see that the binding energy due to gravitation differs from the observed binding energy by about 38 orders of magnitude? Can you not perform a calculation to determine how much mass you would need to make up the difference? Just try it, and you'll see that it doesn't work.

There is no question that the observed binding energies in the nucleus are dozens of orders of magnitude greater than the gravitational binding energy of nuclear particles calculated using classical gravitational laws.

I didn't say anything about forces. I spoke of binding energy.

That's good. I am reluctant to equate the two concepts when speaking about the nucleus. I would like to leave open the possibility that the nuclear force might be viewed as something else - perhaps some kind of inertia in the process by which mass can be 'unraveled' into energy.

Nope. That's why I put them on ice skates.

Hockey players wear skates.

Even if the ball was dropped and rolled away, the skaters would move together, once the exchange had taken place.

I don't follow that. The momentum imparted by a particle is always in the direction of its motion. If the ball was pulled from the other skater and thrown back over the 'grabber's' head, they would move together because the grabbing skater would gain momentum in an equal and opposite direction to the ball. But the ball would depart the scene so it can't be repeated.

I can see how repeated back and forth motion of the same ball via alternating grabs by each skater would move them gradually closer together. The problem, however, is that the 'pull' from each grab is greatest the farther apart they are (so long as the grabber is within 'arms length' of the ball) and gets smaller the closer they get.

Nope again. Look at the law of conservation of momentum. If the exchange is violent enough, the momentum transfer can be as high as you like.

But that momentum lasts only until the ball stops with the grabber. The faster the exchange, the shorter the grabber's motion lasts. Bottom line is that the centre of mass of the grabber and ball cannot change on each grab.

The particles aren't holding onto the quanta with anything. Charged objects are sources of quanta. But as a side note, it is true in the case of the strong force that the quanta are also charged. But that is not necessary for the quanta to mediate the interaction, as demonstrated in the EM case.

I can conceive of virtual particles being exchanged between quarks within a nucleus a lot easier than I can conceive of virtual photons being exchanged over huge distances in an electric/magnetic field. Is it just my lack of imagination? How is it that the force varies as 1/r^2?

Andrew Mason

 

[Andrew Mason]

And if you know this already, then this ring a bell that CLEARLY should have told you that a proton-neutron set up isn't as stable as a proton. Please note that originally, you are indicated that this set up is stable, as if it is the preferred ground state of a nuclei.

That was not what I said nor what I was suggesting. I was using the pairing of a neutron and proton because, while it has a significant binding energy (which makes the configuration endure for long periods of time - hence stable), when it degenerates into separate particles there is a loss of mass. I was suggesting that we might think of the high binding energy of a neutron/proton pair as a 'resistance' to the 'unravelling' of matter rather than as a 'force' times 'distance'.

Andrew Mason

 

[quantumdude]

There is no question that the observed binding energies in the nucleus are dozens of orders of magnitude greater than the gravitational binding energy of nuclear particles calculated using classical gravitational laws.

OK, then what is it that prompts you to ask whether gravity can be responsible for nuclear binding? Clearly, nuclear stability cannot be accounted for by gravity as we know it. So you must be thinking of gravity as we don't know it.

Where does this idea come from?

How is it necessitated by any observational evidence?

Andrew: Or perhaps hockey players grabbing a puck ... but I appreciate your analogy. As I understand your analogy, the skaters would only get closer together if the other was still holding onto the ball.[/color]

Tom: Nope. That's why I put them on ice skates.

Andrew: Hockey players wear skates.

My remark was directed at the blue[/color] part.

Because the "particles" are on ice skates, they continue their motion, even if the ball is dropped after the exchange.

Tom: Even if the ball was dropped and rolled away, the skaters would move together, once the exchange had taken place.

Andrew: I don't follow that. The momentum imparted by a particle is always in the direction of its motion. If the ball was pulled from the other skater and thrown back over the 'grabber's' head, they would move together because the grabbing skater would gain momentum in an equal and opposite direction to the ball. But the ball would depart the scene so it can't be repeated.

But they are on ice skates[/color], so the momentum that was imparted by the exchange does not change, even if no subsequent exchanges take place. And I didn't say anything about throwing the ball after the exchange, I said that even if the ball was dropped, the two would still be attracted. Of course, the exchange cannot be repeated as you note, but that is beside the point. You said that the two skaters would not move towards each other unless one skater was holding the ball, and that is incorrect.

I can see how repeated back and forth motion of the same ball via alternating grabs by each skater would move them gradually closer together. The problem, however, is that the 'pull' from each grab is greatest the farther apart they are (so long as the grabber is within 'arms length' of the ball) and gets smaller the closer they get.

But that momentum lasts only until the ball stops with the grabber. The faster the exchange, the shorter the grabber's motion lasts. Bottom line is that the centre of mass of the grabber and ball cannot change on each grab.

No. The exchange only has to happen once, and the attraction would persist.

Once again, with emphasis: That's why I put them on ice skates.

The two skaters do not stop dead in their tracks once the ball is no longer being exchanged.

I can conceive of virtual particles being exchanged between quarks within a nucleus a lot easier than I can conceive of virtual photons being exchanged over huge distances in an electric/magnetic field. Is it just my lack of imagination?

Yes.

How is it that the force varies as 1/r^2?

The inverse square law is derivable from QFT. In Zee's book, QFT in a Nutshell, he derives it in the first chapter.

However, you need to seek out some education in physics at the basic level first. You said that your next stop is Intro to QFT. That's an admirable sentiment, but you are just not ready for it. That much is apparent from the way you are struggling with the momentum issue with the ice skaters. You should revisit classical mechanics first, and then classical EM theory, then quantum mechanics.

And then try QFT on for size.

 

[marlon]

Andrew,

I suggest you follow the advice of Tom on the QFT-thing here, otherwise it is going to be very hard and even quasi impossible for you to grasp the basic notions of QFT. You not only need to understand the algorithms and "calculus" used in QFT, you also need to see how and why QFT and the fields in particular are introduced. There is a very profound history and reason for this and not knowing this is like studying General Relativity without knowing the concepts of Newtonian physics. It just won't work.

QFT is very hard to learn, yet it is the most interesting and useful model of physics that we have up till now. In my opinion even the LQG and String-theories are Spielerei (this means : "toys") compared to the "maturity" that QFT has reached. Learning this too fast will have the risk of dropping out very soon because you just won't get the point...

just some advice, don't take this the wrong way...
marlon