Gravity in the Nucleus [Archive]

Andrew Mason

Sep27-04, 02:18 PM

Of course... That might be what Nereid meant !

Beside, you cannot only compare an acceleration to a distance, at first it does not make sens.

What is important, it seems to me, is the time required for significant changes in separation to occur. When you are contemplating protons and neutrons in the nucleus, the separation distances are very small. While the earth produces much greater gravitational acceleration at its surface than a proton does at its surface, the time required for signficant changes in separation to be reduced by gravity is much greater:

Example:
For an object that is .01 earth radius above the earth (about 60 km), the time required to return to the surface (ignoring friction) is:

t = \sqrt{2s/g}

g \approx 10 m/sec^2

t = \sqrt{2 \times 120 \times 10^3/10} = 154 seconds

For a proton separated from a neutron by .01 radius of a neutron, the time required to return to the surface of the neutron is:

\therefore t = \sqrt{5 \times 10^{-18}/4 \times 10^{-7}} = 3.5 \times 10^{-5} sec.

(s = .01 \times radius of neutron = 5 \times 10^{-18}m. )

( a \approx 4 \times 10^{-7} m / sec^2)

You need either to know either a speed, or a time scale. Gravity must come out negligible in any case. Besides, the strong interaction is named because it is even larger than the EM interaction.

If we are interested in identifying what keeps the nucleus together (as opposed to what keeps a proton together) the essential question is: what are the forces that work against it?

The magnitude of gravity within the nucleus may be small by comparison to the EM interaction, but how do we know that the EM interaction applies within the nucleus? If it does, then obviously gravity would not be sufficient to keep the nucleus together. But I am not sure that it does. I was hoping someone out there might be able to explain why gravity is not sufficient.

Andrew Mason

humanino

Sep27-04, 02:23 PM

No, gravity by itself is far not sufficient to overcome EM repulsion. It is only the residual strong interaction potential which keeps the nucleons together. Have you calculated EM repulsion ?

Andrew Mason

Sep27-04, 03:34 PM

No, gravity by itself is far not sufficient to overcome EM repulsion. It is only the residual strong interaction potential which keeps the nucleons together. Have you calculated EM repulsion ?
I never said gravity by itself was sufficient to overcome EM repulsion. It is dozens of orders of magnitude smaller. But I am not assuming that it has to overcome EM repulsion in order for the nucleus to stay together (ie. once it is together).

If you put H nuclei together create He, one has to use alot of energy to overcome the EM repulsion (ie. it requires the energy inside a star). But once fusion occurs, do we know that the EM repulsion continues to operate between protons in the nucleus? That would be my question.

EM repulsion certainly doesn't continue when two protons fuse to produce deuterium (and emit a positron). Does EM repulsion continue when an extra proton is added to the nucleus?

Andrew Mason

zefram_c

Sep27-04, 03:41 PM

Andrew Mason

Sep27-04, 04:19 PM

Of course EM repulsion continues within the nucleus. That is why heavier nuclei with a lot of protons are unstable - the repulsion is stronger than the nuclear attractive forces. That's also why no elements with more than 100 protons exist in nature - they're very unstable.
That doesn't necessarily mean that EM repulsion continues within the nucleus of He. It may be that EM forces have a minimum range: they do not apply within a region of space that is larger than a He nucleus but smaller than a nucleus of Einsteinium.

I am not saying this is actually the case. I am wondering if anyone can explain why it is not the case.

Andrew Mason

Nereid

Sep27-04, 04:55 PM

Andrew Mason

Sep28-04, 08:24 AM

I think you've heard of scattering experiments Andrew, in which a beam of electrons (or protons) hits a target of protons (H nuclei); if you read up on those, I think you'll find that there's very clear experimental data to show that the EM force doesn't weaken at short distances; at least to the experimental limit.

Scattering experiments are, for the most part, done with electrons, not protons. The result is that we do not measure electrical repulsion between protons. Perhaps you can explain to me how electron scattering shows that EM force exists within the nucleus and, in particular, that proton-proton repulsion exists within the nucleus.

If electron-proton EM attraction exists to within a very small distance from the nucleus, why is there not some electron energy at which the electron reaches the nucleus but cannot escape it - ie it joins the nucleus? The quantum mechanical explanation of the electrons in the atom based on the uncertainty principle may be quite correct but begs the question: does EM force really have any meaning when elementary charged particles get very close?

Andrew Mason

ZapperZ

Sep28-04, 08:46 AM

If electron-proton EM attraction exists to within a very small distance from the nucleus, why is there not some electron energy at which the electron reaches the nucleus but cannot escape it - ie it joins the nucleus? The quantum mechanical explanation of the electrons in the atom based on the uncertainty principle may be quite correct but begs the question: does EM force really have any meaning when elementary charged particles get very close?

Andrew Mason

Andrew,

When you look at the Schrodinger equation for the hydrogen atom, for example, what do you think that "V" term is?

Furthermore, you don't use the "uncertainty principle" to accurately solve for the energy states, etc. of an atom.

As for electron not reaching the nucleus, who said that? There is a difference between the BOUND STATE solution of an atom, where there is a minium, ground state for an electron-atom system, and free electron colliding with a bare proton/nucleus, which can induce an inverse beta decay! In the latter case, you CAN have an electron capture with the appropriate momentum conservation condition.

Zz.

Nereid

Sep28-04, 08:50 AM

Scattering experiments are, for the most part, done with electrons, not protons. The result is that we do not measure electrical repulsion between protons. Perhaps you can explain to me how electron scattering shows that EM force exists within the nucleus and, in particular, that proton-proton repulsion exists within the nucleus.

If electron-proton EM attraction exists to within a very small distance from the nucleus, why is there not some electron energy at which the electron reaches the nucleus but cannot escape it - ie it joins the nucleus? The quantum mechanical explanation of the electrons in the atom based on the uncertainty principle may be quite correct but begs the question: does EM force really have any meaning when elementary charged particles get very close?

Andrew MasonYou might like to google on 'proton-proton scattering'; it would seem that there have been quite a few such experiments, over a wide range of energies.

I'll leave it to a PF member more familiar with this work than I am to say something about the results of scattering experiments wrt your idea that the EM force may have a different behaviour either in nuclei or over short ranges (or both).

humanino

Sep28-04, 09:24 AM

We use quantum electrodynamics to probe the proton structure at very small distance. For instance at 6 GeV, we are under 2\times10^{-17}m.
If course if you insist in saying "what about thousand times smaller than any actual accessible distance" we would have to give up.

Andrew Mason

Sep28-04, 03:40 PM

Andrew,
When you look at the Schrodinger equation for the hydrogen atom, for example, what do you think that "V" term is?
The Schrodinger equation provides an accurate mathematical model for the quantum mechanical behaviour of the atom, in which EM potential is obviously important. Inside the nucleus may be another matter.

Furthermore, you don't use the "uncertainty principle" to accurately solve for the energy states, etc. of an atom.
I agree. But one does use it to explain why the electron doesn't simply 'fall' into the nucleus due to EM attraction.

As for electron not reaching the nucleus, who said that? There is a difference between the BOUND STATE solution of an atom, where there is a minium, ground state for an electron-atom system, and free electron colliding with a bare proton/nucleus, which can induce an inverse beta decay! In the latter case, you CAN have an electron capture with the appropriate momentum conservation condition.
But it is quite rare and it is not stable. One might think (classically) that the EM attraction would bring it into the nucleus and keep it there, if EM attraction was that strong inside the nucleus.

It is assumed that strong nuclear attractive force works against the coulomb repulsion force that exists between protons. This means that the nuclear force is strong only in the region very close to the nucleus. I am looking for the evidence that this is in fact the case. I am suggesting that the same result would occur if the coulomb force had a minimum range (ie did not operate inside a certain the region close to the proton) so that the force which keeps protons together is something much weaker. I am suggesting, if that is the case, that gravity might actually be the dominant force inside the nucleus.

Andrew Mason

ZapperZ

Sep28-04, 04:31 PM

It is assumed that strong nuclear attractive force works against the coulomb repulsion force that exists between protons. This means that the nuclear force is strong only in the region very close to the nucleus. I am looking for the evidence that this is in fact the case.

I'm sorry, but is this still in doubt? A nucleus consists of a bunch of positively charged protons, and neutral neutrons. If there's nothing else that not only counter the coulombic repulsion, but also is way stronger than the coulombic repulsion, don't you think the nucleus would fly apart?

I am suggesting that the same result would occur if the coulomb force had a minimum range (ie did not operate inside a certain the region close to the proton) so that the force which keeps protons together is something much weaker. I am suggesting, if that is the case, that gravity might actually be the dominant force inside the nucleus.

Andrew Mason

You can suggest anything you like, but without (i) a self-consistent theory and/or (ii) experimental impetus to suggest that, then you might as well propose that bored angels in their spare time pushes the nucleons together. To allow for what you are proposing, you have to rewrite the whole of Maxwell Equations, since the 1/r potential obviously have to be corrected.

What I'm puzzled with is that there ALREADY is a verified, consistent explanation/description for the strong force. What is WRONG with it that is causing you to come up with a whole new speculation on why nucleons can stick together in spite of the coulombic force? Did you find a flaw in the Glashow/Salam/Weinberg model that is causing you to refute their theory? Are you proposing that QCD be dumped in favor of your "gravity"?

Zz.

marlon

Sep28-04, 05:04 PM

Andrew,

I suggest you need to look at the socalled "BETA-FUNCTION" of the strong-force-coupling constant AND the fact it is negative. You know : the famous asymptotic freedom...

As you will know it makes sure that a proton (just like any other baryon) is built out of three quarks with a different colour each so the colour-neutrality is always respected. The EM-processes are much much weaker when looked at some nucleus at quark-scale and this biggest problem you would have is there is no short-range for EM. The mediators will never acquire mass through the Higgs-mechanism.

QCD predicts that the interquark-potential is linear in the long range. So this basically means that in the vacuum state, the quarks will tend to form doublets or triplets...ie mesons and baryons.

I don't really see your problem with a theory that is already well established (apart from the quark-confinement ofcourse) and experimentally verified...

Indeed, in the short range the strong force becomes repulsive, yet this effect is much smaller compared to the i) long range attractive part of the strong force and ii) the attractive residual strong force mediated by the pions...which have mass and thus describe a short range "attractive" interaction. Why attractive? Well, because of the omnipresent colour-neutrality. Just look at how pions are created via the screening-effect in QCD...

regards
marlin

ZapperZ

Sep28-04, 05:08 PM

As you will know it makes sure that a proton (just like any other fermion) is built out of three quarks with a different colour each so the colour-neutrality is always respected.

Er.. I think you mean ".. like any other baryon...", not fermion, especially if you are giving a "three quark" content example.

Zz.

marlon

Sep28-04, 05:17 PM

Er.. I think you mean ".. like any other baryon...", not fermion, especially if you are giving a "three quark" content example.

Zz.

Yes indeed, Zz. Thanks for the correction...my mistake...

marlon

Andrew Mason

Sep28-04, 05:31 PM

You can suggest anything you like, but without (i) a self-consistent theory and/or (ii) experimental impetus to suggest that, then you might as well propose that bored angels in their spare time pushes the nucleons together. To allow for what you are proposing, you have to rewrite the whole of Maxwell Equations, since the 1/r potential obviously have to be corrected.

Well, at some point the 1/r potential breaks down because the proton is not a point charge - it has a finite size. You are not suggesting that the potential goes to - infinity inside the proton are you? I am just suggesting that it might break down in a region that is outside the proton 'surface'.
What I'm puzzled with is that there ALREADY is a verified, consistent explanation/description for the strong force. What is WRONG with it that is causing you to come up with a whole new speculation on why nucleons can stick together in spite of the coulombic force? Did you find a flaw in the Glashow/Salam/Weinberg model that is causing you to refute their theory? Are you proposing that QCD be dumped in favor of your "gravity"?
I am just questioning the existing theory is correct. I am asking what evidence we have that there is a strong nuclear force.

So far the explanation has been, 1. protons repel protons with enormous EM force (\propto 1/r^2) that continues down to the 'surface of the proton'; 2. the nucleus consists of protons having a separation less than the radius of a proton and the nucleus does not fly apart. 3. Therefore there must be a strong nuclear force that is much greater than the coulombic repulsion forces but which operates only in the region of the nucleus.

3 necessarily follows from 1 and 2. I am just asking what evidence we have that 1 is correct.

Andrew Mason

humanino

Sep28-04, 05:39 PM

Well, at some point the 1/r potential breaks down because the proton is not a point charge - it has a finite size.
Even within classical mechanics, this does not imply failure of the EM. Right the 1/r potential of the proton fails, but why would it be so for the point-like constituants ? From far away the proton looks like a point, and when you get near, you can see it is a ball, and if you get near enough, you can actually see the sub-structure. Everything in accordance with EM, or its quantum version necessary to describe the scattering process. But still, it is EM.

humanino

Sep28-04, 05:43 PM

1) is wrong
Please read this thread about nuclear interactions.

ZapperZ

Sep28-04, 06:00 PM

Well, at some point the 1/r potential breaks down because the proton is not a point charge - it has a finite size. You are not suggesting that the potential goes to - infinity inside the proton are you? I am just suggesting that it might break down in a region that is outside the proton 'surface'.

But you just answered your own question. If it is NOT a point charge, then you'll never get to an infinite potential. So what's the problem?

I am just questioning the existing theory is correct. I am asking what evidence we have that there is a strong nuclear force.

Do you even KNOW what the "existing theory" is, i.e. have you studied QFT, QED, and QCD? Or is this questioning simply based on ignorance that you acquired via reading pop-science books?

So far the explanation has been, 1. protons repel protons with enormous EM force (\propto 1/r^2) that continues down to the 'surface of the proton'; 2. the nucleus consists of protons having a separation less than the radius of a proton and the nucleus does not fly apart. 3. Therefore there must be a strong nuclear force that is much greater than the coulombic repulsion forces but which operates only in the region of the nucleus.

3 necessarily follows from 1 and 2. I am just asking what evidence we have that 1 is correct.

Andrew Mason

The discovery of the quarks AND the verification of the hirerchy of the quark model ARE the evidence of the strong force! QCD includes ALL the strong interactions and decay channels that make predicitons on what and where to look in a particle collider.

Zz.

Andrew Mason

Sep28-04, 06:14 PM

Even within classical mechanics, this does not imply failure of the EM. Right the 1/r potential of the proton fails, but why would it be so for the point-like constituants ? From far away the proton looks like a point, and when you get near, you can see it is a ball, and if you get near enough, you can actually see the sub-structure. Everything in accordance with EM, or its quantum version necessary to describe the scattering process. But still, it is EM.
I am not the first to suggest that classical EM theory breaks down at the atomic level (eg. Planck's solution to the ultra-violet catastrophe).

We know that the proton, as with all elementary particles, can be expressed as a wave function. This suggests that when we get down to the regions of the 'surface' of the proton, things get fuzzy.

Within that fuzzy region, we cannot assume that electro-magnetic forces follow classical laws. Since the entire nucleus appears to be within the 'fuzzy region' and since the 'evidence' of the strong force seems to be an inference based on the assumption that enormous EM repulsion exists within the nucleus, I am questioning whether the strong force is real.

So I just ask the question: what evidence do we have for the strong nuclear force that is independent of any assumption that strong EM repulsion forces operate between protons within the nucleus? I am not suggesting it doesn't exist. I am just not aware of it.

Andrew Mason

Andrew Mason

Sep28-04, 06:39 PM

But you just answered your own question. If it is NOT a point charge, then you'll never get to an infinite potential. So what's the problem?
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.

Do you even KNOW what the "existing theory" is, i.e. have you studied QFT, QED, and QCD? Or is this questioning simply based on ignorance that you acquired via reading pop-science books?
I am just asking questions. All questions are based on ignorance. Otherwise, why ask the question?

I don't pretend to have more than a rudimentary grasp of quantum theory. I studied it as an undergraduate in physics but that was many years ago. And I ended up as a lawyer.

The discovery of the quarks AND the verification of the hirerchy of the quark model ARE the evidence of the strong force! QCD includes ALL the strong interactions and decay channels that make predicitons on what and where to look in a particle collider.
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

humanino

Sep28-04, 06:42 PM

I am not the first to suggest that classical EM theory breaks down at the atomic level (eg. Planck's solution to the ultra-violet catastrophe).
How old is that ?We know that the proton, as with all elementary particles, can be expressed as a wave function. This suggests that when we get down to the regions of the 'surface' of the proton, things get fuzzy. It would be meaningless to describe the proton as a wave by itself. This wave is a tensor product of the waves of the constituants.
Within that fuzzy region, we cannot assume that electro-magnetic forces follow classical laws. Since the entire nucleus appears to be within the 'fuzzy region' and since the 'evidence' of the strong force seems to be an inference based on the assumption that enormous EM repulsion exists within the nucleus, I am questioning whether the strong force is real. We will soon be able to provide a snapshot of the interior of the proton. Total information on the content, that is the Wigner pseudo-probability distribution in phase space. The strong force is real there is definitely no doubt.So I just ask the question: what evidence do we have for the strong nuclear force that is independent of any assumption that strong EM repulsion forces operate between protons within the nucleus? I am not suggesting it doesn't exist. I am just not aware of it. Classification of hundreds of particles : the hadrons. The zoo of strongly bound states is far from random. It obeys symmetry. Those symmetries are beautifully interpreted in the standard-model. Besides, there is no such thing as a "strong EM repulsion". It is weak as compared to the strong force.

ZapperZ

Sep28-04, 06:46 PM

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

Sep28-04, 07:20 PM

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 text books 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

Sep28-04, 07:30 PM

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

Sep28-04, 08:53 PM

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

Sep29-04, 12:54 AM

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

Sep29-04, 06:39 AM

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

Sep29-04, 11:17 AM

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

Sep29-04, 11:33 AM

Please read this thread about nuclear interactions.
Andrew, I think you did not take time to read this thread which is short, and which might help you asking questions.

Tom Mattson

Sep29-04, 08:27 PM

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

Sep30-04, 03:00 AM

Andrew, I think you did not take time to read this 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

Sep30-04, 04:52 AM

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 equasions 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 equasions, what are the SI (International Standard) units for \propto_e and \propto_g?

Andrew, based upon these equasions, 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 equasions 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

Sep30-04, 09:10 AM

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.

Tom Mattson

Sep30-04, 10:59 AM

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

Sep30-04, 11:44 AM

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

Sep30-04, 12:11 PM

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) (http://mathworld.wolfram.com/TorsionTensor.html)
T(u,v)=\nabla_uv-\nabla_vu-[u,v]
and it is assumed zero in Einstein's theory :
John Baez on zero torsion (http://math.ucr.edu/home/baez/gr/torsion.html)

Nereid

Sep30-04, 12:11 PM

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. :wink: ), 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

Sep30-04, 03:33 PM

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 :wink: 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

Sep30-04, 04:59 PM

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

Sep30-04, 05:23 PM

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

Sep30-04, 05:50 PM

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

Sep30-04, 06:54 PM

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

Tom Mattson

Sep30-04, 06:55 PM

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

Sep30-04, 07:00 PM

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

Sep30-04, 10:58 PM

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

Oct1-04, 01:12 AM

ZapperZ

Oct1-04, 06:08 AM

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 prefered 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

Oct1-04, 07:51 AM

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 alot 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

Oct1-04, 08:21 AM

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 prefered 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

Tom Mattson

Oct1-04, 03:57 PM

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.

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

Andrew: Hockey players wear skates.

My remark was directed at the blue 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, 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 alot 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

Oct1-04, 05:16 PM

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... :wink:
marlon

Andrew Mason

Oct1-04, 07:23 PM

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?
I am not saying that gravity accounts for nuclear stability. Obviously the force of gravity is not sufficient to account for the binding energy of a proton and neutron (ie. the energy required to unbind them). I am suggesting that the explanation for this energy barrier might not require the existence of a mysterious nuclear force at all. If that is the case, then gravity might be the only force in the nucleus.

An analogy would be to very dense and heavy ball bearings (pretend they are made from neutron star matter) sitting on a level frictionless surface at the bottom of a deep earth well. Their own gravity would attract them to each other and keep them together but is not the force that keeps them in the well. I am suggesting that there may be some energy barrier that keeps the nucleons from leaving the region of the nucleus, but that it is not a force x distance energy barrier (that is where the analogy ends, of course, because the energy barrier that keeps the ball bearings in the well is earth gravity x height of the well).

How is it necessitated by any observational evidence?
It isn't necessitated. The question is whether observational evidence can have an alternate explanation. In case you haven't noticed, I don't like the strong nuclear force.

My remark was directed at the blue part.
I knew that. I was being facetious. I also play hockey.

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

But if the ball drops, it is not moving toward the skater (the grabber). That means the skater has stopped the ball. Since the momentum of the ball in the direction of the skater has to equal to the momentum of the skater in the direction of the ball in the original frame of reference (ie. before the grab), the skater stops moving. The forward momentum of the skater has to equal the backward momentum of the ball. When the ball stops, the skater stops.

But they are on ice skates, so the momentum that was imparted by the exchange does not change, even if no subsequent exchanges take place.
Do you not agree that in the original rest frame, the position of the center of mass of the skater and the ball cannot change? So either the ball keeps moving past the skater's back (ie he throws it behind him) and the skater keeps moving forward toward the other skater, or the ball and the skater stop.

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.
That is what I originally said because, as I explained, I didn't think you were relying on transfer of momentum because unless the ball was very heavy the skater would not move very far toward the other skater.

In my subsequent post I said that the skater would move toward the other skater a little bit and then stop. I said: "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." And I went on to take issue with your suggestion (perhaps I misunderstood) that once the skater began moving as he pulled the ball towards himself, he would continue moving toward the other. I said that he would only continue until the ball reached him and stopped. I still stand by that. I said "But that momentum lasts only until the ball stops with the grabber"

No. The exchange only has to happen once, and the attraction would persist.
I guess I don't know what you mean by 'attraction'. You cannot mean 'motion' because once the ball has stopped, the skater has to stop. (Or are you suggesting that the skaters are skating as well? {that was a joke})

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.
Whatever happens on that frictionless ice surface, the center of mass of the 2 skater and ball system cannot move. I think we have to agree on that.

If one skater grabs the ball and brings it toward him and the ball stops, that skater must stop. If the other skater then grabs that ball and brings it to a stop against his chest, that skater moves a small distance toward the other and then stops.

If the skater who pulls the ball towards him never stops the ball (because the other skater grabs it back before it reaches his chest) and this is kept repeating, the two skaters will continue to move together. But I didn't think that was what you were saying because I thought you said the motion of the first skater to grab the ball would continue even if the ball was dropped (ie. after the first grab).

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.

I am not struggling with momentum at all. I am struggling with your example. I assure you I have no problem with basic physics. I studied physics from 1972-1976. That was a long time ago. I have't heard that the principle of conservation of momentum had changed since then.

You should revisit classical mechanics first, and then classical EM theory, then quantum mechanics.

And then try QFT on for size.
I don't need to revisit classical mechanics. I am rereading my 4th year quantum mechanics text and my 2nd year EM text. I appreciate that you are trying to be helpful, but I think that we are just misunderstanding each other's posts here.

Andrew Mason

Andrew Mason

Oct1-04, 08:42 PM

just some advice, don't take this the wrong way... :wink:
marlon
I do appreciate the advice. I know that the contributors to this board have the best of intentions and I am very grateful for the opportunity to discuss these things. It has spurred my interest in revisiting the difficult areas of physics that 30 years ago drove me into law. Law makes a wonderful career, mind you, but it isn't physics.

Andrew Mason

humanino

Oct1-04, 09:33 PM

I am not saying that gravity accounts for nuclear stability. Obviously the force of gravity is not sufficient to account for the binding energy of a proton and neutron (ie. the energy required to unbind them). I am suggesting that the explanation for this energy barrier might not require the existence of a mysterious nuclear force at all. If that is the case, then gravity might be the only force in the nucleus.

:confused: :bugeye:
The question is whether observational evidence can have an alternate explanation.
The answer seems no, because the strong interaction applications are so numerous, and all in agreement with a very very simple, minimal assumption. Any other explanation would be more complicated. Occam razor.

I don't like the strong nuclear force.
I have contempt for lawyers. But I have no idea what being a lawyer could be so I am polite enough not to tell them, especially when I need their help.

we are just misunderstanding each other's posts here.
I doubt that we misunderstand yours.

humanino

Oct1-04, 10:20 PM

But if the ball drops, it is not moving toward the skater (the grabber). That means the skater has stopped the ball. Since the momentum of the ball in the direction of the skater has to equal to the momentum of the skater in the direction of the ball in the original frame of reference (ie. before the grab), the skater stops moving. The forward momentum of the skater has to equal the backward momentum of the ball. When the ball stops, the skater stops.

This skater analogy has limits, it can not work so well at those scales, and will eventually run into difficulties. It was merely : the exchange can lead to either attraction or repulsion. But digging into it, I am not certain you will gain understanding in the nuclear interaction.

In my opinion, one should accept at some point, the historical developpement : if you are dealing with the proton/neutron structure, the first step is to understand the models which are not derived from fundamental laws. They are "intuited" with
1 classical analogies (as the skater, just closer to the problem)
2 quantum constraints
The reason one has to study those model : they work very well, for an approximative description. At the beginning, we had only innaccurate data, and those models were sufficient. And then, if you want to rigourously derive them, it is impossible. In order to gain understanding in the nuclei, you need to understand the meaning and the values of the parameters of the models.

Do you not agree that in the original rest frame, the position of the center of mass of the skater and the ball cannot change? So either the ball keeps moving past the skater's back (ie he throws it behind him) and the skater keeps moving forward toward the other skater, or the ball and the skater stop.

You are running in the difficulties. This is not a ball exchanged. In fact, there are two parts : a scalar and a vector part. Those are intrisically mixed. They lead to a subtle balance between repuslion and attraction.

I am not struggling with momentum at all. I am struggling with your example. I assure you I have no problem with basic physics. I studied physics from 1972-1976. That was a long time ago. I have't heard that the principle of conservation of momentum had changed since then.

See : this is mean. You know you are playing with an analogy. Do you only want to hear ? : at some point, any analogy fails. It only allows to communicate without equations, in order to emphasize a property. You extract the substance of a calculus, and then you think of an analogous phenomena, and you say "ok, remind in this simple case, it is the same ! :smile: " But you go "yes but also, if the ball is not inflated enough ... :tongue2: " This is not serious this skater thing. You are going nowhere with that.

Chiral Symmetry: Pion-Nucleon Interactions in Constituent Quark Models (http://arxiv.org/abs/hep-ph/9612230)
Nucleon-nucleon potentials from nonlinear field theory (http://smithers.physnet2.uni-hamburg.de/NN/NNnonlin.html)

Andrew Mason

Oct1-04, 11:21 PM

This skater analogy has limits, it can not work so well at those scales, and will eventually run into difficulties. It was merely : the exchange can lead to either attraction or repulsion. But digging into it, I am not certain you will gain understanding in the nuclear interaction.

I will go further than that and say that I am certain I won't.

See : this is mean. You know you are playing with an analogy. Do you only want to hear ? : at some point, any analogy fails. It only allows to communicate without equations, in order to emphasize a property. You extract the substance of a calculus, and then you think of an analogous phenomena, and you say "ok, remind in this simple case, it is the same ! :smile: " But you go "yes but also, if the ball is not inflated enough ... :tongue2: " This is not serious this skater thing. You are going nowhere with that.
I certainly did not intend to be mean and I apologize if it sounded like I was. But the analogy was yours. Conservation of momentum is not a minor detail. In order to have an exchange of gluons creating an enormous attractive force through momentum transfer, one would require some kind of imaginary negative momentum. I am not the only one who has had problems with the concept.

But don't give up on me just yet. I may surprise you with what I really do understand.

Andrew Mason

Orion1

Oct1-04, 11:48 PM

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...

Can Deuterium Binding Energy be described as existing inside a Yukawa Potential Well?

Deuterium Binding Energy:
E_b = ((m_p + m_n) - m_D) E_n

Yukawa Potential Well:
U_y = f^2 \frac{e^{-\frac{r_D}{r_0}}}{r_D}

E_b = U_y

((m_p + m_n) - m_D) E_n = f^2 \frac{e^{- \frac{r_D}{r_0}}}{r_D}

f_D^2 = ((m_p + m_n) - m_D) E_n r_D e^{\frac{r_D}{r_0}}

\boxed{f_D = \sqrt{((m_p + m_n) - m_D) E_n r_D e^{\frac{r_D}{r_0}}}}

Key:
E_n = 931.5 \; Mev*amu^{-1} - mass-energy equivalence
m_p - Proton mass
m_n - Neutron mass
m_D - Deuterium mass
r_D - Deuterium nuclear radius
r_0 - Yukawa nuclear range

humanino

Oct2-04, 12:43 AM

But don't give up on me just yet. I may surprise you with what I really do understand.

I trust you Andrew, you must certainly understand very well parts of physics apart from nuclear theory.

Conservation of momentum is not a minor detail. In order to have an exchange of gluons creating an enormous attractive force through momentum transfer, one would require some kind of imaginary negative momentum. I am not the only one who has had problems with the concept.

This is a very good question indeed, and I have not been able to figure out a satisfactory answer yet. I will work on it, and probably benefit much from the other posts here.

But your argument is not only meant against QCD : two opposite electric charges attract by exchanging photons.

Andrew Mason

Oct2-04, 01:50 AM

The answer seems no, because the strong interaction applications are so numerous, and all in agreement with a very very simple, minimal assumption. Any other explanation would be more complicated. Occam razor.
I'll have to take your word for it at the moment because I really don't understand it in the depth required to have an informed opinion. At this stage I am simply asking questions.

But let me just say that Occam's Razor is a bit of a cop out. Occam's Razor just a generalization, not a principle of science. It does not apply, for example, to the machinery of the cell - the simplest explanation is hardly ever the way it really works. It doesn't apply to General Relativity compared to Newton. Newton was an advocate of it as a principle but as subsequent developments have shown, things were not always as simple as he thought. It applies best in situations where there is more than one possible explanation and one has to determine the most probable one. If it walks like a horse, talks like a horse and smells like a horse then it is probably a horse, not a zebra - (unless, perhaps, you are in the African savannah).

I have contempt for lawyers. But I have no idea what being a lawyer could be so I am polite enough not to tell them, especially when I need their help.
Nelson Mandela is a lawyer. Mahatma Gandhi was a lawyer. Franklin D. Roosevelt and Abraham Lincoln were lawyers. You have contempt for them? Unfortunately, we are bound by our professional oath to defend people and causes that many people dislike. So we learn to develop a thick skin - a very useful thing to have in this forum, I have found.:smile:

Andrew Mason

anti_crank

Oct2-04, 03:38 AM

I'll have to take your word for it at the moment because I really don't understand it in the depth required to have an informed opinion. At this stage I am simply asking questions. You are welcome to ask questions. However, the other question is what to do with the answers you've gotten. This thread has many of the trademarks of the old TD forum; however, the scientific discussion has managed to stay above crackpot material for the time being.

We've given you plenty of reasons, complete with calculations, why gravity is irrelevant in the nucleus and why the strong force is needed. The strong force has been tested in many other ways. It successfully predicts meson structure and lifetimes. The existence of gluons has received huge support from three-jet events. On the theoretical side, the color quantum number is required for anomaly cancellation in the GWS model . Hence without colored quarks, the weak force theory would break down, and a large number of particle physics experiments would be open problems. The italicized terms are terms you might try to do Google searches on, since I don't have the time to explain those things.

Now you suggest that gravity and/or the EM force might behave differently on nuclear scales. Experimental evidence refutes this. For example, if Maxwell's equations were to break down at that scale, EM waves of wavelengths close to the size of nuclei (1fm) should behave differently than other EM waves. This corresponds to an energy of about 1.2 GeV, and experimental data is available that confirms QED to much greater energies (smaller scales).

As for gravity, this may sound like a catch-22 to you. We believe gravity is so weak on nuclear scales that its effects are impossible to measure. We attribute everything we observe to EM/strong/weak interactions. You might say we mistakenly attribute the new gravitational effects to the fictional strong force. But you must keep in mind that when accurate data is available, those models can be tested to great accuracy, and so far they passed every test. Therefore if you want to overthrow the strong force, it is not enough to make broad suggestions about gravity holding the nucleaus together. Rather, you must provide a mathematical description of gravity that reproduces all those effects, then proceed to solve the other problems that the absence of strong nuclear forces would create.

At this point, you might argue that you don't have the training, the knowledge or the skills to make such a model. In that case, and with all due respect, please leave the physics to the physicists. I like to think that no physicist would tell you how to do your job, and I ask for the same professional courtesy from you. We do not have contempt for lawyers; I suspect humanimo was simply offended by your statement that you "don't like" the strong force. More on that later.
But let me just say that Occam's Razor is a bit of a cop out. Occam's Razor just a generalization, not a principle of science. It does not apply, for example, to the machinery of the cell - the simplest explanation is hardly ever the way it really works. It doesn't apply to General Relativity compared to Newton. Newton was an advocate of it as a principle but as subsequent developments have shown, things were not always as simple as he thought. This was a completely different matter. The theories made different predictions, and experiments falsified Newton and supported Einstein.
It applies best in situations where there is more than one possible explanation and one has to determine the most probable one. Is that not exactly what we would have here? Assuming that you somehow manage to construct a model of gravity/EM that does what you want, it will undoubtedly be very complicated. Why should that be the most probable then? The current theories, with the strong force included, are simpler than anything thus constructed. Of all the branches of science, particle physics is especially driven towards greater simplicity, and this drive is what makes for most of today's effort in this field.
In case you haven't noticed, I don't like the strong nuclear force. This is an extremely unscientific attitude. We expect this from creationists who don't like the vast majorities of present theories, classical-physics mindsets who feel their worlds coming apart at the mere thought of relativity or quantum mechanics, and other crackpots who are fixated on one idea and cannot let go of it. In science, personal opinions are a luxury. Furthermore, personal opinions without some valid scientific or mathematical backing are not even science. This is not a good path you're heading down; consider this an attempt to pull you away from the edge.

Andrew Mason

Oct2-04, 10:20 AM

Now you suggest that gravity and/or the EM force might behave differently on nuclear scales. Experimental evidence refutes this. For example, if Maxwell's equations were to break down at that scale, EM waves of wavelengths close to the size of nuclei (1fm) should behave differently than other EM waves. This corresponds to an energy of about 1.2 GeV, and experimental data is available that confirms QED to much greater energies (smaller scales).
I am not suggesting that EM theory breaks down for EM waves with wavelengths smaller than 10^{-15} m.. I was just questioning whether we really have evidence that the coulomb force between protons continues as their 'separation' distance becomes smaller than 10^{-15} m.. That may have more to do with the structure of the proton than EM theory. EM theory assumes point charges and does not explain what creates the coulomb force.
As for gravity, this may sound like a catch-22 to you. We believe gravity is so weak on nuclear scales that its effects are impossible to measure. We attribute everything we observe to EM/strong/weak interactions. You might say we mistakenly attribute the new gravitational effects to the fictional strong force.
Hang on. I am not talking about new gravitational effects. I am not suggesting that some new gravity would account for the huge binding energy of nucleons. I am just questioning whether we have to use traditional concepts of force x distance to account for that binding energy.

At this point, you might argue that you don't have the training, the knowledge or the skills to make such a model. In that case, and with all due respect, please leave the physics to the physicists. I like to think that no physicist would tell you how to do your job, and I ask for the same professional courtesy from you.
Scientific experts try to tell me how to do my job all the time. The problem is that they are frequently wrong and it takes a lawyer (and usually another expert) to make a judge understand that their explanations are wrong.

The current theories, with the strong force included, are simpler than anything thus constructed. Of all the branches of science, particle physics is especially driven towards greater simplicity, and this drive is what makes for most of today's effort in this field.
My interest, and I think it has always been the goal of science, is in trying to understand nature in terms of more fundamental concepts. Biology explains macroscopic life forms in terms of more fundamental units of life: the cell. Molecular biology seeks to explain complex behaviour of cells in terms of fundamental processes (protein synthesis being governed by the machinery of RNA/DNA transcription and translation). Physical chemistry explains the composition of matter in terms of combinations of a finite number of kinds of atoms and the electromagnetic forces between atoms. Physics explains all atoms as combinations of nucleons and electrons.

Nuclear physics tries to explain protons and neutrons in terms of more fundamental particles: quarks. Neutons and protons consist of two different combinatons of three quarks held together by an exchange of virtual particles. At this point we have trouble maintaining a conceptual framework because we have to rely on elaborate mathematical models to describe and predict what is happening. As we probe deeper into the nucleus the models become more elaborate rather than simpler. I am suggesting that perhaps there is a more fundamental principle or some simpler explanation that will tie it all together. I don't think I am alone.

This is an extremely unscientific attitude. We expect this from creationists who don't like the vast majorities of present theories, classical-physics mindsets who feel their worlds coming apart at the mere thought of relativity or quantum mechanics, and other crackpots who are fixated on one idea and cannot let go of it. In science, personal opinions are a luxury. Furthermore, personal opinions without some valid scientific or mathematical backing are not even science.
I do agree with you there. I never said the fact that I didn't 'like' an explanation was a reason to reject the explanation. I just meant it was reason for me to keep trying to see if there is not another more fundamental explanation that fits the evidence.

Andrew Mason

Tom Mattson

Oct2-04, 01:37 PM

I am not saying that gravity accounts for nuclear stability.

If there is no strong force to provide the stability, then something else must do the job. It's as simple as that.

Obviously the force of gravity is not sufficient to account for the binding energy of a proton and neutron (ie. the energy required to unbind them). I am suggesting that the explanation for this energy barrier might not require the existence of a mysterious nuclear force at all. If that is the case, then gravity might be the only force in the nucleus.

There's nothing mysterious about the strong interaction. It's only mysterious to people who have not studied it.

An analogy would be to very dense and heavy ball bearings (pretend they are made from neutron star matter) sitting on a level frictionless surface at the bottom of a deep earth well. Their own gravity would attract them to each other and keep them together but is not the force that keeps them in the well. I am suggesting that there may be some energy barrier that keeps the nucleons from leaving the region of the nucleus, but that it is not a force x distance energy barrier (that is where the analogy ends, of course, because the energy barrier that keeps the ball bearings in the well is earth gravity x height of the well).

This is just silly. The energy well has to correspond to some potential. Don't you see that you're replacing the (well-understood) strong+weak+EM interaction with gravity and a mysterious energy well? Where does this well come from? God?

It isn't necessitated. The question is whether observational evidence can have an alternate explanation. In case you haven't noticed, I don't like the strong nuclear force.

You don't understand the nuclear force. This can be remedied with some study.

But if the ball drops, it is not moving toward the skater (the grabber). That means the skater has stopped the ball. Since the momentum of the ball in the direction of the skater has to equal to the momentum of the skater in the direction of the ball in the original frame of reference (ie. before the grab), the skater stops moving.

No. The if the ball is dropped after the exchange, then the skaters continue moving towards each other. There has to be another exchange to stop the skaters.

The forward momentum of the skater has to equal the backward momentum of the ball. When the ball stops, the skater stops.

Will you please work out the calculation? It's quite simple. You have a skater carrying a ball, and they are both moving at speed v. If the skater drops the ball, the skater does not stop. In fact, the skater moves faster, because of the reduced mass.

Do you not agree that in the original rest frame, the position of the center of mass of the skater and the ball cannot change?

Yes.

So either the ball keeps moving past the skater's back (ie he throws it behind him) and the skater keeps moving forward toward the other skater, or the ball and the skater stop.

No.

Please do a calucation, and you'll see that it does not work like this.

That is what I originally said because, as I explained, I didn't think you were relying on transfer of momentum because unless the ball was very heavy the skater would not move very far toward the other skater.

Please look up the definition of momentum. The ball need not be very heavy, if it is moving fast. I already explained this.

In my subsequent post I said that the skater would move toward the other skater a little bit and then stop. I said: "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." And I went on to take issue with your suggestion (perhaps I misunderstood) that once the skater began moving as he pulled the ball towards himself, he would continue moving toward the other. I said that he would only continue until the ball reached him and stopped. I still stand by that. I said "But that momentum lasts only until the ball stops with the grabber"

And that is wrong.

I guess I don't know what you mean by 'attraction'. You cannot mean 'motion' because once the ball has stopped, the skater has to stop. (Or are you suggesting that the skaters are skating as well? {that was a joke})

Whatever happens on that frictionless ice surface, the center of mass of the 2 skater and ball system cannot move. I think we have to agree on that.

By "attraction" I mean that the skaters continue to move towards each other. While it is true that the center of mass of the entire system must not change its momentum, it is not true that the individual constituents of that system must stop moving. And no, the skaters need not have been skating prior to the exchange.

If the skater who pulls the ball towards him never stops the ball (because the other skater grabs it back before it reaches his chest) and this is kept repeating, the two skaters will continue to move together. But I didn't think that was what you were saying because I thought you said the motion of the first skater to grab the ball would continue even if the ball was dropped (ie. after the first grab).

That is precisely what I was saying.

I am not struggling with momentum at all. I am struggling with your example. I assure you I have no problem with basic physics. I studied physics from 1972-1976. That was a long time ago. I have't heard that the principle of conservation of momentum had changed since then.

You are indeed struggling with the idea of momentum.

First, you believe that only a very massive ball can impart an appreciable momentum transfer. But that is wrong because p=mv, so a lightweight but fast moving ball can also impart such an impulse.

Second, you believe that a system with total momentum of zero implies that the momentum of all the constituents must also be zero. But that is also wrong. Once a momentum transfer takes place (as in, say, the exchange of a ball from one skater to another), the center of mass of the system is motionless despite the fact that each skater continues moving. If a momentum q is imparted to skater 1, and a momentum -q is imparted to skater 2, then the total momentum is conserved and the skaters continue on their merry way, towards each other.

I don't know how to make it any clearer than that.

I don't need to revisit classical mechanics. I am rereading my 4th year quantum mechanics text and my 2nd year EM text. I appreciate that you are trying to be helpful, but I think that we are just misunderstanding each other's posts here.

No, you're understanding my posts just fine. It's just that you need to brush up on basic physics. I'm sorry that you disagree, but it's true.

anti_crank

Oct2-04, 01:38 PM

I am not suggesting that EM theory breaks down for EM waves with wavelengths smaller than 10^{-15} m.. I was just questioning whether we really have evidence that the coulomb force between protons continues as their 'separation' distance becomes smaller than 10^{-15} m.. That may have more to do with the structure of the proton than EM theory. EM theory assumes point charges and does not explain what creates the coulomb force.
My point was that if you change Coulomb's law to act differently on those scales, you have to change Maxwell's equations as well, and so you change the properties of EM waves on those ranges. This consequence cannot be avoided.

If you're suggesting perhaps that protons have a charge distribution that may end up reducing the repulsive force, you may be partially correct. However, since the net charge on both protons remains positive, the net force will necessarily be repulsive, will equal the point-charge repulsion formula to a first order approximation.

Hang on. I am not talking about new gravitational effects. I am not suggesting that some new gravity would account for the huge binding energy of nucleons. I am just questioning whether we have to use traditional concepts of force x distance to account for that binding energy.
The traditional concepts of force x distance are not the right tools for this job. You need quantum mechanics for any analysis of nuclear structure. Let me add here that all alpha decays can be modelled using perturbation theory where the alpha nucleus must tunnel through the Coulomb barrier. This does a very good job of predicting alpha decays; hence the Coulomb force is unchanged for all heavy nuclei undergoing alpha decay. To see how it acts on the scale of smaller nuclei, you can go to scattering experiments. These allow us to probe the structure of protons and even resolve individual quarks. If the EM force behaved differently, those experiments would have to be reinterpreted. The tapestry of science is too tightly woven to hope that you can unravel one thread without having to make massive changes. Scientific experts try to tell me how to do my job all the time. The problem is that they are frequently wrong and it takes a lawyer (and usually another expert) to make a judge understand that their explanations are wrong. I find this quite surprising. What field of law do you practice?
My interest, and I think it has always been the goal of science, is in trying to understand nature in terms of more fundamental concepts... Nuclear physics tries to explain protons and neutrons in terms of more fundamental particles: quarks. Neutons and protons consist of two different combinatons of three quarks held together by an exchange of virtual particles. At this point we have trouble maintaining a conceptual framework because we have to rely on elaborate mathematical models to describe and predict what is happening. As we probe deeper into the nucleus the models become more elaborate rather than simpler. I am suggesting that perhaps there is a more fundamental principle or some simpler explanation that will tie it all together. I don't think I am alone. I'm afraid I fail to see your point. Why exactly are we having trouble maintaining a conceptual framework? The mathematical models may be elaborate, but they do an excellent job of describing Nature. Right now the Standard Model reduces all of physics to some twenty-odd constants, and there are countless plans to reduce their numbers further by making a unified theory. Have you looked at such theories, like SU(5) or supersymmetry? In any event, there is no doubt that the strong force is here to stay, and will have to be included in any such searches. Since we have all done our best to give you reasons for the strong force's existence, the ball is in your court now: what evidence would be strong enough to convince you of its existence?

PS. I just realized that neutron stars might be a good place to see both gravity and strong nuclear forces at work. While I believe that the theories do a good job of predicting the properties of neutron stars (otherwise I'd know about it), I am not an astrophysicist, so I will have to defer this point.

Nereid

Oct2-04, 04:40 PM

PS. I just realized that neutron stars might be a good place to see both gravity and strong nuclear forces at work. While I believe that the theories do a good job of predicting the properties of neutron stars (otherwise I'd know about it), I am not an astrophysicist, so I will have to defer this point.I had a very nice post prepared, ready to post, then my browser crashed :mad: Will Bill give me a refund do you think?

We have modelled neutron stars, compared the results with observations, and the fit is very good. Unfortunately (or not!), we don’t have one in our neighbourhood, so we haven’t been able to perform controlled proton scattering experiments yet :cry: Indeed, the extent to which astronomy provides tests of a great deal of physics can sometimes be difficult to accept … I mean, a faint smudge of light in an image that’s full of other smudges is a gravitational lens, just as Uncle Al predicted?

Some highlights from models, which are based on General Relativity, the Standard Model (particle physics variety), as well as more pedestrian atomic physics, etc:
- there are well observed objects which have masses, densities, temperatures, surface compositions etc that match the models well (and few, if any, which don’t); pulsars (http://imagine.gsfc.nasa.gov/docs/science/know_l1/pulsars.html) are well-known examples
- from stellar evolution theory, we expect neutron stars to form under certain circumstances (e.g. ‘core collapse’ supernovae), and that’s just what we find; the Crab nebula (http://antwrp.gsfc.nasa.gov/apod/ap000711.html) is a good example
- the behaviour of neutron stars when mass is added – e.g. accretion from the other star of a close binary – is reasonably well studied, and observations match predictions closely; an example (http://universe.gsfc.nasa.gov/press/images/rossi2000/)
- research is under way (http://www.physics.ucsb.edu/~piro/research/crust.html) into how waves in neutron stars can provide details of the internal structure of these gigantic nuclei, in a way similar to how seismology gives an idea of the internal structure of the Earth, and helioseismology (http://soi.stanford.edu/results/heliowhat.html) the Sun
- indeed, studying neutron star (http://www.physics.ubc.ca/~heyl/talks/Life%20and%20Times.pdf) may help us answer questions in high energy physics that aren’t possible to address here on Earth

Andrew Mason

Oct2-04, 11:38 PM

If there is no strong force to provide the stability, then something else must do the job. It's as simple as that. I agree.
There's nothing mysterious about the strong interaction. It's only mysterious to people who have not studied it.All of physics is mysterious. We just reduce phenomena to fewer phenomena that we don't understand. That is the essence of all science. One should be cautious about saying these things are resolved. Planck's professor is a good example (he tried to dissuade Planck from taking up physics because all the great questions had been resolved - pre 1900).

How is energy converted to mass? How is it that our entire universe of matter and energy emerged from a point dozens of orders of magnitude smaller than a proton? Everything is mysterious at some level.

This is just silly. The energy well has to correspond to some potential. Don't you see that you're replacing the (well-understood) strong+weak+EM interaction with gravity and a mysterious energy well?I am not replacing anything. I don't have a theory and I am not proposing one. I am just asking questions and suggesting that the nuclear force may not be a correct explanation of reality.

We measure the nuclear force indirectly. We measure energies. The fact is that the separation of a proton and neutron requires the creation of matter. That obviously requires energy. But perhaps the energy is not required to pull against an attractive force for the separation distance. Perhaps it is required to add mass to the separate proton and neutron. I am suggesting that the process of creating matter might be thought of as inhibiting separation.

We don't understand how matter is created. It obviously requires energy. Does it require that a force be applied to the proton for a particular distance? That is what I am asking. So far, no one seems to have come up with a good reason why it has to.

Where does this well come from? God?
Probably the same place that the universe came from. Are you suggesting that God can be ruled out?

Andrew Mason

Andrew Mason

Oct3-04, 12:30 AM

No. The if the ball is dropped after the exchange, then the skaters continue moving towards each other. There has to be another exchange to stop the skaters.
Ok. Let's get this straight. There are two skaters at rest in the frame of reference of the ice, S1 and S2. S2 is holding a ball which is also at rest. S1 grabs the ball from S2 and pulls it toward himself accelerating the ball to a speed v_b. This results in momentum p_b = m_b v_b of the ball and necessarily imparts an equal and opposite momentum to S1 toward S2. S1 moves toward S2 at a speed v_{S1} = p_b/m_{S1}. I think we are agreed to this point.

Then S1 stops the ball at his chest and drops it (if he doesn't stop it, the ball would keep going past him). You seem to be saying that v_{S1} remains the same. I am saying that violates conservation of momentum. The reason is that in order to stop the ball, S1 has to absorb the ball's momentum p_b. Since that is equal and opposite to S1's momentum, this means that S1 stops.

But we don't even have to do that. We know that the momentum of the system is always 0 in the ice's frame of reference. If S2 is not moving and the ball is not moving, S1 can't be moving.

Will you please work out the calculation? It's quite simple. You have a skater carrying a ball, and they are both moving at speed v. Why? How? This is not possible. They cannot possibly be moving at the same speed unless that speed is 0. See above.

If the skater drops the ball, the skater does not stop. He does if he stops the ball and then drops it. I assume you mean by dropping the ball that the ball is stopped in the rest frame of the ice and does not go sailing off behind S1.

In fact, the skater moves faster, because of the reduced mass. Not in this universe. And you think I need help with basic physics?

Please do a calucation, and you'll see that it does not work like this.
Calculations aren't required. It is just simple conservation of momentum. See above.

Please look up the definition of momentum. The ball need not be very heavy, if it is moving fast. I already explained this. I don't need to look up the definition of momentum. Your comment that the skater speeds up if he reduces his mass is astounding really.

By "attraction" I mean that the skaters continue to move towards each other. While it is true that the center of mass of the entire system must not change its momentum, it is not true that the individual constituents of that system must stop moving. And no, the skaters need not have been skating prior to the exchange. Well in your example if S2 and the Ball are not moving then S1 cannot be moving without violating conservation of momentum.

You are indeed struggling with the idea of momentum. I am struggling with your concept of momentum. It is not Newton's.

First, you believe that only a very massive ball can impart an appreciable momentum transfer. I never said that. I said that unless the ball was heavy, the skater would not move very far when the skater pulled it towards him. It has nothing to do with the speed at which he pulls it. It has to do with the location of the center of mass. That point can't change. If the ball is not heavy, the center of mass is very close to the skater. If the ball is very heavy, it is close to the ball, so the skater will move farther.
But that is wrong because p=mv, so a lightweight but fast moving ball can also impart such an impulse. I have never said anything to suggest that I disagreed with that.

Second, you believe that a system with total momentum of zero implies that the momentum of all the constituents must also be zero. But that is also wrong. I never said that. I said the center of mass can't move. The constituents can move all they want. But that point can't change.
You are really twisting what I have said here.
Once a momentum transfer takes place (as in, say, the exchange of a ball from one skater to another), the center of mass of the system is motionless despite the fact that each skater continues moving. If a momentum q is imparted to skater 1, and a momentum -q is imparted to skater 2, then the total momentum is conserved and the skaters continue on their merry way, towards each other. You haven't got Skater 2 moving at all by Skater1 pulling the ball towards himself.

No, you're understanding my posts just fine. It's just that you need to brush up on basic physics. I'm sorry that you disagree, but it's true.This is high school physics, Newton's laws. Your statement that the loss of mass by simply dropping the ball will cause S1 to speed up is patently absurd. I am sure you will realize that when you reread your post.

Andrew Mason

Andrew Mason

Oct3-04, 02:41 AM

I find this quite surprising. Ok, it may be a bit of an exaggeration.
One of the most frequent sources of error (e.g wrongful convictions) is incorrect expert evidence. Experts were found to say that hair and fibre could make a highly probable 'match', that bullets could be matched by their metallugical composition, that babies with heart problems can't have high digoxin levels naturally. A nurse was charged (and later discharged, fortunately) with murdering babies on the basis of bad science. Guy Paul Morin was wrongly convicted of murder on the basis of hair and fibre 'matches' (DNA later exhonerated him). Clayton Johnson was convicted because of experts who said his wife was murdered (later exhonerated by other experts and a whole body of other evidence). Hundreds, possibly thousands of cases based on the now discredited 'science' of bullet matching are now under review in the US.

A good example of 'expert' folly is in the acoustic experts who testified on the JFK assassination before the House Select Committee on Assassinations in 1978 and concluded that there were 4 shots. The fact that no one said that they were sure they heard 4 shots and 90% said there were exactly 3 (most of whom recalled the same pattern of shots 1........2...3) was completely ignored. The National Academy of Sciences later showed the errors. See: http://www.dufourlaw.com/jfk/shot_pattern_evidence.pdf

Even nobel physicist Luis Alvarez wrote an article in Physics Today advocating his new theory of jiggle effect (definitely an unproven technique which, in this case leads to conclusions that are inconsistent with the evidence). He did get some things right, mind you.

What field of law do you practice? mainly litigation. see: http://www.dufourlaw.com/dufwho.htm#AM
For a good example of how an expert messed up a trial I was involved in and later corrected by the court of appeal, see: http://www.lawsociety.sk.ca/judgments/2003/Ca2003/2003skca40.pdf

Since we have all done our best to give you reasons for the strong force's existence, the ball is in your court now: what evidence would be strong enough to convince you of its existence?
Correct me if I am missing something, but the existence of the strong force is inferred from the binding energies that are observed for nuclear particles. Is there any other evidence of the strong force?

PS. I just realized that neutron stars might be a good place to see both gravity and strong nuclear forces at work. While I believe that the theories do a good job of predicting the properties of neutron stars (otherwise I'd know about it), I am not an astrophysicist, so I will have to defer this point.
I think you are right. If neutron star is as dense as the nucleus, then the nucleons in the neutron star are going to be within range of the strong nuclear force, if it exists. Now if we could only get a piece of a neutron star and see if it spontaneously flies apart...

There may be a way of testing the existence of the nuclear force if we could observe a star going through a gravitational collapse. As the density reached nuclear densities, there should be a significant release of energy as the nuclear force takes hold. That may be the theory behind the super nova explosion, I am not sure.
Andrew Mason

Tom Mattson

Oct3-04, 02:40 PM

All of physics is mysterious. We just reduce phenomena to fewer phenomena that we don't understand.

OK fine, I'll amend my statement: The strong force is no more mysterious than any other interaction.

I am not replacing anything. I don't have a theory and I am not proposing one. I am just asking questions and suggesting that the nuclear force may not be a correct explanation of reality.

But that begs the question: Why does the Standard Model agree so well with experiment?

Probably the same place that the universe came from. Are you suggesting that God can be ruled out?

I'm saying that there is no need to introduce it. Likewise, there is no reason to introduce an energy well whose source is unknown.

Tom Mattson

Oct3-04, 02:45 PM

This is high school physics, Newton's laws. Your statement that the loss of mass by simply dropping the ball will cause S1 to speed up is patently absurd. I am sure you will realize that when you reread your post.

I reread my posts, and now I see that I did not write what I intended to say. That's my fault, and I'm sorry for the confusion.

I meant for the exchange to be "sticky". That is, one skater actually rips the ball from the other's grasp, so that a fricitonal force develops between the skaters' hands and the ball. So, a force acts on each skater for a brief time interval. That sets both skaters into motion, and they will continue in that state of motion until acted on by a force.

Sorry, I've written this analogy so many times, in so many posts, that I got sloppy.

Andrew Mason

Oct4-04, 12:45 AM

But that begs the question: Why does the Standard Model agree so well with experiment?
Probably because it was designed to explain experimental data. The Standard model made sense of a vast array of experimental data and reduced great complexity to something simpler. But even those who built the Standard Model don't really believe that it is the ultimate theory.

Murray Gell-Mann writes in his 1994 book "The Quark and the Jaguar" that the Standard Model is an ad hoc theory rather than a theory built on fundamental principles (like General Relativity, Newton's Laws, Maxwell's EM equations). It describes many interactions but does not relate them to each other. It requires arbitrary constants that are experimentally determined and not derivable from principle. And it does not include gravity. Despite all this, it is remarkably successful in explaining and predicting experimental data. It works. But that doesn't mean it is the last word as a theory.

I'm saying that there is no need to introduce it. Likewise, there is no reason to introduce an energy well whose source is unknown.Until we know everything there is to know I wouldn't want to speculate on God's role in all of this.

I didn't introduce an energy well whose source is unknown. Empirically it has been shown to exist. Its source is unknown. I am just trying to understand what it represents. So far I see two possibilities: It could represent a strong force over a very short distance. Or it could represent something else.

Andrew Mason

ZapperZ

Oct4-04, 08:35 AM

Probably because it was designed to explain experimental data. The Standard model made sense of a vast array of experimental data and reduced great complexity to something simpler. But even those who built the Standard Model don't really believe that it is the ultimate theory.

Murray Gell-Mann writes in his 1994 book "The Quark and the Jaguar" that the Standard Model is an ad hoc theory rather than a theory built on fundamental principles (like General Relativity, Newton's Laws, Maxwell's EM equations). It describes many interactions but does not relate them to each other. It requires arbitrary constants that are experimentally determined and not derivable from principle. And it does not include gravity. Despite all this, it is remarkably successful in explaining and predicting experimental data. It works. But that doesn't mean it is the last word as a theory.

Apparently, this thing will simply not go away....

Again, your information is a bit dated like your Feynman text. While the Standard Model came into being as a phenomenological model (as most anything that is new), it has outgrown that status! It has made several predictions that have been verified to be true. There are sound theoretical description that can verify many of its components. You would have known this had you studied something like Perkin's High Energy Physics text. So saying that the Standard Model is nothing more than an "ad hoc" theory in its CURRENT form is like saying the Maxwell equation is nothing more than a collection of experimental observation. Theories and descriptions evolve! 1994 may have been a good year, but man, that is ancient history in many areas of physics.

There are plenty of indications that the Standard Model isn't complete. No high energy physicist in their right mind would even claim that. However, these indications do not come from what you have in mind, nor based on what you have suggested.

May I also suggest that you do not try to counter a physics argument with quotations? Not only are quotations often taken out of context (Einstein's "knowledge-imagination" is a prime example), but also they mean nothing in a physics discussion of CONTENT. Open any physics journals and you will not see a rebuttal or a comment on a physics paper done simply via quotations.

I'm curious. Are you still hanging on to this question: "Can anyone explain to me why gravity would not be a significant force on the 'surface' of a proton or neutron?" In other words, are you still under the false impression that GRAVITY can produce a significant force in nuclear binding?

Zz.

marlon

Oct4-04, 08:55 AM

A very nice question to you Andrew...from our most friendly science advisor, who is always in "the very best of moods"...

marlon

Andrew Mason

Oct4-04, 10:47 AM

I'm curious. Are you still hanging on to this question: "Can anyone explain to me why gravity would not be a significant force on the 'surface' of a proton or neutron?" In other words, are you still under the false impression that GRAVITY can produce a significant force in nuclear binding?
I never said that gravity explained nuclear binding. I just pointed out that gravity on the 'surface' of a nucleon would be sufficient to keep nucleons together - ie. a little separation would cause the nucleons to return together quickly. The nuclear binding energy (of, say, a deuteron) is a different matter. It can't be explained by gravity as we know it, and I never suggested it was.

The atoms of H_2 can escape their EM binding energy 'well' without adding mass to the protons or electrons that comprise those atoms. The nucleons in a nuclear binding energy 'well' can only escape by adding mass to those nucleons. I am suggesting that the nuclear binding energy (or mass creation energy) can be thought of as a kind of barrier to exit rather than a force that keeps the nucleons together. If that is the case, then gravity could be the only force inside that barrier.

Andrew Mason

Tom Mattson

Oct4-04, 11:18 AM

Probably because it was designed to explain experimental data.

OK, now we're getting somewhere. Now all I have to do is get you to accept that fitting the data is a good reason to accept a theory, rather than to reject it.

The Standard model made sense of a vast array of experimental data and reduced great complexity to something simpler. But even those who built the Standard Model don't really believe that it is the ultimate theory.

Of course they don't, and I don't believe that it's the ultimate theory either. But you can be assured that the next step in the process will not involve turning our backs on the Standard Model. In fact, it will most certainly be a requirement of the next theory that it include the Standard Model as a special case.

Murray Gell-Mann writes in his 1994 book "The Quark and the Jaguar" that the Standard Model is an ad hoc theory rather than a theory built on fundamental principles (like General Relativity, Newton's Laws, Maxwell's EM equations). It describes many interactions but does not relate them to each other. It requires arbitrary constants that are experimentally determined and not derivable from principle. And it does not include gravity. Despite all this, it is remarkably successful in explaining and predicting experimental data. It works. But that doesn't mean it is the last word as a theory.

Agreed.

Until we know everything there is to know I wouldn't want to speculate on God's role in all of this.

And there is no need to. Furthermore, there is no need to speculate on the role of a mysterious energy well that comes from nowhere and corresponds to nothing we know of. Do you agree?

I didn't introduce an energy well whose source is unknown. Empirically it has been shown to exist. Its source is unknown. I am just trying to understand what it represents. So far I see two possibilities: It could represent a strong force over a very short distance. Or it could represent something else.

Here's where I see this going.

You think that the strong force doesn't exist and that the EM force mysteriously turns off--for no apparent reason--at subnucleonic distances. In place of that you think that gravity + a potential well is responsible. But the gravitational interaction that is necessary to account for experimental data could not possibly be the same one that we know of classically. Therefore, we are looking at a gravitational interaction that is nothing like Newton's or Einstein's conception of it. We are looking at an interaction that behaves altogether differently from Newton's or Einstein's gravity at a very short distance, that holds the nucleus together, and that accounts for the observed hadron spectra.

I contend that you are simply attaching the label "gravity" to what is more commonly known as "the strong interaction".

ZapperZ

Oct4-04, 11:48 AM

I never said that gravity explained nuclear binding. I just pointed out that gravity on the 'surface' of a nucleon would be sufficient to keep nucleons together - ie. a little separation would cause the nucleons to return together quickly.

But you can only say this if you complete turn OFF the coulomb force. What makes you think it is valid for you to extrapolate gravity up to that scale, and yet you keep complaining that EM fields should not work there? If you accept that gravity works the same way all the way to nuclear scale, then you should be fair and also invoke EM fields too! Then coulombic forces would have severely overwhelm the puny gravitational forces by comparison! Several people have already given you the order of magnitude difference between the two! And don't tell me that we have no experimental data that EM forces should work at that scale, because that argument can be used against your gravity also, which would then make your whole point moot.

The atoms of H_2 can escape their EM binding energy 'well' without adding mass to the protons or electrons that comprise those atoms. The nucleons in a nuclear binding energy 'well' can only escape by adding mass to those nucleons. I am suggesting that the nuclear binding energy (or mass creation energy) can be thought of as a kind of barrier to exit rather than a force that keeps the nucleons together. If that is the case, then gravity could be the only force inside that barrier.

I have no idea what you just said here or why it is even relevant. Atoms of H2 can "escape" from a "binding energy well"? Hello? I was talking about NUCLEAR binding, not MOLECULAR binding. H2 is a MOLECULE, and thus, each H atom have a molecular binding with the other H atom. This is NOT a nuclear binding energy. Please don't change or confuse the subject.

Zz.

Andrew Mason

Oct4-04, 03:31 PM

But you can only say this if you complete turn OFF the coulomb force. What makes you think it is valid for you to extrapolate gravity up to that scale, and yet you keep complaining that EM fields should not work there? If you accept that gravity works the same way all the way to nuclear scale, then you should be fair and also invoke EM fields too!
Gravity appears to be different than the other forces. Nothing in EM or the Standard Model accounts for gravity. Only General Relativity does that. The principle of equivalence says that gravity and inertia can be considered equivalent, so gravity can be looked at as a pseudo-force or a curvature in space time. There is no evidence that space-time has quantum-like discreteness, unlike EM phenomena and the nucleus. So there is no reason to believe that gravity is any different at small distances. EM theory assumes point charges but we know that the proton is not a point charge.

According to the Standard Model, coulomb force is associated with the + and - 2/3 spin of quarks. The model does not explain how the coulomb force results from that and does not predict the nature of force that is created (perhaps my understanding of this is incorrect, in which case I would welcome correction). We don't really know where it begins. We only know that it is associated with quarks and exists in some region outside the proton, perhaps down to the quark level, perhaps not. Gravity, on the other hand, has no such limit because it is not so much a physical property of matter as it is a property of the space-time environment.

Then coulombic forces would have severely overwhelm the puny gravitational forces by comparison! Several people have already given you the order of magnitude difference between the two! And don't tell me that we have no experimental data that EM forces should work at that scale, because that argument can be used against your gravity also, which would then make your whole point moot.
I have explained why gravity would not turn off inside the nucleus. What evidence do we have that EM forces actually exist between protons in a He nucleus?

I have no idea what you just said here or why it is even relevant. Atoms of H2 can "escape" from a "binding energy well"? Hello? I was talking about NUCLEAR binding, not MOLECULAR binding.
Of course H atoms bind with EM molecular binding forces. I said "EM binding energy well" to distinguish it from nuclear binding energy.

My point was that there appears to be a fundamental difference between the two. Molecules can be separated without adding mass to the nuclei or electrons that compose those molecules. Nuculeons cannot be separated without adding mass to the nucleons.

H2 is a MOLECULE, and thus, each H atom have a molecular binding with the other H atom. This is NOT a nuclear binding energy. Please don't change or confuse the subject.
I was not trying to change the subject. I was just pointing out that separation of atoms in a molecule (which derives from a 1/r^2 force and therefore a -1/r energy potential) is very different from the separation of nuclear particles. I point out that one difference is that overcoming the nuclear potential results in adding mass to the separated parts (nucleons) while overcoming the EM potential does not add mass to the separated parts. While the energy required to separate atoms is the due to the need to apply a force over a distance, the energy required to separate nuclei is needed to create mass. Perhaps instead of looking at the nuclear binding energy as a continuous force between nucleons, we could look at it as a 'barrier' to separation. The analogy would be to a can of beans. The can prevents them from escaping. But it does not create a continuous force between the beans.

I am merely suggesting that if we looked at the nuclear potential in these terms, we would not need to invent a force that is enormous within a tiny distance (1 f.) and is effectively 0 outside that distance, and becomes repulsive at very short distances so as not to squish the nucleons.

Andrew Mason

Tom Mattson

Oct4-04, 04:08 PM

There is no evidence that space-time has quantum-like discreteness, unlike EM phenomena and the nucleus. So there is no reason to believe that gravity is any different at small distances.

Yes, there is reason to think that quantum gravity is the correct path. You note general relativity, but fail to note that GR and QFT contradict each other. They can't both be right, and there is little doubt that a correct description of gravity will lead to a quantized theory.

According to the Standard Model, coulomb force is associated with the + and - 2/3 spin of quarks.

The electric charges are +1/3 and -2/3. All quarks have the same spin: 1/2.

The model does not explain how the coulomb force results from that and does not predict the nature of force that is created (perhaps my understanding of this is incorrect, in which case I would welcome correction).

As I noted in an earlier post, Coulomb's law is derivable from QFT. It is reducible to electric charges in motion. What the Standard Model does not explain is the existence of charges. But then again, neither does any other theory.

We don't really know where it begins. We only know that it is associated with quarks and exists in some region outside the proton, perhaps down to the quark level, perhaps not. Gravity, on the other hand, has no such limit because it is not so much a physical property of matter as it is a property of the space-time environment.

I have explained why gravity would not turn off inside the nucleus. What evidence do we have that EM forces actually exist between protons in a He nucleus?

We have yet to see a compelling reason to think that the EM interaction should switch off at any distance. And we have plenty of evidence in favor of the other side, in that all predictions of the Standard Model have been verified.

My point was that there appears to be a fundamental difference between the two. Molecules can be separated without adding mass to the nuclei or electrons that compose those molecules. Nuculeons cannot be separated without adding mass to the nucleons.

That's not true of molecules. There is a slight increase in mass.

I am merely suggesting that if we looked at the nuclear potential in these terms, we would not need to invent a force that is enormous within a tiny distance (1 f.) and is effectively 0 outside that distance, and becomes repulsive at very short distances so as not to squish the nucleons.

So instead, we invent a sourceless potential well to keep the nucleons together because gravity is not strong enough to keep them in, is that it?

I'm sorry, but this idea of yours hasn't a leg to stand on. The simple fact of the matter is that we would have to continue using the Standard Model, because there's not a single quantitative prediction here. And besides, you'd have to tailor the energy well just so that it fit the observed spectrum of hadrons. Whereas with the Standard Model, we have the following few ingredients:

1. Canonical quantization.
2. Special relativity.
3. A gauge group.

From just these 3 things, all the predictions follow.

ZapperZ

Oct4-04, 05:19 PM

Your reply here is laced with ignorance and mistakes, and you seem to somehow have the ability to state all of the statements below as IF you have a complete knowledge of what you are stating. Nowhere in the immediate quote below was even a question or any sense of uncertainty of what you are stating:

Gravity appears to be different than the other forces. Nothing in EM or the Standard Model accounts for gravity. Only General Relativity does that. The principle of equivalence says that gravity and inertia can be considered equivalent, so gravity can be looked at as a pseudo-force or a curvature in space time. There is no evidence that space-time has quantum-like discreteness, unlike EM phenomena and the nucleus. So there is no reason to believe that gravity is any different at small distances. EM theory assumes point charges but we know that the proton is not a point charge.

According to the Standard Model, coulomb force is associated with the + and - 2/3 spin of quarks. The model does not explain how the coulomb force results from that and does not predict the nature of force that is created (perhaps my understanding of this is incorrect, in which case I would welcome correction). We don't really know where it begins. We only know that it is associated with quarks and exists in some region outside the proton, perhaps down to the quark level, perhaps not. Gravity, on the other hand, has no such limit because it is not so much a physical property of matter as it is a property of the space-time environment.

For someone who keeps insisting that all theories are subject to being challenged and being changed, you are quite arrogant in what you think gravitational theory should be and what it should not be. While GR has a number of verifications, in terms of degree of certainty, it has a LESS degree of certainty that classical E&M and QED. Now read that again - GR, which you are putting ALL your faith in, has a lower degree of certainty than classical E&M and QED. Furthermore, and pardon me for saying that, but I do not believe you even have a clue the complex theory of GR, much less are able to judge its validity.

Secondly, + and - 2/3 spin of quarks?!!!!

Thirdly, you have this unhealth obsession with so-called "point charge". If you have studied physics at any considerable level, you would have noticed that the classical electrostatic description has an equivalent form to the classical gravitational description. There is a gauss's law equivalent for gravitational field, for example. Every single description for an electrostatic field, there is an identical equivalent for a gravitational field (except for the absence of an repulsive force). Thus, gravitational force ALSO considers the mass being acted upon as point mass sources! Someone should have clearly pointed this out in your original "derivation" of gravitational force. Don't believe me, go double check what you did... you put ALL of the mass of the proton to be at the radius of another proton. Ignoring the obvious fact that you have two protons overlapping each other already, you yourself are doing the very thing you are criticizing, lumping the mass of an object into a point mass.

So get over this "point charge" problem already.

I have explained why gravity would not turn off inside the nucleus. What evidence do we have that EM forces actually exist between protons in a He nucleus?

What evidence that it doesn't? We have no description or evidence that EM forces can just simply disappear. It is YOUR responsibility to prove that your "new imagination" has a valid impetus to be considered other than just something you thought of out of ignorance.

Of course H atoms bind with EM molecular binding forces. I said "EM binding energy well" to distinguish it from nuclear binding energy.

My point was that there appears to be a fundamental difference between the two. Molecules can be separated without adding mass to the nuclei or electrons that compose those molecules. Nuculeons cannot be separated without adding mass to the nucleons.

I was not trying to change the subject. I was just pointing out that separation of atoms in a molecule (which derives from a 1/r^2 force and therefore a -1/r energy potential) is very different from the separation of nuclear particles. I point out that one difference is that overcoming the nuclear potential results in adding mass to the separated parts (nucleons) while overcoming the EM potential does not add mass to the separated parts. While the energy required to separate atoms is the due to the need to apply a force over a distance, the energy required to separate nuclei is needed to create mass. Perhaps instead of looking at the nuclear binding energy as a continuous force between nucleons, we could look at it as a 'barrier' to separation. The analogy would be to a can of beans. The can prevents them from escaping. But it does not create a continuous force between the beans.

I am merely suggesting that if we looked at the nuclear potential in these terms, we would not need to invent a force that is enormous within a tiny distance (1 f.) and is effectively 0 outside that distance, and becomes repulsive at very short distances so as not to squish the nucleons.

Andrew Mason

This last set of paragraphs are a complete mystery to me. What EXACTLY is this about? Adding mass and not adding mass?? Applying force over a distance? Huh??!! I have to add mass to separate nucleons? Tell me where I am adding mass in an alpha decay? And if you are somehow thinking that separated daughter nucleons ALWAYS have a higher total mass than the original parent nucleus, I have a fission reactor I want you to meet.

You are not suggesting anything. You are trying to ignore a huge part of physics that you have no clue on (QCD), and trying to replace it with your faulty physics knowledge. In the process, you are doing the exact same thing you are finding faults in with EM and QED. But what is even more appaling is that you are STILL insisting there is some validity in your faulty idea inspite of (i) your admission of lack of any serious knowledge in the matter you are talking about and (ii) the REPEATED explanation of those who are more knowledgable than you on why your idea is wrong.

In the end, I find it ironic that you would spend time quoting Gell-Mann for something that suits your needs, and yet you blatantly reject his most significant contribution - the quarks and how they interact. I hate to think that this is how you practice your profession.

Zz.

anti_crank

Oct4-04, 07:14 PM

I find it unbelievable that this thread is still going, and it raises serious doubts in my mind as to whether any progress is being accomplished. For the last time, I will summarize the main points:

-EM waves obey Maxwell's equations at all known ranges, including well past nuclear range. Hence it is contradictory to assume that classical forces from electrodynamics break down at such scales while the Maxwell equations stand since they come from the same mathematics.
-Scattering experiments confirm that the EM force works as expected. In fact, accelerators would not work if classical EM would break down at those scales/energies. In practice, we routinely accelerate particles to high energies (the LHC at CERN can accelerate protons to 100GeV) without any surprises.
-We know from scattering experiments that nucleons (protons/neutrons) have constituents, whose fractional charges can be measured and agree with the quark model
-The strong force model explains the observed spectrum of mesons and baryons beautifully. Heavy quark systems can be solved with the Schroedinger equation and an approximate, single-gluon exchange radial potential. The spectrum of allowed states for such systems matches observations very well.
-A force exists between nucleons and their constituents that allows certain decays which cannot be otherwise explained. The lifetimes are too small for the decays to be either electromagnetic or weak. We know this because lifetimes are inversely proportional to couplings. Gravity - unless strengthened immensly - is nowhere near the right coupling strength to allow such things.
-Gravity is negligible on the nuclear range, unless radical changes are made to it. Such changes are not warranted by any known results, and would make it a downright ugly theory. Try finding the ground state, using either Schroedinger eqn or Bohr quantization rules if Schr. is too hard, for a system of two heavy quarks bound by the gravitational potential. What's the expected radius - anywhere near the size of nuclei?
-Gluons are one prediction of QCD. Their existence, as well as some properties, can be inferred from three-jet events in accelerators and deep nuclear scattering experiments. The latter indicate that over 50% of a proton's momentum is carried by neutral (non-charged) constituents.
-Regarding neutron stars: see Nereid's post.

At this point, let me make an analogy that should make things clear. This problem has already been in the court of science, a court comprised of all scientists, past and present, who have dealt with this problem. Judgement has been passed by a large majority in favor of the strong nuclear force as presently known to provide the best explanation for the structure of nucleons. Since you've not provided any reasons for appeal that carry any merit upon in depth examination, the request for an appeal is denied. It's very possible for forensic experts to be wrong, and it simply goes to show the need to have multiple and independent experts analyse the problem and voice their opinion. Science is not dogmatic or authoritarian as is often believed by the general public, but relies on concrete evidence from experiments. This test was passed by the strong force, and it makes much less difference whether the experiments came before or after the theory than whether the experiments match the theory at all.