What forces are at work when neutrons are ejected?

It may be difficult to separate "theory" from "application," but let''s see if this helps facilitate the discussion.
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Re: What forces are at work when neutrons are ejected?

Post by Frank Sanns »

Before this thread rambles much deeper into conjecture, it might be helpful to look at the situation of the neutron itself. A neutron is not stable outside of an atom and it decays into a proton, electron and an anti-electron neutrino. None of which can exist as themselves inside of the neutron. Banter about forces and uncertainties but even the fundamentals of the fundamentals are poorly understood by mainstream science.

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Re: What forces are at work when neutrons are ejected?

Post by Dennis P Brown »

Frank, I very much used well-understood and clear physics - I did not once use "conjecture". I am somewhat lost at what you are getting at? Please point out where I did not offer clear and exact physics in my answers. Also, I did in fact point out that the neutron is unstable outside a bound system. I have answered the first question Steve and Chris were wondering about and most certainly did not extend this thread in a non-useful manner - I did clarify and expand my answers in response to questions that were asked. Thanks.
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Re: What forces are at work when neutrons are ejected?

Post by Richard Hull »

Let's face it.....

It is all made up as we go along. (Vector force bosons?, gauge particles?, flavors?, color?) It is all dreamed up to explain the unknowable into an acceptable fabric that certain powers have found to explain, to their own and their associates satisfaction. Each new observation that doesn't fit the ever expanding fabric sends the powers scrambling to expand the fabric with new assertions.

It helps that uncertainty is a virtually accepted tenat for within it, what is, in effect, "the hand of God" can make things happen as if by magic or more acceptably the probablistic roll of some nuclear dice that contain an infinite number of sides.

The casting of the bones are read for us by the powers that be, based on their solid,yet, ever expanding logical framework.

Science, for me, ends at the easily demonstrable and perpetually perdictable outcomes. The moment I am told that there is some uncertainty in why something occurs or a possibility of various things will occur due to a single physical setup, Science fades from view and I stop weighing and considering that which is told to me.

In short, I accept the unknowable and find contentment in that which can be expalined fully. I have tried to stop musing over the unknowable, mysterious uncertainies within the nucleus. For at the level of proclaimed uncertainty, the powers that be are, themselves, powerless of persuasion.

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Re: What forces are at work when neutrons are ejected?

Post by Jeff Robertson »

I'm not sure how I feel about the uncertainty principle being used in this context. The uncertainty principle, as the name implies, is intended to explain the fundamental uncertainty in measuring both position and momentum of a particle. The "delta x" and "delta p" are really just standard deviations of their respective variables - it is sloppy notation by many generations of authors that has led people to think they're anything else. So in the rough calculation you did, you solved for the uncertainty in the momentum of the neutron, rather than the momentum itself. I dunno, maybe I'm completely wrong and this is in fact a valid analysis of the nucleus, but this is just my understanding of quantum mechanics. Tunneling and all those fun quantum effects, however, are still in play though (as you mentioned).

I think the ejection of a neutron can be explained with barely, if any, quantum physics. Classically, the energy of a system is defined as the kinetic and thermal energy of a particle, as well as the binding energy holding it. For example, in a proton-electron (hydrogen, if you will) system, the electron is in a stable orbit around the proton. The reason for this is that the total energy of the electron is negative, which is one of the main characteristics of a bound system. The reason for this is that the electron's kinetic/thermal energy is less than the magnitude of the potential binding energy (the potential energy is in fact negative) due to the coulomb force. So E = (KE) + (PE), where PE is a negative number whose magnitude is greater than KE.

It's not that the electron doesn't have kinetic energy, it's just that its kinetic energy isn't enough to overcome the binding force. Kinetic energy can manifest itself in the form of linear motion, rotational motion, and vibration. In the electron-proton system, the electron orbits around the proton in addition to having an intrinsic amount of angular momentum (known as spin), so the kinetic energy reveals itself in this way.

So, back to the main discussion, a neutron bound to the nucleus has kinetic energy (mostly in the form of vibrational I'm assuming), but the strong interaction overwhelms the neutron and forces it to remain confined. If this balance is disturbed, say by the collision of another subatomic particle, the potential energy of the neutron due to the nucleus will be changed. For example, in nuclear fission a neutron collides with a large, barely stable nucleus. The short range of the strong nuclear force cannot hold together such a nucleus anymore, and the neutrons on the outskirts of the nucleus lose a bit of the binding energy they had before. This drop in binding energy is enough for their internal kinetic energy to overcome the energy confining them to the nucleus, so the system becomes unstable and they are free to leave the atom.

I'm not sure if this is a satisfactory explanation, but this is just how I interpret what's going on. Now as to why a neutron is ejected as opposed to a proton, that is an interesting question. I think Frank put it nicely though - the neutron by itself is less stable than a proton, so if one of the two had to go it sort of makes sense that the neutron would be kicked out first.

I don't really see how the magnetic force plays a major role in this. While a neutron does have a small magnetic moment, the other forces at play (coloumb and strong nuclear) are likely much more powerful than any magnetic effects, at least on the atomic scale.

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Re: What forces are at work when neutrons are ejected?

Post by Steven Sesselmann »

Could we explain the neutron ejection as an extreme case Boyles law?

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Re: What forces are at work when neutrons are ejected?

Post by Jeff Robertson »

Steve,

I'm not sure what you mean, could you go into more detail? Boyle's Law, if I remember correctly, is basically the ideal gas law for a constant temperature. The ideal gas law is derived from the basic assumption that a gas is made up of independent moving particles, and the "pressure" in both the ideal gas law and Boyle's Law is due to molecular impacts from gas molecules on the walls of a container. This seems like an entirely different phenomena (and a macroscopic one, compared to the size of the nucleus) than what's responsible for the neutron's behavior.

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Re: What forces are at work when neutrons are ejected?

Post by Dennis P Brown »

Richard, I see you can wax poetic!
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Re: What forces are at work when neutrons are ejected?

Post by Dennis P Brown »

Jeff, I think you miss understand the uncertainty principle and what it is saying - it is not a series of uncertainties in measurements like someone in the lab weighing a tiny bit of chemicals - it is an absolute statement to the limit of knowledge that the universe imposes and says deep things about what atoms and particles can and can not due - it is what keeps electrons from spiraling down into the nucleus.

Remember the electro-static attraction between an electron and proton? Well, have you considered that there are NO significant forces that prevent the electron from reaching the proton? The fact that it does not IS due to the uncertainty principle in action combined with selection rules (take about something pulled out of thin air.)

Calculate the energy an electron would have if bound with a proton on the dimensions of a nucleus and the energy is huge! Far beyond normal nuclear binding energies. And why the electron can not be bound by the electro-magnetic force so close to a proton (the gluon force does not 'see' electrons.

Yes, this principle has no bases in our normal world and appears as a fix but quantum does not ask us to believe anything except experimental results. Until a better theory is developed that explains experimental results better, that is the reality. My calculation is SOP (also, I used the momentum formula to relate KE besides the uncertainty formula) for modern physic courses.

People have spent many decades working on these problems and do you really think they missed something so simple as what you are saying? Maybe but they calculate nuclear constants and predict effects using standard models and this works well - yes, there is a lot not understood but it is pointless to believe they are all wrong and you have a better insight. That is, unless you have experimental data proving that you get better energies? Try calculating the KE of the neutron and see using your idea if it gives a better answer than mine. If so, you are on to something, otherwise, you are not.

Isn't predicting possible results as accurately as possible what we all want a theory to do? Just because we don't like it does not invalidate it.
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Re: What forces are at work when neutrons are ejected?

Post by Dan Tibbets »

Neutron binding is dependent on forces, but it is also dependent on position. I think nuclear chemistry is similar to regular (electron) chemistry. The protons and neutrons are occupying defined orbits (round / elliptical or possibly hyperbolic (if escaping) . And these orbits can be changed- think of fluorescence in regular chemistry, and isomers in nuclear chemistry. If a neutron is near the maximum extent of it's orbit, it will take a relatively small perturbation to escape- or errors of position and or momentum.
Ihave a wak knowledge of the weak force in the picture, except that it is the major consideration in the binding of unlimited neutrons. The competing interactions mediated by the strong force, electromagnetic force and their relative ranges do a good job of predicting proton and electron numbers in a nucleus. But since this interaction is absent with neutrons* so the weak force dominates.

I recall a gravity / planet orbit program ("Newton" I believe) that would calculate the orbits and show a graphic of the orbits of planets around a star, another planet, etc.. If a too long of a time slice was used (speeds up the program) the errors build up and planets will start escaping from their orbits and even the solar system. The problem is that the precision of the calculations suffers. If the time slices are short enough, the orbits are well behaved, and the program is boring as the updates are slow. The uncertainty principle may be considered in the same way, except the highest precision is set by the universe, not by the whim of the computer user. I wonder if time is quantized and if this plays a role. Or, stated another way, the precision is set by God, not by humans. And, a computers precision is based on the number of bits it can handle, A 64 bit processor is better than a 32 bit. The uncertainity of the calculations are always present. Even if the uncertainty principle is ignored, the precision of any calculation is limited by the number of particles (or wave packets if you prefer) in the universe. The numbers are not infinite, so any calculation cannot be absolutely accurate, nor can the calculations always come up with the same answer (perfect precision).

The question becomes not why there is a an Uncertainty Principle, but why it is set at a level that leads to the statistical effects that underlies known quantum physics.

Whether we can understand why is not really relevant. That we can develop a theory that successfully makes predictions is. The standard model does this (so far), and I suspect it does so through multiple levels of assumptions and squirming. It may have little to do with reality, except that it predicts things accurately. String theory derivatives try to answer the question of why, but without testable predictions they are mathematical flights of fantasy that might be elegant and consistent, but are otherwise useless. And, of course they do not answer the question of reality, but only shifts the assumptions to a 'finer' level. If physics is based on strings, what are strings based on?

Sorry, I waxed philosophical.

*The strong force still plays a role with the neutron binding verses the weak force due to the limited range of the strong force in comparison to the volume/ radius of the nucleus. When a neutron is transiting the core of the nucleus it experiences the maximum strong force interaction. But when it is in a more distant portion of it's orbit the strong force interaction has weakened considerably, but the weak force has lessened to a smaller degree (1/r^3 vs 1/r^2?). At some radius the neutron will experience a net repulsive force. The same happens with the protons (?), but since the electromagnetic force is so much stronger than the weak force, it occurs much closer to the nucleus center and the weak repulsion is essentially irrelevant. I have no idea at what radius the weak force becomes stronger than the attractive strong force and what the orbital maximum radius of the neutron may be, or what magnitude of the Uncertainity Principle applies. but the assumption is that as neutrons accumulate, the orbits reach dimensions where there is significant chance of escape. A hint may come from the uncertainty effects on the D-D fusion cross section data (it is very significant).

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Re: What forces are at work when neutrons are ejected?

Post by Frank Sanns »

Dennis,

The conjecture is is from the particle zoo model and other inconsistent explanations. Leaking gluons doesn't really float my boat like many of the other pet theories that are floating around in mainstream science. Actually, the theory for the sake of theory isn't as important than results. Take all the theories that you want and it will be tough to get me interested but give me one experiment with a result and you have my attention.

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Re: What forces are at work when neutrons are ejected?

Post by Frank Sanns »

The Uncertainty Principle is not something mystical or magical. It is not even unique to quantum mechanics. It is applicable to all forms of waves including those in the ocean and sound waves too. It simply is a relationship between trade offs. Usually is it measuring time and precision.

If there is a sound wave with 2 meter wavelength and you only measure for one tenth of the time that the wave goes by, they you will not have a very good estimate on the wavelength or the energy of the wave. If however, you measure for a long time for multiple waves to go by, then you will have a good estimate of the wavelength but it will have taken much more time. Nothing magical here. Just basic wave physics.

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Achiever's madness; when enough is still not enough. ---FS
We have to stop looking at the world through our physical eyes. The universe is NOT what we see. It is the quantum world that is real. The rest is just an electron illusion. ---FS
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Re: What forces are at work when neutrons are ejected?

Post by Dennis P Brown »

No one has said the uncertainty principle is magical and please don't add such nonsense.

Also, the quantum uncertainty principle is not the same as what is used for uncertainty in measurement related to ocean waves - the quantum uncertainty principle puts absolute limits on what can ever be achieved with measurements, and relates this to Planck's constant.

Time does not enter into the measurement equation I gave for the uncertainty principle for position/momentum - what you are using is an invalid analogy. Of course if you limit a time dependent measurement in time, you introduce a finite and well determined error relative to wave position - this has no relevancy to the topic of the neutron energy when it is no longer bound within a nucleus - only quantum principles apply to this situation. Macro measurements based on Newtonian mechanics does not apply - this has been proven by experiment.

When one uses the uncertainty principle, and the standard Newtonian momentum/energy relation corrected for relativistic effects, the exact answer for the neutron KE is obtained for the now free neutron.

Continued arguments on this point are silly - if you desire to believe nonsense, that is your right. If you wish to try and disprove this scientific fact, I'll do my best to help you with your ideas and experiments but if you just want to continue to make verbal arguments, you are wasting everyone's time.
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Re: What forces are at work when neutrons are ejected?

Post by Jeff Robertson »

Hey Dennis,

I know the uncertainty principle very well, and like Frank says (although wrong context) it is not limited to just Heisenberg's famous equation. The formal mathematical model of quantum mechanics contains a famous equation which relates the uncertainty between any two observables. The famous uncertainty principle which you used is just one application of this generalized uncertainty equation, when applied to position and momentum. The "delta x" and "delta p" in Heisenberg's uncertainty principle do in fact represent standard deviations. Not deviations in human error or measurement, as you said, but in the fundamental observables of the particle in question.

I do, in fact, remember the electro-static force between an electron and proton (can't tell if that was intended to be condescending or rhetorical, I'll assume rhetorical). What keeps the electron from spiraling into the nucleus, at least the way I learned it, is the quantization of the electron's angular momentum, which is derived from the wave function of the hydrogen atom. The electron can only take on certain values of angular momentum which are multiples of the reduced planck's constant (excluding the 1/2 spin of the electron itself). Without this quantum law, the electron would radiate as it orbited around the nucleus (as it's constantly accelerating), spiraling down until it collides with the proton. I'm not sure how Heisenberg's uncertainty principle applies to the electron's stable orbit, that's something I'll have to think about. I suppose as the electron travels closer to the nucleus it becomes more defined in its position (as it is stuck in a potential well and there is a less broad region in which it is likely to be), implying a greater uncertainty in the momentum, but I see this as merely a consequence of the wave equation rather than the governing law itself. The quantization of angular momentum is what I learned to be the main governing law which keeps the electron and proton from colliding. An electron's wave function must be in compliance with Schrodinger's equation, and it so happens that virtually every possible solution of the Schrodinger equation for the hydrogen atom allow only discrete values for the electron's angular momentum.

And in the last section of your post, you justify your method of analysis by saying "well, my numbers were close enough." Just because your numbers are in the ballpark of what they should be does not mean you conducted a proper analysis, that's the ends justifying the means. They weren't even in the ballpark at that - you made some comment about neglecting relativity to justify bumping it up an order of magnitude. You act like I'm insinuating that every other physicist is wrong and I think I have discovered the new standard model, when really I just didn't agree with what you did. I'll read up on this tomorrow, it's very possible that I'm just flat out wrong. If your analysis was in fact correct then I'll own up to it, but it doesn't make any of my other points any less valid. My only criticism of your analysis was that the momentum term in the uncertainty principle relates to the standard deviation, rather than the momentum of the particle itself.

In response to what Frank said, the uncertainty principle is an entire different phenomena than what goes on in macroscopic waves (as Dennis said). All quantum mechanic uncertainty principles are derived from the fundamental probabilistic nature of the wave function, which has no analogy towards the macroscopic world. Any uncertainty which exists on a macroscopic level is due entirely to experimental limitations or human error. That said, whether on purpose or accident, Frank stumbled upon an additional uncertainty principle. His analogy to measuring energy and time wasn't invalid at all, in fact it is an actual equation which is perfectly analogous to Heisenberg's uncertainty principle. That is, (delta E)*(delta t) must be greater than or equal to one half the reduced planck constant. This comes directly from the generalized uncertainty principle.

I'm not sure why you're getting so unnecessarily defensive and snippy. This forum is intended to be an open discussion board for ideas and to talk about various physics topics. Frank is not "wasting everyone's time" as you said, he's participating in an intellectual discussion about the underlying physics of nuclear decay. You come across as having a sense of elitism, when in fact you are not the only one who has knowledge on this topic. Telling people they are wasting everyone's time by posting, saying what they believe is nonsense, and that disagreeing with what you say is on par with trying to disprove scientific fact. Your past couple responses seem a touch pompous and just downright impolite. Whether or not someone's thoughts are accurate or not does not give you the right to have such a toxic attitude.

That said, there are some interesting things to think about here. I think tomorrow I'm going to dust off a couple textbooks and read up on some applications of the uncertainty principle.

Cheers,
Jeff
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Re: What forces are at work when neutrons are ejected?

Post by Frank Sanns »

Collapse of a wave function because of observation for example is different than the probability statistics and the uncertainty. The wave function in quantum mechanics is not the same as waves in the ocean as it describes a probability whose form is a wavelike function. However, the Hamiltonian operator (potential and kinetic energies) in the Schroedinger is precisely derived from waves in the ocean. William Hamilton back in the mid 1800s envisioned a planet flooded with water. He imagined what a wave would be like when it passed by its point of origin. Soon it was clear to him that only certain allowed energies (wavelengths) were allowed (constructive interference) while others would not be allowed (destructive interference) as they would quickly die out.

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Re: What forces are at work when neutrons are ejected?

Post by Dennis P Brown »

Jeff,

first off I never said (and this is what you wrote that I said)

"deviations in human error or measurement, as you said, but in the fundamental observable of the particle in question."

I, in fact, said the exact opposite of what you are writing here. I most certainly did point out that the uncertainty principle is not a measure of measurement errors, but an absoult limit on a quatum system.

Also, I did not say my answer was close enough - I carefully pointed out that my approach must be corrected for relativistic effects to get the correct answer.

So here is exactly what I said:

"A rough calculation I did puts it at about 0.1 MeV; this is not corrected for relativistic velocity effects (big effect at these energies) so this value is too low - just as it appears to be - so the real answer would be more like a few MeV. Still, this simple approach shows that the energy of confinement is huge so a neutron (no longer bound by the strong force (read gluon force)) would have a huge KE outside the nucleus."

Note the word 'rough calculation'. And I point out it is not corrected for relativistic effects. Further, at no point and no where did I say my answer was good enough nor even implied that.

Gotta read the post to be accurate Jeff.

I also did add that the other quantum numbers for the electron bound within an atom do in fact come into play but realized this would further confuse someone struggling to understand the uncertainty principle. Glad you decided to expand on that topic but don't imply that I didn't consider this or somehow mised it.

I too have calculated the electron orbits in hydrogen and yes, the other quantum numbers are critical. Also, the orbits are the only solutions. That all said, you really think that will explain the question about why Frank's approch was incorrect? The uncertainty principle does have a factor in this and is a clearer way to explain the issue to him. Maybe not but that is a judgement call. Still, a full hydrogen atom solution would not be helpfull at all.

Relative to the issue of a neutron having KE after leaving the nucleus and why an electron cannot be in a space the size of a nucleus, the uncertainty principle does have a major impact on whether an electron can approch/occupy that volume - do the calculation (and more to the point, most people here can do that calculation using very simple math - that is why I used that example, by the way.)

Yes, spin has a major impact but again, if you notice, the people were just trying to understand why a neutron is 'ejected' with such energy. Your current example does not address this point about the KE of a neutron leaving the atom - my use of the uncertainty principle does and is correct. Frank added a rather simple counter example and I was showing him (without a full blown quantum calculation) that his idea does not work and why. Yes, too simple for people whho have done the real solution of a hydrogen atom. Do you think that would have help Frank understand the error of his counter example?

Jeff, I most certainly did not get "unnecessarily defensive and snippy. " I was polite in my repose after being told by Frank "thread rambles much deeper into conjecture". All I did was point out I used accurate physics. What is defensive about stating a fact?

Or are you referring to when Frank said "Certainty Principle is not something mystical or magical."

This statement is very disrespectful to me and my post and all I answered was "No one has said the uncertainty principle is magical and please don't add such nonsense."

Using the statement 'mystical or magical' relative to my post on a topic that is solid and well know physics was in fact very inappropriate and, frankly, disrespectful to me. So please Jeff, explain where I was silly?

Later I did notice that this thread was continuing to have people add rather far-out attempts to explain away a very simple, straight forward physics concept. Yes, the thread was getting silly and that was said more in jest, but this was true. I will not give the exact example of why I said that because if you read some of the posts, it was not an unreasonable statement. I will not point out its cause out of respect for the poster's honest mistake.

Your knowledge on the subject is obviously good, but others are not so well schooled and I used solid, simple answers to a very good question. Further, I was respectful and handled my posts in a professional manner especially after being told that I was somehow implying that a very clear physics concept was magical - please, my answer to that statement was fully appropiate.

As for the forum, I was in fact doing exactly what it is for - answering an excellent scientific question with well known science. I treated the subject in a manner that I hoped was clear. Please refrain from such incorrect and baseless statements relative to my posts.

Thanks

Aside: your last post, except for the points I don't agree with about my posts, is an excellent explaination and I am glad you added that.
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Re: What forces are at work when neutrons are ejected?

Post by Chris Bradley »

Just to mention - having initiated the thread to see if folks wanted to offer Steven an answer to his original question (as it was already a thread-drift elsewhere), I'd say the matter has now been answered in a few different ways, qualitatively and qualitatively, and we seem to have now exhausted the possible suggestions people want to make.

Of course it's likely that we'll never know for sure. Maybe it is too 'fundamental' for us to ever tease apart, or maybe we will tease it apart one day. Who can say? In any case, I had just wondered if folks had thought on it before.

'The Uncertainty Principle' is a different topic - it may be necessary to mention it as part of an answer to the question, but I think that's been done now.

'People's personal interpretation of "The Uncertainty Principle"' is a yet further topic, so is definitely a thread drift.

Thanks.
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Re: What forces are at work when neutrons are ejected?

Post by Richard Hull »

Hate to go against convention but no one has yet convinced me that a neutron is a particle, existing as such, within the nucleus! (I am certainly open to the idea though)

Anyone been in a nucleus?

The neutron is obviously an extra nuclear particle. We see it exit bulk matter. We can track it and know that it is highly unstable with a relatively short half life in the wild.

Would it be beyond the pale to suggest that it is a nuclear condensate?..... That certain nuclear conditions or instabilities cause it to form suddenly at the edge of the nuclear influence?

We base all of our intra-nuclear knowledge on a framework of guesses and carefully crafted, testable assumptions that work for us and have satisfied and gratify us. I am thinking that the nucleus cares little for what we think or our machinations. We are comfortable with our explanations for the unseeable and unmeasurable within an edifice whose door is forever closed to us. We only see what is dumped as trash or the debris from our frustrated efforts to go inside by blowing the place to bits and assume the debris has meaning. ( A decent assumption, I guess)

We may be too clever by half and may never know the reality of it all.... If there is any at that level.

Just looking for flies in the ointment.

Richard Hull

I have lived with and accepted that the neutron has life in the nucleus. Do not get me wrong here. someone has got to work in gravity boots or be the Devil's advocate, I guess. RH
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Re: What forces are at work when neutrons are ejected?

Post by Chris Bradley »

> Hate to go against convention but no one has yet convinced me that a neutron is a particle, existing as such, within the nucleus! (I am certainly open to the idea though)
I'd point to activation, one that you have done many times yourself, as the experiment that could provide the answer here: If the neutron does not stay as a neutron when it mixes in with a nucleus, then wouldn't we observe prompt beta emissions *during* neutron absorption into a nucleus, rather than sometime *after*?
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Re: What forces are at work when neutrons are ejected?

Post by Richard Hull »

Why should we observe that? (instant beta emission) Because it doesn't fit our theories? Not good enough for me. We have no idea what happens to macro forces in a nucleus. We do make assumptions that they are unchanged, but dream up new forces, both strong and weak to explain away issues based on conteracting the macro extra nuclear forces.

As for beta emission....Another matter entirely! As the balancing act in this process never works out to simple theory, we invent another particle intrinsically massless and, naturally undetectable in any specific beta decay event, but with a kinetic energy that magically rebalances all our probs away. Self consistency retained and testable only over quadrillions of events assumed to be part of past decay events in a vast sea of nuclear events. Good enough for theoretical science and it stands as fact for many. (most)

I have always noted that the neutron decay to a proton and electron, (which are the only actual things ever detected in such decays), might, oddly enough, be what the neutron is composed of, but then there is the retort about spin and magnetic moments, quark balancing, etc., I am told on good authority, kill that concept. Well can't I kill the idea of the neutron being part of an atom? No that is not even to be considered. Give me a break. The neutron in the atom is a must have to make theory work and solve nuclear beta decay issues.

I would be happy with a nucleus of protons and electrons only allowing for but one force that defeats all others and is itself manifested as a variable mass that readily exchanges itself according to a new set of internalized nuclear rules I can't fathom. Certainly better, ocama's razor wise, than a zoo and several extra forces. All other ephemeral particles from neutrons to mesons to quarks can be explained as evaporations from nuclear events so intense and local that different forces and particles exist only in that fleeting environment, but not in the averaged out, current matter/energy cosmic situation. They are not part of the nucleus. They are part of a long dead past belched up from the dead.

Whatever we invent, it is always self-consistent and testable. We call it science because it is all we have to ward off religious dogma and old fashioned superstition. One man's religion is often another man's heresy. Science should not be left out of that mix especially if it starts to tell of things unseen, but imagined even if guided by self consitent theory and is testable. It makes those of us who don't like the other dogmas and disconcerting unknowns feel like we are on solid ground.

For most here, if we are without science, we are truly adrift. I will only let science take me so far afield before I feel a cold breeze up me kilt.

Richard Hull

Why don't neutron stars decay in a few minutes to an hour? Or, in that decay, reform their mass into hydrogen (electron and proton) once clear of the energetic star? Some kind of unknown binding on a scale that is macro, but resembles a nuclear force? Gravity should not slow down decay, or should it? Time dialated decay? RH Too many questions and no good answers.
Progress may have been a good thing once, but it just went on too long. - Yogi Berra
Fusion is the energy of the future....and it always will be
The more complex the idea put forward by the poor amateur, the more likely it will never see embodiment
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Steven Sesselmann
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Re: What forces are at work when neutrons are ejected?

Post by Steven Sesselmann »

Richard,

I have been following this tread, but as I had nothing intelligent to contribute, I stayed out of it. What puzzled my mind was the forces are at play, what causes a neutron to be ejected with 2.4 Mev energy? Not so much where the energy comes from, but what the nature of the force is.

Could it be,

a) ...centripetal force, ie the nuclei spiral rapidly inwards until one particle lets go?

b) ...the neutron is composite and made up from a proton and an electron, where there is a small probability that the electron finds itself on the far side, thereby leaving the two positive faces in contact.?

c) ...thermal energy of the neutron at room temperature being 2.4 Mev higher than the ultra cold state of being in the nucleus.

d) ...that the neutron doesn't move at all, and that we the observers move through time at a relatively different rate than a sub atomic particle, thereby giving us impression that the particle is moving.

Food for thought...

Steven
http://www.gammaspectacular.com - Gamma Spectrometry Systems
https://www.researchgate.net/profile/Steven_Sesselmann - Various papers and patents on RG
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Re: What forces are at work when neutrons are ejected?

Post by Dennis P Brown »

I answered the question on the energy of the neutron 'ejected' from the nucleus very clearly far up in this thread. Whether one wants to believe scientific fact (and that does not mean something else isn't causing the process, only that is how science works) or just go off into speculation is fine. But the answer does not require a force - no more than a photon needs a force to start at the speed of light - that is just as 'counter intuitive’ as the neutron issue yet no one here is asking how that is possible.

So, once more I will answer a question using accepted science which will run against the grain ... but a neutron star does not decay in fifteen minutes for the same reason a neutron does not decay in a nucleus - it is less stable if it decays than if it remains a neutron. A neutron lowers its energy when it is bound. The only way to confirm this is to calculate the energy in each case - trying to use a easy to understand example in our macro world just will not cut it for quantum effects. Can't get around that.
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Re: What forces are at work when neutrons are ejected?

Post by Jeff Robertson »

Another thing to keep in mind when discussing nuclear-related topics such as this is that "force" is a very classical, borderline obsolete term. While it may be appropriate to talk of forces on a macro scale, the concept of forces (in the sense that particles are "pushing" or "pulling" on each other) holds no meaning on the nuclear level (or in the quantum world at all, for that matter). Modern quantum field theory (and most modern quantum theories) states that these "forces" are really the exchange of particles carrying momentum/energy. For the electromagnetic force this is the well known photon, for the strong interaction this is the gluon, and the weak interaction is governed by the exchange of W and Z bosons (some versions of quantum field theory predict an analogous particle for gravity called the "graviton," although gravity is such a small force in comparison that it would be nearly impossible to detect said particle).

That said, if you're trying to explain neutron emission in terms of some force pushing or pulling the neutron out of the atom, you may be left feeling unsatisfied. In fact, if you're trying to reconcile a neutron's behavior with any sort of classical concept of motion or dynamics, chances are good your answer is either wrong or incomplete. Steve's ideas are interesting to think about, but with all the success quantum mechanics has had in explaining subatomic particles, I doubt that neutron behavior can be fully explained with such archaic concepts (although they could very well play a part, this is all just speculation after all).

While I'm more inclined to believe the solutions Dennis or myself proposed (since quantum has been the most successful theory on the nuclear scale to date), I agree with Richard's notion of playing devil's advocate. We can't assume we "know" anything about the neutron. We know nothing for certain about its shape, size, constituents, or behavior. We can't even observe a neutron directly (have you seen a neutron before? because I haven't!), we can only observe its effects on the environment and infer that a neutron must be in play.

All of that said, I'm still a fan of my energy balance proposal . Within every particle there is an internal tug-o-war between the particle's kinetic/thermal energy and the potential energy due to surrounding particles. If the kinetic energy is greater then the particle is free to roam; if the potential energy is greater, then the particle is bound to the system (this is one of the reasons why planets orbit around the sun, and electrons are bound to nuclei). This is how plasma is formed - we give enough energy to the atoms until the nuclei and electrons break free from each other, resulting in a "soup" of ions and electrons freely roaming. Energy conservation is one of the few physics concepts that has survived from classical mechanics all the way through modern quantum theory.

When a neutron is bound to the nucleus, it means that its potential energy due to the strong force is stronger than its internal energy. When some event, such as the collision of an additional neutron with the nucleus, shifts the atom, the neutron is pushed out of equilibrium for a moment. If this shift causes the neutron to temporarily lose enough potential energy so that it drops below its internal energy, the neutron will be able to break free from the nucleus. If it is not enough to upset the neutron's balance, then it'll simply settle back down into equilibrium. These are just my thoughts at least.

Apologies for the long post, this always seems to happen when I start talking about quantum!

Jeff
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Re: What forces are at work when neutrons are ejected?

Post by Frank Sanns »

Jeff Robertson wrote:


> While I'm more inclined to believe the solutions xxxxx or myself proposed (since quantum has been the most successful theory on the nuclear scale to date), I agree with Richard's notion of playing devil's advocate. We can't assume we "know" anything about the neutron. We know nothing for certain about its shape, size, constituents, or behavior. We can't even observe a neutron directly (have you seen a neutron before? because I haven't!), we can only observe its effects on the environment and infer that a neutron must be in play.

Neutrons have no external electric charge so they should and do pass right though solid objects. It is the Coulomb force that keeps normal matter looking and acting the way it macroscopically does. Similarly a visible photon should not interact with a neutron both because of the wavelength of a visible photon would be too long and the lack of charge would make a neutron invisible. A neutron star then should be invisible or should I say transparent to visible light. Gravitational lensing would be expected though as the gravitational forces would still do their work.


> When a neutron is bound to the nucleus, it means that its potential energy due to the strong force is stronger than its internal energy. When some event, such as the collision of an additional neutron with the nucleus, shifts the atom, the neutron is pushed out of equilibrium for a moment. If this shift causes the neutron to temporarily lose enough potential energy so that it drops below its internal energy, the neutron will be able to break free from the nucleus. If it is not enough to upset the neutron's balance, then it'll simply settle back down into equilibrium. These are just my thoughts at least.
>

And what of anti color charge with neutrons? Anti electrical charge is easy to appreciate but when it involves the color charge, all bets are off. Anit matter has its own consequences and ramifications. Is this anti potential energy or positive potential energy or double potential energy? At high packing densities does degeneracy become equivalent to the environment in an atom or vice versa?

Frank Sanns
Achiever's madness; when enough is still not enough. ---FS
We have to stop looking at the world through our physical eyes. The universe is NOT what we see. It is the quantum world that is real. The rest is just an electron illusion. ---FS
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Re: What forces are at work when neutrons are ejected?

Post by Jeff Robertson »

Well I was being facetious with my comment about not being able to "see" neutrons, perhaps I should have made it more obvious. Low energy photons do not interact with neutrons because, as you said, they're electrically neutral. The quarks that make up a neutron (for those who don't know, three "quarks" are what come together to form a neutron) do have electric charges, though, so I believe high energy photons do interact with the quarks inside a neutron. Not completely positive though (no pun intended), if anyone can verify this?

As for anti color charge, I'm not exactly sure what you're getting at. My understanding of anti particles is that they're identical except for having opposite spin and electric charge. If this is true, this would imply that the electromagnetic potential energy of an anti particle would have the opposite sign as the potential energy of its particle counterpart, which is what I think you were saying. The potential we've been talking about in this thread, though, is due to the strong interaction, and as far as I can tell the strong nuclear force interacts with matter and antimatter in the same way.

I'm rather rusty on quarks and all that fun stuff, so take everything I say with a grain of salt.

Jeff
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Re: What forces are at work when neutrons are ejected?

Post by Frank Sanns »

Jeff,

Quantum Chromo Dynamics (QCD) is just one of the many theories related to hadrons which of course includes quarks and obviously then neutrons and protons. I used it as an example because anti matter is not anti coulomb charge matter. Negative and positive electric charges neutralize but they do not annihilate like matter and anti matter (anti color charge). The color charge is an attribute just like a Coulomb charge but with quite different properties. Also spin or mass alone does not cause annihilation .

Quarks, which are the accepted building blocks of hadrons (protons and neutrons in the case of the original thread), have properties that include electric charge of -1/3 or +2/3, mass, spin, and color charge. It is this latter that is quite interesting in the stability or instability of quark composites like protons and neutrons. Call it whatever your will but what is referenced as color charge is what has a profound effect on stabilities, and reactivities. The attributes mentioned above are definitely the smoking gun of ejections from nuclei.

Frank Sanns
Achiever's madness; when enough is still not enough. ---FS
We have to stop looking at the world through our physical eyes. The universe is NOT what we see. It is the quantum world that is real. The rest is just an electron illusion. ---FS
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