Charged or Neutral Strong Nuclear Force Exchanges?
Chalky
the strong nuclear force (now re-christened nuclear force or residual
strong force) is currently illustrated as a neutral pi meson exchange
(itself further complicated by representation of such a neutral meson
exchange as a residual consequence of the standard model)
However, I have a problem with that representation for the following
reason:
IF that force were mediated (let us say 50/50) via neutral meson
exchanges, we might expect to find nuclei containing plural protons
and no neutrons.
Do we observe such nuclei? NO.
IF, on the other hand, that force is mediated only by charged meson
exchanges, we would only expect to find plural nucleon nuclei
containing neutrons, with the statistical proportion of neutrons being
about 50%.
Do we observe an overwhelming preponderance of such nuclei? YES.
Has some subtlety of consideration within the standard model gone over
my head, or is my logic reasonably sound here?
Phillip Helbig---undress to reply
<ef3691d4-22e8-4d8c...@r31g2000vbi.googlegroups.com>,
Chalky <chalk...@bleachboys.co.uk> writes:
> IF that force were mediated (let us say 50/50) via neutral meson
> exchanges, we might expect to find nuclei containing plural protons
> and no neutrons.
>
> Do we observe such nuclei? NO.
>
> IF, on the other hand, that force is mediated only by charged meson
> exchanges, we would only expect to find plural nucleon nuclei
> containing neutrons, with the statistical proportion of neutrons being
> about 50%.
>
> Do we observe an overwhelming preponderance of such nuclei? YES.
>
> Has some subtlety of consideration within the standard model gone over
> my head, or is my logic reasonably sound here?
I think a simpler explanation is that proton-only nuclei are not bound.
Neither are neutron-only nuclei. (IIRC, the dineutron is ALMOST bound.)
Lester Welch
> I note that, athttp://en.wikipedia.org/wiki/Nuclear_force
One cannot neglect the role of Fermi statistics. Related to this
point is the spin of deuterium, which is 1 not 0 - that is the spin of
the nucleons are aligned. Such a state is not allowed for identical
fermions - thus 2He is not possible. The point being that the strong
force has a spin dependent component which can not be ignored.
Chalky
undress to reply) wrote:
> I think a simpler explanation is that proton-only nuclei are not bound.
> Neither are neutron-only nuclei. �
This is obviously true, but the question is why?
(IIRC, the dineutron is ALMOST bound.)
This is ALMOST obvious since there is no net electrostatic repulsion
to overcome :-)
Chalky
> One cannot neglect the role of Fermi statistics. �Related to this
> point is the spin of deuterium, which is 1 not 0 - that is the spin of
> the nucleons are aligned. �Such a state is not allowed for identical
> fermions - thus 2He is not possible. �The point being that the strong
> force has a spin dependent component which can not be ignored.
This would seem to imply that protons can only bind to neutrons and
vice versa in heavy nuclei.
Is that correct?
If so, I guess my original question would resolve into:
How often is this interaction neutral, and how often charged (ie
proton becomes neutron and vice versa)
Lester Welch
>
> This would seem to imply that protons can only bind to neutrons and
> vice versa in heavy nuclei.
> Is that correct?
No. The strong nuclear force causes an attraction between any
combination of p's & n's.
The strength depends of the spin. Some particle/spin combinations are
not allowed because of
Fermi statistics.
Phillip Helbig---undress to reply
<9a114731-974d-40e9...@m38g2000yqd.googlegroups.com>,
Chalky <chalk...@bleachboys.co.uk> writes:
> > (IIRC, the dineutron is ALMOST bound.)
>
> This is ALMOST obvious since there is no net electrostatic repulsion
> to overcome :-)
Not obvious. Hypothetical nuclei consisting of a number of neutrons
larger than 2 are also not bound, but they are not "almost bound".
Chalky
> No. �The strong nuclear force causes an attraction between any
> combination of p's & n's.
> The strength depends of the spin. �Some particle/spin combinations are
> not allowed because of
> Fermi statistics.
OK, but then, why can't you get nuclei containing just two protons of
opposing spins?
Or, for that matter, Deuterium nuclei of no net spin?
Chalky
undress to reply) wrote:
> In article
> <9a114731-974d-40e9-9365-3356d9e60...@m38g2000yqd.googlegroups.com>,
>
> Chalky <chalkys...@bleachboys.co.uk> writes:
> > > (IIRC, the dineutron is ALMOST bound.)
>
> > This is ALMOST obvious since there is no net electrostatic repulsion
> > to overcome :-)
>
> Not obvious. Hypothetical nuclei consisting of a number of neutrons
> larger than 2 are also not bound, but they are not "almost bound".
It seems to me that you may well be delving into realms of speculation
here, which go well beyond the bounds of the scientific method. You
appear to assume that the currently accepted paradigm is accurate to
far beyond the (even theoretical) limits of observational evidence.
I prefer to be more of a scientific sceptic, based on the historical
evidence that every paradigm of natural philosophy hitherto, has been
found to be wanting, in the light of subsequent observational
evidence.
Whilst I appreciate your comments (I assume that you have made these
in the context of total faith in the [current] standard model), I
would prefer to take them with a modicum of salt, given the historical
perspectives that unbiased study of the history of scientific
revolutions teaches us, and the lessons we can learn from that.
However, my response here might possibly be different, if either of
you could provide a credible response to my own prior posting.
Phillip Helbig---undress to reply
<c1379d32-9fc0-4cba...@a32g2000yqm.googlegroups.com>,
Chalky <chalk...@bleachboys.co.uk> writes:
> > > > (IIRC, the dineutron is ALMOST bound.)
> >
> > > This is ALMOST obvious since there is no net electrostatic repulsion
> > > to overcome :-)
> >
> > Not obvious. Hypothetical nuclei consisting of a number of neutrons
> > larger than 2 are also not bound, but they are not "almost bound".
>
> It seems to me that you may well be delving into realms of speculation
> here, which go well beyond the bounds of the scientific method. You
> appear to assume that the currently accepted paradigm is accurate to
> far beyond the (even theoretical) limits of observational evidence.
Nuclear physics is actually a branch of science where essentially all is
known; there are no outstanding puzzles. (Of course, nuclear physics is
ultimately due to particle physics, where there are some big puzzles,
but then so is zoology ultimately due to particle physics although we
don't let that stop us from, say, sequencing the genome of some
species.) According to that theory, which we have no reason to doubt,
nuclei consisting of solely neutrons or protons do not exist.
The burden of proof, of course, is on those who claim that they do.
While it might be impossible to definitively prove that such nuclei
could not exist under any circumstances, it is easy to prove that they
do: just show one to the world. Extraordinary claims demand
extraordinary evidence, and in the case of nuclear physics, where we
have no indication that anything is wrong or even incomplete, that
evidence would have to be strong indeed.
Lester Welch
All we have is experiment to guide us. Put a spin dependence in the
strong force - adjust the parameters so that experiment is agreed with
and you'll find that a deuterium with no spin is not bound and that
the (spin dependent) strong force is not sufficient to overcome the
electrostatic repulsion of two protons. Why is that so? Why are the
parameters the value they are. No one knows - yet. Why does c and e
have the values they do?
Chalky
undress to reply) wrote:
> In article
I think you misunderstood my point here....ie they are not observed,
period, and so, theoretical calculations to determine how extremely
they cannot be observed go beyond the bounds of the experimental
method.
Jonathan Thornburg [remove -animal to reply]
[[dineutrons, diprotons, and other such beasts]]
> I think you misunderstood my point here....ie they are not observed,
> period, and so, theoretical calculations to determine how extremely
> they cannot be observed go beyond the bounds of the experimental
> method.
Consider the question: Why can't a human (unaided by mechanical
equipment or wings) take off from the Earth's surface and fly through
the air by flapping her arms vigorously?
Humans doing this are not observed, period, and so, theoretical
calculations (involving aerodynamics, human muscle power, etc) to
determine how extremely they cannot be observed go beyond the bounds
of the experimental method.
--
-- "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
Dept of Astronomy, Indiana University, Bloomington, Indiana, USA
"If the triangles made a god, it would have three sides." -- Voltaire
Igor Khavkine
<jth...@astro.indiana-zebra.edu> wrote:
> Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> [[dineutrons, diprotons, and other such beasts]]
>
> > I think you misunderstood my point here....ie they are not observed,
> > period, and so, theoretical calculations to determine how extremely
> > they cannot be observed go beyond the bounds of the experimental
> > method.
>
> Consider the question: Why can't a human (unaided by mechanical
> equipment or wings) take off from the Earth's surface and fly through
> the air by flapping her arms vigorously?
>
> Humans doing this are not observed, period, and so, theoretical
> calculations (involving aerodynamics, human muscle power, etc) to
> determine how extremely they cannot be observed go beyond the bounds
> of the experimental method.
With this discussion of what is and is not observed, I wonder if
anyone bothered to check the most accessible place:
http://en.wikipedia.org/wiki/Diproton
http://en.wikipedia.org/wiki/Dineutron
These articles have references which appear to have at least some
discussion of observability, if perhaps only as short lived resonance
states.
Igor
Chalky
<jth...@astro.indiana-zebra.edu> wrote:
> Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> [[dineutrons, diprotons, and other such beasts]]
>
> > I think you misunderstood my point here....ie �they are not observed,
> > period, and so, theoretical calculations to determine �how extremely
> > they cannot be observed go beyond the bounds of the experimental
> > method.
>
> Consider the question: Why can't a human (unaided by mechanical
> equipment or wings) take off from the Earth's surface and fly through
> the air by flapping her arms vigorously?
>
> Humans doing this are not observed, period, and so, theoretical
> calculations (involving aerodynamics, human muscle power, etc) to
> determine how extremely they cannot be observed go beyond the bounds
> of the experimental method.
Your choice of example reads to me uncannily like a justification of
Lord Thompson's late 19th century assertion that heavier than air
flying machines are impossible.
However, you have had the common sense (or political intelligence) to
add additional qualifications, in order to retain a fragment of
credibility to that argument, in the light of a century of additional
unambiguous observational evidence.
Nevertheless, since pure physics assertions (like revenge) are best
served cold, in order to enable considered perspective, I would prefer
to address your argument via allusion to its perceived more general
original context.
IMO Theoretical research physicists should be interested in
identifying and/or discovering universal laws of nature. In this
context, assuming that what we have already learned is everything
that can be learned, is folly in the extreme.
In Lord Kelvin's case, that folly was compounded by the fact that he
should have already known that birds and bees can fly (as can gliding
condors, various spiders, and dandelion seeds, [without flapping any
appendages])
My original point, however, was not that if standard theory asserts
that certain things are impossible (to varying levels of certainty
above 100%), then they must in fact, be possible in inverse
proportion to those asserted levels of certainty. It was that the
specificity of the strong nuclear force might also be explainable
(more simply), if we dump the idea of neutral meson exchange
particles.
AFAICT, no respondent has yet addressed that issue.
Chalky
[[Mod. note -- 23 excessively-quoted lines snipped.
Posters, *please* do this yourselves! -- jt]]
> With this discussion of what is and is not observed, I wonder if
> anyone bothered to check the most accessible place:
>
>
> These articles have references which appear to have at least some
> discussion of observability, if perhaps only as short lived resonance
> states.
>
> Igor
Thanks. The refs seem to indicate that reactions necessarily
attributable to a neutral strong force are statistically pretty rare
in nature, which, I guess is a corollary to what PH said.
Consequently, if I postulated that there were at least 10^4 charged
pion exchanges in nature for every neutral one, this evidence would
not seem to contradict that, AFAICT.
Igor Khavkine
> My original point, however, was not that if standard theory asserts
> that certain things are impossible (to varying levels of certainty
> above 100%), then they must in fact, be possible in inverse
> proportion to those asserted levels of certainty. It was that the
> specificity of the strong nuclear force might also be explainable
> (more simply), if we dump the idea of neutral meson exchange
> particles.
>
> AFAICT, no respondent has yet addressed that issue.
Chalky, your original point has been addressed, precisely by deferring
to the level of confidence in the standard model and other aspects of
nuclear physics. The fact that, as you say, none of the respondents
have demonstrated that your specific idea about meson exchange is
wrong (other than by pointing out that it's different from the
scientifically accepted theory of nuclear forces), is more of a
statement about the collective knowledge of the respondents on the
history of nuclear physics, rather than anything else.
Also, the collective point of the respondents has been that, even
though we don't have the details of history on our finger tips, they
are there to be found in books on nuclear and particle physics. There
for anyone interested to find and study them. *hint, hint* :-)
Now, being more kind to your direct question, here is some historical
information from Abraham Pais's history of particle physics _Inward
Bound_ (OUP, 1986), Chapter 17(f):
"In his first nuclear physics paper [104] Heisenberg had realized that
the approximate equality between the number of protons and neutrons in
nuclei implied that short-range forces between nn could not be very
different from those between pp. Since he did not introduce such
pp-forces at all, he had therefore to assume the nn-force to be weak
compared to np. For the same reason in early models [166] with both
nn- and pp-interactions these two were taken equal. In 1935, it was
shown [167] by variational methods that the difference between the
binding energies of H3 (pnn) and He3 (ppn) can entirely be accounted
for by Coulomb effects---which implies equality of short-range pp- and
nn-forces. 'We know today how hard it is to get good values for [the
binding energies of] these nuclei [168]. However that may be, it was
appreciated early that nuclear pp- and nn-forces, if present, should
be equal, a property since named charge symmetry.
In August 1936 it became clear that such forces actually exist, that
they are strong, and that they are intimately related ta pn-forces.
Three papers received by Physical Review within two days in that month
mark major strides forward in the understanding of nuclear
interactions.
One, by Tuve and co-workers [119], announced evidence from scattering
experiments for strong pp-interactions. Another by Breit, Condon, an
Present [169], contained an analysis of these data, a comparison with
available pn-scattering information and the following conclusion: 'The
pp- and pn-interactions in 1S-states, are found to be equal within th
experimental error. This suggests that interactions between heavy
particles are equal also in other states.'
The third paper [170], by Cassen and Condon, deals with a formalism
that incorporates equality of pp-, pn-, and nn-forces in 1S-states, a
property called charge independence. Their tools are Heisenberg's
matrices tau, Eq. (17.11), wich they called the characteristic
vector. In 1937 Wigner [171] gave tau the name 'isotopic spin', not a
fortunate choice. Today the common temm is 'isospin'."
Here are the relevant references.
104. W. Heisenberg, Zeitschr. f. Phys. v78, p156, 1932.
119. W. H. Wells, Phys. Rev. v47, p591, 1935;
M. G. White, Phys. Rev. v49, p309, 1936;
M. A. Tuve, N. Heydenberg, and L. R. Hafstad,
Phys. Rev. v.50, p.806, 1936.
166. K. Guggenheim, J. de Phys. et le Rad. v5, p475, 1934;
L. A. Yaung, Phys. Rev. v47, 972, 1935.
167. E. Feinberg and J. G. Knipp, Phys. Rev. v48, p906, 1935.
168. R. Peierls, in Nuclear physics in retrospect, Ed. R. H. Stuewer
(Univ. Minnesota Press, Minneanapolis; 1979), p189.
169. G. Breit, E. U. Condon. and R. D. Preent. Phys. Rev. v5O, p825,
1936.
170. B. Cassen and E. U. Condon, Rhys. Rev. v5O, p84, 1936.
171. E. P. Wigner, Phys. Rev. v51, p106, 1937.
The evidence for charge independence and notions of isospin later lead
to the postulation of the three pions, which are now known to fit all
the data at appropriate energies, hence their membership in the
standard theory of nuclear forces. For more information, you can start
by looking up "pion" in Pais's book and chasing the references.
Hope this helps.
Igor
CarlB
stability is that the nuclear force is a Yukawa potential and falls
off exponentially, while the electric force is carried by the massless
photon and falls off 1/r^2. Consequently, at long enough distances,
the electric potential dominates. From that, it's a matter of QM to
compute how far "long enough" is. Mostly it will depend on the
proton / neutron masses.
> Consider the question: Why can't a human (unaided by mechanical
> equipment or wings) take off from the Earth's surface and fly through
> the air by flapping her arms vigorously?
In fact I can fly, but only for brief periods of time. (About 1 second
or so.)
Chalky
undress to reply) wrote:
> In article
> Nuclear physics is actually a branch of science where essentially all is
> known; there are no outstanding puzzles.
In that case, perhaps the combined wisdom of the respondents can also
be used to establish whether there are any further blunders in my
(admittedly simplistic) modelling of the strong nuclear force as a
vacuum fluctuation. For this I take the measured massenergy of the
pion as del E in Heisenberg's uncertainty relationship and interpret
the resultant del T as both the spatial and temporal radius of any
such vacuum fluctuation (when expressed in natural units with c=1).
Now, for my purposes I get a far more credible figure for R by
plugging in the charged pion mass than by using the neutral one.
This leads me to ask next: what is the typical radius of a strong
nuclear vacuum fluctuation according to nuclear physics?
While we are on this subject, I recall (I think from Wiki) that the
composition of the neutral pion is not actually known.
If not, why can't we just model it as a quantum superposition of two
oppositely charged pions.....
And, If so, shouldn't the 3 vacuum fluctuations then all have the same
radius?
Igor Khavkine
> On Nov 30, 5:34 pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig---undress to reply) wrote:
> > In article
> > <c1379d32-9fc0-4cba-857d-eec883c9a...@a32g2000yqm.googlegroups.com>,
> > Nuclear physics is actually a branch of science where essentially all is
> > known; there are no outstanding puzzles.
>
> In that case, perhaps the combined wisdom of the respondents can also
> be used to establish whether there are any further blunders in my
> (admittedly simplistic) modelling of the strong nuclear force as a
> vacuum fluctuation.
And why would you do this? This kind of reasoning is so inaccurate
that the only sensible way to interpret it is as a heuristic way of
arranging various constants (hbar, speed of light, pion mass) into an
expression with units of length or time. It is just as easy to look at
the internucleon Yukawa potential, which has the trifecta of nice
features that it serves as the basis for some semi-realistic nuclear
models, it justifiable from pion field theory, and has an explicit
length parameter determining its range as a function of the pion mass.
> For this I take the measured massenergy of the
> pion as del E in Heisenberg's uncertainty relationship and interpret
> the resultant del T as both the spatial and temporal radius of any
> such vacuum fluctuation (when expressed in natural units with c=1).
>
> Now, for my purposes I get a far more credible figure for R by
> plugging in the charged pion mass than by using the neutral one.
Umm, these charged and neutral pion masses are within 5% of each
other. I don't see how using one mass or the other would give you
significantly different ballpark estimates, which is the only kind of
estimate that you can obtain this way.
> This leads me to ask next: what is the typical radius of a strong
> nuclear vacuum fluctuation according to nuclear physics?
I'm guessing that what you are really after here is the range of the
internucleon force. This length is set by the explicit length
parameter from the Yukawa potential. This parameter is just the
Compton wavelength of the pion: hbar / m_pion c ~ 1 fermi, in
agreement with the folk wisdom in nuclear physics.
> While we are on this subject, I recall (I think from Wiki) that the
> composition of the neutral pion is not actually known.
> If not, why can't we just model it as a quantum superposition of two
> oppositely charged pions.....
Lets see, in this Wikipedia article and section
http://en.wikipedia.org/wiki/Pion#endnote_quarkcontent
there is a footnote next to the listed composition of the neutral
pion, (u u-bar - d d-bar)/sqrt(2). The footnote says: "Make-up inexact
due to non-zero quark masses." If this is what you mean, then you are
reading way too much uncertainty into that statement. While I'm no
expert, the most likely reason for this remark is that the given
expression excludes corrections to it that are of order m_quark/
m_pion. The masses of up an down quarks are of the order of 1-5 MeV,
while the pions sit around 135 MeV. Therefore, the missing corrections
should also be of order 5%. But you don't have to believe me, there's
a reference to a standard particle physics book next to that footnote
(Griffiths), where you can read much more about it.
So, the answer to your "Why not?" question is that there is actually
quite a bit known about the make up of the neutral pion, and what is
known is not along the lines of what you suggest. On a completely
separate track, your suggestion does not work because (at least to my
knowledge) there has never been an observation of a coherent quantum
superposition of states of different electric charge. In fact, the
absence of such superpositions follows from standard quantum mechanics
and is known as "charge superselection".
Hope this helps.
Igor
Chalky
> On Dec 9, 9:21 pm, Chalky <chalkys...@bleachboys.co.uk> wrote:
>
> > On Nov 30, 5:34 pm, hel...@astro.multiCLOTHESvax.de (Phillip Helbig---undress to reply) wrote:
> > > In article
> > > <c1379d32-9fc0-4cba-857d-eec883c9a...@a32g2000yqm.googlegroups.com>,
> > > Nuclear physics is actually a branch of science where essentially all is
> > > known; there are no outstanding puzzles.
>
> > In that case, perhaps the combined wisdom of the respondents can also
> > be used to establish whether there are any further blunders in my
> > (admittedly simplistic) modelling of the strong nuclear force as a
> > vacuum fluctuation.
>
> And why would you do this?
In order to facilitate a direct derivation of the strong nuclear
gravity interaction at the big bang, via direct logical deduction from
the axiomatic foundations of GR theory. IE vaguely along the lines
suggested by Misner, Thorne, and Wheeler, in the final chapter of
GRAVITATION.
> This kind of reasoning is so inaccurate
> that the only sensible way to interpret it is as a heuristic way of
> arranging various constants (hbar, speed of light, pion mass) into an
> expression with units of length or time.
I would suggest it is also statistically relevant, both for my
purposes and for yours.
> It is just as easy to look at
> the internucleon Yukawa potential, which has the trifecta of nice
> features that it serves as the basis for some semi-realistic nuclear
> models, it justifiable from pion field theory, and has an explicit
> length parameter determining its range as a function of the pion mass.
It is nice to receive your direct confirmation that my simplistic
approach does, in fact, have an exact parallel in the standard model
(see further comment below)
> > For this I take the measured massenergy of the
> > pion as del E in Heisenberg's uncertainty relationship and interpret
> > the resultant del T as both the spatial and temporal radius of any
> > such vacuum fluctuation (when expressed in natural units with c=1).
>
> > Now, for my purposes I get a far more credible figure for R by
> > plugging in the charged pion mass than by using the neutral one.
>
> Umm, these charged and neutral pion masses are within 5% of each
> other. I don't see how using one mass or the other would give you
> significantly different ballpark estimates, which is the only kind of
> estimate that you can obtain this way.
Actually, the resultant predicted value of mean matter density now,
turns out to be proportional to the sixth power of that massenergy,
which makes a huge difference when comparing resultant predictions
with the 5 year WMAP data (published this year).
> > This leads me to ask next: what is the typical radius of a strong
> > nuclear vacuum fluctuation according to nuclear physics?
>
> I'm guessing that what you are really after here is the range of the
> internucleon force. This length is set by the explicit length
> parameter from the Yukawa potential. This parameter is just the
> Compton wavelength of the pion: hbar / m_pion c ~ 1 fermi, in
> agreement with the folk wisdom in nuclear physics.
I think you will find that this is EXACTLY the same as what I
suggested, numerically.