# B World made up of 2nd & 3rd Generation particles

1. Aug 30, 2016

### cube137

This is a common question asked but I can't find a good reference.

Supposed the first generation particles (electron, quark up, down) vanish and we only had STABLE second generation particles composing of muon, strange and charm quarks that suddenly became stable (don't decay).. would they form stable molecules and objects (Or stable third generation set of particles like the tau, bottom & top quarks that don't decay, what kind of molecules and objects would they form?)

2. Aug 30, 2016

### ChrisVer

what kind of answer are you searching for? If the world was not the world we live in, then it would be something else...
Since you want to remove the 1st generation completely, then the ones you are talking about could form "atoms" or "molecules" (they are the lightest particles and so cannot decay), but they would be totally different... for example the atomic orbitals of muons are different to those of electrons, something that can affect the chemical behavior of atoms and so molecules.

3. Aug 30, 2016

### cube137

How different.. for example.. can strange and charm quarks form into nucleus and the muons become orbitals of them.. could we have periodic table too of this 2nd gen particles (if somehow they became stables and some vacuum symmetry unbreaking removed the first gen particles).

Just pondering on the purpose of the 2nd and 3rd gen particles and thinking what kinds of molecules would be produced if they were stable.

4. Aug 30, 2016

### ChrisVer

strange and charm quarks can form hadron states which are of course unstable because they can decay (their mass permits it) into something else. If you don't give them a path to decay into, then those hadrons could be stable ones (as the protons are).

the muons are like heavier electrons, so yes- given the right charges they can occupy atomic orbitals... they already do that http://www.physics.umd.edu/courses/Phys401/bedaque08/homework_5_solution.pdf ...

However, as physics is already built as it is, you can't really answer such questions which at the end of the day come down to "if physics is not what we know it is, then what ... ". First of all taking generations out of the scope can result in several misfunctionings of the theory (going the other way around to how the charm quark was predicted).

Molecules or atoms won't be so affected by the nature of the nuclei.... they will be mostly affected by the heavier muons.

5. Aug 30, 2016

### Staff: Mentor

It depends on how the first generation vanishes - you cannot easily remove one generation from the standard model. If we ignore things like CP violation (which needs at least three generations), then you just get heavier particles everywhere.

2 charm + 1 strange give a "heavier proton", 1 charm + 2 strange give a "heavier neutron".
Now the "proton" is much heavier than the "neutron", and will decay quickly to a "neutron" plus antimuon plus muonneutrino. The energy difference is so large that not even nuclei can stabilize a "proton", so no stable nuclei can form. Atoms are extremely short-living in that world.

6. Aug 30, 2016

### cube137

I'm just wondering when I came across chapter 8 "Problems of the Standard Model" in Peter Woit "Not Even Wrong". Quoting briefly:

"*Why do the quark and leptons of each generation come in a certain pattern? In mathematical terms, the quarks and leptons come in certain representations of the SU(3)xSU(2)XU(1) symmetry group. Why these specific representations and not others? This includes the questons of why the weak interactions are chiral, with only one-handedness of particles experiencing the SU(2) gauge field force
*Why three generations? Could there be more with highest masses that we have not seen?"

Anyone know more than Woit?

7. Aug 30, 2016

### ChrisVer

No, since we don't know what is beyond the Standard Model.
However the last questions don't imply the vanishing of a generation, but ask whether there is a 4th generation... Searches looking for 4th generation fermions are done, and up to now nothing has been found.
Similarily searches for Left Right symmetry (eg a Model like SU(3) x SU(2) x SU(2) x U(1) ) theories have also been conducted.

8. Aug 30, 2016

### cube137

Of course I know a generation doesn't just vanish.. but I found the following information from Lubos Motl...

"Otherwise, the fact that there are 3 generations in a particular Universe can be derived from deeper properties of string theory (half of the Euler character of the Calabi-Yau shape, assuming a conventional heterotic compactification for a while), and as I have hinted, even at this very point, it might be possible to show that the number of generations cannot be one, among other forbidden values. While three-generation models are known, it's not fully known at this moment whether 3 generations is a unique solution to some conditions or whether it's a coincidence, as the anthropic reasoning wants us to immediately believe."

9. Aug 30, 2016

### Staff: Mentor

It can be derived in speculative extensions of the SM. Which just means "if we find a 4th generation, that specific extension approach is wrong". There are other extensions with more generations.

10. Aug 30, 2016

### ChrisVer

Well, yes there are such string models which give the number of generations as the Euler character of the Calabi-Yau shape. However, I cannot really help you there because I am not a string theorist (neither a string believer), since string theory has 2 major problems: it lacks any measurable prediction and it has problems within its own framework (eg it's not 1 theory)... but I think that if you manipulate a theory too much it can give you what you ask for, but that is not actually a prediction (eg what is special about half of the Euler character?) . Maybe someone else could help.

11. Aug 31, 2016

### ohwilleke

This is not a common question at all. But, it is a rather obvious one and props to you for taking the time to think about it because asking it is a good frame for exploring parts of the Standard Model and particle physics that most people don't spend much time thinking about in an enjoyable way.

It is non-trivial to figure out just what such a world would look like, but we have almost all of the information we need to figure it out from what we know about the Standard Model so long as the assumptions are clarified. Certainly, hadrons with third generation quarks would be as scarce as they are in our world in a world with second but not first generation quarks. And, only a very small number of hadrons containing only charm and/or strange quarks would be stable. The trick would be to first figure out which hadrons were stable or metastable. This, in turn, would depend upon whether charm quarks could decay to strange quarks in the same manner that down quarks can decay to up quarks in our world (with the reverse also occurring in each respective scenario when not barred by conservation of energy), or whether charm and strange quarks simply could not decay further even though that is not strictly analogous to the first generation particle behavior. I'll focus on the former, because it is closer to the concept that I think you're trying to explore and involves dynamics that are pretty easy to make sense of. Even a slight deviation from the Standard Model's weak force makes the question a lot more foreign and complicated when simply eliminating the option of first generation fundamental fermions without specifically specifying that they are stable creates the kind of stability that I think you are envisioning.

Electromagnetically, a spin-1/2 baryon with two charm quarks and a strange quark, or with one charm quark and two strange quarks, have the same basic chemical properties as protons and neutrons respectively. But, the proton and neutron have almost identical masses, while the second generation equivalent would not, so baryons with charm quarks would be strongly prone to decay, if possible.

The charmed omega baryon which is analogous to the neutron has a mass of about 2.7 GeV, and the doubly charmed omega baryon which is analogous to the proton has an unknown mass which would be at least about 3.9 GeV. Several spin-3/2 baryons would be lighter than the spin-1/2 baryons made up only of charm and strange quarks. The omega baryon is spin-3/2 and has a mass of about 1.7 GeV and an electromagnetic charge of -1. A spin-3/2 charmed omega baryon (analogous to the positively charge delta baryon in our world) has a mass of about 2.8 GeV. The mass of a spin-3/2 double charmed omega baryon (analogous to the neutral delta baryon in our world) has an unknown mass but would be at least 4.0 GeV. The mass of a triply charmed spin-3/2 omega baryon (analogous to the charge +2 delta baryon in our world) has an unknown mass but would be at least 5.2 GeV.

If charmed baryons were allowed to decay to less charmed baryons in this world, almost all matter would decay to uncharmed omega baryons, atypical isotypes of atomic nuclei would be very rare, and every atomic nucleus would have a negative electric charge. Assuming that the W+ bosons produced in charm quark ultimately decayed to produce mostly antimuons and muon neutrino pairs (plausible given conservation of charge), the resulting atoms would have chemistry a lot like atoms in our own world but with only one isotype of each periodic table element and atomic masses a bit more than twice as great.

The fact that baryons in atoms would usually have spin-3/2 rather than spin-1/2 would also tweak the chemistry, in ways that would be predictable but subtle. No spin-3/2 particles in our world have mean lifetimes of more than a fraction of a second, so we don't have a lot of experimental data on how chemistry would differ in a world of spin-3/2 rather than spin-1/2 particles. But, since both are fermions, and most chemistry involves the leptons associated with an atom, rather than direct interaction of their nuclei, the fact that the nucleons were spin-3/2 rather than spin-1/2 would probably mostly affect the stability of certain elements. Instead of many isotypes of multi-baryon elements in our world that differ by number of neutrons, the variation among atoms with the same number of nucleons would depend on the extent to which the spin-3/2 nucleons had aligned or not aligned spins (and there would be only one kind of pseudo-hydrogen).

The mix of elements by atomic number might be different, however, depending mostly upon what the analog to the pion (which has a mass of less than 1/6th of the proton) which transmits the non-fundamental nuclear binding force in our world looked like. The mesons which have only second generation quark content are the 3 spin-0 pseudoscalar: charmed eta meson (mass ca. 3.0 GeV and no electric charge) and strange D meson (mass ca. 2.0 GeV and charge +1), the anti-strange D meson (mass ca. 2.0 GeV and charge -1); the spin-1 vector mesons: the Phi meson (electrically neutral and about 1.0 GeV), the electrically neutral J/Psi meson (electrically neutral and about 3.1 GeV), the vector strange D meson (electrically +1 charged and about 2.1 GeV) and the vector anti-strange D meson (electrically -1 charged and about 2.1 GeV).

Since all of these mesons have greater masses relative to the uncharmed omega meson that would be the predominant nucleon in this world than the mass of the pion relative to nucleons in our world, the nuclear binding force carrier boson in a second generation quark only world would be much heavier than in our world relative to the hadrons bound in a nuclei which would tend to make heavier elements in the periodic table much less stable than they are in our world (in part because heavier carrier bosons have a shorter range and in part because heavier carrier bosons relative to the hadrons would be produced less frequently).

Also, the Phi meson is by far the lightest, so it would be the most obvious primary carrier for the non-fundamental nuclear binding force (and would have a very long lifetime eventually decaying to photons or to muon and anti-muon pairs), but since it is a vector meson, rather than a pseudo-scalar meson, this would significantly alter the character of the interaction, making it more like a spin-1 weak force boson or gluon with dynamical mass, and less like the pseudo-scalar meson which acts as the carrier boson in our world. In particular, a spin-1 nuclear binding force meson would probably give rise to more internal structure on pseudo-atom nuclei than in nuclei of our world where the nuclear binding force is carried by a spin-0 pseudoscalar meson. The nuclei of pseudo-atoms might have something of a molecular-like structure.

And keep in mind that about 99.93% of stuff in the universe is made up of only ten chemical elements; just four (Hydrogen, Helium, Oxygen and Carbon) make up more than 99%. Two of those ten chemical elements (helium and neon) moreover, are chemically inert. Our world only seems complex chemically because we live in an extremely atypical little corner of it.

Basically, in a second generation only world, the chemical equivalent of hydrogen would make up a much larger percentage of all matter relative to pseudo-helium and other heavier pseudo-elements, than in our world. And, since pseudo-helium in this world, like helium in our world, would be chemically inert, the share of molecules that were simple two atom hydrogen molecules would be much greater, and most other molecules would be one or more pseudo-hydrogen atoms bound to a single heavier than pseudo-helium element (e.g. pseudo-water and pseudo-methane).

In particular, nuclear fusion, which powers stars and arises from the reduced binding energy per nucleon in heavier elements relative to lighter elements, would be far less potent in this world than in our own. This would mean that stars would produce fewer heavy elements, that stars would emit fewer photons relative to their mass, and that stars would collapse into black holes are far lower luminosities. Supernovas would be less energetic. My intuition isn't clear on the question of whether these dimmer stars would burn out faster or slower, but it certainly isn't particularly likely that they would live the same length of time as stars of comparable mass in our own world.

This doesn't mean that there couldn't be enclaves in this universe of heavier elements in planets, just as the tiny percentage of non-light elements in the universe are concentrated in places like terrestrial planets like Earth, and they could have interesting and Earth-like chemistry. But, with less stable heavy elements, this world wouldn't be one that is almost indistinguishable from our own.

A third-generation particle only world where top quarks could decay into bottom quarks would be quite similar. A third-generation particle only world where top quarks were stable, in contrast, might have more heavy elements because pion equivalents containing only bottom quarks and anti-bottom quarks would be light relative to hadrons containing any top quarks.

Last edited: Aug 31, 2016
12. Aug 31, 2016

### cube137

Wow. you are really very good about nucleons and hadrons. You can make the above a good sci-am magazine contribution :) I'd like to ask a question. Do you know of any beyond the standard model or papers where the quarks or a particular quark is made up of other particles (subquarks or monopoles) that is bonded by string-like excitations of the superconducting Higgs vaccum? And where the nuclear forces holding the nucleons together arise from the residual coupling between their strings. Have you heard anything like this? And furthermore, if there is a vacuum domain activated (localize symmetry breaking), an effective reduction takes place in the degree of colour-shade symmetry that can separate the quarks (complementing or nullifying asymptotic freedom)?

13. Aug 31, 2016

### ohwilleke

AFAIK, the fact that the particle content of the Standard Model is represented by the SU(3)xSU(2)xU(1) symmetry group tells you that you should have four different kinds of fundamental fermions but does not itself explain why there have to be three generations of them.

There are some plausible reasons of internal consistency that are suggestive of reasons why there are three generations rather than more than three generations (which in turn implies that our roster of non-supersymmetric fundamental fermions is probably complete).

For example, the only way that particles change generations in the Standard Model is via W boson interactions. Generally, the heavier a particle is, the shorter its mean lifetime. The top mean lifetime of the heaviest quark, the top quark, is only a bit longer than the mean lifetime of the W boson. It could be that the decay rate of the W boson imposes a ceiling on the rate at which particles decay and that this ceiling, indirectly, imposes a limit on the heaviest mass that a fundamental particle can have. And, the mathematical structure of the Standard Model is such that fundamental fermions must come in sets of four (an up-type quark, a down-type quark, a charged lepton and a neutrino), which is how the charm quark, the top quark, the muon neutrino, and the tau neutrino were predicted.

The Yukawa (i.e. Higgs field coupling strength) of the top quark is almost 1, and there are suggestive reasons to think that a particle with a Yukawa of more than 1 is not possible.

The CKM matrix in the Standard Model which governs the probability of weak force transitions from one quark type to another imply that the probability of any given type of up quark quark transforming into one of the three down type quarks is almost exactly 100% in each of the three cases, and visa versa (for down type quarks transitioning to up type quarks). So, any mixing between higher order quarks and Standard Model quarks would have to be very small. Very small mixings, in turn, are generally associated with very large mass differences. But very heavy mass differences put pressure on the weak force decay bounds.

Of course, there are direct exclusions of fourth generation fermions from the LHC, but those bounds aren't large enough in and of themselves to rule out plausible SM4 theories (which is what theories with a fourth generation of Standard Model fermions that differ only in mass are called).

The fact that W and Z boson decays and neutrino oscillation data and cosmology data all strongly favor the hypothesis that there are only three rather than four species of "fertile" neutrinos with masses of 45 GeV or less (and accumulating Higgs boson decay data is close to pushing that bound to 62.5 GeV or less) means that if there were a fourth generation left handed neutrino mass eigenstate, that it would have to have a mass of more than 45 GeV, when the heaviest of the three neutrino mass eigenstates which are known is not more than about 0.1 eV (i.e. 0.0000000001 GeV) and all three of those masses are within a dozens of meV of each other, makes it highly unlikely that there is a fourth generation neutrino, and if there is not a fourth generation neutrino, then there can't be a fourth generation of any Standard Model fermion for consistency reasons.

Last edited: Aug 31, 2016
14. Aug 31, 2016

### ChrisVer

well you can still have sterile neutrinos no?
and also I think that the heaviest neutrino bound is some MeV ...?

15. Aug 31, 2016

### ohwilleke

I'll only address this briefly as it really belongs in the Beyond the Standard Model forum, while the original question which is really just a fun way to elicit the Standard Model properties of higher generation fermions does not.

Models in which some or all of the Standard Model fundamental particles are actually composite particles made up of something more fundamental are generically called "Preon" models after terminology which, if I recall correctly, was the terminology used by Pati and Salam in their 1974 paper which was one of the earliest preon model proposals (also "Technicolor" models propose a composite substitute for the Higgs boson). I've contributed to and in several cases been the original author of many of the articles on Wikipedia related to preon models and a number of notable preon papers are cited in the footnotes to the Preon article at Wikipedia and in other articles linked to it. So far, the LHC and prior colliders have not seen any sign of compositeness in the fundamental leptons and quarks, and have placed extremely strict bonds on the energy scales at which such compositeness could arise in the context of fairly naive and straightforward versions of preon models. Nobody has found any evidence distinguishing fundamental particles from the point particle representation that they have in the Standard Model at any scale we can probe.

Also, only a pretty small minority of preon models explain more than one generation of fundamental fermions or provide any insight into how preons give rise to the masses of the fundamental particles of the Standard Model which in that analysis are actually composite.

There are also quite a few papers that explore the idea that leptons and quarks are more similar than they seem in a scheme in which leptons are possible because there are really four colors rather than three colors, which one of those colors, or certain combinations of those colors, giving rise to leptons that don't interact via the strong force rather than quarks (in much the same way that nobel gases are chemically inert despite being composite particles made up of things that do interact chemically when found in other configurations). Indeed, Pati and Salam's original 1974 preon paper advanced this hypothesis. The paper was Pati, J.C.; Salam, A. (1974). "Lepton number as the fourth "color"". Physical Review D. 10: 275–289.

I am not aware of any peon models that specifically looks at bonds in the nature of "string-like excitations of the superconducting Higgs vacuum" and the connection between string excitation modes and particular fundamental particles in the Standard Model is much less direct and determinate than popularizations of string theory have implied.

The nuclear force holding nucleons together is established to be a residual force derived from the strong force carried by gluons that binds quarks together, and no one is publishing alternatives to this (i.e. basically QCD which is part of the Standard Model) although some physicists who have proposed preon models have considered the possibility that gluons may actually be composite bosons which are a residual force of the force that binds preons together.

16. Aug 31, 2016

### ohwilleke

Cosmology bounds establish that sterile neutrinos have to be heavy enough that they don't constitute hot dark matter and that they must have small mixing angles with ordinary neutrinos. But, a number of experiments in the last couple of years have pretty much ruled out the possibility of a light sterile neutrino ca. 1 eV +/- and the evidence from reactor anomalies that initially suggested this possibility has turned out to be a statistical fluke as more data has been collected and better analysis has been done. Now, sterile neutrinos are pretty much restricted to hypothetical heavy right handed particles that oscillate with ordinary neutrinos in some sort of see saw mechanism and dark matter candidates. Since sterile neutrinos would not interact via the weak force they would not leave traces in W and Z boson decays.

The neutrino bound you are referencing is from direct measurements of the mass of individual neutrinos based upon things like the lack of an experimentally discernible difference in neutrino speed from the speed of light for neutrinos of a given energy (mostly kinetic). But, these kinds of measurements are hopelessly crude.

Cosmology measurements place the upper bound on the heaviest neutrino mass much lighter (ca. 0.1 eV) and this is corroborated (1) by the small known mass differences between the three neutrino mass eigenstates as precisely determined in neutrino oscillation experiments and (2) by the absence of neutrinoless double beta decay observations to date which implies an upper bound on the Majorana component of neutrino mass. Given the small known mass differences between the three neutrino mass eigenstates, if the absolute mass of the neutrino mass eigenstates were even tens of eVs or more, let alone of MeV scale, the neutrino mass eigenstates would be virtually degenerate and a degenerate neutrino mass hierarchy is pretty strongly disfavored at this point.

17. Aug 31, 2016

### cube137

Please go to https://www.physicsforums.com/threads/preon-quark-models-excluded-by-lhc.883777/ for this Beyond the Standard Model topic. Thanks.

18. Sep 1, 2016

### Staff: Mentor

Good point about the negatively charged omega baryon, but I'm not so sure about the chemistry part. Can you get stable nuclei (apart from the antihydrogen equivalent) with only negatively charged baryons? I also don't see how the nuclear spin would influence chemistry. As far as I understand this would only influence the hyperfine structure, which is irrelevant for chemistry.

If we want to include cosmology, then the whole big bang nucleosynthesis would have looked differently because all the reaction channels and rates are completely different. Even earlier: a missing first generation would probably influence baryogenesis and maybe also leptogenesis, and I have no idea how.

In our universe, the first stars were made out of huge amounts of hydrogen+helium, and they only could cool down effectively once the temperature was sufficient to ionize hydrogen molecules. To ionize a pseudo-hydrogen/muon molecule needs much more energy - I'm not sure if that temperature can be reached at all without a star, so star formation might be impossible. On the other hand, the muons bring the omegas close together - if pseudo-helium is stable, it can be produced directly in the interstellar gas via muon-catalyzed fusion.

19. Sep 1, 2016

### ohwilleke

Only if the leptons around them are antimuons, as I suggest, rather than muons. Anti-muons around negatively charged nuclei would behave chemically pretty much the same as electrons around positively charged nuclei. Basically, a second generation particle universe would have the opposite electrical charges for most kinds of matter.

Assuming that the overall universe was neutral in electric charge, the other possibility is that not enough antimuons can be generated to match omegas 1 to 1. I could imagine that charge conservation might make it impossible for some charm quarks to decay to strange quarks, and that you might even get queer structures with doubly charmed omegas and uncharmed omegas bound electromagnetically to each other in the absence of leptons. Ideally, one would run some sort of Monte Carlo simulation and see what happens.

I don't know precisely how nuclear spin would influence chemistry, because all real world chemistry involves spin-1/2 baryons and spin-1/2 leptons. But, I can imagine that it could have some sort of influence, in a manner similar to the way that different isomers of a chemical can have different properties.

While it would be over the top to calculate all of the reaction channels and rates from scratch, I have little doubt that I am right in terms of general trends deriving from a weaker and shorter range nuclear binding force. Still, somehow or other, I have little doubt that star formation would be possible, albeit different. It is honestly pretty remarkable how much of cosmology is a function of pion mass in our real world because pion mass is pivotal in determining the effective properties of the nuclear binding force that drives fusion reactions.

Your point about the spacing of the omegas due to having antimuons rather than electrons associated with them is well taken and a good insight.

As far as baryogenesis and leptongenesis go, I agree that it would be wildly different (by assumption), but I think there is no choice in this hypothetical other than to take that as an assumed given. Honestly, we don't really have any real credible consensus explanations for either baryogenesis or leptogensis now anyway. I would implicitly assume that aggregate baryon number and aggregate lepton number would be more or less the same, simply for want of any good theory pointing in any other direction.

20. Sep 1, 2016

### Staff: Mentor

I'm not talking about the (anti)muons. I am talking about the nuclei. I can imagine smaller nuclei - no beta decay available and the strong interaction should win over the electric repulsion for a while. But without the equivalent of neutrons available and with a weaker residual strong interaction, the nuclei can't get too large. And I don't know where the limit is.
Why?
That doesn't make sense. Different isomers are different arrangements of atoms, with absolutely no connection to the hyperfine structure.
Also, most nuclei have spin values different from 1/2 already.

21. Sep 2, 2016

### ohwilleke

This is a good point. I agree that nuclei with more charged nucleons (i.e. higher numbered pseudo-elements) would be less stable and hence absent or more rare in nature, and that the lack of the equivalent of neutrons would contribute to that effect. Perhaps the heaviest element that would be stable in nature in this world might be pseudo-iron or pseudo-aluminum, or something like that.

Conservation of charge dictates that it ought to be possible to have almost all charged particles confined in electromagnetically neutral composite systems analogous to atoms, because for every baryon with a negative integer charge that is something else out there with an equal and opposite charge and the strength of the EM force will tend to sort them out.

But, suppose that we have a whole lot of negatively charged omega nuclei. This means that we need an equal and opposite number of positively charged components. And, all the lepton sector has to offer that is stable are anti-muons.

Now, if a charm quark decays to a strange quark, it has to emit a W+ boson. But, while some of the time the W+ boson will decay to an antimuon and a muon neutrino, or to something else that produces an antimuon and muon neutrino and other stuff, sometimes the W+ boson will decay hadronically, leaving us without the antimuon that we need to be perfectly analogous to electrons orbiting protons. If one runs out of antimuons, then some positively charged hadron will have to step in to neutralize the electric charge of the atomic system, and because of conservation of charge, any time that there are insufficient antimuons, there will be sufficient hadrons to balance things out.

My first wild, off the cuff guess was that charmed omegas might be those hadrons and if that was the case they could substitute for antimuons if they existed; but you are right that since they are not stable, it doesn't make a lot of sense for charmed omegas to fulfill that role unless there was some counterprocess continually replacing the ones that decayed as there is in the case of a bound neutron. On second thought, uncharmed anti-omegas might do the job of end products of hadronic W+ boson decays that maintain charge conservation even better than charmed omegas, as they could annihilate with ordinary uncharmed omegas, eliminating the antimuon-less omegas completely in order to balance the charge imbalance. So, on second thought, maybe charmed omegas aren't needed.

We've never seen how spin-3/2 particles behave when bound in nucleus-like clumps bound by the strong force (because in our world they aren't stable for long enough for such nuclei to assemble). So, we really don't know what impact that might have. Spin-3/2, in principle, can have somewhat more complicated properties than spin-1/2 objects.

In the absence of any better analogy I can think of, isomer chemistry seems to me to be the best available analogy for something that has a minor lesser order impact but isn't necessarily something that can be completely ignored in every circumstance. Obviously, there is guesswork involved, and I'd welcome another analogy. But, I am not comfortable that a nucleus made of spin-3/2 particles will necessarily act exactly like a nucleus made of spin-1/2 particles in all observable respects, even though my intuition is that the impact of having a nucleus made of spin-3/2 rather than spin-1/2 particles would be subtle.

My intuition is that the greater complexity of spin-3/2 interaction would giving rise to greater structure within the nucleus or would change the global spin properties of a nucleus in a way that might conceivably have some subtle chemical consequences.

For example, I wouldn't be surprised if different spin alignments in a nucleus made of five omega baryons could require different amounts of binding energy, giving rise to slightly different masses for different spin alignment types (usually mixed together in a way that averaged out, but separable chemically in principle much like isotypes but with smaller differences in mass between types). Maybe it wouldn't, and obviously I'm only considering gross qualitative features of these nuclei rather than a rigorous quantitative analysis, but that kind of order of magnitude type effect is plausible.

Last edited: Sep 2, 2016
22. Sep 2, 2016

### Staff: Mentor

I'm not even sure about pseudo-helium. The nuclear potential is a Yukawa potential, and to estimate its strength we can evaluate it at the size scale of the nuclei. For protons and neutrons, with pions as mediators, the exponential in the Yukawa term ís about $-\frac 1 \hbar m_\pi \cdot r_p = -0.6$, and e-0.6=0.5 is a reasonable number. If we replace the pion by a phi meson we get -4.1 in the exponent, e-4.1 is 0.017. Unless the nuclei sizes go down (possible), the residual strong interaction gets weaker by a factor ~30.

2He decays in our world already (into two protons), and you want to see it with a nuclear force of just 3% of its current strength?

The W+ is virtual and it does not have hadronic decay channels available if we discuss the decay of a charmed hadron. c->s+antimuon+neutrino is the only available option (maybe also anti-tau + neutrino in some cases, but that anti.-tau then always produces an antimuon), and this option is always available. There are not "insufficient antimuons", the antimuons are produced in the decay.
It does not matter. We also have never observed how humans behave if you move a grain of dust on the moon by exactly 1 cm. We still know that it won't have any relevant effect.

Different isomers usually have completely different chemical reactions. Comparing this to effects of the nuclear spin is just ridiculous. Do you really think ethanole and dimethyl ether are similar just because they both have the sum formula C2H6O? Hint: One is a liquid at room temperature, the other one is a gas. And only one of them will make you drunk.
Do you mean isotope effects? That is a better analogy, although isotope effects are still much larger than the hyperfine structure.

23. Sep 5, 2016

### nikkkom

What part of SM requires that leptons and quarks must have the same number of generations?

24. Sep 5, 2016

Staff Emeritus
If you want your theory to be anomaly free, you need to use complete representations.

25. Sep 6, 2016

### ohwilleke

A famous variation on the kind of speculation found in this thread is a 2006 paper discussing what the universe would like like without the weak force, which concludes that it wouldn't have to be very much different than our world.

http://arxiv.org/abs/hep-ph/0604027