# Two new heavy baryons Discovered

1. Nov 1, 2006

### neutrino

http://physicsweb.org/articles/news/10/10/17/1

2. Nov 1, 2006

### selfAdjoint

Staff Emeritus
Thanks for the update, neutrino!

So these are B-particles (as I understood it that means particles containing a bottom quark)? Has any work been done on how they behave under CP symmetry? Other B-particles violate it.

(Added) From the article by Belle Dunne' (fortunate name!)

Exciting times. My Particle Data Group handbook just came and already it's out of date!

Last edited: Nov 1, 2006
3. Nov 1, 2006

### Kea

Damn! Can't these guys slow down a little bit. We're trying to catch up...

4. Nov 1, 2006

### CarlB

If I ever get done with the lepton masses, I'm going to start working on the meson masses. I tried the quarks, but unfortunately they are very model dependent and the models disagree.

I went to another string theory seminar. Saw a fascinating lecture on "spin chains". They use arithmetic that is vaguely similar to mine.

The whole reason for solving the $$\rho^2=\rho$$ equation is to get a non perturbative answer to a nonlinear interaction problem. It seems to me that the techniques I use could be used also over on their problem, but I haven't seen enough of it.

But I really liked the idea, strings as sequences of spin states (at least in my limited understanding):
http://www.phys.hawaii.edu/indico/c...?contribId=194&amp;sessionId=168&amp;confId=3

Carl

5. Nov 1, 2006

### Kea

We have to do quarks first, because that's the way I see sorting out the theory. It might all be craziness, but we don't think it's just a model. This might take a while if no one else gets interested in this...

6. Nov 1, 2006

### CarlB

The problem is that my version of the theory is a "heavy quark" theory. I'll get infinity for all their masses. I don't know how to split that off.

And when I looked up the data on the PDG I found that no one really knows what the correct model for mass is.

But the meson masses are very well measured to high accuracy. And there are some well known formulas that work a bit better than anyone would have expected. So I think there's room in there for a fit.

The other place to take it is to calculate the gauge boson couplings. As far as I can tell, the gauge bosons are composed of 6 preons each, which means that instead of 3x3 matrices like the Koide formula uses, you will have to mess around with 6x6 matrices. And of course the entries are not complex numbers but are (not quite primitive) idempotents taken from the complex Clifford algebra C(4,1). The "not quite" is from the fermions which appear to need two primitive idempotents in each preon oriented in a particular direction (which I call a "snuark").

I wasted 3 months of my life trying to get the quarks to work before I gave up and began working on the neutrinos and "solved" that problem in a day or two. I went to the quarks first because their masses are better known than the neutrinos, I thought. But the neutrinos turned out to be easier.

Carl

7. Nov 1, 2006

### Kea

I scrolled through a few data sites but they are such a mess I realised it was going to be too much trouble. The quark CKM matrix is at least almost a circulant. Wiki puts it at

0.9753 0.221 0.003
0.221 0.9747 0.040
0.009 0.039 0.9991

Does anyone have errors for these numbers?

8. Nov 2, 2006

### CarlB

Yes, the CKM matrix is a good place to go. But the MNS matrix isn't a circulant so I'm not sure what it will buy you.

If I had to guess, I would say that multiplying the CKM matrix by a matrix of the circulant eigenvectors could be interesting. I did that in the MASSES2 paper. If you pick the right order of eigenvectors, the result is that the CKM matrix becomes a 24th root of unity (or various other roots of unity).

The theoretical reason for doing this is that the CKM and MNS matrices compare mass eigenstates to flavour eigenstates. To get the nice matrix, what you want is something that compares apples to apples.

The whole difference between the neutrinos and electrons is that 12th root of unity that makes their masses so much different. If we were in the same room this would be a lot easier to explain. Basically, the neutrino mass circle has to have 24 states in it. One left, one right, and 22 "sterile". The electron mass circle has two states in it, one left, one right. Mass consists of running around the circle. Gauge bosons consist of running one step around the circle but jumping off the circle you are on and jumping on to the other circle. That means that you have your state wrong by whatever the difference is between the (non sterile) left and right neutrino states. It could be just about any nth root of unity where n divides 24, if I recall.

Carl

P.S. I'm starting to get more interesting DNS hits on my websites. I have to admit that I laughed when I saw this one:

Of course the President of UCB is a physicist:
http://www.universityofcalifornia.edu/president/bio.html

But the office of the President is probably huge and has lots of employees who might be wasting time surfing on their office computer.

9. Nov 2, 2006

### arivero

There is a minor mistery on meson masses. The scales where mass lives are roughly well stablished:

electromagnetic breaking, where electromagnetic mass differences are.

chiral (or QCD) are, where the pion is.

QCD (or so) area, where the proton is

Electroweak area, for W, Z etc

Mistery is, three of these areas is also inhabited by a fundamental lepton: electron, muon, tau, and the for one got another peculiar fermion, the top.

There is absolutly no reason, AFAIK (AFAWK?), for the muon and tau to live so near of QCD-marked areas.

10. Nov 2, 2006

### arivero

Except if quarks are supersymetrical to diquarks

11. Nov 2, 2006

### Kea

Last edited: Nov 2, 2006
12. Nov 2, 2006

### CarlB

I've had a fascinating series of conversations with physicists in the last day or so. As an amateur, you know you're being appreciated when the pros check to see if you referenced their work.

If you want to be the person who figures out the Koide formula for the meson masses I would think about taking some time out to begin on it sooner rather than later. Of course this could be a recipe to go waste a lot of time.

13. Nov 2, 2006

### arivero

:rofl:

Problem is, due to susy breaking (lets suposse it, for the sake of discussion) we have not three but six mases to fit: $$\pi^+,K^+,D,D(s),B^+,B(c)$$ I would expect that whatever that breaks susy is also breaking Koide.

EDITED: a first way to look into this is to "restore" part of the symmetry by putting the masses of the (d,s,b) quarks to be small enough. Then we get two triplets of charged bosons, the (Ud)(Us)(Ub) at the level of the muon, and the (Cd),(Cs),(Cb) at the level of the tau. We can say, if mass of the down sector are zero, that the flavour symmetry is completely restored in the (dsb) sector but still broken in (uct). Now, this is not enough to pair the mesons with the leptons; we need two mesons to be near zero, not near muon or tau. Ideally one combination of (U*) and another of (C*) should be zero and pair with the two spin states of the electron. These combinations could happen to be the Goldstone boses of some spontaneusly broken symmetry.

EDITED again: Having determined the pairings for charged leptons, the neutral ones follow by "isospin", ie by substracting a W particle: if Ud has charge -1, we produce a charge 0 going to Dd or to Uu. We could fix a criteria of producing the "neutrino" always from the particle, or always from the antiparticle, but I do not feel sure.

Last edited: Nov 3, 2006
14. Nov 2, 2006

### arivero

Now, seriously thinking, the funny think is that Koide original derivation of the formula was a model of preons. Thus it seems feasible to derive a generalised formula for mesons, where we are supposed to know what the "preons" are. But most probably the colour force mix everything.

Last edited: Nov 3, 2006
15. Nov 2, 2006

### Kea

Hurrah! Progress!

16. Nov 2, 2006

### CarlB

I don't know my mesons, but what I would think that the formula would have to predict 21 masses, that is (u,d,s,t,b,c)^2 = 36 with the antiparticles having the same mass (u/s = s/u, so subtract 15). Part of the problem is that it is hard to get masses for the higher massed particles. Another part is that it is hard to get quantum numbers. And the final problem is that the quarks involved are supposedly mixtures.

I suspect that if you write the thing in preons the mixture problem will go away, so what I would be inclined to ignore the purported mixtures. But the parity, charge and spin quantum numbers should be okay.

Carl

17. Nov 3, 2006

### arivero

And substract any combination having the top. This is about 25 counting particles and antiparticles. But separate in charge 0 or charge +-1, only the charged ones should have some relation with the formula for charged leptons.

EDITED: Really one needs to predict:
Code (Text):

2 diquarks of charge +4/3    (uu)(cc)
6 mesons of charge +1        (uD)(uS)(uB)(cD)(cS)(cB)
6 diquarks of charge +2/3    (DD)(SS)(BB)(DS)(DB)(SB)
6 diquarks of charge +1/3    (ud)(us)(ub)(cd)(cs)(cb)
13 mesons of charge 0        (uU)(dD) (uC)(dS) (dB) (cU)(sD) (cC)(sS) (sB)  (bD)(bS)(bB)
6 diquarks of charge -1/3    (UD)(US)(UB)(CD)(CS)(CB)
6 diquarks of charge -2/3    (dd)(ss)(bb)(ds)(db)(sb)
6 mesons of charge -1        (dU)(dC)(sU)(sC)(bU)(bC)
2 diquarks of charge -4/3    (UU)(CC)

Charges +-1 can surely be related to the formula for charged leptons. Charge 0 would need the removal of a U(1) subgroup to get the 12 degrees of freedom of the massive neutrinos (as when going down from U(3) to SU(3); lets say that here we are looking at SU(5)_flavour). Or perhaps just to remove or isolate (bB), that can not come from beta decay of any charged meson.

In the diquark side, everything comes nicely in packs of 6 degrees of freedom except these pesky 4/3

But probably everything is broken because of SU(3) colour plus some extra breaking from flavour asymmetry, so one can not really to hope see further than the theoretists working with lattice QCD do.

Last edited: Nov 3, 2006
18. Nov 3, 2006

### arivero

Incidentally,
$$(\sqrt{m_\tau}+\sqrt{m_\mu}+\sqrt{m_e})^2 \sim (\sqrt{m_u}+\sqrt{m_c})(\sqrt{m_d}+\sqrt{m_s}+\sqrt{m_b})$$

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