I’ve got a paper on the hadrons ready to submit to Phys Math Central. This is a fairly new peer reviewed open access journal for which I have a “pass” that allows me to avoid having to pay the $1500 submission fee, so long as I submit before January 31. This is a big deal and I want to do it right, so I’m looking for advice from readers.
The paper as it stands is here:
Koide mass formulas for the hadrons, 49 pages, LaTeX.
The subject is the extension of Koide’s lepton mass formula to the neutrinos and then to the hadrons. I’ve written the background section so it should be accessible to typical grad students in physics.
I’ve put this together as an example of applying quantum information theory to the practical problem of the hadron masses. This all is fairly simple stuff and it uses very basic ideas in quantum mechanics.
Quantum information treats the information contained in quantum states. For a colored particle, that information is “red”, “green” or “blue”. But the usual method of modeling the color force is instead to approximate the color force as a modification to the Coulomb force.
On looking at the problem as a color force problem, we find that the excited states of a color bound state, in the quantum information approximation, must come in triplets and these triplets are related by the discrete Fourier transform on 3 variables.
So we apply the discrete Fourier transform to triplets of particles and what do we find. Well we get a simple equation relating the charged lepton masses which implies Koide’s mass equation. And we get an implication for a neutrino version. These two equations tell us how leptons look when they are excited, as if the leptons were composite particles.
The mesons and hadrons are composite particles and also have excitations. So, naturally, we plug those excitation mass numbers into the discrete Fourier transform and what do we get. Yes, we get more copies of the lepton mass equations.
The paper includes 33 mass equations that fit triplets of mesons, and 6 more that fit triplets of baryons.
45 Comments
January 14, 2009 at 12:21 pm
Excellent news. Hopefully the reviewers will be reasonable people who don’t think the physics content of one’s paper depends on the hardness of one’s hats.
January 16, 2009 at 2:55 pm
Quite a tour de force, Carl. Have you sent this to Tommaso? I’m sure he’ll have some great suggestions concerning the overall organization, as the experimental particle physics community is likely your primary audience. I, for one, think some images to help with the Berry-Pancharatnam topological phase discussion would be helpful.
A few small typos I found:
“The pure denstiy matrix…” pg. 12 first paragraph
“FOne would then write…” pg. 15 third paragraph
“problem of findint” pg. 22 first paragraph
Overall, excellent work though. Your paper adds much support for the importance of circulants in particle physics. Any nonperturbative unified theory of particle physics should have something to say about such circulants, especially if such a theory ends up being a matrix theory.
January 16, 2009 at 3:55 pm
Okay, I put the thing through a spell chequer, and now it’s got about two dozen improvements. I also like the idea about a picture showing BP phase and I’ve got those from other stuff I’ve written, in particular my book on density matrices.
By the way, I’m not sure whether to add it, but there is a subtlety in the comment where it notes that two degrees of freedom could be either spin-1/2 and get BP phase, or spin-0 and not. And that is that it isn’t that easy to get these kinds of things to come out. If you take two spin-1/2 objects with BP phase charge n=1 and combine them aligning, the result is a spin-1 object and it gets a BP phase “charge” of n=2.
So it messes things up to say that neutrinos and electrons are made from similar things. To fix it, my instinct is that you have to go to braid statistics. Along that line, note equation (11) of this:
New Kinds of Quantum Statistics
Frank Wilczek
http://arxiv.org/abs/0812.5097
Basically, another way of describing the BP phase is as one of these new kinds of quantum statistics. I’m thinking I need to add this as a reference.
January 17, 2009 at 10:08 am
Yup, sounds like you need to write down some BP charge “fusion rules” for various particle species (spin 0, spin 1/2, spin 1, etc.), as fusion rules specify the possible value of the charge when you combine two charged particles in anyon theory.
Some good references are:
Preskill’s KITP talk
and especially page 5 of
Non-Abelian Anyons and Topological Quantum Computation – Freedman et al.
It appears you have a non-abelian anyon model with multiple fusion channels.
January 17, 2009 at 2:44 pm
Obviously y’all are going to have to write the papers connecting this up as non abelian anyons.
The difference between what I’m doing and what is going on in with anyons or braids is that I’m primarily concerned with what happens to one quantum object. So to me, the various RG, RB, BR, etc., objects are components to a single path. The path loops back on itself, but it is the path of a single particle. With the braid theory, they’re keeping track of three (or more) particles simultaneously. So these are inherently multiparticle quantum mechanics.
The non Abelian stuff Wilczek talks about in
http://arxiv.org/abs/0812.5097
is algebraically related to the non Hermitian pure density matrices in a fairly simple way.
We’re considering braids of 3 objects, use x, y, and z for the Pauli spin matrices (or Clifford algebra generators). The braid is small enough that no two braids commute. Then the remaining braid rules are:
and Wilczek’s solution is
which defines a swap between the x and y states.
Now I’m using R, G, and B which are related to x, y, and z by
R = (1+x)/2 or x = 2R-1,
G = (1+y)/2 or y = 2G-1,
B = (1+z)/2 or z = 2B-1.
and you might want to rewrite i using i=xyz, but either way you can write a braid operator in terms of the projection operators for MUB spin.
January 19, 2009 at 10:52 am
Hi Carl,
On pg. 11, I think there is a typo somewhere in here…
exp(i tan−1(3 −√8)) ≈ exp(i 0.28599378)
January 20, 2009 at 12:19 am
Thanks Dave,
It’s fixed. I screwed up the final calculation. The number 0.2859… should be 0.16991845…. This is the angle whose cosine and sine are
Carl
January 20, 2009 at 8:08 am
You probably knew this, but it’s also (arcsin(1/3))/2.
Or, half the angle (19.47deg) where the lower points of an inscibed tetrahedron fall on the unit sphere.
January 20, 2009 at 8:53 am
…and (arccos(1-2/9))/4=.16991845
January 20, 2009 at 9:28 am
even simpler… (cos(19.47))/2=sqrt(2/9)
January 20, 2009 at 1:56 pm
Dave,
No, I’m stunned. That’s very much appreciated news to me. I’m going to add something into the paper to that effect.
The reason it’s interesting to me (and also no doubt to Marni Sheppeard) is because the number 2/9 appears as the mysterious charge or amplitude later on, see the text just below equation (67). Also known as “that damned number”.
If you give me a real name and I’ll add a note to the effect that you pointed this out. Otherwise, I’ll attribute it to “Dave”.
January 20, 2009 at 3:47 pm
Cool, Dave! I’m sure you realise just how cool.
January 20, 2009 at 7:04 pm
Hi Carl,
I would be very honored to be mentioned in your amazing paper, Thank You!
Looking at pg.14, 1st paragraph, I noticed tan(19.47)=sqrt(1/8).
maybe relevant?…
tan(90-19.47)=sqrt(8), and tan((90-19.47)/2)=sqrt(1/2)
January 21, 2009 at 3:25 am
I saw that you sugested at physics forums that in order to find the 3 generations of Garrett Lisi’s model, one should use your method. But you haven’t yet applied your method to find the havier quarks. So, how will you find them?
Carl: I suggested one way to expand E8 to 3 generations was to look at the discrete Fourier transform as described in the paper. The first applications are with heavy quarkonium. These are mesons made with heavy quarks. Of course quarks themselves are not observable so I can’t find them.
January 21, 2009 at 12:27 pm
Pretty good Dave, you probably also recognized this as part of the geometry of the star hexagon in a two dimensional drawing.
And Carl, guess this would connect with one of your previous papers on the Star of David.
Are you trying to “keep it simple” for this paper,
or have you revised the more “precise value”?
Carl: Mike, the value 2/9 is not as exact enough for measurement. But there always seem to be corrections to qm formulas and 2/9 is pretty close. So I decided it’s better to stick with the 2/9. Part of the reason for this is that it’s at least possible that the error could appear not as a correction to delta, but instead as a correction to the tau mass in particular.
January 22, 2009 at 7:50 am
I got it now. The Top quark is an excited state in E8.
January 22, 2009 at 8:05 am
Hi Carl, Then for future reference you might also be interested in the approximation cos (77.16) = 2/9.
77.16 is close to the apex angle of the large triangle in the heptagon or ~ sin (90/7) = 2/9.
Both the heptagon and the hexagon fit in the
Cosmological Circle from which alpha,
the fine structure constant, is derived; and a very close approximation to your latest value for delta is on the fqxi essay page…
We noted alpha^3, pi, and 1.8 in our essay. 1.8 = 9/5.
((9/5 + (alpha^3 x pi^1/3))^-1 – 3/9 = 0.2222220466 ~ 2/9
Also as an historical aside and “confirmation”
an even simpler formulation from harmonic proportion was found from the Foundation Stone of ancient geometry. This was the Greek template for the formation of matter. The star tetrahedron is related to Metatron’s Cube which is an aspect of the Foundation Stone. Basically the proportions involved include 144/648 = 162/729 = 2/9.
January 22, 2009 at 2:24 pm
Hmm, interesting….
If you have a sphere of radius 1 with an inscribed tetrahedron, then a circle placed within a face will have a radius of sqrt(2/9), exact.
January 23, 2009 at 8:09 am
I think we’re barking up the wrong tree focusing on 2/9 rad. If you take the cube root of 2/3 rad, you get 2/9 rad, (and the other 2 roots, of course.)
January 23, 2009 at 10:08 am
And that makes the extra neutrino term(pi/12), actually pi/4, before taking the cube root.
January 23, 2009 at 10:29 pm
Dave,
You’re quite right that before taking the cubed root the value is 2/3 instead of 2/9, and the pi/12 becomes pi/4. And pi/4 is the correct Berry-Pancharatnam or topological phase. But it turns out that I don’t have any better explanation of 2/3 than 2/9.
In the paper, the reason that the pi/12 gets a cubed root while the 2/9 does not is that the pi/12 only appears when you go around the color space and return back where you started. It’s a topological phase. On the other hand, the 2/9 is supposed to come from a QFT correction to the vertex. So it applies to the vertices individually; you don’t have to string together a sequence that gets you back where you started.
As far as what this has to do with various geometric objects, I have no idea. The basic problem is that there are a lot of ways of getting 2 and 3 out of mathematics.
January 30, 2009 at 10:54 am
The binding energy of the quarks is commonly listed as 313MeV, (proton 938.27MeV), and is “1″ the natural lepton mass scale (where tau is 5.6617).
Or, where the angle is pi/2.
Since most of the nucleon mass is binding energy, this might be the “ground state” energy of a single up-up bond.
February 2, 2009 at 8:14 pm
I have a very simple formula for the top mass, which involves a recognition that in natural units, the top quark and electron are nearly equidistant from “unity”…
(cos(2(arctan(3-sqrt8))))^2=8/9 exact
..or just take (2/9)*4, or 2/3+2/9
electron in “natural” units=.00162812978=e_n
1/(e_n/(8/9))=545.957
545.957 * 313.8563Mev=171.352GeV
(313Mev converts natural units to MeV)
PDG top mass=171.2 (/171.352=.999113)
Don’t know if it’s a coincidence, but that flip around unity might illustrate the essential difference between leptons and quarks.
February 4, 2009 at 11:32 am
Interesting Dave, more connections… first, thanks for posting this,
another curious connection.
Dave
January 22, 2009 at 2:24 pm
“Hmm, interesting….
If you have a sphere of radius 1 with an inscribed tetrahedron, then a circle placed within a face will have a radius of sqrt(2/9), exact.”
***
As you probably know, 8/9 is the whole tone ratio in classical harmonics and is also represented by the ancients as a squaring of the circle. A circle with circumference 8 is “squared” (nearly the same area) by a square with a perimeter of 9. This is the “main” construction in the Cosmological Circle. The perfect fifth 3/2, or here 2/3 was represented by a triangle in a circle. 2/3 = 6/9 = 486/729 (a tri-circle with total circumference of 729 is circumscribed by a circle of 486 circumference). 8/9 – 6/9 = 2/9. And from the Foundation Stone 8/9 = 648/729. 144 = 12^2 360/12 = 30. Ratios common to both the Foundation Stone and the Cosmological Circle.
Carl’s now “blessed number” delta again,
((144 – (alpha^3 / 30)^1/2))/648 = 0.2222220466 ~ 2/9
Same value as above post:
January 22, 2009 at 8:05 am
((9/5 + (alpha^3 x pi^1/3))^-1 – 3/9 = 0.2222220466 ~ 2/9
February 5, 2009 at 9:33 pm
Here is a very convincing description of the arrangement of quarks in a baryon…
http://www.unclear2nuclear.com/shape.php
February 7, 2009 at 7:37 pm
I found a simple formula for the neutron-proton mass difference.
2006 CODATA neutron=1.00866491597
proton=1.00727646677
/ .33693868676 = 2.9936156209 = Nc
/ .33693868676 = 2.9894948439 = Pc
(.3369.. converts amu to natural units based on the Carl’s marvelous lepton mass formula)
(1/N^.5)+15=15.5779656=Nt
1/(Nt^2)=.00412077137
Nc – .00412077137 = 2.9894948495 = Pt
Pt / Pc = 1.00000000189187
It might be a coincidence, but it popped up as soon as I put the N-P difference into my natural mass values spreadsheet this morning.
February 8, 2009 at 7:30 am
Ok Dave, Thanks for the link on “triangular” quarks,
and where is Carl?
15?
http://www.earthtech.org/michael/bio/puzzling.wav
February 8, 2009 at 11:28 am
Hi Mike,
Since the Neutron & proton are composed of three quark-bonds, the real number is probably 15/3.
February 9, 2009 at 5:32 am
Wow!
When 1/15 , as input to Carl’s lepton formula is 48.7026950deg, and it’s cos^2=0.43555555556!
February 9, 2009 at 12:58 pm
Dave, you should write these cool things down carefully in a small paper, and try to publish it.
February 9, 2009 at 1:32 pm
The way to distinguish numerology from physics is that physics uses multiple examples of a formula working while numerology typically gives one only a single coincidence. That said, I just noticed that the 15.577xxx in David Lackey’s comment of February 7, is that the .577 part is the first 3 digits of the Euler-Mascheroni constant. I suspect that having a good memory for this sort of numerical coincidences is not a good thing.
February 10, 2009 at 7:18 am
Well anyway, numero neato Dave… 15 again!
(7/5)^2 x 2/9 = (2 x 7^2)/15^2 = 0.435555555…
February 10, 2009 at 10:26 am
The neutron has 3 electric charge bonds:
2/3*-1/3=-2/3, of which there are 2 = -4/9
and one of -1/3 * -1/3 = +1/9 (Total is -3/9)
The proton also has 3: +2/3*-1/3=-2/9 of which there are 2 = -4/9
and one +2/3*+2/3=+4/9 (Total is 0/9)
Since the difference between the is -1/3, and 15 is that 3rd, the total would be 45 in reciprocal natural units, 1/45=.022222…
February 11, 2009 at 9:40 pm
OK, I figured out a possible solution to where 15 comes from…
down -> up + W-
neutron -> proton + e + v
The W boson is 256.162 (sqrt=16.005) in natural units, and since the Carl’s lepton equation always adds a unit just before taking the square to obtain the “real” mass value, let’s do the same here… 15+1.005=16.005.
Since we must conserve mass, perhaps that extra “unit” is used up in the creation of the 2 leptons.
Perhaps it IS that “+1″ in the lepton formula.
When the quite massive W decays into electron & anti-neutrino, this is where the mass might go.
But the weird thing is, wouldn’t this be negative mass, since we are subtracting it from the neutron to create the proton? Is “negative mass” positive charge”?
February 16, 2009 at 7:08 pm
Thought I’d quickly mention something really cool, see http://arxiv.org/abs/hep-ph/0609131
In it, Pestieau & Castro describe the standard relationship between the W, Z, and B bosons (Mw^2 + MB^2 = Mz^2), but the B (carrier of weak hypercharge) turns out to be 137.08 in Brannen units! (Carl, hope you don’t mind if I call them that)
BTW, the weak mixing angle is 28.1547 deg, or acos(W/Z), using 2006 PDG for W & Z , and
(atan 1.5)/2 = 28.154966!
March 2, 2009 at 1:05 pm
e=.302775638, 1/e=3+e
(1-e)/(1+e)=.535183758 = tan(28.154966)
e*h-bar/2Mp=.050639933136=nuclear magneton in natural units=un (h-bar=1, Mp is in “natural” lepton mass units)
Proton mag moment/un=2.792847356 (2006 CODATA)
Actual Proton mag. moment in natural units=
0.1414296033=up
up/sqrt(2)=0.1000058316
March 3, 2009 at 6:36 am
Using the same method as above, the muon magnetic moment (including muon g-factor) is
0.45021923962
Proton mag moment / muon mag moment=
.1414296033 / .45021923962 = .31413496078
.31413496078 / pi = 0.099992263612
April 8, 2009 at 7:26 am
at the end of
http://theory.fnal.gov/seminars/slides/2009/CSturm.pdf
they give a charm mass of 986MeV
986MeV/313.8564MeV=3.141564
(where 313MeV is the same “lepton unity” value as the Feb 2 post above)
3.141564/pi=0.99999091232(!)
April 8, 2009 at 3:34 pm
Dave, that’s a nice coincidence, but when they only give three digits accuracy for the 986 MeV, we only end up with 3 digit accuracy in ratios involving it.
April 8, 2009 at 8:16 pm
You’re absolutely right, I’ve wished there was a way to edit posts more than once!
April 17, 2009 at 11:06 am
I was wondering about the geometric quantity which the 3 leptons are the cube roots of:
(1+sqrt2*cos(3*12.73239))*313.85637 MeV =662.6804 Mev
When the 3 down-type quarks are converted using 662.6804 MeV as the new conversion constant the latest PDG values yield sqrt (geometric) masses of:
d=.0872094 s=.396155 b=2.517518
Amazingly, not only do they sum to 3.00088 (as the leptons do), but after taking away the “1+” term, (d+s)/b = -0.999418854, just like (elec+muon)/tau = -1.
April 17, 2009 at 1:12 pm
Sorry, if (d-1+s-1)/(b-1)=-1, then d+s+b=3
~ so both can’t be significant.
May 4, 2009 at 9:14 am
Hi Carl,
Since the electron mass is the magic threshold for creating stable matter from light, I was looking at the relationship between it’s gravitational and charge force numerators. Check this out…
From http://findarticles.com/p/articles/mi_m1200/is_n6_v151/ai_19123470/
“From their data, the researchers obtained a value of the fine structure constant, a number that characterizes the inherent strength of the electromagnetic force. As expected theoretically, the newly obtained value of 1/128.5 is significantly larger than the 1/137 observed for a fully screened electron.”
May 4, 2009 at 9:18 am
Sorry, the latex html I put in after “check this out” didn’t appear above. It is
ke^2 / (Gme^2)^1/3 = 1/128.5^2
May 12, 2009 at 6:02 pm
In geometric units, the e, u, t masses sum to “6″, or 3.3569994E-027 kg = M_leptons
M_leptons * c / 2pi = 1.6017399 E-019 = e_lep
e_lep /elem. chg = .999728
If this is more than a coincidence, that means charge has dimensions of momentum.