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	<title>Comments on: Three exquisite exclusive charmonium signals</title>
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	<link>http://dorigo.wordpress.com/2008/12/29/three-exquisite-exclusive-charmonium-signals/</link>
	<description>private thoughts of a physicist and chessplayer</description>
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		<title>By: Louise</title>
		<link>http://dorigo.wordpress.com/2008/12/29/three-exquisite-exclusive-charmonium-signals/#comment-103290</link>
		<dc:creator>Louise</dc:creator>
		<pubDate>Wed, 31 Dec 2008 14:39:01 +0000</pubDate>
		<guid isPermaLink="false">http://dorigo.wordpress.com/?p=1883#comment-103290</guid>
		<description>HI Tommaso:  Off-subject, but I would like to draw your attention to the blog of our friend Kea.  She has a new job in Oxford, and was due to speak there January 7.  The authorities have taken her passport until further notice and left her to wash dishes.  You may recall my experiencing similiar problems with UK immigration.  This is madness.</description>
		<content:encoded><![CDATA[<p>HI Tommaso:  Off-subject, but I would like to draw your attention to the blog of our friend Kea.  She has a new job in Oxford, and was due to speak there January 7.  The authorities have taken her passport until further notice and left her to wash dishes.  You may recall my experiencing similiar problems with UK immigration.  This is madness.</p>
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		<title>By: Namit</title>
		<link>http://dorigo.wordpress.com/2008/12/29/three-exquisite-exclusive-charmonium-signals/#comment-103286</link>
		<dc:creator>Namit</dc:creator>
		<pubDate>Wed, 31 Dec 2008 04:05:36 +0000</pubDate>
		<guid isPermaLink="false">http://dorigo.wordpress.com/?p=1883#comment-103286</guid>
		<description>Thanks a lot Tommaso and Igor for the links and references.</description>
		<content:encoded><![CDATA[<p>Thanks a lot Tommaso and Igor for the links and references.</p>
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		<title>By: Michael Schmitt</title>
		<link>http://dorigo.wordpress.com/2008/12/29/three-exquisite-exclusive-charmonium-signals/#comment-103284</link>
		<dc:creator>Michael Schmitt</dc:creator>
		<pubDate>Wed, 31 Dec 2008 03:48:27 +0000</pubDate>
		<guid isPermaLink="false">http://dorigo.wordpress.com/?p=1883#comment-103284</guid>
		<description>Hi Tommaso,  very nice post!

I like this analysis as well.  For me, the differences between the production mechanism for the vector mesons (J/psi and Psi(2s)) and the scalar (chi_c) are very important.  As you alluded to, the chi_c is a stand-in for the Higgs boson.  For the chi_c, the charm quark plays the main role, while for the Higgs boson, it is the top quark.  If I understand correctly, the CDF results confirm the moderate-upper end of the theoretical predictions for exclusive Higgs production at the LHC.   Mike Albrow wrote a superb exposition of this physics - see arXiv:0812.0612

regards,
Michael</description>
		<content:encoded><![CDATA[<p>Hi Tommaso,  very nice post!</p>
<p>I like this analysis as well.  For me, the differences between the production mechanism for the vector mesons (J/psi and Psi(2s)) and the scalar (chi_c) are very important.  As you alluded to, the chi_c is a stand-in for the Higgs boson.  For the chi_c, the charm quark plays the main role, while for the Higgs boson, it is the top quark.  If I understand correctly, the CDF results confirm the moderate-upper end of the theoretical predictions for exclusive Higgs production at the LHC.   Mike Albrow wrote a superb exposition of this physics &#8211; see arXiv:0812.0612</p>
<p>regards,<br />
Michael</p>
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		<title>By: carlbrannen</title>
		<link>http://dorigo.wordpress.com/2008/12/29/three-exquisite-exclusive-charmonium-signals/#comment-103275</link>
		<dc:creator>carlbrannen</dc:creator>
		<pubDate>Tue, 30 Dec 2008 22:57:35 +0000</pubDate>
		<guid isPermaLink="false">http://dorigo.wordpress.com/?p=1883#comment-103275</guid>
		<description>Tommaso,

I did a Monte Carlo to find the distributions and learned some things about ratios of random variables doing this. A little puzzle that an experimentalist might enjoy:

Pick three numbers from a uniform [0,1] distribution (or any other uniform distribution, same result). Rearrange them so that they are ordered a&lt;b&lt;c. Then the ratio (b-a)/(c-a) has what distribution?

With 6 masses, there are 20 ways of splitting them up into two distinct groups of three so the best fit is far better. The simulation is easy. I use java.

By the way, the easiest way to characterize a triplet of three numbers a&lt;b&lt;c is by that ratio (b-a)/(c-a). For example, if the three numbers happen to be the masses of the three lowest excitations of a hydrogen-like atom (i.e. some atom stripped to one electron), then they follow a law
$latex E_n = s - t/n^2$ where s and t are constants. Uh, s is the mass of the bare nucleus plus the mass of the electron. For such a triplet, we have:
(b-a)/(c-a) = 27/32.
Looking at ratios is a good way of picking triplets out of data.

I did the distribution calculations for heavy quarkonium only, as it is unique in having the masses so accuratelyl measured that the fits are worse than the experimental measurements. This makes the calculation easy. The paper includes 36 mass formlas for other hadrons (i.e 120 counting heavy quarkonium). Also there are some heavy mesons whose masses are accurately known but not the quantum numbers and the paper predicts the quantum numbers for them.</description>
		<content:encoded><![CDATA[<p>Tommaso,</p>
<p>I did a Monte Carlo to find the distributions and learned some things about ratios of random variables doing this. A little puzzle that an experimentalist might enjoy:</p>
<p>Pick three numbers from a uniform [0,1] distribution (or any other uniform distribution, same result). Rearrange them so that they are ordered a&lt;b&lt;c. Then the ratio (b-a)/(c-a) has what distribution?</p>
<p>With 6 masses, there are 20 ways of splitting them up into two distinct groups of three so the best fit is far better. The simulation is easy. I use java.</p>
<p>By the way, the easiest way to characterize a triplet of three numbers a&lt;b&lt;c is by that ratio (b-a)/(c-a). For example, if the three numbers happen to be the masses of the three lowest excitations of a hydrogen-like atom (i.e. some atom stripped to one electron), then they follow a law<br />
<img src='http://l.wordpress.com/latex.php?latex=E_n+%3D+s+-+t%2Fn%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='E_n = s - t/n^2' title='E_n = s - t/n^2' class='latex' /> where s and t are constants. Uh, s is the mass of the bare nucleus plus the mass of the electron. For such a triplet, we have:<br />
(b-a)/(c-a) = 27/32.<br />
Looking at ratios is a good way of picking triplets out of data.</p>
<p>I did the distribution calculations for heavy quarkonium only, as it is unique in having the masses so accuratelyl measured that the fits are worse than the experimental measurements. This makes the calculation easy. The paper includes 36 mass formlas for other hadrons (i.e 120 counting heavy quarkonium). Also there are some heavy mesons whose masses are accurately known but not the quantum numbers and the paper predicts the quantum numbers for them.</p>
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	<item>
		<title>By: Carl Brannen</title>
		<link>http://dorigo.wordpress.com/2008/12/29/three-exquisite-exclusive-charmonium-signals/#comment-103273</link>
		<dc:creator>Carl Brannen</dc:creator>
		<pubDate>Tue, 30 Dec 2008 21:49:57 +0000</pubDate>
		<guid isPermaLink="false">http://dorigo.wordpress.com/?p=1883#comment-103273</guid>
		<description>Tommaso,

I&#039;ve been very busy on this and want to do it right. There are a total of 38 hadron equations like the ones for heavy quarkonium, but most hadron resonance masses are not known to great accuracy so it is only in heavy quarkonium where the fits are much worse than the experimental result. So I did the statistics for heavy quarkonium fits only.

In the process of doing the Montecarlo for random ratios I found a result that surprised me a little. Take three samples from a uniformly distributed random variable. Rearrange the three samples so that a&lt;b&lt;c. Then what is the distribution of the ratio:
(b-a)/(c-a)? Clearly the ratio is between 0 and 1. To calculate it from first principles, the rearrangement is most of the trouble as the inequalities break up the triple integral from 0 to 1. I found the result by simulation and then proved it because it&#039;s easier if you already know the answer...

Ratios are useful for analyzing structure in data that is organized in triplets. For example, if the mass formula were like that for hydrogen like atoms (i.e. nuceli plus one electron):
M_n = M - k/n^2
for k a constant and M the sum of the nuclei and electron masses, then the ratio (b-a)/(c-a) for the three lowest masses, n=1,2,3 is always 27/32.</description>
		<content:encoded><![CDATA[<p>Tommaso,</p>
<p>I&#8217;ve been very busy on this and want to do it right. There are a total of 38 hadron equations like the ones for heavy quarkonium, but most hadron resonance masses are not known to great accuracy so it is only in heavy quarkonium where the fits are much worse than the experimental result. So I did the statistics for heavy quarkonium fits only.</p>
<p>In the process of doing the Montecarlo for random ratios I found a result that surprised me a little. Take three samples from a uniformly distributed random variable. Rearrange the three samples so that a&lt;b&lt;c. Then what is the distribution of the ratio:<br />
(b-a)/(c-a)? Clearly the ratio is between 0 and 1. To calculate it from first principles, the rearrangement is most of the trouble as the inequalities break up the triple integral from 0 to 1. I found the result by simulation and then proved it because it&#8217;s easier if you already know the answer&#8230;</p>
<p>Ratios are useful for analyzing structure in data that is organized in triplets. For example, if the mass formula were like that for hydrogen like atoms (i.e. nuceli plus one electron):<br />
M_n = M &#8211; k/n^2<br />
for k a constant and M the sum of the nuclei and electron masses, then the ratio (b-a)/(c-a) for the three lowest masses, n=1,2,3 is always 27/32.</p>
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		<title>By: Igor Ivanov</title>
		<link>http://dorigo.wordpress.com/2008/12/29/three-exquisite-exclusive-charmonium-signals/#comment-103268</link>
		<dc:creator>Igor Ivanov</dc:creator>
		<pubDate>Tue, 30 Dec 2008 15:12:29 +0000</pubDate>
		<guid isPermaLink="false">http://dorigo.wordpress.com/?p=1883#comment-103268</guid>
		<description>Dear Namit,

the field these results belong to is called &quot;diffraction at hadronic colliders&quot;. Tommaso cited a few theory papers with some particular examples of calculations. There is in fact a series of workshops devoted to diffraction, and the last one, Diffraction-2008, took place last September, the website is http://www.cs.infn.it/diff2008/
There you can find many more results and theoretical calculations. 

There is also a project for the LHC, called FP420, which aims at detecting both protons with very forward roman pots (220+420 meters away from the main detector), which together with the information from central detector will make a very clean study of such processes possible at the LHC.

Cheers,
Igor</description>
		<content:encoded><![CDATA[<p>Dear Namit,</p>
<p>the field these results belong to is called &#8220;diffraction at hadronic colliders&#8221;. Tommaso cited a few theory papers with some particular examples of calculations. There is in fact a series of workshops devoted to diffraction, and the last one, Diffraction-2008, took place last September, the website is <a href="http://www.cs.infn.it/diff2008/" rel="nofollow">http://www.cs.infn.it/diff2008/</a><br />
There you can find many more results and theoretical calculations. </p>
<p>There is also a project for the LHC, called FP420, which aims at detecting both protons with very forward roman pots (220+420 meters away from the main detector), which together with the information from central detector will make a very clean study of such processes possible at the LHC.</p>
<p>Cheers,<br />
Igor</p>
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		<title>By: dorigo</title>
		<link>http://dorigo.wordpress.com/2008/12/29/three-exquisite-exclusive-charmonium-signals/#comment-103263</link>
		<dc:creator>dorigo</dc:creator>
		<pubDate>Tue, 30 Dec 2008 11:13:28 +0000</pubDate>
		<guid isPermaLink="false">http://dorigo.wordpress.com/?p=1883#comment-103263</guid>
		<description>Hi Carl,

good to know you are working hard. I have sent you the voucher, you should receive it in a week or so. The coincidences you mention seem striking to me, but I think one would have to complement such an analysis with an estimate of the &quot;trial factor&quot; of putting together constants and functions to fit those numbers. That may give the reader some perspective.

Namit, the link I had provided was not html&#039;ed correctly, now I have fixed it. If you visit the web site of the analysis you will find more information. In any case I paste below a few theoretical references:

A. Szczurek, Proc. Blois 2007 (Hamburg), 231, arXiv:0709.0224.
S. Klein and J. Nystrand, Phys. Rev. Lett. 92:142003 (2004).
V.P. Goncalves and M.V.T. Machado, Eur. Phys. J. C40, 519 (2005).
L. Motyka and G. Watt, arXiv:0805.2113.
F. Yuan, Phys. Lett. B510, 155 (2001).

Cheers,
T.</description>
		<content:encoded><![CDATA[<p>Hi Carl,</p>
<p>good to know you are working hard. I have sent you the voucher, you should receive it in a week or so. The coincidences you mention seem striking to me, but I think one would have to complement such an analysis with an estimate of the &#8220;trial factor&#8221; of putting together constants and functions to fit those numbers. That may give the reader some perspective.</p>
<p>Namit, the link I had provided was not html&#8217;ed correctly, now I have fixed it. If you visit the web site of the analysis you will find more information. In any case I paste below a few theoretical references:</p>
<p>A. Szczurek, Proc. Blois 2007 (Hamburg), 231, arXiv:0709.0224.<br />
S. Klein and J. Nystrand, Phys. Rev. Lett. 92:142003 (2004).<br />
V.P. Goncalves and M.V.T. Machado, Eur. Phys. J. C40, 519 (2005).<br />
L. Motyka and G. Watt, arXiv:0805.2113.<br />
F. Yuan, Phys. Lett. B510, 155 (2001).</p>
<p>Cheers,<br />
T.</p>
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	<item>
		<title>By: Namit</title>
		<link>http://dorigo.wordpress.com/2008/12/29/three-exquisite-exclusive-charmonium-signals/#comment-103262</link>
		<dc:creator>Namit</dc:creator>
		<pubDate>Tue, 30 Dec 2008 06:14:03 +0000</pubDate>
		<guid isPermaLink="false">http://dorigo.wordpress.com/?p=1883#comment-103262</guid>
		<description>Dear Tomasso,
                 Could you provide a pointer/link to the theory paper with which the results are compared with.
Regards</description>
		<content:encoded><![CDATA[<p>Dear Tomasso,<br />
                 Could you provide a pointer/link to the theory paper with which the results are compared with.<br />
Regards</p>
]]></content:encoded>
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	<item>
		<title>By: carlbrannen</title>
		<link>http://dorigo.wordpress.com/2008/12/29/three-exquisite-exclusive-charmonium-signals/#comment-103261</link>
		<dc:creator>carlbrannen</dc:creator>
		<pubDate>Tue, 30 Dec 2008 01:55:28 +0000</pubDate>
		<guid isPermaLink="false">http://dorigo.wordpress.com/?p=1883#comment-103261</guid>
		<description>Nice bumps. It seems to me that you could change the gluon number problem by using a ggg interaction, aren&#039;t they allowed?

The paper I&#039;m planning on sending to Phys Math Central revolves around the heavy quarkonium states. In short, it says that the six J/psi mesons and the six Upsilon mesons are in direct correspondence with each other, and also with the six leptons; they are corresponding excitations.

To see the correspondence, you have to split the sixes up into two triplets each, with formulas similar to each other and to the charged leptons and the neutrinos. Let&#039;s see if I can write the mass formulas here using the local LaTeX. Since there is no preview, this is an exercise in perfection. Which reminds me, the heavy snow in Seattle has kept me from the chess club for 2 weeks but now the roads are clear.

Then the three charged leptons have masses (in MeV) given by
$latex \sqrt{m_{eg}} = 25.054 309 435(\sqrt{2}+\cos(2/9 + 2g\pi/3)),$
where g is the generation number. The above constant was chosen so that it gives the masses to within experimental error. The three neutrino masses (in eV) are given by
$latex \sqrt{m_{\nu g}} = 0.1414(\sqrt{2}+\cos(2/9 + 2g\pi/3 + \pi/12)).$
This gives the right values for the neutrino oscillation measurements, which are of the absolute values of the differences in the squares.

Thus the leptons can be written as two slightly different triplets. The claim is that the J/psi and Upsilon also show the same triplets. Here&#039;s the J/psi mass formulas (using the same overall constant $latex \mu = 25.054 ...$, as in the charged lepton formula):
$latex \sqrt{m_{\psi 1 g}} = \mu(2.44247 - 0.25002\cos(2/9+2g\pi/3+\pi/12)),$
$latex \sqrt{m_{\psi 2 g}} = \mu(2.51049 - 0.08943\cos(2/9+2g\pi/3)),$
and for the Upsilon:
$latex \sqrt{m_{\Upsilon 1 g}} = \mu(3.99332 - 0.12667\cos(2/9+2g\pi/3+\pi/12)),$
$latex \sqrt{m_{\Upsilon 2 g}} = \mu(4.13723 - 0.07754\cos(2/9+2g\pi/3)).$
Note that each equation uses two arbitrary constants to give three masses. And some of the arbitrary constants are suggestively close to rational numbers like 4 or 1/4.

Unlike the leptons, the heavy meson fits are not exact, but have errors of magnitude around the isospin splitting (which ought to be how inaccurate a color calculation should be). For the J/psi you get 
[3096.916, 3775.2, 4421.1] versus experiment
[3096.916(.011), 3775.2(1.7), 4421(4)]
and
[3686.08, 4040, 4150] versus
[3686.093(.034), 4039(1), 4153(3)].

The Upsilon mesons 
[9456, 10035, 10554] versus
[9460.30(26), 10023.26(31), 10579.4(1.2)]
and 
[(10355.2, 10864.4, 11019.5] versus
[10355.2(.5), 10865(8), 11019(8)].
Note that the really bad fit is that of the lighter Upsilon mesons. It may not be a coincidence that these are the states that are most easily modeled using non relativistic QM and have now been renamed $latex \Upsilon(1S), \Upsilon(2S), \Upsilon(4S)$ if I recall.

Anyway, I am still busily improving the paper. Phys Math Central requires a background section to familiarize non specialists which is a very good idea.</description>
		<content:encoded><![CDATA[<p>Nice bumps. It seems to me that you could change the gluon number problem by using a ggg interaction, aren&#8217;t they allowed?</p>
<p>The paper I&#8217;m planning on sending to Phys Math Central revolves around the heavy quarkonium states. In short, it says that the six J/psi mesons and the six Upsilon mesons are in direct correspondence with each other, and also with the six leptons; they are corresponding excitations.</p>
<p>To see the correspondence, you have to split the sixes up into two triplets each, with formulas similar to each other and to the charged leptons and the neutrinos. Let&#8217;s see if I can write the mass formulas here using the local LaTeX. Since there is no preview, this is an exercise in perfection. Which reminds me, the heavy snow in Seattle has kept me from the chess club for 2 weeks but now the roads are clear.</p>
<p>Then the three charged leptons have masses (in MeV) given by<br />
<img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7Bm_%7Beg%7D%7D+%3D+25.054+309+435%28%5Csqrt%7B2%7D%2B%5Ccos%282%2F9+%2B+2g%5Cpi%2F3%29%29%2C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{m_{eg}} = 25.054 309 435(\sqrt{2}+\cos(2/9 + 2g\pi/3)),' title='\sqrt{m_{eg}} = 25.054 309 435(\sqrt{2}+\cos(2/9 + 2g\pi/3)),' class='latex' /><br />
where g is the generation number. The above constant was chosen so that it gives the masses to within experimental error. The three neutrino masses (in eV) are given by<br />
<img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7Bm_%7B%5Cnu+g%7D%7D+%3D+0.1414%28%5Csqrt%7B2%7D%2B%5Ccos%282%2F9+%2B+2g%5Cpi%2F3+%2B+%5Cpi%2F12%29%29.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{m_{\nu g}} = 0.1414(\sqrt{2}+\cos(2/9 + 2g\pi/3 + \pi/12)).' title='\sqrt{m_{\nu g}} = 0.1414(\sqrt{2}+\cos(2/9 + 2g\pi/3 + \pi/12)).' class='latex' /><br />
This gives the right values for the neutrino oscillation measurements, which are of the absolute values of the differences in the squares.</p>
<p>Thus the leptons can be written as two slightly different triplets. The claim is that the J/psi and Upsilon also show the same triplets. Here&#8217;s the J/psi mass formulas (using the same overall constant <img src='http://l.wordpress.com/latex.php?latex=%5Cmu+%3D+25.054+...&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\mu = 25.054 ...' title='\mu = 25.054 ...' class='latex' />, as in the charged lepton formula):<br />
<img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7Bm_%7B%5Cpsi+1+g%7D%7D+%3D+%5Cmu%282.44247+-+0.25002%5Ccos%282%2F9%2B2g%5Cpi%2F3%2B%5Cpi%2F12%29%29%2C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{m_{\psi 1 g}} = \mu(2.44247 - 0.25002\cos(2/9+2g\pi/3+\pi/12)),' title='\sqrt{m_{\psi 1 g}} = \mu(2.44247 - 0.25002\cos(2/9+2g\pi/3+\pi/12)),' class='latex' /><br />
<img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7Bm_%7B%5Cpsi+2+g%7D%7D+%3D+%5Cmu%282.51049+-+0.08943%5Ccos%282%2F9%2B2g%5Cpi%2F3%29%29%2C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{m_{\psi 2 g}} = \mu(2.51049 - 0.08943\cos(2/9+2g\pi/3)),' title='\sqrt{m_{\psi 2 g}} = \mu(2.51049 - 0.08943\cos(2/9+2g\pi/3)),' class='latex' /><br />
and for the Upsilon:<br />
<img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7Bm_%7B%5CUpsilon+1+g%7D%7D+%3D+%5Cmu%283.99332+-+0.12667%5Ccos%282%2F9%2B2g%5Cpi%2F3%2B%5Cpi%2F12%29%29%2C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{m_{\Upsilon 1 g}} = \mu(3.99332 - 0.12667\cos(2/9+2g\pi/3+\pi/12)),' title='\sqrt{m_{\Upsilon 1 g}} = \mu(3.99332 - 0.12667\cos(2/9+2g\pi/3+\pi/12)),' class='latex' /><br />
<img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7Bm_%7B%5CUpsilon+2+g%7D%7D+%3D+%5Cmu%284.13723+-+0.07754%5Ccos%282%2F9%2B2g%5Cpi%2F3%29%29.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{m_{\Upsilon 2 g}} = \mu(4.13723 - 0.07754\cos(2/9+2g\pi/3)).' title='\sqrt{m_{\Upsilon 2 g}} = \mu(4.13723 - 0.07754\cos(2/9+2g\pi/3)).' class='latex' /><br />
Note that each equation uses two arbitrary constants to give three masses. And some of the arbitrary constants are suggestively close to rational numbers like 4 or 1/4.</p>
<p>Unlike the leptons, the heavy meson fits are not exact, but have errors of magnitude around the isospin splitting (which ought to be how inaccurate a color calculation should be). For the J/psi you get<br />
[3096.916, 3775.2, 4421.1] versus experiment<br />
[3096.916(.011), 3775.2(1.7), 4421(4)]<br />
and<br />
[3686.08, 4040, 4150] versus<br />
[3686.093(.034), 4039(1), 4153(3)].</p>
<p>The Upsilon mesons<br />
[9456, 10035, 10554] versus<br />
[9460.30(26), 10023.26(31), 10579.4(1.2)]<br />
and<br />
[(10355.2, 10864.4, 11019.5] versus<br />
[10355.2(.5), 10865(8), 11019(8)].<br />
Note that the really bad fit is that of the lighter Upsilon mesons. It may not be a coincidence that these are the states that are most easily modeled using non relativistic QM and have now been renamed <img src='http://l.wordpress.com/latex.php?latex=%5CUpsilon%281S%29%2C+%5CUpsilon%282S%29%2C+%5CUpsilon%284S%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Upsilon(1S), \Upsilon(2S), \Upsilon(4S)' title='\Upsilon(1S), \Upsilon(2S), \Upsilon(4S)' class='latex' /> if I recall.</p>
<p>Anyway, I am still busily improving the paper. Phys Math Central requires a background section to familiarize non specialists which is a very good idea.</p>
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