Here's a mass formula for the pions that I hope the reader will find amusing. Begin with the mass formula for the charged leptons (since I want to use "n" for radial excitations of mesons, I will use "g" for the quantum number that gives the generation):
\sqrt{m_{e g}}/25.0544 = \sqrt{0.5} + \cos(2/9 + 2g\pi/3)
where 25.0544 \sqrt{\textrm{eV}} is a mass scaling constant that makes the leptons nice and we will use for the mesons as well.
The basic idea is to think of the mesons as having radial excitations "n" with quantum numbers like the spherical harmonics of the hydrogen atom, but also having color excitations following Koide's formula. This means that we get the hydrogen wave functions, but tripled.
Guess that the three lowest mass pions = (\pi, \pi(1300), \pi(1800)) with masses (139.57, 1300, 1812) are the n=1, l=0 states with three Koide generations distinguishing them. Let's use "g" for the generation number. Then the formula for these three pion masses turns out to be:
\sqrt{m_{\pi 10g}}/25.0544 = 1.196797 -0.743543\cos(2/9 + 2g\pi/3).
Make the same guess about the three lowest mass J=1 pions = (\pi_1(1400), \pi_1(1600), \pi_1(2015)). Their masses are (1376, 1653, 2013). Since J=1, these have n=2, l=1. Note the last of these, along with the last two in the next group of three, is hard to find in the PDG. It's listed on the page titled "further states":
http://pdg.lbl.gov/2007/listings/m300.pdf
The Koide formula for these j=1 pion masses is:
\sqrt{m_{\pi 21g}}/25.0544 = 1.6313 + 0.1792986\cos(2/9 + 2g\pi/3 + \pi/12).
The extra angle pi/12 means that these have a formula like the neutrinos.
And the three lowest J=2 pions are the (\pi_2(1670), \pi_2(1880), \pi_2(2005)) with masses (1672.4, 1880, 1990). These presumably have n=3, l=2. They also have a Koide formula:
\sqrt{m_{\pi 32g}}/25.0544 = 1.71477 + 0.0864566\cos(2/9 + 2g\pi/3 + 2\pi/12).
The above 3 equations give 9 masses and they work pretty well.
For the hydrogen excitations, the total energy (or mass) of the hydrogen atom is approximately:
m_{H n} = 10^9 - 13.6 /n^2 eV.
The second term, the binding energy, is very small compared to the total energy of the hydrogen atom (about 1GeV) which is mostly due to the rest mass of the proton. Consequently, if I rewrite this as a formula for the square root of the mass of the hydrogen atom, the square root energy levels will still follow a 1/n^2 law:
\sqrt{m_{H n}} = 10^{9/2} - (10^{-9/2}\times 13.6/2) /n^2 \sqrt{eV}.
So based on the energies of the hydrogen atom, one might look for 1/n^2 dependency in the three Koide mass formulas given above, in order to unify the three equations. The three "v" terms need to be: (1.19680, 1.6313, 1.71477). These have n=(1,2,3). They can be fairly well approximated with the formula v_n = 16/9-\sqrt{1/3}/n^2. The three "s" terms need to be (-0.743543, +0.1792986, +0.0864566 ). The sign change can be accounted for by making all three signs negative, but taking an extra phase of pi for the n=1 and n=2 case. This can be accomplished by adding a phase of nj\pi/2=n(n-1)\pi/2 to the phase angle. Then the numbers are fairly closely approximated by s_n = -0.75/n^2. The resulting formula for nine pion masses is fairly compact:
\sqrt{m_{\pi nlg}}/25.0544 = 16/9 - \sqrt{3}/n^2 -(3/4)\cos(2/9 + 2n\pi/3 + jn\pi/2)/n^2
The above simplifications give an approximations of the v's as follows:
1.19680 -> 1.2004275
1.6313 -> 1.63344
1.71477 -> 1.71362
And the s's are approximated by:
-0.743543 -> -3/4
+0.1792986 -> 0.1875,
+0.0864566 -> 0.08333
which is fairly close.
The nine computed masses are as follows:
137.99 1271.2 1834.1
1365.5 1659.6 2032.4
1676.1 1883.0 1977.3
The nine actual masses are:
136.5(2.5) 1300(100) 1812(14)
1376(17) 1652(21) 2014(?)
1672.4(3.2) 1876-2003(?) 1974-2005(?)
Note: The error in the pi mass is due to the mass difference between the pi+ and pi0. And the masses labeled with (?) do not have mass spreads in the PDG. If more than one measurement is given, the range of measurements are given.
