The Koide Lepton Mass Formula and Geometry of Circles

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by  Jerzy Kocik

The Koide Lepton Mass Formula and Geometry of Circles
Jerzy Kocik
Department of Mathematics, Southern Illinois University, Carbondale, IL 62901∗
AremarkableformalsimilaritybetweenKoide’sLeptonmassformulaandageneralizedDescartes
circle formula is reported.
Keywords: Lepton mass, Koide formula, Descartes circle theorem, geometry.
I. INTRODUCTION formula. Descartes – in his 1654 letter to the princess
of Bohemia, Elizabeth II – showed that the curvatures
Quarksandleptonsarebelievedtobethefundamental of four mutually tangent circles (reciprocal of radii), say
2 particles of matter yet their nature is still far from being a,b,c,d, satisfy the following “Descartes’s formula”:
1 understood. One of the exciting puzzles is a formula
Theorem II.1 (Descartes’s Circle Formula, 1654 [2]).
0 involving the masses of the three leptons, discovered by
The curvatures of four circles in Descartes’s configura-
2
Yoshio Koide [6, 7]:
tion satisfy this equation:
n
2 √ √ √
Ja me+mµ+mτ = 3( me+ mµ+ mτ)2. (I.1) (a+b+c+d)2 =2(a2+b2+c2+d2) (II.1)

5 SeeTable1forthecorrespondingnumericalvalues(from where a = 1/r , b = 1/r , etc., denote reciprocals of
1 2
] [10]). 2. Fromra Ddeisiccaartlelse dto bKeoniddes , which are signed curvatures. In Fig.1
h in the middle, circle D is a boundary of an unbounded
p electron m =0.510998910 MeV/c2 In this dnoistek w(er ewgainotn too cualtls aidtteentDion) thoe an ccuerioituss fboernmdal issimnileagriatyti ovfe K.oCidier’cs lfeormula to
- e Descartes circle formula. Descartes – in his letter to the princess of Bohemia, Elizabeth
D in the right figure has bend equal zero.
n =1me II, showed in 1645 that the curvatures of four mutually tangent circles (reciprocal of
√
e m =1 radii), say a,b,c,d, satisfy the following “Descartes formula”:
g e

s. muon mµ =105.658367 MeV/c2
c =206.768282 m B A A D
e
ysi √mµ =14.37943957 me1/2 A C D C B C
h tau mτ =1776.84 MeV/c2 B

p =3477.1894 m
[ √ e
mτ =58.97
1 FIG.Fi1g:urFeo 1u:r Fcoirucr lceiricnlesv ainr ivoaurisouDs eDsecsacratretess ccoonnffiigguurraatitoinosn s.

v
TABLE I: Lepton masses. Theorem 1 (Descartes Circle Formula, 1654 [Des]): The curvatures of four circles in
7
Descartes Tcohnefigsuirmatiiolan rsiattyisfoy fthDis eesqcuaatritoens: ’s circle formula to Koide’s
6
Quite remarkably, Koide used his formula to predic t lepton mass formula is striking: represent masses as re-
0 (a + b + c + d)2 = 2 (a2 + b2 + c2 + d2) (2)
2 the mass of the tau lepton with surprising accuracy: ciprocals of disk areas, equivalently, their curvatures as
1. old measured mτ =1784±4 [1970’s] wcuerrvea tausd=rq1eiffsu/r. ea1,Ir rnebe n=F1trig/oruc2oro, ete seffi1t ocic.fn,i edmtnheenta osmftseoie drrsde,tlcehi,ap erncoidsrccuallmews eDoof gfirsea stdaqi iub( IcoaIaur.lnel1eds)da: rybe2 exoncifdne sap,s nttw euhafniodcbrho ouaanfrdee dsi gdnieskd
0 predicted by Koide m =1776.97 [1982] (region2 o/u3ts.idTe Dh)e hfeonucer tiths bteenrdm is mnegaaytivbee. Ctairkcleen Da isn tzheer roig,hdt fi=gur0e, hcaos rb-end equal
2 τ zero.
1 new measured mτ =1776.99±.3 [2002] responding geometrically to a line as a special circle.
v: newer measured mτ =1776.84±.17 [2011] The siTmhileariftay cotfo Dreoscfa2rt/es3 ciirncvlea floidrmautleas tot hKeoiedxe’asc ltepctoonr rmeassps ofonrdmeunlac ies striking:
represewnt imthassDese assc raerctiperso’csalfso orfm diuskla a.reaHs,o ewqueivvaelre,nttlyh tehepirr ceusrevnatturaesu atsh soqruare roots
i
X (all in MeV/c2). Some interesting formal associationosf mashseas,s afnodu wned gaet f(o2r) mexucleapt ffoorr ac idricffleersenitn cogeeffniceireantl fporo sthitei osunmt ohfa stquares: 2
r have been noticed since [1, 4, 9, 11], but after almost 3i0nsteadg oef n2e/3r.a l Tizhees foDuretshc taerrmte ms’asy. be taken as zero, d =0, corresponding geometrically
a to a line as a special circle.
years the consensus is that the “mystery of the lepton

Theorem II.2 (General Circle Formula [5]). The radii
mass formula” [8] remains unsolved. The factor of 2/3 invalidates the exact correspondence with Descartes’ formula.
Howevoerf, nthe+ p2re(snen−t a1u)th-osrp hhaesr efsouCnd1 ,a. .fo.r,mCulna+ f2ori ncirgcleens eirna glepnoersailt ipoonsition that
generaliinzesR Dnescsaartteiss’f.y a quadratic equation

II. FROM DESCARTES TO KOIDE
Theorem 2 (General Circle FormuBla F[jkB])T: T=he0 radii of n+2 (n–1)-sph(eIreIs. 2C)1,…, Cn+2
in general position in Rn satisfy a quadratic equation
In this note we want to call attention to a curious for - or simply BFBT = 0 (3)
mal similarity of Koide’s formula to Descartes’s circle
or simply
F b b =0
ijFibjb = 0
ij i j

where Bw =h e[bre1,…B,b=n] i[sb t1h,e .r.o.w, mbnat]riixs (tchoveecrtoowr) omf athter cixurv(actouvreesc otof rth)e ospfhtehrees and F =
∗Electronicaddress: [email protected] f –1 is thceu inrvveartsue roef sthoe f“ctohnefigsuprahteiorne smaatrnixd” Ff wi=th efn−tr1iesi sdetfihneedi bnyv erse of

(d – r – r)/2rr in general
ij i j i j (4)
fij =
cos , if C and C intersect
i j

Here d = distance between the centers of ith and jth spheres, r = radius of the ith sphere.
ij i
For spheres that intersect,  is the angle of their intersection.
2
Consider the special case where the product for every pair is the same, say p. Then one
calculates the inverse of the corresponding matrix f:

12p p p p 
1 p p p
 
f = p 1 p p  F = 1  p 12p p p 
p p 1 p 3p22p1  p p 12p p 
   
p p p 1  p p p 12p

Solving (3) gives the following geometric result:

Proposition: Four circles of curvatures a, b, c, d, respectively, intersecting pairwise at
the angle  = arcos p satisfy the quadratic equation

  p
a2 b2 c2 d2  (abcd)2 (5)
3p1

Consider the special case where the product for every pair is the same, say p. Then one

calculates the inverse of the corresponding matrix f:
For circles mutually tangent we have p = 1 and the above formula reduces to

Descartes f1orm2pula. pWe arep howepver interested in the value 2/3 for the coefficient on the
1 p p p
f = pp p1 p1 pp  F = r i3gph2t1 2spide1.  pp 1p2p 1p2p pp 
p p p 1 Corollary: pKoidep’s leptopn m1ass2 pformula may be interpreted as a Descartes-like circle
formula (3) for p = 2/3, where the squared curvatures correspond to the masses of the

Solving (3) gives the following geometthrirce ree sluelpt:t ons. One mass is assumed zero and corresponds to a straight line. Each pair of
circles intersects under the same angle:
Proposition: Four circles of curvatu res a, b, c, d, respectively, intersecting pairwise at
the angle  = arcos p satisfy the quadratic equation  = arccos(2/3)  0.841 rad  15/56   48.2

  p
a2 b2 c2 dI2nterestingly(, atheb racdii,d r)e2c iprocals of squ(5a)r e roots of masses, are close to integer values,
3p1
namely, r = 1, r  4.10, r  58.97, but not close enough to warrant further attention.
  e

2
For circles mutually tangent we have p = 1 and the above formula reduces to
Descartes formula. We are however i nterested in the value 2/3 for the coefficient on the
right side. electron
the “configuration matrix” f with entries defined by electron

(cid:40) Corollary: Koide’s lepton mass for mula may be interpreted as a Descartes-like circle
f = (dij −ri−rj)/2rirj in genfoerrmaulla (3) for p = 2/3, where the s quared curvatures correspond to the masses of the
ij cosϕ, if Ci tahnrede lCepjtoinns.t e Orsneec mt.ass is assumed z ero and corresponds to a straight line. Each pair of muon
circles intersects under the same angle:
(II.3)
 = arccos(2/3)  0.841 rad  15/56   48.2
Here dij = distance between the centers of ith and
jth spheres, r = radius of the ith spInhteerreest.ingFlyo,r thsep rhadeiri,e rseciprocals of square roots of masses, are close to integemr vuaolune s,
i namely, r = 1, r  4.10, r  58.97, but not close enough to warrant further attention.
thatintersect,ϕistheangleoftheirintersection. (Quite e

interestingly, the above theorem can be derived from the tau tau

fact that circles in the Euclidean plan e may be regarded electron
as vectors in the Minkowski space [5] .) electron

Consider the special case where th e product for every
Figure 3: Lepton masses via circle configuration (on the left a detail)
pair is the same, say p. Then one calculates the inverse
muon
of the corresponding matrix f:

 
−1 p p p
 p −1 p p  muon
f = 
 p p −1 p  tau tau
p p p −1 3
FIG.2: Leptonmassesviacircleconfiguration(detailonbot-
 
1−2 p p Figupre 3: Lepton mtaossmes) .via circle configuration (on the left a detail)
1  p 1− 2 p p 
⇒F =  
3p2+2p−1 p p 1−2 p  that may be a candidate for geometry of generalized
p p p 1−2
Koide formula:

√ √ √
Solving (II.2) gives the following geomet ric result: m +m +m +...=κ·( m + m + m3 +...)2.
1 2 3 1 2 3
Proposition II.1. Four circles of curvatures a, b, c,
The relation is thus
d, respectively, intersecting pairwise at the angle ϕ =
arccosp satisfy the quadratic equation
1 2 2
p p= and κ= ⇒p= .
(a2+b2+c2+d2)= (a+b+c+d)2. (II.4) n+1−1/κ 3 2n−1
3p−1 (II.6)
For circles mutually tangent we have p = 1 and the
above formula reduces to Descartes’s formula. We are,
III. CONCLUSIONS
however,interestedinthevalue2/3forthecoefficienton
the right side.
We conclude with some remarks:
Corollary II.1. Koide’s lepton mass formula may be
1. Leptons and segments. Ifoneconsidersthreeseg-
interpreted as a Descartes-like circle formula (II.2) for
mentsonalineratherthancirclesinplane,formula
p = 2/3, where the squared curvatures correspond to the
(II.2) with n=1, becomes
masses of the three leptons. One mass is assumed zero
and corresponds to a straight line. Each pair of circles p
intersects under the same angle: a2+b2+c2 = (a+b+c)2.
2p−1
ϕ=arccos(2/3)≈0.841 rad ≈15/56π ≈48.2◦. To make the coefficient in Koide’s formula equal
κ = 2/3 we would need p = 2. The segments
Interestingly, the radii, reciprocals of square roots of for e, µ, τ would be (−1.40,1.40), (−1.49,1.68),
masses, are close to integer values, namely, rτ = 1, (−1.42,1.47), respectively.
r ≈ 4.10, r ≈ 58.97, but not close enough to warrant
µ e
further attention. 2. Neutrinos. Recently, Brannen observed [1] that
thesamepatternisfollowedbytheneutrinomasses
Since the Koide formula might be applicable for other undertheconditionthatthesquarerootofthefirst
particle families, let us provide a solution to the configu- mass is taken to be negative.
ration of n+2 spheres in n-dimensional space:
With his data, m = 0.000388eV, m =
1 2
p 0.00895eV,m =0.0507eV,thefactorof2/3isob-
b2+b2+...+b2 = (b +...+b )2 (II.5) 3
1 2 n+2 (n+1)p−1 1 n+2 tained with the very good precision of 0.01% and
Since the Koide’s formula might be applicable for other particle families, let us
provide a solution to the configuration of n+2 spheres in n-dimensional space:

p
b2 b2 ...b2  (b ...b )2 (5)
1 2 n2 (n1)p1 1 n2

that may be a candidate for geometry of generalized Koide formula:

m  m Since mthe K..o.ide’s form(ulam might bme applicmable f.o.r. )o2t her particle families, let us
1provid2e a so3lution to the configura1tion of n2+2 spher3es in n-dimensional space:

The relation is thus p
b2b2...b2  (b ...b )2 (5)
1 1 2 n2 (n1)p1 21 n2
p and  2  p (6)
n11/ 3 2n1
that may be a candidate for geometry of generalized Koide formula:

3. Conclusions m1m2m3...  ( m1  m2  m3 ...)2

The relation is thus
We conclude with some remarks:
1 2
p and 2  p (6)
1. Leptons and segments. If one considner1s th1r/ee segments on a lin3e rather th2ann c1ircles in
plane, formula (3) wi th n=1, becomes
3. Conclusions
p
a2 b2 c2  (abc)2
We conclude with some remarks:
2p1

1. Leptons and segments. If one considers three segments on a line rather than circles in
To make the coefficpileannte , ifno rKmuoliad (e3’)s w fiothr mn=u1l,a b eecqoumaels  = 2/3 we would need p = 2. The
segments for e, ,  would be (–1.40, 1.40), (1.49, 1.68), (1.4p2, 1.47), respectively.
a2b2c2  (abc)2
2p1
2. Neutrinos. Recently, Brannen observed [Bra] that the same pattern is followed by the

neutrino masses undTero tmhaek ec othned ictoioefnfi ctiheantt tinh eK soqiduea’sr ef orromoutl ao fe qtuhael fi r=s t2 m/3 awsse iwso tualdk enne etdo pb e= 2. The
negative. segments for e, ,  would be (–1.40, 1.40), (1.49, 1.68), (1.42, 1.47), respectively.

2. Neutrinos. Recently, Brannen observed [Bra] that the same pattern is followed by the

n e u=t r0in.0o0 m03a8ss8e sM uenVde r =th e1 con dition that the s quar e =ro o–t1 o f the first mass is taken to be
n1 e g=a 0ti.v0e0. 8 95 MeV = 23.106701  1 = 4.8028
2 1 2
 = 0. 0507 MeV = 130.6701   = 11.4311
3  = 0.000388 MeV 1 = 1  3  = –1 3
1 1 1
Tab le2 2 =: 0N.0e0u8t9r5in oM meVa s s e=s 23.06701 1 2 = 4.8028
the res u l3t i =n g0. 0e5q0u7 a MtieoVn = 130.6701 1 3 = 11.4311 be that of six 4-spheres in a five-dimensional space
The factor of 2/3 is obtained with the very goToabdl e p2r: eNceiustiroinno moafs s0es. 01%. The resulting with p = 2/9, giving an intersection angle of ϕ ≈
equation Teqhuea tmfiaocn t +o mr mo1f ++2/ 3mm is2 =o+ b2tmai n(3e–d= mwi32t h+( −them√ vm e+ry1  +gmoo√)d2 m pr2ec+isio√n mof 30).201%. The( 7r)e sulting r(3a/p7id)πly≈to707..12%◦.wWheennoontelytthhaetltahset tphrreeceisqiounargkrso,wcs,
1 2 3 1 2 3
poses no pr3mob +le mm +t mo o=u 2r (g–eomm +e trmic +i ntmer)p2 retation, (7) b and t, are considered.
poses no problem to our geoamsentreicg aitnitveerprc1eutarvtia2otnu, ra3es ins3egaastsiovec1 iacutervd2attuore u3ins baossuoncdiaetded to
√
unbounded disk definpeosde sb yno t hdperio srbkelegdmioe ntfio n ooeuudtrs gibdeyoem ate hctreiircc rilenegt.e iroprnetaotuiotns, iadse naegactiirvce lceu.rvature is associated to up m =1m m =1
u u u
unbounded disk defined by the region outside a circle. down m =12m √m =3.464
d u √ d
strange ms =210mu ms =14.491
√
charm m =2500m m =50
c u √ c
bottom m =9000m m =94.868
b u √ b
top mt =348000mu mt =589.915

TABLE II: Quark masses

Figure 2: Neutrino masses via disk configuration (√on left a detail)
Figure 2: NeutrinFo ImGa.s3s:esD viisak dciosnk ficgounrfaigtiuornatifoonr a(onne lgeaftt iav edetamil).
4. Coda. Whether this intriguing connection with ge-
4
But estimations of neutrino masses change rapidly ometry will contribute to an understanding of the
4 with new experiments and only upper bounds are masses of leptons remains an interesting question
measured. Most recent estimations are: electron andrequiresfurtherinvestigation. Theanalogyde-
neutrino: m < 2.2eV, muon neutrino: m < 170 scribed above may turn out to be merely superfi-
keV, and tau neutrino: m < 15.5 MeV [3]. Thus, cial, but given the current state of understanding
taking the highest value for the muon neutrino about the matter, any interesting structural paral-
would suggest a rather low value of 2.4 MeV for lels are worthy of our consideration in the effort of
the tau neutrino if Koide’s formula with ratio 2/3 reconstructing deeper patterns.
were to be satisfied.
Acknowledgements
3. Quarks. Quark mass estimations taken from [11]
are presented in the table below. An extended ver-
sion of Koide’s equation would suggest 2/3 with I am grateful to Philip Feinsilver for his encouraging
a precision of 5% [11]. The geometry here would interest and priceless comments.
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[4] J.M.G´erard,F.Goffinet,andM.Herquet,ANewLook able from Particle Data Group at the Internet site
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