All the lepton masses from G, pi, e

Jay R. Yablon

If one needs 5 terms to approximate the 0.511.. MeV mass of the electron
to 0.509 MeV then there's no more "numerical coincidence" value left in it.
Jay will find out that there are many other combinations that will lead to
equal or better results.

Hans, this is Jay. I am not fixed for sure on the 5 terms for the electron, but what is very nice about this result is that the electron mass is then characterized in leading order by only four-loop terms and that all five of the four loop terms go into the electron mass. So, the fact that every one of these terms has a common Feynman diagram interpretation as a four-loop term seems to be based on some physics beyond coincidence and beyond picking and choosing terms

Still the approach makes sense and the most intriguing hint may be the anomalies.
To bring them up one more time:

0.00115869 = muon / Z mass ratio
0.00115965 = electron magnetic anomaly
0.00000635 = electron / W mass ratio
0.00000626 = difference between muon and electron magnetic anomaly


Our friend arivero has now got me looking at these anomalies using the terms I have developed. I will let you know what I find.

Jay

arivero

A small corrected improvement. While mm/mZ is 99.76% of the Schwinger correction, the following approximation
[tex]{m_\mu\over m_Z}={\alpha \over 2 \pi} - 0.5 ({\alpha\over\pi})^2[/tex]
is 99.9982% accurate

Still I hope the 0.5 comes from addition of multiple Feynman diagrams, then being approximate itself. I am afraid than an exact -1/2 coefficient in second order QED comes not from a finite term, but associated to the infrared divergence.

arivero

Hmm but at second order of alpha, there is a vacuum polarisation term due to pairs electron positron, and it is not negligible. It is [tex]0.015687 \alpha^2/\pi^2[/tex], ie .0000000846.

Er, wait... lets supposse this term is not in your ratio. Add it!
0.0011586922+.0000000846=.0011587768

Ok it does not seem very much of an improvement. But if we add also my term (1/2) (mm/mW)^2 we have
0.0011586922+.0000000846+.0000008608=.0011596376
to be compared with experimental 0.0011596521. No bad.

I would conclude that our series (well, two terms) on masses does not approach to the experimental magnetic moment, but to the two loop QED (or full electroweak, does not matter) vertex correction, excluding the vacuum polarisation loop. On first examination, it seems that this loop is recovered when we introduce the product of electron and tau masses, but I have not examined the expansion for this anomalous moment.

Just for the record, the two loop correction for electron is 0.5 a/pi - 0.3284794 (a/pi)^2. In our case, disregarding the vacuum polarisation amounts to a second term coefficient -.3441636 We have
[tex]a_e^{\mbox{2qed-v.p.}}=.00115955280[/tex]
[tex]{m_\mu \over m_Z}+ \frac12 {m_\mu^2 \over m_W^2}=.001159553[/tex]
agreement about, well, dammn, it is already inside the experimental error for Z0... if you want, respective to central values it is of Z and W, it is 99.99997%.

If one feels bad about having the 2 m_W^2 in the denominator, you can always use the square of an unknown neutral mass about 114 GeV... the LHC start will wait some time yet.

Hans de Vries

That's just the value you want it to have! Do you have you any more
data like that. Like the explicit value of the same term in the Muon?
Really interesting would also be the relation with the muon/electron
mass with these terms in the formula's (which are known analytically.
I suppose that we can ignore the same second order terms for virtual
muon-anti muon pairs for the time being.


0.000 006 263 813 : Difference between electron and muon anomaly
0.000 000 084 639 : First vacuum polarization term of the electron
----------------------------------------------------------------------
0.000 006 348 452 : Sum of the vacuum polarization terms (the above)

0.000 006 353 732 : (+/-3 000) mass ratio of electron and the W boson.
0.000 000 002 998 : uncertainty from the Z mass


Regards, Hans.

arivero

That was the point. We have hit experimental precision in both expressions.

Sunday night I mailed you a separate email, the bibliography tells where to find such terms. You would not like me to write the closed expression for the "vacuum polarised" terms of the muon, it is about four lines in the Phys Rev. Moreover, the next order terms are also relevant.

Alejandro

arivero

Hmm, I had tried the same with the data truncated at two loops for electron and three for the muon and I got a somehow weaker result. It seems one needs to add the three loop data for the electron, but it is a mess because then we have three new diagrams to exclude.

Hans de Vries

The other interesting observation is that the missing term
becomes equal in both cases to within experimental value.


0.00115877693 : mu/mZ + VP2
0.00115965218 : electron anomaly
0.00000087525 : missing

0.00116504602 : mu/mZ + me/mw (+ VP2 - VP2)
0.0011659208_ : muon anomaly
0.00000087478 : missing

0.00000000047 : missing1 - missing2
0.00000002668 : uncertainty due to Z (cancels if missings are subtracted)
0.00000000299 : uncertainty due to W mass


So there may be a single missing term.


Regards, Hans

PS: VP2 = 0.00000008464 : First vacuum polarization term of the electron
which is a second order term.

PPS: I'm not entirely sure if the term 0.00116504602 : mu/mZ + me/mw (+ VP2 - VP2)
should indeed not include VP2.

me/mW - VP2 = difference between muon and electron anomaly.
mu/mZ + VP2 = all self energy terms + first vacuum polarization term of the electron anomaly

arivero

Making it in a reverse way: as the missing term is already in the electron anomalous moment, assume it is sort of square of the first term. So it is mu^2/X^2. Solve for X:
X=sqrt(mu^2/.000000875015)=112.95 GeV

The quantity is interesting in two ways. We suspect of a neutral scalar H0 at 115 GeV, and it could have this role. But also mw*sqrt(2) is 113.87 GeV, so we can use the W particle, which was my first attempt for the missing term. Lacking of more theory, both are equally suitable: values up to 114.5 GeV are covered by the Z indeterminacy. The first has the advantage of not using an arbitrary 1/2 coefficient and it is neutral as the Z, but it has not been discovered (yet?), the second is an already discovered particle but we have used it for the "vacuum polarised graphs", and it is surprising to have it here too, even if squared.

arivero

For the sake of completeness, references. The industry of calculation of the anomalous moment seems to be based in Cornell, around a veteran named T. Kinoshi-ta. Other group does exist in North America around A. Czarnecki

http://arxiv.org/abs/hep-ph/9810512, from Czarnecki and Marciano, is the main entry point for the calculation up to order alpha^4. It is regretly a short preprint and it does not separate loop by loop, so one is referred to more detailed bibliography, which is not in the arxiv

The five diagrams for order alpha^2 appear well separated in Levine and Wright Phys. Rev. D 8, 3171-3179 (1973) http://prola.aps.org/abstract/PRD/v8/i9/p3171_1. I got from here the specific value we were using above.

Also some sums for 40 diagrams of order alpha^3 are presented there. Note that of these, 12 diagrams are vacuum polarisation loops, amounting perhaps to a contribution of 0.37 (alpha/pi)^3

The alpha^2 "polarisation loop", depending of the mass quotient of the external and internal lepton, is studied both analytic and numerically by Li Mendel and Samuel, Phys. Rev. D 47, 1723-1725 (1993) http://prola.aps.org/abstract/PRD/v47/i4/p1723_1

Also Samuel Li and Mendel provide a calculation of tau lepton, Phys. Rev. Lett. 67, 668-670 (1991) http://prola.aps.org/abstract/PRL/v67/i6/p668_1 up to order alpha^3. For this lepton, the contributions for quarks are already noticeable at this scale.

Jay R. Yablon

Alejandro"

Do you and Hans have any close fits relating the tau magnetic moment with any lepton-to-electroweak-boson ratio?

Jay.

arivero

Hmm the answer past yesterday was yes, the answer today is more towards no. On one side the quantity [tex]m_e m_\tau/m_W^2[/tex] in of the right order of magnitude to do further corrections in our calculations, but we do not need it anymore, giving the uncertainty in the mass of Z. On other hand, and more concretely answering your question, the difference between anomalous moment of mu and tau can only be covered with a new quotient [tex]m_e/m_{X^+}[/tex], and the mass of the new X+ particle should be around 68-70 GeV. At these energy range, the LEP2 presented an slight "statistical" deviation, but no particle

arivero

80425

So I was getting a discrepance between calculations at home and calculations at work.

Jay R. Yablon

Dear Alejandro and Hans:

I just posted to my web site http://home.nycap.rr.com/jry/FermionMass.htm, a Gordon-like decomposition of the Fermion mass into a term due to electromagnetic charge, and a term due to magnetic moment. This is still an early working draft.

I hope this can help you in your efforts by providing a covariant field theory context for your efforts to characterize the magnetic moments. I know that your efforts have helped me recognize that consideration of magnetic moments is likely to be a critical aspect of what I am attempting to do.

Best,

Jay.

arivero

Probably Hans is also exploring this way, but I am not so optimistic about a direct connection; perhaps a semiclassical effect, could be. But even that is strange to manage. To me, it seems more as if the symmetry breaking mechanism of the electroweak group (and its vacuum value) were needing of the lepton masses is some misterious way.

From our quadratic formulae we can get rather intriguing equations. For instance this one:

[tex]{m_\tau\over m_Z} + {m_\mu\over m_W}=
{m_\tau\over m_\mu} a_\mu^{s.e.} + {m_\mu\over m_e} a_\mu^{v.p.}
[/tex]

Where [tex]a_\mu^{s.e.},a_\mu^{v.p.}[/tex] are the self-energy and vacuum polarisation parts of the muon anomalous magnetic moment; note that the v.p. part depends internally of lepton mass quotients, while the s.e. is mass independent, in QED (in the full electroweak theory new dependences appear).

arivero

Some of the development of the thread uploaded at http://arxiv.org/abs/hep-ph/0503104

Hans de Vries

Hmm,

I was doing something else and ran just incidently into this one:

[tex]\sqrt{ \ 2 \ \frac{m_V}{m_Z} \ \frac{m_{\tau}}{m_e}} \ = \ 137.038 (12)[/tex]

m_V is the vacuum expectation value of 246.22046 GeV (according to Jay)
The biggest uncertainty is from the tau mass.


Regards, Hans


[itex]\ \ \alpha \ \ \ [/itex] = 1/137.03599911
m_V = 246220.46
m_Z = 91187.6 (+-2.1)
m_τ = 1776.99(+0.29-0.26)
m_e = 0.51099892 (+-0.00000004)

arivero

Also this vacuum should have a high uncertainness. I wonder where did Jay got so many digits from.