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Magnetic moment of the Muon and Tau

by zincshow
Tags: magnetic, moment, muon
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zincshow
#1
Nov5-13, 11:24 AM
P: 93
For the electron, wiki lists its magnetic moment as −1.00115965218076(27) μB. It does not list it for the Muon or Tau. Wiki does show for the Muon the Anomalous magnetic dipole moment of 0.00116592089. Is it correct to assume the magnetic moment of the Muon is -1.00116592089 μB?

For the Tau, wiki does not show either a magnetic moment or an anomalous magnetic dipole moment. What is the magnetic moment of the Tau?

A google search produces results like http://arxiv.org/pdf/hep-ph/0702027v1.pdf but unfortunately, the article is beyond my pay grade for understanding.
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Bill_K
#2
Nov5-13, 11:49 AM
Sci Advisor
Thanks
Bill_K's Avatar
P: 4,160
Quote Quote by zincshow View Post
For the Tau, wiki does not show either a magnetic moment or an anomalous magnetic dipole moment. What is the magnetic moment of the Tau?
Regarding the anomalous magnetic moment of the tau, this paper says,

Theoretically, the standard model predicts aτ = 1.1769(4) x 10−3
(Since the tau has such a short lifetime, aτ can't be measured by the usual means, so the paper is mainly devoted to clever, indirect ways to do it.)
mfb
#3
Nov5-13, 12:16 PM
Mentor
P: 11,631
Muon: Right

The MEG experiment is much more sensitive than the upper limit on the branching ratio used in the paper, so the measurements improved since then (>2 orders of magnitude).

zincshow
#4
Nov5-13, 02:56 PM
P: 93
Magnetic moment of the Muon and Tau

Great thanks, wonder why wiki doesn't just say:

Electron: −1.00115965218076(27) μB
Muon: -1.00116592089 μB
and Tau: -1.0011769 μB
lpetrich
#5
Nov19-13, 09:56 PM
P: 518
Strictly speaking, it's only the electron that has a magnetic moment close to 1 Bohr magneton. The other charged leptons (electron flavors) have much smaller magnetic moments. That's because the Bohr magneton is the Dirac value of the electron's magnetic moment. In general, the Dirac value is
q/(2m)

for charge q, mass m, hbar = c = 1

The anomalous magnetic moment a is defined be

(magnetic moment) = (Dirac magnetic moment) * (1 + a)

So (MM, muon) ~ 1/200 * (MM, electron) and (MM,tau) ~ 1/1800 * (MM, electron)
zincshow
#6
Nov20-13, 02:01 PM
P: 93
Thanks for the message, it is different from the others and makes sense to me. You are suggesting the better estimates would be:

Electron: about −1.0 μB
Muon: -0.005 μB
and Tau: -0.0005 μB

Also, given the confusion in the answers, do you know any online references I could use? Would not it make more sense for the Muon to be 200 * (MM electron) rather then 1/200 * (MM electron)? Ie. wouldn't it be a lot harder to flip a Muon then it is to flip an Electron?

Much appreciate the help with this.
lpetrich
#7
Nov23-13, 02:35 AM
P: 518
Another case of macroscopic intuitions causing trouble. I'll give a hand-waving argument for an elementary particle's magnetic moment.

Angular momentum J = m*((r)x(v))
Magnetic moment mu = (1/2)*q*((r)x(v))
mu = (q/(2m))*J
for (charge distribution) ~ (mass distribution)

Strictly speaking, mu = g*(q/(2m))*J

For orbital angular momentum, g = 1
For an elementary fermion's spin, g = 2


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