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The muon is produced with suitable particle accelerators, to yield a large spin polarization, nearly 100%. It is then implanted as a probe in solids, liquids and gasses. It generally ends up inside the sample in few equivalent, most energetically-convenient sites, retaining its initial spin polarization. Parity violation allows the detection of the muon spin coherent behaviour from the direction of its decay-positron (electron), averaged over an ensemble of muons which are implanted one by one, or in bunches, at subsequent times.

The muon is a lepton with the following properties

Charge	{$\pm$}
Mass	{$m_\mu\,=\,0.1126\,m_p\,=\,206.7684(6)\,m_e \,= \,105.65849(4)\, \mbox{Mev/c}^2$}
Spin	{$I_\mu\,=\,\frac 1 2$}
Magnetogyric ratio	{$\frac {\gamma_\mu} {2\pi}\,=\,135.53875(6)\,\mbox{MHz/T}$}
Mean lifetime	{$\tau_{\mu^+}\,=\,2.19703(4)\,\mu\mbox{s}$}

The most convenient way to produce intense muon beams suitable for implantation is to exploit the decay of charged pions. We now specialize to positive muons. In the rest frame of the pion that decays:

{$ (1) \qquad\qquad \pi^+ \rightarrow \mu^+ + \nu $}

the outcoming muon and neutrino have opposite linear momentum, equal to 29.79 MeV/c, as can be obtained by imposing also the conservation of the relativistic energy (see here for a summary of relativistic kinematics):

{$ (2) \qquad\qquad \sqrt{m_\pi^2c^4+p_\pi^2c^2}= \sqrt{m_\mu^2c^4+p_\mu^2c^2}+\sqrt{m_\nu^2c^4+p_\nu^2c^2} $}

with {$p_\pi=0,\quad p_\nu=-p_\mu, \quad m_\nu=0 $} (effectively) and {$m_\pi=139.57\, \mbox{MeV/c}^2$}.

The muon kinetic energy is then

{$ (3) \qquad\qquad E_k = \sqrt{m_\mu^2c^4+p_\mu^2c^2} - m_\mu c^2 = 4.119\,\mbox{Mev/c^2} $}

Comparing the terms in Eq. (2) is is easy to see that, whereas the neutrino is entirely relativistic (ignoring its mass, it would travel at the speed of light), the muon is not in the fully relativistic limit, having {$\beta=0.272$}, hence a very small {$\gamma=1.039$}. Equation (3) sets the energy that a muon originating from a pion at rest has to spend to get through thin vacuum windows along the beamline and to implant inside tha sample.

Note:

{$ m_e=9.1093826(16),\ 10^{-31}\,\mbox{kg}=0.51099907(15)\,\mbox{MeV} $}

{$ c=299792458\,\mbox{m/s},\qquad\qquad e=1.60217653(14)\,10^{-19}\,\mbox{C}$}

{$ h=6.626 0693(11)\,10^{-34}\,\mbox{Js}$}

{$ \gamma_\mu=\frac e {m_\mu}(1+\frac{g_\mu-2} 2) $}, where the measured quantity is {$1+\frac {g_\mu-2} 2 = 1.001165923(8)$}.

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