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MagneticPowders

< Magnetic single crystal | Index | Knight shift >

If, with the same local field of the previous page, we have a polycrystal, since the spectrum of precession frequencies is unaffected by the orientation of the individual crystallite, we must simply take the podwer average of Eq. 2 and 3,

{$ \overline P_{z,y}(t) = \frac 1 {4\pi} \int_0^\pi\sin\theta d\theta\int_0^{2\pi} d\phi P_{z,y}(t;\Theta,\theta,\phi) $}

All terms linear in {$\sin\phi$} and {$\cos\phi$} average to zero and it is straightforward to check that

{$ (1) \qquad\qquad \begin{eqnarray} \overline P_{z}(t) &=& - \frac 1 3 \cos\Theta \left[1+ 2 \cos\gamma_\mu B_\mu \right]\\ \overline P_{y}(t) &=& - \frac 1 3 \sin\Theta \left[ 1 + 2 \cos\gamma_\mu B_\mu \right]\end{eqnarray} $}

Intuitively, the 1 to 2 proportion between the longitudinal and the transverse terms is due to the fact that precessions take place around field components orthogonal to the initial muon spin direction, with two such Cartesian components, while the constant term arises from field components parallel to the initial muon spin direction, and there is only one such component. It is simpler to recognize the result of Eq. 1 without spin rotation ({$\Theta=0$}). In this case the U-D detectors measure zero polarization and the F-B detectors see {$\frac 1 3 $} of polarization longitudinal and {$\frac 2 3$} transverse. as shown in the figure (with total asymmetry {$A=0.27$}). The example in the figure also shows that, quite typically, both the transverse precessing component and the longitudinal one relax in time, and that, furthermore, the two relaxation rates are not the same.

Longitudinal relaxation in this context can only be due to time dependent interactions and it is a {$T_1$} process, whereas transverse relaxation will be probably dominated by inhomogeneity of the local field (a so-called {$T_2^*$} process). For this distinction we refer to an NMR primer?.

< Magnetic single crystal | Index | Knight shift >