3.2.1 The Raw Asymmetry for a Pair of Counters

Consider a pair of opposing counters--e.g. L and R in Fig. 3.1 with a magnetic field applied parallel to the ${\bf \hat{z}}$ axis. The ``raw asymmetry'' of the histograms N_L (t) and N_R (t) is defined as  

$$A_{\rm raw} (t) \equiv A_{LR}(t) = \frac{ \left[ N_{L}(t)-B_... ...circ} \right] + \left[ N_{R}(t) - B_{R}^{\circ} \right] } \, ,$$ (7)

where

 

$$N_{L}(t) = N_{L}^{\circ}e^{-t/ \tau_{\mu}} \left[ 1+A_{L}^{\... ... \, P_{L}(t) \right] + B_{L}^{\circ} \,\,\, \hbox{\rm and},\\$$ (8)

 

$$N_{R}(t) = N_{R}^{\circ}e^{-t/ \tau_{\mu}} \left[ 1+A_{R}^{\circ} \, P_{R}(t) \right] + B_{R}^{\circ} \, ,$$ (9)

so that

 

$$A_{\rm raw}(t) = \frac{ N_{L}^{\circ} \left[ 1+A_{L}^{\circ}... ...] + N_{R}^{\circ} \left[ 1+A_{R}^{\circ} P_{R}(t) \right]} \, .$$ (10)

The reason for introducing $A_{\rm raw} (t)$ is to eliminate the muon lifetime (which is a well known quantity) and the random backgrounds $B_{L}^{\circ}$and $B_{R}^{\circ}$. Ideally, the two counters in question are identical to one another, so that the histograms recorded by the two counters differ only by a phase. In this idealistic situation the difference in phase between the histograms is due solely to the geometry of the positron counters with respect to the sample. The polarization of the muon spin which is seen by each counter is

$$P_{L}(t) = \cos(\omega_{\mu} t + \psi) \, ,$$ (11)

$$P_{R}(t) = \cos(\omega_{\mu} t + \phi) \, ,$$ (12)

where $\psi$ and $\phi$ are the initial phases of the muon spin polarization vector in counters L and R, respectively. If the counters are aligned precisely opposite one another, the difference between these phases is $\mid \psi - \phi \mid = 180^{\circ}$, so that

$$P_{L}(t) = \cos[\omega_{\mu} + \, (\phi + \pi)] = -\cos(\omega_{\mu} + \phi ) = -P_{R}(t) \, .$$ (13)

The counters L and R measure the projection of the muon polarization on the ${\bf \hat{x}}$ axis so that

$$P_{L}(t) \equiv P_{x}(t) \; \; \; \; \hbox{\rm and}, \; \; \; \; P_{R}(t) = -P_{L}(t) \equiv -P_{x}(t) \, .$$ (14)

Thus Eq. (3.7) becomes  

$$A_{\rm raw}(t) = \frac{ \left( N_{L}^{\circ} - N_{R}^{\circ}... ...L}^{\circ} - N_{R}^{\circ} A_{R}^{\circ} \right) P_{x}(t)} \, .$$ (15)

If $N_{i}^{\circ} = N_{j}^{\circ}$ and $A_{i}^{\circ} = A_{j}^{\circ} \equiv A^{\circ}$ then Eq. (3.15) reduces to  

$$A_{\rm raw}(t) = A^{\circ} P_{x}(t)$$ (16)

If $B_{i}^{\circ} = B_{j}^{\circ}$, then using Eqs. (3.7), (3.8), (3.9) and (3.16) the x-component of the polarization can be written as

$$P_{x}(t) = \frac{ N_{i}(t) - N_{j}(t) }{ N_{i}^{\circ} + N_{j}^{\circ}} e^{ t / \tau_{\mu}} \, .$$ (17)