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# MulabTwo

Better strategy for global fit

Free notes

GPS Spin rotation on for simultaneous fit of ZF on 2-1, 3-4 groups

Assume one has a good calibration set: high T with full asymmetry and no internal field, measured in low Transverse Field (e.g. 50G).

This data set can be fitted as decay four independent histograms:

$N_k(t)=N_{0k} e^{-t/\tau_\mu} [1+A_k\cos(\omega_\mu t + \phi_k)] + B_k\quad\quad k=1,\cdots,4$

yielding four replica of the two relevant histogram-dependent parameters, $N_k, A_k$

With these eight parameters one can do the following on any new data set of a subsequent ZF temperature scan:

1) determine the two following quantities for each group:

$\alpha_{21}=\frac {N_{02}} {N_{01}} \qquad \qquad \beta_{21}=\frac {A_2}{A_1}$

$\alpha_{34}=\frac {N_{03}} {N_{04}} \qquad \qquad \beta_{34}=\frac {A_3}{A_4}$

2) Fit the following data compositions:

$N_2(t)+ \alpha_{21}\beta_{21}N_1(t) \qquad \mbox{to} \qquad N_2(1+\beta_{21})$

But this amounts to simply checking the individual earlier fits.

Then, assuming $\alpha_{21}$ and $\beta_{21}$ constant with temperature one could use in the subsequent temperature scan

$\frac{N_2(t)- \alpha_{21}N_1(t)}{2N_2 e^{-t/\tau_\mu}}=\frac{A_2+A_1} 2 G(t)$

which is an average asymmetry. The important parameter however is $\alpha_{21}$, to eliminate constant terms from the nominator of the fraction in the above expression. The other parameter, $\beta_{21}$, simply produces a different average asymmetry. It is important to use the Fit values in the denominator, and not the values determined from Data.

Now in the double fit of Mulab-2.0 each pair has its own average asymmetry and the UD/FB ratio in the fit parameters takes that into account.

Background subtraction

For a new muasymmetry.m

1. determine $B_F^0,B_B^0$ from the pre-prompt counts as in older muasymmetry.m
2. determine total $N_0$ rate by fitting the sum $N_F(t)-B_F^0 + \alpha(N_B(t)-B_B^0)$ to $N_0 e^{-t/\tau_\mu}$; the experimental quantity contains also a little residual contamination from the polarization that does not cancel exactly if the F and B counters have different muon asymmetries.
3. determine $A^0(t)=\frac{N_F(t)-B_F^0-\alpha(N_B(t)-B_B^0)}{N_0 e^{-t/\tau_\mu}}$; this is an approximation to the true asymmetry $A(t)$; moreover the background correction may not be exact.
4. In a next iteration optimize the parameters $N_{F0},\beta_F$ and $B_F$ to minimize the square deviations from zero of the following combination of experimental data: $N_F(t)-N_{F0}(1+\beta_F A^0(t))e^{-t/\tau_\mu}-B_F$
5. redo the same for the backward counter(s) $N_{B0},\beta_B$ and $B_B$
6. the refined values are $\alpha=\frac {N_{F0}} {N_{B0}}$ and $B_F, B_B$ with which one can compute a corrected asymmetry with the new parameters

$A(t)=\frac{N_F(t)-B_F-\frac {N_{F0}} {N_{B0}}(N_B(t)-B_B)}{2 N_{FO}e^{-t/\tau_\mu} }$

Unfortunately does not seem to converge...