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Correcting for pion/electron veto inefficiency

At this point we turn again to the figures (Fig. 3.4 and Fig. 3.5) that illustrate the difference in the approaches to kaon identification in different spectrometer settings. As was already discussed in Section  3.4, a Cherenkov veto ($\bar{C}$) was used to obtain kaon-enhanced data samples. The price of the high cleanliness of a PID with $\bar{C}$ is its high inefficiency: any event resulting in a firing of the vetoing Chereknov's counter (a passage of any charged particle of sufficient velocity through the counter's aperture) may result in a loss of a kaon. In particular, such can be the outcome of a coincidence of the kaon with a pion or an electron in the counter's acceptance. To evaluate this inefficiency, we take the following steps:

1. On-line: obtain a sample without the $\bar{C}$ veto and a sufficient statistics of kaons.
2. Off-line: in the sample without veto, find a way of counting the kaons with and (non-trivial !) without the veto.

The counting of the rejected kaons can be possible in presense of a device sensitive to the difference between kaons and lighter particles. Presense of a device capable of identifying kaons independently of the Cherenkovs is ideal. Such an ideal case is realized in the weak magnetic field setting, (see Fig. 3.4) where the $m^2$ measurement (due to the good time-of-flight resolution) allows one to tell kaons from lighter particles even in the absence of the veto. The inefficiency correction factor is therefore easily calculable:

$$1 + \frac{N(K)_{lost}}{N(K)_{detected}}$$ (32)

The case of a strong field setting (see Fig. 3.5) is complicated by the fact that a particle lost due to the veto is not necessarily a kaon, but a kaon or a lighter particle. The lighter particle is most likely a pion and will be called so in the following discussion.

Figure: Correcting for the Cherenkov veto inefficiency in the strong field case, 4% most central events. The number of rejected kaons is evaluated by subtracting the clean pion $m^2$ line shape scaled by a proper multiplier $\Upsilon$. + = all vetoed tracks $\,dN_{lost}/\,dm^2$; $\Diamond$ = ratio of the pure pion line $\,dN_{clean, light}/\,dm^2$ to the ``all vetoed tracks'' distribution, $\circ$ (also in the insert) = $\,dN(K+p)_{lost}/\,dm^2$ obtained as ``all vetoed tracks'' minus $\Upsilon$ - scaled pion line (see Eq.  4.31). The shaded histogram shows the $m^2$ distribution of $K/p$ tracks which were not vetoed.

It is possible to count the lost kaons without identifying them, taking advantage of the fact that the kaon and pion peaks in the $m^2$ spectrum of vetoed tracks (see Fig. 4.10) are separated, even though they overlap. In order to subtract the pions from all vetoed tracks, we select a clean sample of undoubtedly light tracks ($\pi$+$e$) from a sample of good events with a single hit in hodoscopes $H2$ and $H3$ and only one reconstructed track, by requiring a large (characteristic of a pion) signal in $C1$. Since this requirement is highly selective, we use the shape of the distribution $\,dN_{clean, light}/\,dm^2$ so obtained, but not the magnitude of the probability it represents. Therefore, we seek a multiplier $\Upsilon$ that allows us to subtract the light particle distribution without remainder to obtain the number of the vetoed $K$ and $p$ only. We do it by dividing the ``$+$'' histogram on Fig. 4.10 by the clean light track distribution (not shown). Flatness of the ratio so obtained (shown as $\Diamond$) is to be expected in the range of $m^2$ where the light particle line shape selected is representative of the sample we want to subtract. We fit the ratio in the flat region to obtain the required multiplier $\Upsilon$, and subtract the light particle line shape:

$$\frac{\,dN(K+p)_{lost}}{\,dm^2} = \frac{\,dN_{lost}}{\,dm^2} - \Upsilon \frac{\,dN_{clean, light}}{\,dm^2}$$ (33)

The histogram plotted on Fig. 4.10 by $\circ$ represents the $m^2$ distribution of vetoed $K$ and $p$ only, $\,dN(K+p)_{lost}/\,dm^2$. To get $N(K)_{lost}$, we pick an $m^2$ window (vertical lines on Fig. 4.10). By solving the standard problem of interpreting a sum of two Gaussian peaks as ``signal'' + ``background'', $K$ being the signal and $p$ the background, one finds that the ``lost'' kaons in the window are 98-99% clean of protons.