Impact Parameter Resolution

Impact parameter resolution $\sigma_\delta$ is calculated as

$$\begin{array} {lll} \sigma_\delta^2 = \left( E'^{r\phi} \right)_{11},\end{array}$$ (105)

where $E'^{r\phi}$ is the full error matrix obtained from $E^{r\phi}$ given by Eq.3.2.48 by moving the pivot to the interaction point. In the pivot moving we can include the effect of thin-layer multiple scattering, for instance, at the beam pipe wall, according to the method explained in Section 2.4. If the material thickness is non-negligible, we need to use the method in Subsection 3.2.b instead. The effect of external coordinate information can also be implemented as explained there.

Let us consider again a tracking device having (n+1) equally sapced sampling points with an idential spatial resolution $\sigma_x$.For simplicity, we further assume that multiple scattering in the tracking volume is negligible. Then the error matrix is given by Eq.B.7, where the pivot is located at y₀ = 0 which is the -th hit position. When the -th hit is at r = R_in, we need to transform the error matrix by $\Delta y = - R_{in}$, using Eq.3.2.19. Ignoring the material between the -th hit and the interaction point (by setting E_S = 0), we obtain Fig.3.2 which plots $\sigma_\delta/\sigma_x$ as a function of the ratio of the extrapolation length and the lever arm length: R_in/(R_out-R_in).

 

Figure 3.2:   The impact parameter resolution normalized by the coordinate error plotted as a function of the ratio of the distance from the innermost hit to the interaction point to the lever arm length for $N_{sample} = 10, 20, \cdots, 100$. No multiple scattering and no external curvature information are assumed.

Notice that the impact parameter resolution here is much worse in general than that expected from a straight-line approximation. This is because of the curvature error. If we have an external tracking device which provides additional curvature information, we can thus improve the impact parameter resolution significantly. We can do this by modifiying $E^{r\phi}$, according to

$$\begin{array} {lll} \left( E^{r\phi} \right)^{-1} \righta... ...cr 0 & 0 & 1/\left(\sigma_\kappa^{ext}\right)^2 },\end{array}$$ (106)

and by following the same steps as above.