Inclusion of Geometrical Constraints

In the last subsection, we tried to determine the common vertex of given n_track tracks without touching the track parameter vectors themselves: these helical tracks do not necessarily pass through the common vertex. Here, we require these tracks to originate from a common vertex:

$$\begin{array} {lll} \nonumber {\bf x}_v = (x_v,y_v,z_v)^T\end{array}$$

so that i-th track, for instance, can be parametrized as

$$\begin{array} {lll} \left\{ \begin{array} {lllll} x & = & x... ...tan\tilde{\lambda}^i \cdot \phi .\end{array}\right. \end{array}$$ (38)

Notice that, for a given trial vertex ${\bf x}_v$, the track parameter vector has now only three components:

$$\begin{array} {lll} \nonumber \tilde{{\bf a}}_i = (\tilde{\phi}_0^i,\tilde{\kappa}^i,\tan\tilde{\lambda}^i)^T,\end{array}$$

since the track has to pass through the pivot ${\bf x}_v$ and therefore $\tilde{d}_\rho = \tilde{d}_z = 0$.

In order to calculate the $\chi^2$, we need to transform the pivot of the i-th track from the trial vertex ${\bf x}_v$ to the pivot of the corresponding measured track. This induces the following change of the helix parameters:

$$\begin{array} {lll} \begin{array} {lll} d_\rho^i & = & \le... ...\lambda^i & = & \tan\tilde{\lambda}^i , \end{array}\end{array}$$ (39)

which gives the helix parameter vector:

$$\begin{array} {lll} \nonumber {\bf a}(\tilde{{\bf a}}^i, {\... ...\rho^i, \phi_0^i, \kappa^i, d_z^i, \tan\lambda^i )^T\end{array}$$

to be compared with the corresponding measured track parameter vector ${\bf a}_a$.Thus, we arrive at the $\chi^2$ definition:

$$\begin{array} {lll} \chi^2 = \sum_i \Delta{\bf a}_i^T \cdot E_{{\bf a}_i}^{-1} \cdot \Delta{\bf a}_i\end{array}$$ (40)

with

$$\begin{array} {lll} \nonumber {\bf a}_i = {\bf a}_i - {\bf a}(\tilde{{\bf a}}^i, {\bf x}_v) ,\end{array}$$

the minimization of which determinies the parameter vector:

$$\begin{array} {lll} \nonumber {\bf A} \equiv \pmatrix{ {\bf... ...}_i \cr \vdots \cr \tilde{{\bf a}}_{n_{track}} }\end{array}$$

containing $3 \cdot \left( n_{track} + 1 \right)$ components. Notice that the necessary calculations of the first and the second derivatives of the $\chi^2$ require only the evaluation of the transformation matrix:

$$\begin{array} {lll} \nonumber \frac{ \partial {\bf a}(\tilde{{\bf a}}^i, {\bf x}_v) }{ \partial {\bf A} }\end{array}$$

and what then follows is the same as with the last subsection.