DV Analyze Recon

Calculating kinematic variables in Moller Lab Frame

Vector Magnitude

The pythagorean theorem is used to take the 3 cartesean components in the Lab frame to find the magnitude of the Moller momentum vector, [math]|\vec{p}_2\ '|[/math].

[math]|\vec{p}_2\ '|=\sqrt{p_x^2+p_y^2+p_z^2}[/math]

Finding the correct kinematic values starting from knowing the momentum of the Moller electron, [math]p^'_{2}[/math] , in the Lab frame,

xz Plane

Figure 1: Definition of Moller electron variables in the Lab Frame in the x-z plane.

Using [math]\theta '_2=\arccos \left(\frac{p^'_{2(z)}}{p^'_{2}}\right)[/math]

[math]\Longrightarrow {p^'_{2(z)}=p^'_{2}\cos(\theta '_2)}[/math]

Checking on the sign resulting from the cosine function, we are limited to:

[math]0^\circ \le \theta '_2 \le 60^\circ \equiv 0 \le \theta '_2 \le 1.046\ Radians[/math]

Since,

[math]\frac{p^'_{2(z)}}{p^'_{2}}=cos(\theta '_2)[/math]

[math]\Longrightarrow p^'_{2(z)}\ should\ always\ be\ positive[/math]

xy Plane

Figure 2: Definition of Moller electron variables in the Lab Frame in the x-y plane.

Similarly, [math]\phi '_2=\arccos \left( \frac{p^'_{2(x) Lab}}{p^'_{2(xy)}} \right)[/math]

where [math]p_{2(xy)}^'=\sqrt{(p_{2(x)}^')^2+(p^'_{2(y)})^2}[/math]

[math](p^'_{2(xy)})^2=(p^'_{2(x)})^2+(p^'_{2(y)})^2[/math]

and using [math]p^2=p_{(x)}^2+p_{(y)}^2+p_{(z)}^2[/math]

this gives [math](p^'_{2})^2=(p^'_{2(xy)})^2+(p^'_{2(z)})^2[/math]

[math]\Longrightarrow (p'_{2})^2-(p'_{2(z)})^2 = (p'_{2(xy)})^2[/math]

[math]\Longrightarrow p_{2(xy)}^'=\sqrt{(p^'_{2})^2-(p^'_{2(z)})^2}[/math]

which gives[math]\phi '_2 = \arccos \left( \frac{p_{2(x)}'}{\sqrt{p_{2}^{'\ 2}-p_{2(z)}^{'\ 2}}}\right)[/math]

[math]\Longrightarrow p_{2(x)}'=\sqrt{p_{2}^{'\ 2}-p_{2(z)}^{'\ 2}} \cos(\phi)[/math]

Similarly, using [math]p_{2}^2=p_{2(x)}^2+p_{2(y)}^2+p_{2(z)}^2[/math]

[math]\Longrightarrow p_{2}^{'\ 2}-p_{2(x)}^{'\ 2}-p_{2(z)}^{'\ 2}=p_{2(y)}^{'\ 2}[/math]

[math]p_{2(y)}'=\sqrt{p_{2}^{'\ 2}-p_{2(x)}^{'\ 2}-p_{2(z)}^{'\ 2}}[/math]

[math]p_{x}[/math] and [math]p_{y}[/math] results based on [math]\phi[/math]

Checking on the sign from the cosine results for [math]\phi '_2[/math]

We have the limiting range that [math]\phi[/math] must fall within:

[math]-\pi \le \phi '_2 \le \pi\ Radians[/math]

Examining the signs of the components which make up the angle [math]\phi[/math] in the 4 quadrants which make up the xy plane:

[math]For\ 0 \ge \phi '_2 \ge \frac{-\pi}{2}\ Radians[/math]

px=POSITIVE

py=NEGATIVE

[math]For\ 0 \le \phi '_2 \le \frac{\pi}{2}\ Radians[/math]

px=POSITIVE

py=POSITIVE

[math]For\ \frac{-\pi}{2} \ge \phi '_2 \ge -\pi\ Radians[/math]

px=NEGATIVE

py=NEGATIVE

[math]For\ \frac{\pi}{2} \le \phi '_2 \le \pi\ Radians[/math]

px=NEGATIVE

py=POSITIVE

Analysis.groovy

From a GEMC run of solenoid field strength of 0T, the eg12_rec.0.evio output file of the reconstruction is analyzed. The different kinematic variables are displayed as shown:

Using the phythagorean theorm to construct the Generated Event momentum vector length, we find:

[math]|\vec{p}_2\ '|=\sqrt{1.23162997^2+1.49707997^2+3.6786499^2}=4.15819983964[/math]

Euler Angles

We can use the Euler angles to perform the rotations.

For the rotation about the y axis.

And the rotation about the z axis.

Transformation Matrix

The Euler angles can be applied using a transformation matrix

[math]\left( \begin{array}{ccc} \cos (\theta ) & 0 & -\sin (\theta ) \\ 0 & 1 & 0 \\ \sin (\theta ) & 0 & \cos (\theta ) \\ \end{array} \right).\left( \begin{array}{c} x \\ y \\ z \\ \end{array} \right)[/math][math]=\left( \begin{array}{c} x \cos (\theta )-z \sin (\theta ) \\ y \\ z \cos (\theta )+x \sin (\theta ) \\ \end{array} \right)[/math]

For event #29, in sector 3, the location of the first interaction is given by

Converting -25 degrees to radians,

[math]\theta =-0.436332[/math]

which is the rotation the detectors are rotated from the y axis.

[math]\left( \begin{array}{ccc} \cos (\theta ) & 0 & -\sin (\theta ) \\ 0 & 1 & 0 \\ \sin (\theta ) & 0 & \cos (\theta ) \\ \end{array} \right).\left( \begin{array}{c} -15.76 \\ 0 \\ 237.43 \\ \end{array} \right)[/math][math]=\left( \begin{array}{c} 86.0588 \\ 0. \\ 221.845 \\ \end{array} \right)[/math]

Finding [math]\phi =\frac{120\ 2 \pi }{360};[/math] since "sector -1" =3-1=2*60=120 degrees

[math]\left( \begin{array}{ccc} \cos (\phi ) & -\sin (\phi ) & 0 \\ \sin (\phi ) & \cos (\phi ) & 0 \\ 0 & 0 & 1 \\ \end{array} \right).\left( \begin{array}{c} 86.0588 \\ 0. \\ 221.845 \\ \end{array} \right)[/math][math]=\left( \begin{array}{c} -43.0294 \\ 74.5291 \\ 221.845 \\ \end{array} \right)[/math]

This shows how the coordinates are transformed and explains the validity of using the TBTracking information to obtain a phi angle in the lab frame.