Difference between revisions of "Quality Checks"

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<math>\phi_{\gamma} = tan^{-1}(\frac{p_{e x}^{\prime}}{p_{e z}^{\prime}})</math><br>
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<math>\phi_{\gamma} = tan^{-1}(\frac{p_{e y}^{\prime}}{p_{e x}^{\prime}})</math><br>
  
  

Revision as of 23:12, 17 May 2008

Quality Checks

Run Summary Table

The table below uses a characteristic DST file to try and estimate the sample size for a semi-inclusive analysis of pion electroproduction. The column marked "cuts" below indicates the number of events kept when the standard EC based electron identification cuts, described above, are used: [math]EC_{tot}\gt 0.2*p [/math] and
[math] EC_{inner}\gt 0.08*p[/math]. The next step will be to compare unpolarized pion production rates in order to evaluate the CLAS detectors efficiencies for measuring charged pions with different torus polarities. The question is whether you get the same rates for negatively charged pions in one torus polarity to positively charged pions using the opposite torus polarity.

Beam Energy Torus Current Target Begin Run End Run file used # trig([math]10^6[/math]) events remaining after [math]e^-[/math] cuts(%) expected # evts([math]10^6[/math]) events remaining after [math]e^-[/math] and [math]\pi^+[/math] cuts(%) expected # evts([math]10^6[/math]) events remaining after [math]e^-[/math] and [math]\pi^-[/math] cuts(%) expected # evts([math]10^6[/math])
4239 2250 NH3 28205 28277 /cache/mss/home/nguler/dst/dst28205_05.B00 1108.72 60.8 674.1 8.3 92.02 3.24 35.92
ND3 28074 28190 /cache/mss/home/nguler/dst/dst28187_05.B00 1117.87 59.6 666.25 7.99 89.32 3.3 36.9
-2250 NH3 28407 28479 /cache/mss/home/nguler/dst/dst28409_05.B00 1013.57 24.2 245.28 0.12 1.22 0 0
ND3 28278 28403 /cache/mss/home/nguler/dst/dst28400_05.B00 1556.04 23.9 371.89 0.02 0.31 0.05 0.51
5735 2250 NH3 27074 27195 /cache/mss/home/nguler/dst/dst27095_05.B00 1442.25 57.7 832.18 9.3 134.13 3.8 59.13
ND3 27116 27170 /cache/mss/home/nguler/dst/dst27141_05.B00 624.55 59.1 369.10 9.53 59.52 3.9 24.36
-2250 NH3 26911 27015 /cache/mss/home/nguler/dst/dst26988_05.B00 900.93 80.7 727.05 7.14 64.33 9.9 89.19
ND3 27022 27068 /cache/mss/home/nguler/dst/dst27055_05.B00 711.53 80 569.22 6.97 49.59 10.1 71.86

Rates

Unpolarized Pion electroproduction

Rates from other experiments in our Kinematic range

Center of Mass Frame Transformation

We have proton and electron. In the Lab frame electron is moving along the x-axis with momentum ;[math]\vec{p_e}[/math] and proton is at rest. The 4-vectors are:

Lab Frame
[math]P_e=[/math]([math]E_e[/math],[math]p_e[/math],0,0) and for proton :[math]P_p=[/math]([math]m_p[/math],0,0,0)
CM Frame
:[math]{P_e}^{\prime}=[/math]([math]{E_e}^{\prime}[/math],[math]{p_e}^{\prime}[/math],[math]0[/math],[math]0[/math]) and for proton :[math]{P_p}^{\prime}=[/math]([math]{E_p}^{\prime}[/math],[math]{p_p}^{\prime}[/math],[math]0[/math],[math]0[/math])
Find [math] \beta_{CM} [/math] such that [math]P_{tot}^{CM}=0 =p_e^{\prime} + {p_p}^{\prime}[/math]
[math]\left ( \begin{matrix} {E_e}^{\prime} \\ p_e^{\prime} \\ 0 \\ 0 \end{matrix} \right )= \left [ \begin{matrix} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 & 1 \end{matrix} \right ] \left ( \begin{matrix} E_e \\ p_e \\ 0 \\ 0 \end{matrix} \right )[/math]


[math]\left ( \begin{matrix} {E_p}^{\prime} \\ p_p^{\prime} \\ 0 \\ 0 \end{matrix} \right )= \left [ \begin{matrix} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &1\end{matrix} \right ] \left ( \begin{matrix} m_p \\ 0 \\ 0 \\ 0 \end{matrix} \right )[/math]

Using the last two equations we will get the following for x component:

[math]{p_e}^{\prime}=-\gamma_{cm}(\beta_{cm} E_e-p_e)[/math]
[math]p_p^{\prime} = - \gamma_{cm} \beta_{cm} m_p[/math]
[math] \gamma_{cm}(p_e - \beta_{cm} E_e)= \gamma_{cm} \beta_{cm} m_p [/math]
[math]\beta_{cm} = \frac {p_e}{m_p + E_e}[/math]
Example of the Missing Mass Calculation for the following reaction [math]e^- p^+ \rightarrow (e^-)^{\prime} \pi^{-} X [/math]

Event number 3143292 1.jpg Event number 3143292 2.jpg Event number 3143292 3.jpg

[math]p_e = 5.736 Gev \sim E_e[/math] : electron mass is neglibible
[math]m_p = 0.938 GeV[/math] : Mass of a proton
[math]\beta_{cm} = \frac{5.736}{6.674} = 0.859 \lt 1[/math]
[math]\gamma_{cm} = \frac{1}{\sqrt{1 - \beta_{cm}^2}} = \frac{1}{\sqrt{1 - 0.859^2}} = 1.9532[/math]
[math]\left ( \begin{matrix} {E_e}^{\prime} \\ p_{ex}^{\prime} \\ p_{ey}^{\prime} \\ p_{ez}^{\prime} \end{matrix} \right )= \left [ \begin{matrix} \gamma_{cm} & 0 & 0 & -\gamma_{cm} \beta_{cm} \\ 0 & 1 &0 &0 \\ 0 &0 &1 &0 \\ -\gamma_{cm} \beta_{cm} &0 &0 &\gamma_{cm}\end{matrix} \right ] \left ( \begin{matrix} E_e=4.4915 \\ -0.549 \\ 0.974 \\ 4.3501 \end{matrix} \right )[/math]
[math]\left ( \begin{matrix} {E_p}^{\prime} \\ p_{px}^{\prime} \\ p_{py}^{\prime} \\ p_{pz}^{\prime} \end{matrix} \right )= \left [ \begin{matrix} \gamma_{cm} & 0 & 0 & -\gamma_{cm} \beta_{cm} \\ 0 & 1 &0 &0 \\ 0 &0 &1 &0 \\ -\gamma_{cm} \beta_{cm} &0 &0 & \gamma_{cm} \end{matrix} \right ] \left ( \begin{matrix} E_p = 1.4563 \\ 0.3697 \\ -0.3447 \\ 0.9924 \end{matrix} \right )[/math]
[math]\left ( \begin{matrix} {E_{\pi^-}}^{\prime} \\ p_{\pi^- x}^{\prime} \\ p_{{\pi^-} y}^{\prime} \\ p_{{\pi^-} z}^{\prime} \end{matrix} \right )= \left [ \begin{matrix} \gamma_{cm} & 0 & 0 & -\gamma_{cm} \beta_{cm} \\ 0 & 1 &0 &0 \\ 0 &0 &1 &0 \\ -\gamma_{cm} \beta_{cm} &0 &0 & \gamma_{cm} \end{matrix} \right ] \left ( \begin{matrix} E_{\pi^-} = 0.5757 \\ 0.1052 \\ -0.4394 \\ 0.3282 \end{matrix} \right )[/math]
Electron
[math]\left ( \begin{matrix} {E_e}^{\prime} \\ p_{ex}^{\prime} \\ p_{ey}^{\prime} \\ p_{ez}^{\prime} \end{matrix} \right )= \left ( \begin{matrix} 4.4915 \gamma_{cm} - 4.3501 \gamma_{cm} \beta_{cm} \\ -0.549 \\ 0.974 \\ -4.4915 \gamma_{cm} \beta_{cm} + 4.3501 \gamma_{cm} \end{matrix} \right ) = \left ( \begin{matrix} 1.4742 \\ -0.549 \\ 0.974 \\ 0.96078 \end{matrix} \right )[/math]
Proton
[math]\left ( \begin{matrix} {E_p}^{\prime} \\ p_{px}^{\prime} \\ p_{py}^{\prime} \\ p_{pz}^{\prime} \end{matrix} \right )= \left ( \begin{matrix} 1.4563\gamma_{cm} - 0.9924 \gamma_{cm} \beta_{cm} \\ 0.3697 \\ -0.3447 \\-1.4563 \gamma_{cm} \beta_{cm} + 0.9924\gamma_{cm} \end{matrix} \right ) = \left ( \begin{matrix} 1.17939 \\ 0.3697 \\ -0.3447 \\ -0.505023 \end{matrix} \right )[/math]
[math]\pi^-[/math]
[math]\left ( \begin{matrix} {E_{\pi^-}}^{\prime} \\ p_{\pi^- x}^{\prime} \\ p_{{\pi^-} y}^{\prime} \\ p_{{\pi^-} z}^{\prime} \end{matrix} \right )= \left ( \begin{matrix} 0.5757\gamma_{cm} - 0.3282 \gamma_{cm} \beta_{cm} \\ 0.1052 \\ -0.4394 \\ -0.5757 \gamma_{cm} \beta_{cm} + 0.3282\gamma_{cm} \end{matrix} \right ) = \left ( \begin{matrix} 0.5738 \\ 0.1052 \\ -0.4394 \\ -0.324868 \end{matrix} \right )[/math]

[math]\vec {P_{tot}}^{\prime} = (p_{ex}^{\prime} + p_{px}^{\prime} + p_{\pi^-x}^{\prime})\hat{x} + (p_{ey}^{\prime} + p_{py}^{\prime} + p_{\pi^-y}^{\prime})\hat{y} + (p_{ez}^{\prime} + p_{pz}^{\prime} + p_{\pi^-z}^{\prime})\hat{z} = - 0.0741 \hat{x} + 0.1899 \hat{y} + 0.13 \hat{z} [/math]

Missing Mass
Conservation of the 4-momentum gives us following

[math](P_e)^\mu + (P_p)^\mu = ({P_e}^{\prime})^\mu + ({P_X}^{\prime})^\mu + ({P_{\pi^-}}^{\prime})^\mu[/math]
[math](P_e)_\mu + (P_p)_\mu = ({P_e}^{\prime})_\mu + ({P_X}^{\prime})_\mu + ({P_{\pi^-}}^{\prime})_\mu[/math]

Solving it for the final proton state

[math]{M_x}^2 = ({P_X}^{\prime})_\mu({P_X}^{\prime})^\mu = [(P_e)_\mu + (P_p)_\mu - ({P_e}^{\prime})_\mu - ({P_{\pi^-}}^{\prime})_\mu][(P_e)^\mu + (P_p)^\mu - ({P_e}^{\prime})^\mu - ({P_{\pi^-}}^{\prime})^\mu][/math]

In our case 4-vectors for particles are

[math](P_e)_\mu = ( 5.736, 0, 0, 5.736 GeV)[/math]
[math](P_p)_\mu = (m_p, 0, 0, 0)[/math]
[math]({P_e}^{\prime})_\mu = (4.4914861, -0.549, 0.974, 4.3501 )[/math]
[math]({P_{\pi^-}}^{\prime})_\mu = (0.575721, 0.1052, -0.4394, 0.3282)[/math]

Plug and chug

[math]{M_x}^2 = [( 5.736, 0, 0, 5.736 GeV) + ( m_p, 0, 0, 0 ) - (4.4914861, -0.549, 0.974, 4.3501 ) - (0.575721, 0.1052, -0.4394, 0.3282)[/math]] [[math]\left (\begin{matrix} 5.736 \\ 0 \\ 0 \\ 5.736 \end{matrix} \right )[/math] + [math]\left (\begin{matrix} m_p \\ 0 \\ 0 \\ 0 \end{matrix} \right) [/math] - [math]\left (\begin{matrix} 4.4914861 \\ -0.549 \\ 0.974 \\ 4.3501 \end{matrix} \right) [/math] - [math]\left ( \begin{matrix}0.575721 \\ 0.1052 \\ -0.4394 \\ 0.3282 \end{matrix} \right) ] = 0.981198614 GeV^2[/math]

[math]M_x = 0.9905547 GeV[/math]

Example of the Missing Mass Calculation for the following reaction [math]\vec{e}p \rightarrow (e^-)^{\prime} n \pi^+[/math]

Used file dst27095_05.B00. Target [math]NH_3[/math], Beam Energy 5.735 GeV and Torus Current +2250. Event_number=3106861

Event number 3106861 sector 1 4.jpg Event number 3106861 sector 3 6.jpg Event number 3106861 calorimeter.jpg Event number 3106861 particle momentums sectors phi angle.jpg

[math]p_e = 5.736 Gev \sim E_e[/math] : electron mass is negligible
[math]m_p = 0.938 GeV[/math] : Mass of a proton
[math]m_n = 0.939566 GeV[/math] : Mass of a neutron
[math]\beta_{cm} = \frac{5.736}{6.674} = 0.859 \lt 1[/math]
[math]\gamma_{cm} = \frac{1}{\sqrt{1 - \beta_{cm}^2}} = \frac{1}{\sqrt{1 - 0.859^2}} = 1.9559[/math]
[math]\left ( \begin{matrix} {E_e}^{\prime} \\ p_{ex}^{\prime} \\ p_{ey}^{\prime} \\ p_{ez}^{\prime} \end{matrix} \right )= \left [ \begin{matrix} \gamma_{cm} & 0 & 0 & -\gamma_{cm} \beta_{cm} \\ 0 & 1 &0 &0 \\ 0 &0 &1 &0 \\ -\gamma_{cm} \beta_{cm} &0 &0 &\gamma_{cm}\end{matrix} \right ] \left ( \begin{matrix} E_e=2.7165 \\ 0.8769 \\ -0.117 \\ 2.5684 \end{matrix} \right )[/math]
[math]\left ( \begin{matrix} {E_n}^{\prime} \\ p_{nx}^{\prime} \\ p_{ny}^{\prime} \\ p_{nz}^{\prime} \end{matrix} \right )= \left [ \begin{matrix} \gamma_{cm} & 0 & 0 & -\gamma_{cm} \beta_{cm} \\ 0 & 1 &0 &0 \\ 0 &0 &1 &0 \\ -\gamma_{cm} \beta_{cm} &0 &0 & \gamma_{cm} \end{matrix} \right ] \left ( \begin{matrix} E_n =2.0218 \\ -0.4811 \\ -0.9008 \\ 1.4704 \end{matrix} \right )[/math]
[math]\left ( \begin{matrix} {E_{\pi^+}}^{\prime} \\ p_{\pi^+ x}^{\prime} \\ p_{{\pi^+} y}^{\prime} \\ p_{{\pi^+} z}^{\prime} \end{matrix} \right )= \left [ \begin{matrix} \gamma_{cm} & 0 & 0 & -\gamma_{cm} \beta_{cm} \\ 0 & 1 &0 &0 \\ 0 &0 &1 &0 \\ -\gamma_{cm} \beta_{cm} &0 &0 & \gamma_{cm} \end{matrix} \right ] \left ( \begin{matrix} E_{\pi^+} =2.3431 \\ -0.6918 \\ 0.8242 \\ 2.0764 \end{matrix} \right )[/math]


Electron
[math]\left ( \begin{matrix} {E_e}^{\prime} \\ p_{ex}^{\prime} \\ p_{ey}^{\prime} \\ p_{ez}^{\prime} \end{matrix} \right )= \left ( \begin{matrix} 2.7165 \gamma_{cm} - 2.5684 \gamma_{cm} \beta_{cm} \\ 0.8769 \\ -0.117 \\ -2.7165\gamma_{cm} \beta_{cm} + 2.5684 \gamma_{cm} \end{matrix} \right ) = \left ( \begin{matrix} 0.99569 \\ 0.8769 \\ -0.117 \\ 0.45728 \end{matrix} \right )[/math]
Neutron
[math]\left ( \begin{matrix} {E_n}^{\prime} \\ p_{nx}^{\prime} \\ p_{ny}^{\prime} \\ p_{nz}^{\prime} \end{matrix} \right )= \left ( \begin{matrix} 2.0218\gamma_{cm} - 1.4704 \gamma_{cm} \beta_{cm} \\ -0.4811 \\ -0.9008 \\-2.0218 \gamma_{cm} \beta_{cm} + 1.4704\gamma_{cm} \end{matrix} \right ) = \left ( \begin{matrix} 1.4828 \\ -0.4811 \\ -0.9008 \\ -0.52256 \end{matrix} \right )[/math]
[math]\pi^+[/math]
[math]\left ( \begin{matrix} {E_{\pi^+}}^{\prime} \\ p_{\pi^+ x}^{\prime} \\ p_{{\pi^+} y}^{\prime} \\ p_{{\pi^+} z}^{\prime} \end{matrix} \right )= \left ( \begin{matrix} 2.3431\gamma_{cm} - 2.0764 \gamma_{cm} \beta_{cm} \\ -0.6918 \\ 0.8242 \\ -2.3431 \gamma_{cm} \beta_{cm} + 2.0764\gamma_{cm} \end{matrix} \right ) = \left ( \begin{matrix} 1.09257 \\ -0.6918 \\ 0.8242 \\ 0.1226 \end{matrix} \right )[/math]

[math]\vec {P_{tot}}^{\prime} = (p_{ex}^{\prime} + p_{nx}^{\prime} + p_{\pi^+x}^{\prime})\hat{x} + (p_{ey}^{\prime} + p_{ny}^{\prime} + p_{\pi^+y}^{\prime})\hat{y} + (p_{ez}^{\prime} + p_{nz}^{\prime} + p_{\pi^+z}^{\prime})\hat{z} = -0.296 \hat{x} + -0.1936 \hat{y} + 0.0573 \hat{z} [/math]

Missing Mass Calculation
Below is the conservation of the 4-momentum

[math](P_e)^\mu + (P_p)^\mu = ({P_e}^{\prime})^\mu + ({P_X}^{\prime})^\mu + ({P_{\pi^+}}^{\prime})^\mu[/math]
[math](P_e)_\mu + (P_p)_\mu = ({P_e}^{\prime})_\mu + ({P_X}^{\prime})_\mu + ({P_{\pi^+}}^{\prime})_\mu[/math]

Solving it for the final neutron state

[math]{M_x}^2 = ({P_X}^{\prime})_\mu({P_X}^{\prime})^\mu = [(P_e)_\mu + (P_p)_\mu - ({P_e}^{\prime})_\mu - ({P_{\pi^+}}^{\prime})_\mu][(P_e)^\mu + (P_p)^\mu - ({P_e}^{\prime})^\mu - ({P_{\pi^+}}^{\prime})^\mu][/math]

The 4-vectors for the particles in this event

[math](P_e)_\mu = ( 5.736, 0, 0, 5.736 GeV)[/math]
[math](P_p)_\mu = (m_p, 0, 0, 0)[/math]
[math]({P_e}^{\prime})_\mu = (2.7164, 0.8769, -0.117, 2.5684 )[/math]
[math]({P_{\pi^+}}^{\prime})_\mu = (2.3431, -0.6918, 0.8242, 2.0764)[/math]

[math]{M_x}^2 = [( 5.736, 0, 0, 5.736 GeV) + ( m_p, 0, 0, 0 ) - ( 2.7164, 0.8769, -0.117, 2.5684 ) - ( 2.3431, -0.6918, 0.8242, 2.0764 )[/math]] [[math]\left (\begin{matrix} 5.736 \\ 0 \\ 0 \\ 5.736 \end{matrix} \right )[/math] + [math]\left (\begin{matrix} m_p \\ 0 \\ 0 \\ 0 \end{matrix} \right) [/math] - [math]\left (\begin{matrix} 2.7164 \\ 0.8769 \\ -0.117 \\ 2.5684 \end{matrix} \right) [/math] - [math]\left ( \begin{matrix} 2.3431 \\ -0.6918 \\ 0.8242 \\ 2.0764\end{matrix} \right) ] = 0.882724889 GeV^2[/math]

[math]M_x = 0.9395344 GeV[/math]

[math]\phi_{diff}^{LAB} = \phi_e^{LAB} - \phi_{\pi^+}^{LAB} = (120 + 35.3) -26.4 = 128.9[/math]

[math]{\phi_{\pi}}^{CM}[/math] transformation from LAB frame to CM frame

Used file dst27095_05, event_number=3106861

File dst27095 event number 3106861 phi gamma theta x and phi pi angles in cm frame.jpg

[math]\beta_{cm} = \frac{5.736}{6.674} = 0.859 \lt 1[/math]
[math]\gamma_{cm} = \frac{1}{\sqrt{1 - \beta_{cm}^2}} = \frac{1}{\sqrt{1 - 0.859^2}} = 1.9559[/math]
[math]E_e = \sqrt{{p_e}^2 + {m_e}^2} = \sqrt{0.8769^2 + 0.117^2 + 2.56842^2} = 2.7165[/math]
Calculation of [math]{{\phi}_{\gamma}}^{CM}[/math]
[math]{\phi_{\gamma}}^{CM} = tan^{-1}(\frac{{p_{e z}}^{CM}}{{p_{e z}}^{ CM \prime}})[/math]
[math]{p_{e z}}^{CM} = -\gamma_{CM} \beta_{CM}{p_{e z}}^{LAB} + \gamma_{CM} {p_{e z}}^{LAB} = [/math]
[math] = -1.9559 \times 0.8594 \times 5.736 + 1.9559 \times 5.736 = 1.5773 [/math]
[math]{p_{e z}}^{CM \prime} = -\gamma_{CM} \beta_{CM} E_e + \gamma_{CM} {p_{e z}}^{LAB \prime} = [/math]
[math]= -1.9559 \times 0.8594 \times 2.7165 + 1.9559 \times 2.5684 = 0.4574 [/math]
[math]\phi_{\gamma} = tan^{-1}(\frac{1.5773}{0.4574}) = 73.83 [/math]
Calculation of [math]{\theta_x}^{CM}[/math]


[math]{\theta_x}^{CM} = cos^{-1}(\frac{p_{e z}^{CM \prime} - {p_{e z}}^{CM}}{\sqrt{{p_{e x}^{CM \prime}}^2 + {p_{e y}^{CM \prime}}^2 + ({p_{e z}^{CM \prime} - {p_{e z}}^{CM}})^2}}) = [/math]
[math] = cos^{-1}(\frac{0.4574 - 1.5773}{\sqrt{0.8769^2 + (-0.117)^2 + (0.4574 - 1.5773)^2}}) = 141.69 [/math]


Calculation of [math]{\phi_{\pi}}^{CM}[/math]
[math]E_{\pi} = \sqrt{{p_{\pi}}^2 + {m_{\pi}}^2} = [/math]
[math] = \sqrt{(-0.6918)^2 + 0.8242^2 + 2.0764^2 + 0.13969^2} = 2.3428 [/math]


[math]{p_{\pi z}}^{CM} = -\gamma_{CM} \beta_{CM} E_{\pi} + \gamma_{CM} {p_{\pi z}}^{LAB} = [/math]
[math]= -1.9559 \times 0.8594 \times 2.3428 + 1.9559 \times 2.0764 = 0.1232 [/math]


[math]{\phi}_{\pi}^{CM} = tan^{-1}(\frac{{p_{\pi y}}^{CM \prime}}{{p_{\pi x}}^{CM \prime}}) = [/math]


[math] = tan^{-1}(\frac{-sin{{\phi}_{\gamma}} \times p_{\pi x}^{CM} + cos{{\phi}_{\gamma}} \times p_{\pi y}^{CM}}{sin{\theta_x}\times(cos{{\phi}_{\gamma}} \times p_{\pi x}^{CM} + sin{{\phi}_{\gamma}} \times p_{\pi y}^{CM}) + cos{\theta_x} \times p_{\pi z}^{CM}}) = [/math]


[math] = tan^{-1}(\frac{-sin(73.83) \times (-0.6918) + cos(73.83) \times 0.8242}{sin(141.69)\times(cos(73.83) \times (-0.6918) + sin(73.83) \times 0.8242) + cos(141.69) \times 0.1232} = 72.92 [/math]
Missing_Mass(experimental data)

W missing mass no cuts 27095 experimental data.gif

The mean value of the missing mass is around 2.056 GeV.

[math]\phi[/math] angle
[math]\phi[/math] angle for electrons and pions ([math]\pi^+[/math]) in lab frame [math]\phi_{e^-}^{LAB}[/math], [math]\phi_{\pi^+}^{LAB}[/math]
Electron fiducial cut at electron momentum range : 2.15 < [math]p_e[/math] < 2.53 GeV for sector 1. The histograms on the right show the < [math]\phi_e[/math] distributions at two values of [math]\theta_e[/math]. The highlighted area in the center indicates the selected fiducial range. [1]


[2]

Phi angles

Below is shown histograms before changing angles by sectors for pions and electrons.
E sector vs e phi angle without cuts before change file dst27095.gif Pion sector vs pion phi angle without cuts before changing phi angles by sector file dst27095.gif

ELECTRON SECTOR vs ELECTRON [math]\phi[/math] ANGLE
dst27095_05.B00, B>0, target material NH3, Target Polarization (78.19) and Beam energy 5.7 GeV dst26988_05.B00, B<0, target material NH3, Target Polarization (-67.82) and Beam energy 5.7 GeV dst27109_05.B00, B>0, target material NH3, Target Polarization (-69.81) and Beam energy 5.7 GeV
E sector vs e phi angle positive torus after change file dst27095 without cuts.gif E sector vs e phi angle negative torus file dst27095 without cuts.gif

E sector vs e phi angle positive torus file dst27109 without cuts.gif


[math]\phi[/math] ANGLE DISTRIBUTION FOR ELECTRONS AND PIONS(in each event we have just two particles eelctron and positive pion), FILE dst27095_05
ELECTRONS PIONS(prt_id=4)
Electron phi angle without cuts in Lab Frame file dst27095.gif Pion phi angle without cuts in Lab Frame file dst27095.gif


SECTOR_VS_[math]\phi[/math] FOR ELECTRONS AND PIONS, FILE dst27095_05
ELECTRONS PIONS(prt_id=4)
Electron sector vs electron phi angle without cuts in Lab Frame file dst27095.gif Pion sector vs pion phi angle without cuts in Lab Frame file dst27095.gif


Eelectron sector vs phi angle difference of electron and pion without cuts in Lab frame File dst27095 05.gif

Graph [math]\phi_{\pi}^{CM}[/math] for Pions hitting paddle #7.  The y-axis should be pion counting rate in units of pions per nanCoulomb.
[math]\phi[/math] angle in the Center of Mass Frame

The variables below are in Lab Frame:

From [math]\vec{e}p \rightarrow n \pi^+[/math] from CLAS

Kinematics of single [math]\pi^+[/math] electroproduction



From the above picture we can write down the momentum x,y and z components for pion in terms of angle and total momentum.

[math]{p_{\pi x}}^{LAB} = p_{\pi}^{LAB} {sin {\theta}}_{\pi}^{LAB} cos {\phi}_{\pi}^{LAB}[/math]

[math]{p_{\pi y}}^{LAB} = p_{pi}^{LAB} {sin {\theta}}_{\pi}^{LAB} sin {\phi}_{\pi}^{LAB}[/math]

[math]{\phi}_{\pi}^{LAB} = arctg(\frac{{p_{\pi y}}^{LAB}}{{p_{\pi x}}^{LAB}})[/math]

where [math]{p_{\pi x}}^{LAB}[/math] and [math]{p_{\pi y}}^{LAB}[/math] are the x and y components of the pion momentum.

[math](P_e)^{\mu}[/math] - Initial electron 4-momentum
[math](P_N)^{\mu}[/math] - Target Nucleon 4-momentum
[math](P_e^{\prime})^{ \mu}[/math] - Scattered electron 4-momentum
[math](P_h)^{\mu}[/math] - Hadron final state 4-momentum
[math](P_m)^{\mu}[/math] - Meson final state 4-momentum

[math]h=n[/math], [math]m=\pi^+[/math] for [math]\vec{e} (\vec{p},\vec{e}^{\prime}) \pi^+ n[/math]


In Inclusive [math]\vec{e} (\vec{p},\vec{e}^{\prime}) X [/math]

Then The Missing Mass [math]W = (E_x ^2 - p_x ^2)[/math]


In Exclusive [math]\vec{e} (\vec{p},\vec{e}^{\prime}) \pi^+ X[/math]

Then Missing Mass [math] M = (E_h ^2 - p_h ^2)[/math]


Conservation of 4-momentum gives

[math](P_e)^\mu + (P_N)^\mu = ({P_e}^{\prime})^\mu + {P_h}^\mu + ({P_m})^\mu[/math]

[math]{P_h}^\mu = (P_e)^\mu - ({P_e}^{\prime})^\mu + (P_N)^\mu -({P_m})^\mu[/math]


[math]q^\mu[/math] - 4-momentum of the exchanged virtual photon([math]\gamma[/math])

[math]q^\mu = (P_e^{\prime})^\mu - P_e ^{\mu} = (E_e ^{\prime}, {\vec{p}_e} ^{\prime}) - (E_e, {\vec{p}_e}) = [/math]


[math] = (E_e ^{\prime} - E_e,{\vec{p}_e} ^{\prime} - {\vec{p}_e} ) = (0,{\vec{p}_e} ^{\prime} - {\vec{p}_e}) = (0, p_{e,x}^{\prime}\hat{x} + p_{e,y}^{\prime}\hat{y} + (p_{e,z}^{\prime} - p_{e,z})\hat{z})[/math]


[math]\phi_{\gamma} = tan^{-1}(\frac{p_{e y}^{\prime}}{p_{e x}^{\prime}})[/math]


[math]\theta_x = cos^{-1}(\frac{p_{e z}^{\prime} - p_{e z}}{\sqrt{{p_{e x}^{\prime}}^2 + {p_{e y}^{\prime}}^2 + ({p_{e z}^{\prime} - p_{e z}})^2}})[/math]

[math]p_{e x}^{\prime} = p_{e x}^{\prime CM}[/math]

[math]p_{e y}^{\prime} = p_{e y}^{\prime CM}[/math]

[math]p_{e z}^{\prime CM} = -E_e \gamma_{CM} \beta_{CM} + p_{e z }^{\prime LAB} \gamma_{CM}[/math]

[math]p_{e z}^{CM} = -p_{e z}^{LAB} \gamma_{CM} \beta_{CM} + p_{e z}^{LAB} \gamma_{CM}[/math]

[math]p_{e z}^{LAB} = Beam Energy [/math]

axis rotation

First the coordinate system is rotated around z-axis by [math]\phi_{\gamma}[/math] angle and then around y-axis by [math]\theta_x[/math] angle. Below is presented the transformation matrix.

Rotation around phi gamma angle.gifRotation around theta x angle.gif

[math]\left ( \begin{matrix} p_{\pi x}^{LAB{\prime}} \\ p_{\pi y}^{LAB{\prime}} \\p_{\pi z}^{LAB{\prime}} \end{matrix} \right )= \left [ \begin{matrix} cos {\theta}_x & 0 & -sin {\theta_x} \\ 0 & 1 &0 \\ sin {\theta_x} &0 & cos {\theta_x} \end{matrix} \right ] \left [ \begin{matrix} cos {\phi_{\gamma}} & sin {\phi_{\gamma}} & 0 \\ -sin {\phi_{\gamma}} & cos {\phi_{\gamma}} &0 \\ 0 &0 & 1 \end{matrix} \right ] \left ( \begin{matrix} p_{\pi x}^{LAB} \\p_{\pi y}^{LAB} \\ p_{\pi z}^{LAB} \end{matrix} \right ) = [/math]

[math]= \left ( \begin{matrix} cos {\theta}_x (cos {\phi_{\gamma}} p_{\pi x}^{LAB} + sin {\phi_{\gamma}} p_{\pi y}^{LAB}) - sin {\theta}_x p_{\pi z}^{LAB} \\ -sin {\phi_{\gamma}} p_{\pi x}^{LAB} + cos {\phi_{\gamma}} p_{\pi y}^{LAB} \\ sin {\theta}_x (cos {\phi_{\gamma}} p_{\pi x}^{LAB} + sin {\phi_{\gamma}} p_{\pi y}^{LAB}) + cos {\theta}_x p_{\pi z}^{LAB} \end{matrix} \right ) [/math]


[math]{\phi}_{\pi}^{LAB{\prime}} = tan^{-1}(\frac{{p_{\pi y}}^{LAB{\prime}}}{{p_{\pi x}}^{LAB{\prime}}}) = tan^{-1}(\frac{-sin{\phi_{\gamma}} \times p_{\pi x}^{LAB} + cos{\phi_{\gamma}} \times p_{\pi y}^{LAB}}{sin{\theta_x}\times(cos{\phi_{\gamma}} \times p_{\pi x}^{LAB} + sin{\phi_{\gamma}} \times p_{\pi y}^{LAB}) + cos{\theta_x} \times p_{\pi z}^{LAB}})[/math]

[math]{\phi}_{\pi}^{CM} = tan^{-1}(\frac{{p_{\pi y}}^{CM}}{{p_{\pi x}}^{CM}}) = tan^{-1}(\frac{-sin{\phi_{\gamma}} \times p_{\pi x}^{CM} + cos{\phi_{\gamma}} \times p_{\pi y}^{CM}}{sin{\theta_x}\times(cos{\phi_{\gamma}} \times p_{\pi x}^{CM} + sin{\phi_{\gamma}} \times p_{\pi y}^{CM}) + cos{\theta_x} \times p_{\pi z}^{CM}})[/math]


[math]p_{\pi x}^{CM} = p_{\pi x}^{LAB}[/math]

[math]p_{\pi y}^{CM} = p_{\pi y}^{LAB}[/math]

[math]p_{\pi z}^{CM} = -E_{\pi} \gamma_{CM} \beta_{CM} + p_{\pi z}^{LAB} \gamma_{CM}[/math]

Phi rate

Pion sector vs pion phi angle without cuts in CM Frame file dst27095 after change.gif

Pion phi angle in CM Frame without cuts using theta x and phi gamma angles file dst27095.gif

Pion phi angle in CM Frame without cuts using theta x and phi gamma angles weighted by luminosity file dst27095 1.gif

Pion phi angle in cm frame vs differential cross section.jpg |[3]Differential cross section vs [math]{\phi_{\pi}}^*[/math] in the [math]\Delta(1232)[/math] region at fixed [math]cos \theta_{\pi}^* = -0.1[/math] for different bins in [math]Q^2[/math]


Q sqrd without cuts dst27095.gif


Insert Above graph but for 1.6 < Q^2 < 1.9

30 should be added to the phi angle in cm frame

Pi sector vs phi angle in cm frame without any cuts file 27095 without angle change by pi sectors.gif


Pion phi angle in CM Frame without cuts using theta x and phi gamma angles weighted by luminosity cut on q sqrd 1 6 and 1 9 file dst27095 05 30 added.gif

30 should be added to the phi angle in cm frame

phi angle in cm frame for different runs
Pion phi angle in CM Frame without cuts using theta x and phi gamma angles weighted by luminosity cut on q sqrd 0 9 and 1 6 file dst27095 1.gif 


 start calculating cross -section
Luminosity Calculation for NH3(Ammonia) target

The target materials are located in the target cells. They are made out of polychlorotrifluoroethylene with a thickness of 0.2 mm. The cells itself are in diameter 15 (15.7) mm and in length 10 (12.7) mm, with 0.025 mm aluminum entrance windows and 0.05 mm kapton exit windows. [4]

[math]L = \frac{i_{scattered}}{\sigma} \sim i_{beam} \rho_{target} l_{target}[/math]
[math]{\rho}_{ammonia} = 0.971 \frac{g}{cm^3}[/math][5]
Avogadro Number: [math]N_A = 6.02214 \times\ 10^{23} {mol}^{-1} [/math]
For file dst27095_05.B00 beam current is
[math]i_{beam} = 6 nAmp [/math]
[math]1 Amp = 6.242x10^{18}[/math] electrons per second
[math]L = 6 nAmp \times 0.971 \frac{g}{cm^3} \times l_{target} \times cm =[/math]
[math]= 6 \times 10^{-9} \times 6.242 \times 10^{18} \times 0.971 \times l_{target} \times \frac{ [\# of electrons]}{[second]} \times \frac{g}{cm^3} \times cm =[/math]
[math]= 3.6366 \times 10^{10} \times l_{target} \times \frac{[ \# of electrons]}{[second]} \times \frac{g}{cm^2} =[/math]
[math]= 3.6366 \times 10^{10} \times \frac{1 mol}{17} \times l_{target} \times \frac{[ \# of electrons]}{[second]} \times \frac{1}{cm^2} =[/math]
[math]= 3.6366 \times 10^{10} \times \frac{6.02214 \times 10^{23}}{17} \times l_{target} \times \frac{[ \# of electrons]}{[second]} \times \frac{1}{cm^2} =[/math]
[math]= 1.288 \times 10^{33} \times l_{target} \times \frac{[ \# of electrons]}{second \times cm^2}[/math]
[math]= 1.288 \times 10^{33} \times 1 \times \frac{[ \# of electrons]}{second \times cm^2}[/math]
[math]= 1.288 \times 10^{33} \frac{[ \# of electrons]}{second \times cm^2}[/math]


[math]1 nAmp = 309[/math] counts on Faraday cup.[6]
[math] L = 1.288 \times 10^{33} \frac{1572}{309 \times 6} \frac{[ \# of electrons]}{second \times cm^2} =[/math]
[math]= 1.09209 \times 10^{33} \frac{[ \# of electrons]}{second \times cm^2}[/math]

The luminosity of the continuous electron beam was [math]10^{34}[/math] [math]cm^{-2} sec^{-1}[/math] [7] [8]

why are we a factor of 2 different from reference 8 above?  Ref 8 says at 4 nA L = 15 x 10^{33}.  The above says 1 x 10 ^{33} when I=6 nA.


Cross Section Calculation
[math]\frac{d\sigma}{d\Omega} = \frac{1}{L} \frac{d^{2}N}{d\Omega dt}[/math]

where

[math]L[/math] is the Luminosity.
[math]N[/math] is the number of interactions.
[math]\sigma[/math] is the total cross section.
[math]d\Omega[/math] is the differential solid angle.
[math] \frac{d\sigma}{d\Omega}[/math] is the differential cross section.

F cup file dst27095.gifF cup integral file dst27095.gif

Conservation of Momentum [math]\Rightarrow[/math] :

[math]\vec{p}_e = \vec{p}_{e^{\prime}} + \vec{p}_{\gamma}[/math] [math]\Rightarrow[/math] [math]\vec{p}_{e^{\prime}} = \vec{p}_e - \vec{p}_{\gamma}[/math]
[math]{p_{e^{\prime}}}^2 = {p_e}^2 - 2 p_e p_{\gamma}cos {\theta_x} + {p_{\gamma}}^2[/math]
[math]cos {\theta_x} = \frac {{p_e}^2 + {p_{\gamma}}^2 - {p_{e^{\prime}}}^2}{2 p_e p_{\gamma}}[/math]

Conservation of Energy [math]\Rightarrow[/math] :

[math] E_e = E_{e^{\prime}} + E_{\gamma}[/math]
[math]\sqrt {(m_e c^2)^2 + (p_e c)^2} = \sqrt {(m_e c^2)^2 + (p_{e^{\prime}} c)^2} + p_{\gamma} c[/math]
[math]p_{\gamma} c = \sqrt {(m_e c^2)^2 + (p_e c)^2} - \sqrt {(m_e c^2)^2 + (p_{e^{\prime}} c)^2}[/math]
[math]p_{\gamma} = \sqrt {(m_e)^2 + (p_e)^2} - \sqrt {(m_e)^2 + (p_{e^{\prime}})^2}[/math]

The [math]cos {\theta_x}[/math] can be written in the following way
[math]cos {\theta_x} = \frac {{p_e}^2 + ( \sqrt {(m_e)^2 + (p_e)^2} - \sqrt {(m_e)^2 + (p_{e^{\prime}})^2})^2 - {p_{e^{\prime}}}^2}{2 p_e ( \sqrt {(m_e)^2 + (p_e)^2} - \sqrt {(m_e)^2 + (p_{e^{\prime}})^2}) }[/math]

Pion Rates -vs- Paddle for opposite sign Torus fields

using all events in which the first particle (the one which caused the trigger) is defined as an electrons and passes the

above electron cuts.

sc_paddle vs X_bjorken 5.7 GeV Beam Energy
no cuts cuts no cuts cuts
Electrons B > 0 B<0
Electrons sc paddle vs X dst 27095 without cuts.gif Electrons sc paddle vs X dst 27095 with cuts.gif Electrons sc paddle vs X dst 26988 without cuts.gif Electrons sc paddle vs X dst 26988 with cuts.gif
[math]\pi^-[/math] B > 0 B<0
Pions^- sc paddle vs X dst 27095 without cuts.gif Pions^- sc paddle vs X dst 27095 with cuts.gif Pions^- sc paddle vs X dst 26988 without cuts.gif Pions^- sc paddle vs X dst 26988 with cuts.gif
[math]\pi^+[/math] B > 0 B<0
Pions^plus sc paddle vs X dst 27095 without cuts.gif Pions^plus sc paddle vs X dst 27095 with cuts.gif Pions^plus sc paddle vs X dst 26988 without cuts.gif Pions^plus sc paddle vs X dst 26988 with cuts.gif
sc_paddle vs X_bjorken with cuts 5.7 GeV Beam Energy(number of events=2)
[math]\pi^-[/math] [math]\pi^-[/math]
B>0 B<0
Pions^- sc paddle vs X dst 27095 with cuts num events 2.gif Pions^- sc paddle vs X dst 26988 with cuts num events 2.gif
[math]\pi^+[/math] [math]\pi^+[/math]
B>0 B<0
Pions^plus sc paddle vs X dst 27095 with cuts num events 2.gif Pions^plus sc paddle vs X dst 26988 with cuts num events 2.gif
sc_paddle vs Momentum 5.7 GeV Beam Energy
There is a curvature problem.  When B > 0 then I expect the high momentum electrons to hit the lower
paddle numbers   (inbending).   I can see this when I look at the B>0 plot for electrons with cuts.  
When B < 0 then the electrons  are bending outwards which makes me expect the the higher momentum
electrons will high the higher numbered paddles.  I do not see this for B>0 with electron cuts.
no cuts cuts no cuts cuts
Electons B > 0 B<0
Electrons sc paddle vs momentum dst 27095 without cuts.gif Electrons sc paddle vs momentum dst 27095 with cuts.gif Electrons sc paddle vs momentum dst 26988 without cuts.gif Electrons sc paddle vs momentum dst 26988 with cuts.gif
[math]\pi^-[/math] B > 0 B<0
Pions^- sc paddle vs momentum dst 27095 without cuts.gif Pions^- sc paddle vs momentum dst 27095 with cuts.gif
 Pions^- sc paddle vs momentum dst 26988 without cuts.gif Pions^- sc paddle vs momentum dst 26988 with cuts.gif
[math]\pi^+[/math] B > 0 B<0
Pions^plus sc paddle vs momentum dst 27095 without cuts.gif Pions^plus sc paddle vs momentum dst 27095 with cuts.gif
 Pions^plus sc paddle vs momentum dst 26988 without cuts.gif Pions^plus sc paddle vs momentum dst 26988 with cuts.gif
sc_paddle vs Momentum with cuts 5.7 GeV Beam Energy(number of events=2)
[math]\pi^-[/math] [math]\pi^-[/math]
B>0 B<0
Pions^- sc paddle vs momentum dst 27095 with cuts num events 2.gif Pions^- sc paddle vs momentum dst 26988 with cuts num events 2.gif
[math]\pi^+[/math] [math]\pi^+[/math]
B>0 B<0
Pions^plus sc paddle vs momentum dst 27095 with cuts num events 2.gif Pions^plus sc paddle vs momentum dst 26988 with cuts num events 2.gif


Used file dst26988_05.B00(Energy=5.7GeV and Torus=-2250)

F cup dst26988 05.gif F cup int dst26988 05.gif Number of pions dst26988 05.gif


Paddle 7 Rates and statistics

The number of events per trigger is measured for the respective DST file above and then the Total number events in the data set is estimated from that.

[math]X_{bj}[/math] [math]\pi^-[/math](B>0) [math]\pi^+[/math](B<0)
Total Number Events [math](10^{3})[/math] Number events per [math]10^6[/math] triggers Total Number Events [math](10^{3})[/math] Number events per [math]10^6[/math] trigger
0.1 5.1 71 24.6 547
0.2 6.9 96 13.7 305
0.3 3.7 51 6.2 137
0.4 3.3 45 2.7 60
0.5 0.9 13 0.99 22
Paddle 17 Rates and statistics
[math]X_{bj}[/math] [math]\pi^-[/math](B<0) [math]\pi^+[/math](B>0)
Total Number Events [math](10^{3})[/math] Number events per [math]10^{6}[/math] trigger Total Number Events [math](10^{3})[/math] Number events per [math]10^{6}[/math] trigger
0.1 6.2 137 4.6 64
0.2 3.5 79 4.9 67
0.3 1.7 39 2.6 36
0.4 0.3 7 2.1 29
0.5 0.1 2 0.6 8
Paddle 5 and 8 Rates and statistics for electrons
[math]X_{bj}[/math] [math]e^-[/math] sc_paddle=5 (B>0) [math]e^-[/math] sc_paddle=8 (B<0)
Total Number Events [math](10^{3})[/math] Number events per [math]10^{6}[/math] trigger [math](10^{3})[/math] Total Number Events [math](10^{3})[/math] Number events per [math]10^{6}[/math] trigger [math](10^{4})[/math]
0.1 384.9 5.314 1665.2 3.706
0.2 382.5 5.282 977.8 2.176
0.3 264.9 3.657 567.1 1.262
0.4 159.5 2.202 328.6 0.7313
0.5 99 1.367 218.2 0.4856
Histograms for 5.7 GeV Beam Energy
Electron energy/momentum Electron Theta ([math]\theta[/math]) Electron Qsqrd Electron X_bjorken
B>0 and sc_paddle=5
Electrons energy momentum dst 27095 with cuts.gif Electrons theta dst 27095 with cuts.gif Electrons Qsqrd dst 27095 with cuts.gif Electrons X bjorken dst 27095 with cuts.gif
B<0 and sc_paddle=8
Electrons energy momentum dst 26988 with cuts.gif Electrons theta dst 26988 with cuts.gif Electrons Qsqrd dst 26988 with cuts.gif Electrons X bjorken dst 26988 with cuts.gif
Normalized X_bjorken for electrons
B>0 and sc_paddle=5 B<0 and sc_paddle=8
X bjorken electrons with cuts sc paddle 5 dst27095.gif X bjorken electrons with cuts sc paddle 8 dst26988.gif

Asymmetries

Systematic Errors

Media:SebastianSysErrIncl.pdf Sebastian's Writeup


[Back to Analysis]

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