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Studies on reconstructing the Z0
invariant
mass through the decay:
0
Z   

Syropoulos Vasileios
Supervisor: Nicola Serra
2
Reconstructing
The specific channel is interesting because:
• Check lepton universality.
• It is an experimentally challenging channel.
• Ideal control channel for .
(BR~10-7
,within the SM, more expected in models Beyond SM)
0
Z   

Motivation
SB   

3
Reconstructing 0
Z   

Motivation
High Branching Ratio, 9.31 ± 0.06% (PDG)
More than 2 charged particles in the final state defines the decay point of tau.
0
Z        
 
Many neutrinos in the final state.
0
Z          
 
   
     
We are also interested in final states with:
0
3    

0
3    

4
Reconstructing 0
Z   

The problem
•Merge the 3 pion vertices into one.
•Define transverse momentum as along tau flight direction.
Before going further it is necessary to clarify:
Solving the problem involves developing a Kinematic Fitter:
•Calculating the neutrinos missing momenta.
•Making a selection between the possible configurations.
•Reconstructing the Z mass.
5
Reconstructing 0
Z   

Calculating neutrino momenta
4. Calculate
1. Move from the Lab frame to the tau rest frame.
 3
3 3,P E p
   ,P p p
 

  ,0P M

2. From momentum conservation we have:
   
2 23 3
P P P P P P     
    
3
2 2
3
3
2
E p M E
M M
E
M
   
 


   

  


3. Moving back to the Lab frame :
 3 3 3
L Llab
P P E  
     
 3 3 3
Llab
E E P  
     
 3 3
T Tlab
P P 

5. Repeat steps for the second neutrino.
 3 3 3 3
2 2
3
lab L L lab
T
E E P P
M P
   






Notice the sign
3 pion lab 4 momenta: Experimentally obtained
Rest frame components: Calculated
6
Reconstructing 0
Z   

Quadratic ambiguity.
There are two compatible decay configurations in the tau rest frame.
There is no way to obtain the correct configuration without knowing the tau momenta.
For each neutrino there are two beta vectors along the tau flight direction.
Two beta vectors x Two neutrinos = Four possible combinations.
Selection criteria: Optimum Z mass reconstruction.
Be aware that the background is modified by this selection criteria.
7
Tests
Compare calculated beta vectors with the toy MC.
The calculated beta vectors are
perfectly aligned.
The calculated magnitude is correct up to 3
decimal digits.
Testing the second b vector produced same order of magnitude results.
Reconstructing 0
Z   

8
•There is probably a systematic error.
•Needs further clarification.
Tests
Compare neutrino energy in the tau rest frame with toy MC.
Testing the second neutrino produced same order of magnitude results.
Reconstructing 0
Z   

9
Results
Reconstructing 0
Z   

0.4 GeV RMS
Toy MC peaks at 91.19 GeV.
True values should produce a spike
10
Results
Reconstructing 0
Z   

Residual neutrino momentum.
11
Always correct choice, by identifying the beta vectors.
The residual momentum followed similar behavior.
Last test
Reconstructing 0
Z   

Our calculations peaks at 91.2 GeV.
Toy MC peaks at 91.19 GeV.
12
More cross checking of the algorithm’s consistency will definitely enable us to
obtain a kinematic fitter.
Conclusions
Reconstructing 0
Z   

Most likely, all the deviations are due to either systematic or rounding errors.
The MC values that have been used were not smeared.
Smearing is under studding.
The results are promising. The algorithm is capable of finding the correct Z mass.
We need to understand the nature of this slight deviation.
13
Thank you for your attention

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Z_tautau[Syropoulos Vasileios]

  • 1. Studies on reconstructing the Z0 invariant mass through the decay: 0 Z     Syropoulos Vasileios Supervisor: Nicola Serra
  • 2. 2 Reconstructing The specific channel is interesting because: • Check lepton universality. • It is an experimentally challenging channel. • Ideal control channel for . (BR~10-7 ,within the SM, more expected in models Beyond SM) 0 Z     Motivation SB    
  • 3. 3 Reconstructing 0 Z     Motivation High Branching Ratio, 9.31 ± 0.06% (PDG) More than 2 charged particles in the final state defines the decay point of tau. 0 Z           Many neutrinos in the final state. 0 Z                       We are also interested in final states with: 0 3      0 3     
  • 4. 4 Reconstructing 0 Z     The problem •Merge the 3 pion vertices into one. •Define transverse momentum as along tau flight direction. Before going further it is necessary to clarify: Solving the problem involves developing a Kinematic Fitter: •Calculating the neutrinos missing momenta. •Making a selection between the possible configurations. •Reconstructing the Z mass.
  • 5. 5 Reconstructing 0 Z     Calculating neutrino momenta 4. Calculate 1. Move from the Lab frame to the tau rest frame.  3 3 3,P E p    ,P p p      ,0P M  2. From momentum conservation we have:     2 23 3 P P P P P P           3 2 2 3 3 2 E p M E M M E M                   3. Moving back to the Lab frame :  3 3 3 L Llab P P E          3 3 3 Llab E E P          3 3 T Tlab P P   5. Repeat steps for the second neutrino.  3 3 3 3 2 2 3 lab L L lab T E E P P M P           Notice the sign 3 pion lab 4 momenta: Experimentally obtained Rest frame components: Calculated
  • 6. 6 Reconstructing 0 Z     Quadratic ambiguity. There are two compatible decay configurations in the tau rest frame. There is no way to obtain the correct configuration without knowing the tau momenta. For each neutrino there are two beta vectors along the tau flight direction. Two beta vectors x Two neutrinos = Four possible combinations. Selection criteria: Optimum Z mass reconstruction. Be aware that the background is modified by this selection criteria.
  • 7. 7 Tests Compare calculated beta vectors with the toy MC. The calculated beta vectors are perfectly aligned. The calculated magnitude is correct up to 3 decimal digits. Testing the second b vector produced same order of magnitude results. Reconstructing 0 Z    
  • 8. 8 •There is probably a systematic error. •Needs further clarification. Tests Compare neutrino energy in the tau rest frame with toy MC. Testing the second neutrino produced same order of magnitude results. Reconstructing 0 Z    
  • 9. 9 Results Reconstructing 0 Z     0.4 GeV RMS Toy MC peaks at 91.19 GeV. True values should produce a spike
  • 10. 10 Results Reconstructing 0 Z     Residual neutrino momentum.
  • 11. 11 Always correct choice, by identifying the beta vectors. The residual momentum followed similar behavior. Last test Reconstructing 0 Z     Our calculations peaks at 91.2 GeV. Toy MC peaks at 91.19 GeV.
  • 12. 12 More cross checking of the algorithm’s consistency will definitely enable us to obtain a kinematic fitter. Conclusions Reconstructing 0 Z     Most likely, all the deviations are due to either systematic or rounding errors. The MC values that have been used were not smeared. Smearing is under studding. The results are promising. The algorithm is capable of finding the correct Z mass. We need to understand the nature of this slight deviation.
  • 13. 13 Thank you for your attention