Energy loss improvements and tracking
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Energy loss improvements and tracking. Niels van Eldik, Peter Kluit, Alan Poppleton, Andi Salzburger, Sharka Todorova Common Tracking Meeting 4 July 2013. Back tracking and the impact of the Landau.

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Energy loss improvements and tracking

Energy loss improvementsand tracking

Niels van Eldik, Peter Kluit, Alan Poppleton, Andi Salzburger, Sharka Todorova

Common Tracking Meeting 4 July 2013


Back tracking and the impact of the landau

Back tracking and the impact of the Landau

If one has a Landau distribution for the Eloss one can calculate the distribution at the IP – so after backtracking- by the following convolution:

N(p’) = Integral Landau(E,EMOP,σL)*Gaus(p’-E, σp) dE

Note that also the impact of the track variation in the Barrel calorimeter can be written in this form where σp = EMOPdL/L.

The point is that the MOP value of the distribution N(p’) is not

EMOP anymore. If σL is very small the shift will be very small. However for larger values of σL the shift will depend on σp .

This means that in the backtracking the new EMOP value after the convolution should be used.

The proposal is to calculate the correction analytically using per track the σp.


Numerical issues with the landau distribution and convolutions

Numerical issues with the Landau distribution and convolutions

The second issue is that one has to use numerical integration and FFT to get the Landau convoluted with a Gaussian distribution. It is far from easy to reach the required numerical precision.

To circumvent these problems I decided to move to a set of analytical functions that allow convolution and fast robust fitting.

The Landau will be approximated by:

Landau(x) = x/(a2+x2) 2 for x>0

x = E + E0

a = 3.59524 σL

EMOP = E0 + a/√3

Gaussian(σreso) ≈ Breit-Wigner 2 ~ 1/(b2+x2) 2

b ≈ 0.87388 σreso

These functions can be convoluted analytically and fitted to the data.


Approximation to the landau distribution

Approximation to the Landau distribution

In red the chosen approximation to the Landau function

Landau(x) = x/(a2+x2) 2 for x>0 compared to the root implementation


Data fitted to the model

Data fitted to the model

Here a fit to some extreme cases.

First a Barrel fit where the distribution is mostly gaussian.

p0 = normalization

p1 = mean

p2 = a ~ Landau width

p3 = b = BW2 width

So b ~ sigma resolution


Data fitted to the model1

Data fitted to the model

  • The numerical analytical functions are in place. They are able to describe the data. For quite extreme cases from the Gaussian to the Landau and the Resolution dominated. More examples and fits are in the backup

  • Next slides:

  • Work out the formulae to give corrections to EMOP

  • after convolution with a Gaussian. Give the mathematics for Landau distributions and 68% (95%) CL intervals.

  • Finally some implications for MS and CB track fitting are discussed.


Modeling the shift in e mop

Modeling the shift in EMOP

  • Problem: suppose we have a MOP value and we have a Landau error σL and a gaussian sigma σreso what is the shift ΔEMOP?

  • Here we use the parameters

    • a = 3.59524 σL

    • b = 0.87388 σreso

  • ΔEMOP = function (a,b)

  • It has a special form ΔEMOP = a f(b/a)

    • f(x=b/a) is shown on the next page

    • It has an asymptotic behaviour where f->1 for x->inf

    • The function can be approximated by

    • f(x) = (p0*x*x+p1*x*x*x)/(1+p2*x+p3*x*x+p4*x*x*x)

    • p0 1.75479e+00 p1 5.88545e-01

    • p2 4.28703e+00 p3 4.36873e+00

    • p4 6.05171e-01


Modeling the shift in e mop1

Modeling the shift in EMOP


Typical example in the calorimeter the shift in e mop

Typical example in the Calorimeter: the shift in EMOP

Barrel a = 200 MeV

EC a = 450 MeV

Assume a dp/p of 2%

This gives the curves on the left*.

So at high momenta one gets the full shift of a. At low momenta the shift is smaller.

That is why the Jpsi pT > 5 GeV is very different from the Z pT > 20 GeV

where the MOP shift starts to rise.

* For simplicity it is assumed that a= 3.595 σL does not change with p


Typical example in the muon spectrometer the shift in e mop

Typical example in the Muon Spectrometer: the shift in EMOP

Barrel a = 20 MeV

EC a = 45 MeV

Assume a dp/p of 2%

This gives the curves on the left.

So at high momenta one gets the full shift of a. At 3 GeV (EC 6 GeV) the shift is already 8 (20) MeV.

The dependence is rather small but should be accounted for…


Convoluting landau distributions and the track fit

Convoluting Landau distributions and the track fit

  • Problem: suppose we have several EMOP values and errors. As discussed in the Muon-PUB-2008-002 note,

  • the formula to combine these values is:

*

This has very special features:

The MOP value shifts depending on the MOP errors (7)

The errors do not add up quadratically but linearly (8)

This has implications for track fitting.

It means that Eloss has to be combined using these formula.

For the MS track fit I would propose to collect the total

material between the measurements and use the formulae

above to aggregate the Eloss.

* I prefer the form: (σ1+σ2) ln (σ1+σ2) - σ1 ln σ1- σ2 ln σ2


Landau distributions and cl intervals

Landau distributions and CL intervals

We want to derive analytic expression for the different

CL intervals. For a pure Landau distribution we define σ+

and σ- as containing 68% of the events on the upper

lower side of the MOP Eloss value.

68% CL σ- = 1.02 σL σ+ = 4.65 σL

95% CL 2σ- = 1.82 σL 2σ+ = 21.86 σL

After convolution with a Gaussian with σResothe formulae

become (checked using the Landau-BW approximations):

σ- = √ (1.02 σL )2 + (σReso) 2

σ+ = √ (1.82 σL )2 + (σReso) 2

2σ- = √ (4.65 σL )2 + (2 σReso) 2

2σ+ = √ (21.86 σL )2 + (2 σReso) 2

These expressions we need for Calorimeter object

in the CB trackfit.

Big!


Treatment of the calorimeter eloss

Treatment of the Calorimeter Eloss

  • Proposal: from the Calorimeter Tracking Geometry we obtain:

  • The EMOP (from ID to MS)

  • The σL on the Eloss

  • The dL/L: uncertainty on track length in Calorimeter

  • NB in the calculation of 1) and 2) the Landau formalism of

  • slide 11(adding errors linearly) should be used.

  • From this we calculate:

  • EMOP corrected using the formalism on slide 7 (dL/L gives a

  • smearing and therefore shifts the mean).

  • We also need σ+ and σ- from slide 12.

  • This allows to make the comparison of the measured

  • Eloss in the Calorimeter with EMOP corrected+ 2 σ.


Treatment of the calorimeter eloss1

Treatment of the Calorimeter Eloss

  • Next step is to determine the Eloss and errors that should be used in case no Calorimeter measurement is used.

  • Now we have to calculate:

  • EMOP corrected using dp/p inside MS and dL/L using the

  • formalism on slide 7. We also need σ+ and σ- from slide 12.

  • The value of the EMOP corrected will depend on whether

  • we looked at the measured Eloss or not. NB if we did look the

  • Landau tail is reduced. Currently I work on the mathematics

  • for this case.

  • This Eloss object can be used for two purposes:

  • Back extrapolation of the MS track to the IP

  • -> SA parameters

  • Perform the Combined fit

  • -> CB parameters


Treatment of the muonspectrometer eloss

Treatment of the MuonSpectrometer Eloss

  • To do a perfect job inside the Muon Spectrometer

  • One should firstly aggregate according to the formalism for Landau distributions

  • Secondly one should reiterate using the uncertainty on the momentum to update EMOP corrected and then refit. This is depicted on slide 10.

  • Question: what is done in the Step Propagator? Is the MOP value returned? How is the aggregation of the Eloss done (is the Landau formalism used)?

  • One could use the Landau formalism in the track fit.

  • However one needs a re-iteration propagating the error on the momentum into the EMOP value.

  • For the calorimeter Eloss object one could make the object taking into account the momentum error…


Backup slides

Backup slides


Introduction

Introduction

  • Currently the Energy loss description for muons passing the calorimeter has a precision of about Et 100 MeV.

  • For mass measurements using the combined and or muon (standalone) measurements it is important to improve significantly the precision of the E loss description.

  • The target that we want to put here is a description that is accurate at e.g. the 10 MeV level.

  • This we want to achieve on a track-track basis.

  • This project will need improvements in several areas:

  • the Tracking Calorimeter description: E loss MOP and sigma modeling

  • 2) (analytic) computing and reconstruction


Introduction physics

Introduction: Physics

  • Concerning the physics aspects:

  • Firstly one can get rid of part of the Landau tail. In particular for isolated muons. This is done by the Muid Combined algorithm.

  • Secondly, we cannot fully get rid of the Landau tails; but we can give the right (well-centered) MOP value. This means that the MOP of di-muon resonances like Jpsi and Z will be (about) well centered.

  • Thirdly, we cannot get the “mean” of the momentum or mass distribution well-centered. This has to do with the fact that a Landau distribution does NOT have a mean. So depending on the cuts, the “mean” will change. The MOP value is however well defined.


Calorimeter tracking geometry

Calorimeter Tracking geometry

  • The current situation is the following.

  • There is a layer based description of the calorimeter to describe the E loss and its error.

  • This description is not used in Muid because it is not precise enough. The Muid description is basically an eta-phi-p map that gives MOP Eloss and its error.

  • There are two possible roads that improve the situation:

  • Improve the E loss description of the Calorimeter

  • Improve the precision of the Muid Eloss map

  • It is hard to conceive that a map with a precision of 10 MeV can be made. We think however it is possible to achieve a very precise description of the E loss in the calorimeter.

  • This would require a volume based TG description of the calorimeter.


Calorimeter tracking geometry1

Calorimeter Tracking geometry

In the Muon system a volume based TG description gives a very accurate description of the material traversed.

For the Muon system this is the best we can achieve.

A similar description of the calorimeter would give on a track-by-track basis the E loss. And this is - in my opinion - the ‘ultimate’ description one wants to have.

Both Sharka and Andi are willing to work on this topic.

Requirements

Eloss: for a given track one wants to know the MOP Eloss value and its error. Secondly one needs the track length in the Calorimeter and its error.

Identification: crossed cells and the MOP E loss per cell and its error (for use in Muon Calorimeter Identification algorithms)


Energy loss modeling

Energy loss modeling

Energy loss modeling is described by a Landau distribution that is characterized by a MOP value and a width or sigma value.

The MOP energy and the sigma depend on the momentum of the muon and consists of a linear and a logarithmic term. The tracking geometry takes uses dependencies. To obtain a precise description Geant4 simulations and TG should agree on these underlying parametrisations and the material characteristics (such as X0).

A study of the Eloss shape in the MC in different regions of the detector was performed.

It was found out that the Eloss shape in the Barrel does NOT have a Landau shape. In the Endcap it can be described by a Landau distribution.


Energy loss modeling1

Energy loss modeling

Fit: frac*Landau + (1-frac) Gaussian

Puzzle

Barrel has 83% Gaussian! Endcap only 4%

Only 17% Landau as expected 96% Landau


Energy loss ec plots for one eta bin and a 2 gev p range

Energy loss EC plots forone eta bin and a 2 GeV p range

Here eta = 2.0-2.1; p 12-14; 14-16 and 16-18 GeV

Enough stats in Endcap from 14 GeV onwards: observe a large Landau tail (as expected)


Energy loss barrel plots for one eta bin and a 2 gev p range

Energy loss Barrel plots forone eta bin and a 2 GeV p range

Barrel eta 0-0.1 same momentum range observe a larger Gaussian component


Energy loss barrel plots for one eta bin and a 2 gev p range1

Energy loss Barrel plots forone eta bin and a 2 GeV p range

Barrel eta 0-0.1 to lower momenta: one observes an increase in the Gaussian component


Barrel energy loss modeling

Barrel: Energy loss modeling

Barrel has more Gaussian shape. Why is this?

First observed by Alan and Kostas. Their explanation:

The distribution has two components: one of Landau component and another track length component.

If the track length varies due to multiple scattering the mean of the Landau will be smeared with a Gaussian.

Here a model that indeed does the job for 10% track length smearing


Energy loss modeling2

Energy loss modeling

Conclude:

Energy loss modeling is described by a Landau distribution that is characterized by a MOP value and a width or sigma value.

Need also to know the track length variations dL/L in the Calorimeter to account for the effect observed in the Barrel.

That is why on slide this is added as a requirement to the TG description.

Reconstruction and computing issues:

- Back tracking and the impact of the Landau tail.

- Numerical precision and the Landau distribution, convolutions.

- Approximation to the Landau distribution.


Numerical issues with the landau distribution

Numerical issues with the Landau distribution

The first approach was to use the formula on slide 10 and compute it.

This turns out to be not that trivial: the Landau distribution itself using e.g. the root implementation has a problem. The MOP value you put in is not the maximum of the distribution…

It is shifted by about 40 MeV.

MOP value of 3 GeV input

The maximum of the curve is shifted by 40 MeV

The point is that we need a precision of << 10 MeV on the MOP value


Back tracking and the impact of the landau1

Back tracking and the impact of the Landau

Suppose we have a track measured in the MS its q/p distribution will be rather Gaussian.

The momentum resolution dp/p in the Muon System is given below:


Can the data be fitted with the current model

Can the data be fitted with the current model?

Here a fit to some extreme cases.

An Endcap fit where the distribution is mostly Landau with small 0.02 dp/p smearing.

p0 = normalization

p1 = mean

p2 = a ~ Landau width

p3 = b = BW2 width

So b ~ sigma resolution


Can the data be fitted with the current model1

Can the data be fitted with the current model?

Here a fit to some extreme cases.

An Endcap fit where the distribution is mostly Landau with large 0.10 dp/p smearing.

p0 = normalization

p1 = mean

p2 = a ~ Landau width

p3 = b = BW2 width

So b ~ sigma resolution

Note the large b value

As expected the MOP

value increases wrt the small smearing case


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