sPHENIX tracking discussion

["Huang, Jin" <jhuang AT bnl.gov>]

Dear John and tracking team

 

Interesting question. Just to share few thoughts, and hope to be useful.

 

1. For azimuthal angular separation, it is 3mrad for the TPC outer pad row, which is worse but not too far from MVTX pixels. And TPC separate tracks in 3D (charge/p, eta, phi), as you also pointed out
2. We probably need 2-3 pixel/pad separation to cleanly separate two track’s clusters due to charge sharing among pixels/pads in both MVTX and TPC.
3. In the end, the limiting factor may be the tracking pattern recognition and fake rejection algorithms. And they may need a special tune for fine separations.

 

Therefore, a more defendable answer could come from a quick plot from the current MDC2 HIJING simulation, to plot track-pair reconstruction efficiency as function of the 3D separation in (charge/p, eta, phi) space, normalized by the product of two stand-alone single track reconstruction efficiency in the same centrality and momentum bin.

 

Cheers

 

Jin

 

 

______________________________

 

Jin HUANG

 

Physicist, Ph.D.

Brookhaven National Laboratory

Physics Department, Bldg 510 C

Upton, NY 11973-5000

 

Office: 631-344-5898

Cell:   757-604-9946

______________________________

 

From: sPHENIX-tracking-l <sphenix-tracking-l-bounces AT lists.bnl.gov> On Behalf Of Lajoie, John G [PHYSA] via sPHENIX-tracking-l
Sent: Saturday, July 30, 2022 1:00 PM
To: sphenix-tracking-l AT lists.bnl.gov
Subject: [Sphenix-tracking-l] check my numbers....

 

Dear sPHENIX Tracking folks:

 

Just to make sure I’m not going off the rails here, check my numbers for me….

 

I’m interested in the *minimum* separation between two tracks that sPHENIX can resolve, as part of this effort to use N-particle correlations to look at jet substructure.  I can get a start on estimating this by looking at the parameters of the MVTX.  If I assume that the tracks are not allowed to share *any* MVTX hits, and the MVTX has a 28um pixel, then the best I can do is:  

 

~28um/24mm = 1.2 e-3

 

Assuming they have to be separated at the first MVTX layer. If they can share hits in the MVTX, but I require the last hit to be unique, I get

 

~28um/39mm = 0.7e-3

 

This is a very back-of-the-envelope estimate, and basically assumes to equal momenta particles.  If they are not equal mpomenta then they will separate based on momentum in both the MVTX, and certainly in the TPC, which would allow you to go to even smaller angles. 

 

Of course, this depends a bit on how the tracking is handled.   Are multiple TPC tracks allowed to be matched to MVTX tracks?

 

Sorry if these questions are somewhat naïve, just trying to think through this….

 

John

 

 

John Lajoie

he, him, his

Harmon-Ye Professor of Physics

Iowa State University

 

(515) 294-6952

lajoie AT iastate.edu