Home » Groups » STAR Protected

Collins and IFF Angles at STAR/COMPASS/HERMES

Updated on Sun, 2013-05-05 19:27. Originally created by yuxip on 2012-10-18 21:58.

This is the extended version of the message that I posted to the spin hypernews

 Hi All,

 
Sorry for digging out the old thread. But I've spent the past few days looking
at the definitions of Collins angles at STAR/Hermes/Compass and I seem to
have found a way relating one to another. We can also resolve the sign issue between
our Collins and IFF asymmetries.
 
Here is the short conclusion.
 
a). In our Collins analysis phi_s is different from Trento/Hermes convention while phi_h is
the same. As a result, our Collins angle phi_collins(star) is shifted w.r.t Trento as follows,
 
        phi_collins(trento) = pi - phi_collins(star) = pi - (phi_s(star) - phi_h(star))
 
with    phi_s(star) = pi - phi_s(trento)
        phi_h(star) = phi_h(trento)
 
But in terms of the sine modulation our definition gives the same moments <sin(phi_collins))
as Hermes' results.
 
b). In our IFF analysis phi_s and phi_R is defined in the same way as phi_s(star) and phi_h(star).
But that doesn't mean the sign of A_{UT}^{sin(phi_RS)} from IFF is opposite to A_{N}^{sin(phi_c)}
from Collins once we corrected the spin patterns. The correction depends on how you calculated the
symmetries with the wrong spin states.
 
 
c). The spin rotation around y axis mentioned by Anselm exists in both pp and
SIDIS in the same way. The reason that Compass has negative Collins asymmetry
is simply because of a different choice in defining the Collins angle. The physics
remains the same --when the spin of the 'outgoing' quark is pointing up, pi+ will be
produced preferentially to the left if looking down to this quark's momentum.
 
----

Argument for a).
 
  In our Collins analysis phi_s and phi_h are defined in the same way(please confirm?)
  as D'Alesio's paper Phys.Rev.D 83.034021(2011) where you firstly build a hadronic c.m.
  frame on the scattering plane in such a way that (p_jet)_x > 0, then the helicity frame
  of the fragmenting parton follows from rotating the hadronic c.m. frame around its
  y-axis so that the z-axis is in the same direction as the outgoing parton. In this
  way phi_s(star) is simply the azimuthal angle of proton's spin in hadronic c.m. frame,
  phi_h(star) is the azimuthal angle of hadron's momentum in the helicity frame of the
  fragmenting parton.
 
  The above scheme is different from Trento convention, but they are connected by
 
        phi_s(star/initial) = pi - phi_s(trento/initial)
        phi_h(star) = phi_h(trento)
 
 
  Of course phi_s(trento/initial) is in terms of the spin of initial quark (target). You can make sense
  out of these relations by transforming fig.1 of D'Alesio's paper into a frame where incoming and outgoing
  partons are on the same z-axis, therefore the picture of the new frame is similar to fig.1 of Trento paper
  (arxiv: hep/ph0410050v2).
 
  However, in going from the initial to final hard scattered parton,
 
        phi_s(star/final)   = phi_s(star/initial)
        phi_s(trento/final) = pi - phi_s(trento/initial)
 
  because the spin transfer from inital to final generates a spin rotation around
  y-axis (by pi) due to the opposite momentum directions of incoming/outgoing quarks
  in gamma-nucleon(as in sidis) or partonic helicity(as in pp) frames. However
  in STAR/D'Alesio's scheme this process becomes hidden because the x-axis of the
  partonic helicity frame is flipped instead of changing phi_s. Therefore,
 
       phi_s(star/final) = phi_s(trento/final)
 
  and the Collins angle,

  phi_collins(trento) = phi_s(trento/initial) + phi_h(trento)
                           = pi - phi_s(trento/final) + phi_h(trento)
                           = pi - phi_s(star/final) + phi_h(star)
                           = pi - [phi_s(star/initial) - phi_h(star)]
                           = pi - phi_collins(star)
 
  So in terms of sine modulation sin(phi_collins(trento)) = sin(phi_collins(star)), and
  our collins asymmetry should have the same sign as Hermes' result.
Argument for b).
 
  The way to figure out how phi_s(iff) is related to Trento is similar as above, by
  Transforming fig.1 of Bacchetta & Radici's paper (arxiv: hep/ph0409174v2) into a
  partonic c.m. frame where the new P_B and P_C are lying on the new z-axis, in order
  to mimic the gamma-nucleon frame as outlined by fig.1 of the Trento paper. Notice that
  you can say P_C -> q only in that frame.
 
  Still we are looking at the moments of sin(phi_s(star) - phi_R(star)) as our Collins analysis
  with phi_h(star) replaced by phi_R(star). There is no problem with phi_R(star) and phi_h(star),
  they are the same as phi_h(trento).
 
  Here is my opinion regarding how we should correct the sign of A_{UT}^{sin(phi_RS)}.
 
  b1.) Correct phi_s(star) to account for the wrong spin pattern, where
 
        phi_s(star/right) = pi + phi_s(star/wrong).
 
       So what's previously been plotted as a function of phi_RS(wrong) ( = phi_s(star/wrong) - phi_R(star) )
       should now be changed to pi + phi_RS(wrong) = phi_RS(right) = pi + phi_s(star/wrong) - phi_R(star).
       For example, when extracting A_UT from raw yield asymmetries vs phi_RS Anselm plotted
 
        asy.(wrong) = N(up/wrong)-N(down/wrong) / [N(up/wrong)+N(down/wrong)]
 
       vs phi_RS(wrong) and the amplitude of sin(phi_RS(wrong)) modulation is A_UT (positive).
       Now with the right spin states the x-axis should be shifted by pi, which will give you negative
       A_UT.
 
  b2). Correct the sign of raw yield asymmetry.
 
        The asy. plotted on the y-axis should also have its sign flipped in order to have
 
        asy.(right) = N(up/right)-N(down/right) / [N(up/right)+N(down/right)] = -asy.(wrong)
 
        This will flip the sign of A_UT again.

 Therefore the sign of A_{UT}^{sin(phi_RS)} will be flipped twice to account for the corrections to the spin
 pattern. This still produces a positive asymmetry for IFF as what it is now.
 
 As a comparsion, since in Collins analysis one does not have to calculated asy. of the raw yield between spin states
 and everything can be done with a single spin state,in principle we only need to flip the sign of A_{N}^{sin(phi_s - phi_h)}
 once. But as the original amplitude was plotted in terms of sin(phi_h - phi_s), you need a second flip from there.
 
 
Argument for c).
 
 No arugment for now, but fig.1 of arxiv: hep/ex-1111.0869v1 is a nice plot showing the definition of anlges at
 COMPASS. Both of phi_s and phi_h follow Trento convention, except that they have chosen a different collins
 angle phi_c = phi_s - (pi - phi_h)

 

Groups: