[b1_ana] Follow up on Re: Fwd: Re: A_PV with Deuteron Target

[Elena Long]

Good afternoon,

A minor point, but doesn't it make more sense to look at 
sigma_pol/sigma_u instead of the difference of them, since we're 
looking at (Npol/Nu - 1)? I don't think that really changes your 
conclusions much, though. This probably isn't terribly helpful yet, but 
I LaTeX'ed the extra asymmetries into our method here: 
https://hallcweb.jlab.org/wiki/index.php/Elong-13-05-20

The Arenhovel paper you're mentioning is Phys. Rev. C 43, 1022–1037 
(1991), yeah? Just so I understand that I'm looking at the same points 
as you, for Fig. 14a and 19 are you looking at q_cm^2 ~ 20?

Take care,
Ellie

Elena Long, Ph.D.
Post Doctoral Research Associate
University of New Hampshire
elena.long at unh.edu
ellie at jlab.org
http://nuclear.unh.edu/~elong
(603) 862-1962

On Sun 19 May 2013 07:03:20 PM EDT, O. A. Rondon wrote:
> Hi,
>
> After thinking a bit about Wally's target-only A_PV asymmetry, I
> recalled that actually there is an electromagnetic target-only
> asymmetry, per Arenhoevel's Inclusive electrodisintegration paper that
> we were discussing when we explored b_1(x~1), see eq. (25).
>
> So, really, the difference pol. - unpol., with P_b.ne.0, is
>
> sig_U*(1 + D*A1*P_b*P_z + A_zz*P_zz + A_v^d*P_z) - sig_U*(1 + APV*P_b),
>
> Since the field is longitudinal, A_v^d may not go away, it depends on
> sin(phi_d)*sin(theta_d)/sqrt(2). phi_d and theta_d are the angles of the
> field relative to the q vector, so theta_d = theta_q, see fig. 1.
>
> Since the field will be in the horizontal plane, A_V^d would cancel
> exactly if the acceptance were symmetric about the plane, but given the
> deflection of the scattered electrons by the field, it may not go
> entirely away without some acceptance cuts.
>
> The good news is that it is suppressed by the double sine, and since the
> angles are small, the contribution is very suppressed: the deflection
> for E' = 7.3 GeV is 0.35 deg., which projects to phi_d = 1.6 deg. for
> the worst case (HMS at 12.5 deg, E' = 7.3 GeV, theta_q = 22.3 deg.) so
> the sine factors are 8E-3, assuming there is NO cancellation between
> events with phi_d > 0 and phi_d < 0.
>
> This A_V^d, and any target-only A_PV, can be further canceled by
> flipping P_z from one period to the next, and adding the events,
> weighted by P_z. So it looks like we'll need to take an equal number of
> pol. periods with opposite charge-weighted P_z anyway.
>
> I tried to estimate the magnitude of the kinematic factor rho_{LT} and
> the form factor F_{LT}^{1,-1} that enter in eq. (25). For example, from
> fig. 14a, I see that at W = Delta(1230), Q^2 = 0.8 GeV^2, the dominant
> F_T = 0.038 fm, while fig. 19 shows F_{LT}^{1,-1}(Delta, 0.8 GeV^2) ~
> 0.9E-4 fm, and it has a decreasing trend with Q^2, which is good for us.
>
> So it looks like the contribution of A_V^d(F_{LT}^{1,-1}) would be about
> 2E-3 of sigma_U, suppressed by the sine factors ~8E-3, but may be
> magnified by rho_{LT}, eq. (5), that seems to be ~ 300, dominated by
> 1/eta = 1/tan^2(12.5/2) and Q^2 = 3.8, or A_V^d ~ 5E-3, before any
> cancellations from adding the data for opposite P_z, and cutting the
> acceptance in phi to be symmetric about phi = 0.
>
> I encourage everyone to review these points carefully.
>
> Cheers,
>
> Oscar
>
>
> O. A. Rondon wrote:
>> Hi,
>>
>> The classical deuteron PV experiment by Charlie Prescott at SLAC was
>> polarized beam on unpolarized deuterons, i.e. Qweak on deuterons.
>>
>> To see a PV with unpolarized beam, we would need to make a difference of
>> target P_z's. But we aren't making any, and we shouldn't. We can only
>> change the P_z sign at the end of every pol. - unpol. cycle. To observe
>> any P_z dependent PV asymmetry we would have to subtract
>> pol. P_z - pol. (-P_z).
>>
>> This would not give us any useful information at the PV level, or even
>> at the b1 level. Remember that we are doing pol. - unpol. for b1. In
>> fact, if P_b is not exactly zero, in the difference pol. - unpol. there
>> are two extra terms:
>>
>> sigma_U*(1 + D*A1*P_b*P_z + A_zz*P_zz) - sigma_U*(1 + APV*P_b),
>>
>> so Dave must be thinking of the APV*Pb term, which is another reason for
>> asking for true unpol. beam. With P_b not zero, this term is always
>> there in pol. - unpol., no matter how we change P_z.
>>
>> The E155x paper indicates  APV = A_EW ~ 8E-5*Q^2, (for protons or
>> deuterons,) so for our proposal, with max. Q^2 <~ 5 GeV^2, A_EW = 4E-4,
>> so P_b <~ 20% would be needed to keep it < 1E-4. Again, it would be best
>> to measure just
>>
>> sigma_U*(1 + A_zz*P_zz) - sigma_U
>>
>> At the A_1 level, changing P_z to -P_z might let us see if the A_1 due
>> to any residual P_b is not changing from period to period (i.e. if it's
>> a real A_1, not a false one, independent of P_z,) but that is not very
>> useful: Any P_b related background (A_1 related or not) would need to be
>> subtracted from our pol. - unpol. difference. I rather stay with the
>> P_z sign that gives the highest P_zz (e.g. +P_z polarizes faster than
>> -P_z, or something).
>>
>> But someone could come up with additional quantitative estimates that
>> may help.
>>
>> Cheers,
>>
>> Oscar
>>
>> Narbe Kalantarians wrote:
>>> I'll look further in to this myself. But, so far, it seems that it may
>>> not be too worrisome.
>>>
>>> Narbe
>>>
>>>
>>> -------- Original Message --------
>>> Subject:     Re: A_PV with Deuteron Target
>>> Date:     Fri, 17 May 2013 11:21:31 -0400 (EDT)
>>> From:     Wally Melnitchouk <wmelnitc at jlab.org>
>>> To:     Narbe Kalantarians <narbe at jlab.org>
>>>
>>>
>>>
>>> Hi Narbe,
>>>
>>> It's a nice proposal and I hope you get it approved.
>>>
>>> The paper where (with Tim Hobbs) we discussed PVDIS on polarized
>>> nucleons is PRD 77, 114023 (2008), Sec. V.  There we considered
>>> only scattering from spin-1/2 targets (in which the deuteron was a
>>> proton + neutron), so did not encounter the b1 structure function
>>> explicitly.  For unpolarized electrons, taking differences between
>>> cross sections with different target polarizations produces a PV
>>> asymmetry.
>>>
>>> Correct me if I'm wrong, but my understanding of b1 is that this is
>>> parity-conserving, so it's not clear to me (without writing down the
>>> full expressions for the cross sections in terms of the lepton and
>>> hadron tensors) how the P-conserving and P-violating contributions
>>> mix.  To measure b1 in your experiment, do you need to polarize the
>>> electron at all?
>>>
>>> In any event, the PV contribution is very small and can be calculated
>>> in terms of spin-dependent PDFs, so if one is in the DIS region the
>>> uncertainty from this would be expected to be small.
>>>
>>> Unfortunately I leave on a trip to Europe tomorrow so don't have time
>>> to look deeper into this, but I'm very curious about this myself.
>>> Maybe let me know what you find, and we can discuss further over email,
>>> if you like.
>>>
>>> Cheers
>>> Wally
>>>
>>>
>>> On Thu, 16 May 2013, Narbe Kalantarians wrote:
>>>
>>>> Hi Wally,
>>>>
>>>> A few of us have proposed a measurement of the deuteron tensor spin
>>>> structure function b1 for this upcoming PAC.  This would involve a
>>>> (longitudinally) polarized deuteron target and unpolarized (or as low as
>>>> possible) beam. I've attached a copy of it, in case you want to see it.
>>>> The kinematic range we are considering is
>>>> 0.16 < x < 0.49,
>>>> 0.8 < Q^2 5.0 GeV^2
>>>>
>>>> With 11 GeV beam.
>>>>
>>>> The basic gist of it is to get the tensor asymmetry A_zz and determine
>>>> b1 from it.
>>>>
>>>> We are undergoing the TAC review. The chair of that committee, Dave
>>>> Mack, mentioned that there is a PV asymmetry with polarized deuteron
>>>> (target).  I recall seeing some calculations you and a student (T.
>>>> Hobbes?) did some time a go for this.
>>>>
>>>> Could you send that write up and let me know if this could be a sizeable
>>>> and/or significantly contributing asymmetry to take in to account?
>>>>
>>>> Thanks,
>>>> Narbe
>>>>
>>>
>>>
>>>
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