Elong-13-05-01

Cross-Section Issue - Fixed

From looking at the dilution factor, f, I found a problem with how the cross-sections were calculated. We know that, naively,
$f=\frac{N_{D}}{(N_{D}+N_{N}+N_{He})} = \frac{t\cdot R_{D}}{(t\cdot R_{D}+t\cdot R_{N}+t\cdot R_{He})} = \frac{R_{D}}{(R_{D}+R_{N}+R_{He})} \approx \frac{N_{D}}{N_{D}+N_{N}}\approx \frac{3\sigma_{D}}{3\sigma_{D}+\sigma{N}}\approx \frac{3\cdot 2}{3\cdot 2 + 14}\approx 6/20=0.3$.

Originally, I was (wrongly) using $\frac{d^2\sigma^u}{d\Omega dE'} = \left( \frac{d\sigma}{d\Omega} \right) _{\mathrm{Mott}} \left[ \frac{2\cdot \left(\frac{F_1^{X}}{A_X} \right)}{m_{X}}\tan^2\left( \frac{\theta_{e'}}{2} \right) + \frac{\left( \frac{F_2^X}{A_X}\right) }{\nu} \right]$

where X = nucleus and $\left( \frac{d\sigma}{d\Omega} \right) _{\mathrm{Mott}}=\frac{Z^2 \alpha^2 \hbar^2 c^2}{4E^2\sin^4\left( \frac{\theta}{2} \right)}\cos^2\left( \frac{\theta}{2} \right)$.
This gave $f \approx 0.062$.

The problem came from using a nuclear Mott cross-section but a nucleon structure cross-section. To fix this, I first found the Mott cross-section for a proton,
$\left( \frac{d\sigma}{d\Omega} \right) _{\mathrm{Mott}_{\mathrm{p}}}=\frac{1^2 \alpha^2 \hbar^2 c^2}{4E^2\sin^4\left( \frac{\theta}{2} \right)}\cos^2\left( \frac{\theta}{2} \right)$,

and then multiplied that by the structure cross-section at the total number of nucleons,

$\frac{d^2\sigma^u}{d\Omega dE'} = A_X \left( \frac{d\sigma}{d\Omega} \right) _{\mathrm{Mott}_{\mathrm{p}}} \left[ \frac{2\cdot \left(\frac{F_1^{X}}{A_X} \right)}{m_{p}}\tan^2\left( \frac{\theta_{e'}}{2} \right) + \frac{\left( \frac{F_2^X}{A_X}\right) }{\nu} \right]$.

This gave $f \approx \frac{3\sigma_{D}}{3\sigma_{D}+\sigma{N}} \approx 0.29$, as expected.

Plugging all this into the rates code gave (NOTE: The error on b₁^d is wrong, as discussed below):

Matching HERMES F₁^d, dA_zz^d, and db₁^d

I wanted to double-check the code against the HERMES data, but this required a few extra steps. First, using the values for $A_{zz}^d$ and $b_{1}^d$ that they measured, along with the formula $b_1^d = -\frac{3}{2}A^d_{zz}F^d_1$, I found what they were using for $F^d_1$. The $F^d_1$ they're using is per nucleon, and for HERMES' two highest x points (the rest are inaccessible to the Bosted code because for 0.012<x<0.063, 16.75<W²<42.87, which is outside the W range that the code can handle) they match the code relatively well:

<x>	Q 2	HERMES $F_1^d$	Bosted $\frac{F_1^d}{A_d}$
0.128	2.33	1.018	0.620 (Outside Bosted "Good" range)
0.248	3.11	0.496	0.486
0.452	4.69	0.161	0.171

This means that the way I was calculating the error $\delta b_1^d=\frac{3}{2}\delta A_{zz}^d F_1^d$ was off by a factor of $A_d=2$. This was corrected.

I also played a bit with the statistics, trying to match the HERMES results. However, that's a very quick estimate of trying to use the SMHS to copy the HERMES detector, and so the entire link should be taken with a huge grain of salt.

Fixed Rates Code

Now that everything is working as it should, I've replotted the lowest three <x> points where we can spend 4 weeks taking data. Since the HMS can't get into the range we need for $F_1^d$, I've removed those calculations. The results (on a re-worked scale to emphasize their value) are:

Where

Target Material	ND3
Target Length	3 cm
fdil	Calculated ~ 0.3
pf	0.65
Pzz	20%
E0	11 GeV
xBjorken	0.1, 0.3, 0.5
Q2	Variable
θq	Calculated
θe'	Calculated
E'	Calculated
W	Calculated

A_zz Formalism

Just for clarification, I'm using the formalism as described below.

Target Material = ND3

$z_{\mathrm{tgt}} = 3\mathrm{ cm}$

$p_f = 0.65$

$P_{zz} = 20\%$

$N_{A} = 6.0221413\cdot 10^{23}$

$\rho_{\mathrm{He}} = 0.1412 \mathrm{g/cm}^3$

$M_{\mathrm{He}} = 4.0026 \mathrm{g/mole}$

$\rho_{\mathrm{ND}_3} = 1.007 \mathrm{g/cm}^3$

$M_{\mathrm{ND}_3} = 20 \mathrm{g/mole}$

$I_{\mathrm{beam}} = 0.115 \mathrm{\mu A}$

$\delta F_1^d = 5\%$

(1)	$R_{\mathrm{Total}} = \mathcal{A}\left[ \mathcal{L}_{\mathrm{He}}\left( \frac{d^2\sigma_{\mathrm{He}}^u}{d\Omega dE'}\right) + \mathcal{L}_{\mathrm{N}}\left( \frac{d^2\sigma_{\mathrm{N}}^u}{d\Omega dE'}\right) + \mathcal{L}_{\mathrm{D}}\left( \frac{d^2\sigma_{\mathrm{D}}}{d\Omega dE'}\right) \right]$
(2)	$R_{\mathrm{Total}} = \mathcal{A}\left[ \mathcal{L}_{\mathrm{He}}\left( \frac{d^2\sigma_{\mathrm{He}}^u}{d\Omega dE'}\right) + \mathcal{L}_{\mathrm{N}}\left( \frac{d^2\sigma_{\mathrm{N}}^u}{d\Omega dE'}\right) + \mathcal{L}_{\mathrm{D}}\left( \frac{d^2\sigma_{\mathrm{D}}^u}{d\Omega dE'}\left[ 1 + \frac{1}{2}P_{zz}A_{zz}^d \right]\right) \right] $
(2a)	$R_{\mathrm{Total}} = \mathcal{A}\left[ \mathcal{L}_{\mathrm{He}} \sigma_{\mathrm{He}}^u + \mathcal{L}_{\mathrm{N}} \sigma_{\mathrm{N}}^u + \mathcal{L}_{\mathrm{D}} \sigma_{\mathrm{D}}^u\left( 1 + \frac{1}{2}P_{zz}A_{zz}^d \right) \right] $

where

(3)	$\mathcal{A} = \left( \Delta\Omega \Delta E' \right)$
(4)	$\mathcal{L}_{\mathrm{He}} = \left[ \mathcal{N}_A \frac{\rho_{\mathrm{He}}}{M_{\mathrm{He}}}\left(1 - p_f\right) \right] \cdot \left( \frac{I_{\mathrm{beam}}}{e} \right) \cdot z_{\mathrm{tgt}}$
(5)	$\mathcal{L}_{\mathrm{N}} = \left[ \mathcal{N}_A \frac{\rho_{\mathrm{ND}_3}}{M_{\mathrm{ND}_3}} p_f \right] \cdot \left( \frac{I_{\mathrm{beam}}}{e} \right) \cdot z_{\mathrm{tgt}}$
(6)	$\mathcal{L}_{\mathrm{D}} = 3\left[ \mathcal{N}_A \frac{\rho_{\mathrm{ND}_3}}{M_{\mathrm{ND}_3}} p_f \right] \cdot \left( \frac{I_{\mathrm{beam}}}{e} \right) \cdot z_{\mathrm{tgt}}$
(7)	$\sigma^u_X = \frac{d^2\sigma^u_X}{d\Omega dE'} = A_X \left( \frac{d\sigma}{d\Omega} \right) _{\mathrm{Mott}_{\mathrm{p}}} \left[ \frac{2\cdot \left(\frac{F_1^{X}}{A_X} \right)}{m_{p}}\tan^2\left( \frac{\theta_{e'}}{2} \right) + \frac{\left( \frac{F_2^X}{A_X}\right) }{\nu} \right]$
(8)	$\left( \frac{d\sigma}{d\Omega} \right) _{\mathrm{Mott}_{\mathrm{p}}}=\frac{1^2 \alpha^2 \hbar^2 c^2}{4E^2\sin^4\left( \frac{\theta}{2} \right)}\cos^2\left( \frac{\theta}{2} \right)$
(9)	$\sigma_D = \frac{d^2\sigma_{\mathrm{D}}}{d\Omega dE'} = \frac{d^2\sigma_{\mathrm{D}}^u}{d\Omega dE'}\left[ 1 + \frac{1}{2}P_{zz}A_{zz}^d \right] $

Then

(10)	$R^+_{\mathrm{Total}} - R^-_{\mathrm{Total}} = \mathcal{A}\left[ \mathcal{L}_{\mathrm{He}} \sigma_{\mathrm{He}}^u + \mathcal{L}_{\mathrm{N}} \sigma_{\mathrm{N}}^u + \mathcal{L}_{\mathrm{D}} \sigma_{\mathrm{D}}^u\left( 1 + \frac{1}{2}P_{zz}^+ A_{zz}^d \right) \right] - \mathcal{A}\left[ \mathcal{L}_{\mathrm{He}} \sigma_{\mathrm{He}}^u + \mathcal{L}_{\mathrm{N}} \sigma_{\mathrm{N}}^u + \mathcal{L}_{\mathrm{D}} \sigma_{\mathrm{D}}^u\left( 1 + \frac{1}{2}P_{zz}^- A_{zz}^d \right) \right] $
(10a)	$R^+_{\mathrm{Total}} - R^-_{\mathrm{Total}} = \mathcal{A}\mathcal{L}_{\mathrm{D}}\sigma_{\mathrm{D}}^u \frac{1}{2}A_{zz} \left[ P_{zz}^+ - P_{zz}^- \right]$
(11)	$R^+_{\mathrm{Total}} + R^-_{\mathrm{Total}} = \mathcal{A}\left[ \mathcal{L}_{\mathrm{He}} \sigma_{\mathrm{He}}^u + \mathcal{L}_{\mathrm{N}} \sigma_{\mathrm{N}}^u + \mathcal{L}_{\mathrm{D}} \sigma_{\mathrm{D}}^u\left( 1 + \frac{1}{2}P_{zz}^+ A_{zz}^d \right) \right] + \mathcal{A}\left[ \mathcal{L}_{\mathrm{He}} \sigma_{\mathrm{He}}^u + \mathcal{L}_{\mathrm{N}} \sigma_{\mathrm{N}}^u + \mathcal{L}_{\mathrm{D}} \sigma_{\mathrm{D}}^u\left( 1 + \frac{1}{2}P_{zz}^- A_{zz}^d \right) \right] $
(11a)	$R^+_{\mathrm{Total}} + R^-_{\mathrm{Total}} = 2\mathcal{A}\left[ \mathcal{L}_{\mathrm{He}} \sigma_{\mathrm{He}}^u + \mathcal{L}_{\mathrm{N}} \sigma_{\mathrm{N}}^u + \mathcal{L}_{\mathrm{D}} \sigma_{\mathrm{D}}^u \right] + \mathcal{A}\mathcal{L}_{\mathrm{D}} \sigma_{\mathrm{D}}^u\left[ \frac{1}{2}P_{zz}^+ A_{zz}^d + \frac{1}{2}P_{zz}^- A_{zz}^d \right] $

If we assume that $P_{zz}^+ = -P_{zz}^- = P_{zz}$, then

(12)	$R^+_{\mathrm{Total}} - R^-_{\mathrm{Total}} = \mathcal{A}\mathcal{L}_{\mathrm{D}}\sigma_{\mathrm{D}}^u A_{zz} P_{zz}$
(13)	$R^+_{\mathrm{Total}} + R^-_{\mathrm{Total}} = 2\mathcal{A}\left[ \mathcal{L}_{\mathrm{He}} \sigma_{\mathrm{He}}^u + \mathcal{L}_{\mathrm{N}} \sigma_{\mathrm{N}}^u + \mathcal{L}_{\mathrm{D}} \sigma_{\mathrm{D}}^u \right] $
(14)	$A_{\mathrm{meas}} = \frac{R^+_{\mathrm{Total}} - R^-_{\mathrm{Total}}}{R^+_{\mathrm{Total}} + R^-_{\mathrm{Total}}}$
(14a)	$A_{\mathrm{meas}} = \frac{\mathcal{L}_{\mathrm{D}}\sigma_{\mathrm{D}}^u A_{zz} P_{zz}}{2\left[ \mathcal{L}_{\mathrm{He}} \sigma_{\mathrm{He}}^u + \mathcal{L}_{\mathrm{N}} \sigma_{\mathrm{N}}^u + \mathcal{L}_{\mathrm{D}} \sigma_{\mathrm{D}}^u \right]}$
(14b)	$A_{\mathrm{meas}} = \left( \frac{\mathcal{L}_{\mathrm{D}}\sigma_{\mathrm{D}}^u}{\mathcal{L}_{\mathrm{He}} \sigma_{\mathrm{He}}^u + \mathcal{L}_{\mathrm{N}} \sigma_{\mathrm{N}}^u + \mathcal{L}_{\mathrm{D}} \sigma_{\mathrm{D}}^u} \right) \frac{ A_{zz} P_{zz}}{2}$
(14c)	$A_{\mathrm{meas}} = f \frac{ A_{zz} P_{zz}}{2}$
(15)	$A_{zz} = \frac{2 }{f \cdot P_{zz}}A_{\mathrm{meas}}$

In order to get the uncertainty, we'd use

(16)	$\delta A_{zz} = \sqrt{\left( \frac{\partial A_{zz}}{\partial A_{\mathrm{meas}}} \delta A_{\mathrm{meas}} \right)^2 + \left( \frac{\partial A_{zz}}{\partial f} \delta f \right)^2 + \left( \frac{\partial A_{zz}}{\partial P_{zz}} \delta P_{zz} \right)^2 }$
(16a)	$\delta A_{zz} = \sqrt{\left( \delta A_{zz}^{\mathrm{Stat}} \right) ^2 + \left( \delta A_{zz}^{\mathrm{Dil}} \right) ^2 + \left( \delta A_{zz}^{\mathrm{Pol}} \right) ^2 }$
(16b)	$\delta A_{zz} = \sqrt{\left( \delta A_{zz}^{\mathrm{Stat}} \right) ^2 + \left( \delta A_{zz}^{\mathrm{Sys}} \right) ^2 }$

Ignoring $\delta A_{zz}^{\mathrm{Sys}}$ for now (and in all of the plots I'm showing), then

(17)	$ \delta A_{zz}^{\mathrm{Stat}} = \frac{\partial A_{zz}}{\partial A_{\mathrm{meas}}} \delta A_{\mathrm{meas}} = \frac{2}{f\cdot P_{zz}} \delta A_{\mathrm{meas}} = \frac{2}{f\cdot P_{zz}}\frac{1}{\sqrt{N_{\mathrm{Total}}}} $
(17a)	$ \delta A_{zz}^{\mathrm{Stat}} = \frac{2}{f\cdot P_{zz}\sqrt{t\cdot R_{\mathrm{Total}}}} $

Using the same formalism that HERMES used (which defines $F_{1_{\mathrm{HERMES}}}^d = \frac{(1 + Q^2/\nu^2)F_2^d}{2x(1+R)}$ with $F_2^d=\frac{F_2^p + F_2^n}{2}$ as a per nucleon quantity, which corresponds to the Bosted that uses per nucleus by $F_{1_{\mathrm{HERMES}}}^d = \frac{F_{1_{\mathrm{Bosted}}}^d}{A_{\mathrm{D}}} = \frac{F_1^d}{2}$ -- as described above), we can extract $b_1^d$ and its uncertainty by

(18)	$ b_1^d = - \frac{3}{2}A_{zz} \left( \frac{F_1^d}{A_{\mathrm{D}}} \right)= - \frac{3}{2}A_{zz} \left( \frac{F_1^d}{2} \right)$
(19)	$ \delta b_1^d =\sqrt{ \left(\frac{\partial b_1^d}{\partial A_{zz}} \delta A_{zz} \right)^2 + \left(\frac{\partial b_1^d}{\partial F_1^d} \delta F_1^d \right)^2 }$
(19a)	$ \delta b_1^d =\sqrt{ \left[ - \frac{3}{2} \left( \frac{F_1^d}{2} \right)\delta A_{zz} \right]^2 + \left[ - \frac{3}{2} A_{zz} \left( \frac{1}{2} \right)\delta F_1^d \right]^2 }$

Playing with 30 Days

Note that for everything below, I'm using

<x>	Q 2	W	E'	θe'	θq	cos(θq)
0.1	1.01	3.16	5.62	7.33	7.52	0.99
0.3	1.50	2.09	8.34	7.33	21.3	0.93
0.452	2.58	2.00	7.96	9.85	23.31	0.92
0.5	3.10	2.00	7.70	10.98	23.05	0.92

The most interesting point seems to be at <x> ~ 0.5. We can focus the majority of our time there (28 days for <x>=0.5, and 1 day for <x>=0.1 and 0.3), and still get reasonable results for two more points.

We can do even better if we sit directly on <x> = 0.452 (28 days for <x>=0.452, and 1 day for <x>=0.1 and 0.3), since that lets us pull the angle down a bit and thus get a higher rate:

If we wanted to do better than HERMES for all points, we could do 24 days for <x>=0.5, and 3 days for <x>=0.1 and 0.3

If we wanted to do a lot better than HERMES for all points, we could do 16 days for <x>=0.5, and 7 days for <x>=0.1 and 0.3

If we wanted to split our time up evenly, we could do 10 days for <x>=0.5, 0.3, and 0.1

Finally, to get a sense of how this changes, for each frame we take 2 days away from <x>=0.5 and give one day each to <x>=0.1 and 0.3

--E. Long 20:31, 1 May 2013 (UTC)