TF InclusiveDeltaDoverD

Delta_D_over_D

[math] q_i(x) \equiv q_i^{\parallel}(x) + q_i^{\perp}(x)[/math]

[math] \Delta q_i(x) \equiv q_i^{\parallel}(x) - q_i^{\perp}(x)[/math]

[math] F_1(x) \equiv \frac{1}{2} \sum_q e_i^2 q_i(x) [/math]

using the above definition to define the proton and neutron unpolarized structure function :

[math] F_1^p(x) \equiv \frac{1}{2} \sum_q e_i^2 q_i^p(x) = \frac{1}{2}\left [ \left( \frac{2}{3} \right)^2 u^p(x)+ \left( \frac{-1}{3} \right)^2 d^p(x)\right ] =\frac{1}{2} \left [\frac{4}{9}u^p(x) + \frac{1}{9}d^p(x)\right ][/math]

[math] F_1^n(x) \equiv \frac{1}{2} \sum_q e_i^2 q_i^p(x) = \frac{1}{2}\left [ \left( \frac{2}{3} \right)^2 u^n(x)+ \left( \frac{-1}{3} \right)^2 d^n(x)\right ] =\frac{1}{2} \left [\frac{4}{9}u^n(x) + \frac{1}{9}d^n(x)\right ][/math]

The above is true within the framework of the constituent quark model when in the valence quark region [math]\left ( x_bj\gt 0.5 \right )[/math] where the more massive quarks are ignored as well as anti-quarks

Using Isospin symmetry

[math]u(x) \equiv u^p(x)\equiv d^n(x) \;\;\;\;\;[/math] and [math]\;\;\;\;\;d(x) \equiv d^p(x)\equiv u^n(x) [/math]

The unpolarized structure functions for the proton and neutron may be written as

[math] F_1^p(x) =\frac{1}{2} \left [\frac{4}{9}u(x) + \frac{1}{9}d(x)\right ] \;\;\;\;\;[/math] [math] F_1^n(x)=\frac{1}{2} \left [\frac{4}{9}d(x) + \frac{1}{9}u(x)\right ][/math]

similarly for the polarized structure function

[math] g_1(x) \equiv \frac{1}{2} \sum_q e_i^2 \Delta q_i(x) [/math]

[math] g_1^p(x) =\frac{1}{2} \left [\frac{4}{9} \Delta u(x) + \frac{1}{9} \Delta d(x)\right ] \;\;\;\;\;[/math] [math] g_1^n(x)=\frac{1}{2} \left [\frac{4}{9} \Delta d(x) + \frac{1}{9} \Delta u(x)\right ][/math]

[math] A_1(x,Q^2) \equiv \frac{\sigma_{1/2}^T - \sigma_{3/2}^T}{\sigma_{1/2}^T - \sigma_{3/2}^T} = \frac{g_1(x,Q^2) - \frac{Q^2}{\nu^2} g_2(x,Q^2)}{F_1(x,Q^2)} \approx \frac{q_1(x,Q^2)}{F_1(x,Q^2)}[/math]

In the non-relativistic constituent quark model

[math] A_1^p = \frac{4\Delta u + \Delta d}{4u+d} \;\;\;\;\ A_1^n = \frac{\Delta u + 4\Delta d}{u+4d}[/math] <ref>https://arxiv.org/abs/hep-ph/9809255 PHYSICAL REVIEW D, VOLUME 59, 034013 Valence quark spin distribution functions Nathan Isgur , https://arxiv.org/abs/hep-ph/0411005 The Spin Structure of the Proton Steven D. Bass Rev.Mod.Phys.77:1257-1302,2005</ref>

One can use the two equations above and solve for the polarized quark distributions assuming the unpolarized are known and get , for example

[math] \frac{\Delta d}{d} = \frac{4}{15} \left ( 4 + \frac{u}{d}\right ) A_1^n + \frac{1}{15} \left ( 1 + 4\frac{u}{d}\right ) A_1^p[/math]

Then using the above approximation for A_1

[math] \frac{\Delta d}{d} = \frac{4}{15} \left ( 4 + \frac{u}{d}\right ) \frac{q_1^n}{F_1^n} + \frac{1}{15} \left ( 1 + 4\frac{u}{d}\right ) \frac{q_1^p}{F_1^p}[/math]

[math]g_1^d \approx \left ( 1 - 1.5 \omega_D \right ) \left ( g_1^n + g_1^p \right )[/math]<ref> Eq. 28 from https://arxiv.org/abs/1505.07877 which is based on https://arxiv.org/abs/0809.4308</ref>

<references />

Delta_D_over_D