TF InclusiveDeltaDoverD: Difference between revisions
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using the above definition to define the proton and neutron unpolarized structure function: | 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{2}{9}u^p(x) + \frac{1}{9}d^p(x)</math> | <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{2}{9}u^n(x) + \frac{1}{9}d^n(x)</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> | ||
<math> F_1^n(x) \equiv \frac{1}{2} \sum_q e_i^2 q_i^n(x) </math> | <math> F_1^n(x) \equiv \frac{1}{2} \sum_q e_i^2 q_i^n(x) </math> | ||
Revision as of 16:12, 22 September 2018
[math]\displaystyle{ q_i(x) \equiv q_i^{\parallel}(x) + q_i^{\perp}(x) }[/math]
[math]\displaystyle{ \Delta q_i(x) \equiv q_i^{\parallel}(x) - q_i^{\perp}(x) }[/math]
[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ 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]
[math]\displaystyle{ F_1^n(x) \equiv \frac{1}{2} \sum_q e_i^2 q_i^n(x) }[/math]
Using Isosping symmetry
[math]\displaystyle{ u(x) \equiv u^p(x)\equiv d^n(x) \;\;\;\;\; }[/math] and [math]\displaystyle{ d(x) \equiv d^p(x)\equiv u^n(x) }[/math]
[math]\displaystyle{ g_1(x) \equiv \frac{1}{2} \sum_q e_i^2 \Delta q_i(x) }[/math]
[math]\displaystyle{ 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>
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