- #1
TheMan112
- 43
- 1
I've been banging my head against the wall for days now trying to figure out how one determines the number of quarks inside the nucleon.
I understand it comes from the fact that the Gross-Llewellyn-Smith sum rule is equal to three:
[tex]\int ^1 _0 F_3 ^N (x) = \int ^1 _0 (u_V (x) + d_V (x)) = 3[/tex]
This is based on neutrino-nucleon scattering data, a professor at my uni told me that it is derivable from these two equations:
[tex]\frac{d \sigma^{\nu N}}{dx} = \frac{G^2 M E}{\pi} x (q(x) + \bar q (x) /3)[/tex]
[tex]\frac{d \sigma^{\bar \nu N}}{dx} = \frac{G^2 M E}{\pi} x (\bar q(x) + q (x) /3)[/tex]
Where N is your average nucleon.
I don't know if this something that should follow naturally but I feel rather lost. :tongue:
I understand it comes from the fact that the Gross-Llewellyn-Smith sum rule is equal to three:
[tex]\int ^1 _0 F_3 ^N (x) = \int ^1 _0 (u_V (x) + d_V (x)) = 3[/tex]
This is based on neutrino-nucleon scattering data, a professor at my uni told me that it is derivable from these two equations:
[tex]\frac{d \sigma^{\nu N}}{dx} = \frac{G^2 M E}{\pi} x (q(x) + \bar q (x) /3)[/tex]
[tex]\frac{d \sigma^{\bar \nu N}}{dx} = \frac{G^2 M E}{\pi} x (\bar q(x) + q (x) /3)[/tex]
Where N is your average nucleon.
I don't know if this something that should follow naturally but I feel rather lost. :tongue: