1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Number of quarks inside a nucleon

  1. Dec 1, 2008 #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:
     
  2. jcsd
  3. Dec 1, 2008 #2

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Think about as two equations in two unknowns: get qV(x) on one side and integrate it.
     
  4. Dec 2, 2008 #3

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    I was unclear - the two unknowns I was speaking of are q_V(x) and qbar(x). (i.e. the valence and sea distributions). Pull q_V to one side, integrate it, and G&L-S tell you it's 3.
     
  5. Dec 2, 2008 #4
    Am I supposed to get [tex]q_V (x)[/tex] from [tex]q(x)[/tex] but keep [tex]\bar q(x)[/tex]? I tried to extract [tex]q(x)[/tex] from the system of equations, but I got a rather messy expression:

    [tex]q(x) = \frac{9 \pi}{8 G^2 M E} \frac{1}{x} \left(\frac{1}{3} \frac{d \sigma^{\bar \nu N}}{dx} - \frac{d \sigma^{\nu N}}{dx}\right)[/tex]

    Not least when one would be integrating this, by parts and all. ;)

    Edit: I suppose a better formulation might be; how do I get [tex]F_3 ^N (x)[/tex] from the last two equations (in the first post).
     
    Last edited: Dec 2, 2008
  6. Dec 2, 2008 #5

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    qV = q - qbar.
     
  7. Dec 2, 2008 #6

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    It's probably also worth pointing out that there are many oversimplifications that go into the G&L-S sum rule. Experimentally, it's around two-and-a-half, but what was surprising wasn't that it wasn't exactly three, but that it wasn't even farther off.
     
  8. Dec 2, 2008 #7
    I agree, but would like to point out that the gluon radiation (or any name you give to higher orders in the evolution) come in excellent agreement with the data.

    Sum_GLS(3 GeV^2) = 2.50 +/- 0.018 (stat) +/- 0.078 (syst) (CCFR collaboration Fermilab, Phys Lett B331 1994 655)
    compared to 2.66 +/- 0.04 expected from CTEQ95 for instance.
     
  9. Dec 2, 2008 #8

    atyy

    User Avatar
    Science Advisor

    Does (syst) stand for systematic error? How do they estimate it?
     
  10. Dec 3, 2008 #9
    In analogue with my last post I come to the expression for the antiquarks:

    [tex]\bar q (x) = \frac{9 \pi}{8 G^2 M E} \frac{1}{x} \left(\frac{1}{3} \frac{d \sigma^{\nu N}}{dx} - \frac{d \sigma^{\bar \nu N}}{dx} \right)[/tex]

    The valence quarks can be calculated by subtracting [tex]q(x)[/tex] and [tex] \bar q (x)[/tex] as Vanadium suggested:

    [tex]q_V (x) = q (x) - \bar q (x) = \frac{\pi}{G^2 M E} \frac{1}{x} \frac{3}{2} \left( \frac{d \sigma^{\bar \nu N}}{dx} - \frac{d \sigma^{\nu N}}{dx}\right)[/tex]

    I'm not sure how one reaches an integer answer when integrating this expression using GLS.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Number of quarks inside a nucleon
Loading...