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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

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

    Vanadium 50

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    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

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

    Vanadium 50

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    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

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    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.
     
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