Number of quarks inside a nucleon

[TheMan112]

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:

$$\int ^1 _0 F_3 ^N (x) = \int ^1 _0 (u_V (x) + d_V (x)) = 3$$

This is based on neutrino-nucleon scattering data, a professor at my uni told me that it is derivable from these two equations:

$$\frac{d \sigma^{\nu N}}{dx} = \frac{G^2 M E}{\pi} x (q(x) + \bar q (x) /3)$$

$$\frac{d \sigma^{\bar \nu N}}{dx} = \frac{G^2 M E}{\pi} x (\bar q(x) + q (x) /3)$$

Where N is your average nucleon.

I don't know if this something that should follow naturally but I feel rather lost.

 

[Vanadium 50]

Think about as two equations in two unknowns: get qV(x) on one side and integrate it.

 

[Vanadium 50]

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.

 

[TheMan112]

Am I supposed to get $$q_V (x)$$ from $$q(x)$$ but keep $$\bar q(x)$$? I tried to extract $$q(x)$$ from the system of equations, but I got a rather messy expression:

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

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

Edit: I suppose a better formulation might be; how do I get $$F_3 ^N (x)$$ from the last two equations (in the first post).

 

[Vanadium 50]

qV = q - qbar.

 

[Vanadium 50]

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.

 

[humanino]

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.

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.

 

[atyy]

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.

Does (syst) stand for systematic error? How do they estimate it?

 

[TheMan112]

In analogue with my last post I come to the expression for the antiquarks:

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

The valence quarks can be calculated by subtracting $$q(x)$$ and $$\bar q (x)$$ as Vanadium suggested:

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

I'm not sure how one reaches an integer answer when integrating this expression using GLS.