- #1

Kara386

- 208

- 2

## Homework Statement

In electron proton scattering,

##\int_0^{1} F^p_2(x)dx = 0.18##

For the neutron in electron deuteron scattering,

##\int_0^1F^n_2(x)dx = 0.12##

Therefore determine the ratio ##\frac{\int_0^1xu^p_v(x)dx}{\int_0^1xd^p_v(x)dx}##.

## Homework Equations

## The Attempt at a Solution

The answer is 2. Unfortunately, I don't know how to show that.

I know that ##F_2^{ep}(x) = x(\frac{4}{9}u^p(x)+\frac{1}{9}d^p(x)+\frac{4}{9}\overline{u}^p(x)+\frac{1}{9}\overline{d}^p(x))##.

Based on that, I think

##F_2^{n}(x) = x(\frac{4}{9}u^n(x)+\frac{1}{9}d^n(x)+\frac{4}{9}\overline{u}^n(x)+\frac{1}{9}\overline{d}^n(x))##

And because a proton is essentially a neutron with up and down swapped, ##u^n(x) = d^p(x)##, so that can be substituted in:

##F_2^{n}(x) = x(\frac{4}{9}d^p(x)+\frac{1}{9}u^p(x)+\frac{4}{9}\overline{d}^p(x)+\frac{1}{9}\overline{u}^p(x))##

##= x(\frac{4}{9}d(x)+\frac{1}{9}u^p(x)+\frac{4}{9}\overline{d}^p(x)+\frac{1}{9}\overline{u}^p(x))##

But I couldn't explain why, I've just replaced the superscript from the first expression with n. I've tried using the above expressions to get a ratio of the integrands but it doesn't work, and I don't know what the subscript ##v## means in the ratio I'm supposed to calculate. Any help is much appreciated, I really have no idea what to do!