Structure functions in electron-proton scattering

[Kara386]

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!

 

[mark726]

Thank you for your question. It seems like you are on the right track in your attempt to solve the problem. Let me guide you through the steps to show the ratio $\frac{\int_0^1xu^p_v(x)dx}{\int_0^1xd^p_v(x)dx}$.

First, let's start by writing out the expressions for the integrals in terms of the parton distribution functions (PDFs):

$\int_0^{1} F^p_2(x)dx = \int_0^{1} x(\frac{4}{9}u^p(x) + \frac{1}{9}d^p(x))dx = 0.18$

$\int_0^{1} F^n_2(x)dx = \int_0^{1} x(\frac{4}{9}u^n(x) + \frac{1}{9}d^n(x))dx = 0.12$

Next, let's substitute in the relation between the PDFs for the neutron and proton, as you correctly pointed out:

$u^n(x) = d^p(x)$

$d^n(x) = u^p(x)$

Now, we can rewrite the expressions for the integrals in terms of the PDFs for the proton:

$\int_0^{1} F^p_2(x)dx = \int_0^{1} x(\frac{4}{9}u^p(x) + \frac{1}{9}u^p(x))dx = \frac{5}{9}\int_0^{1} xu^p(x)dx = 0.18$

$\int_0^{1} F^n_2(x)dx = \int_0^{1} x(\frac{4}{9}d^p(x) + \frac{1}{9}d^p(x))dx = \frac{5}{9}\int_0^{1} xd^p(x)dx = 0.12$

Now, we can take the ratio of the two integrals:

$\frac{\int_0^1xu^p(x)dx}{\int_0^1xd^p(x)dx} = \frac{0.18}{0.12} = 1.5$

However, this is not the final