There's somethin from Thomson (Modern Particle Physics) that I am a little mythed by.

Section 8.4.3, eq. 8.3 is given as

$$ \frac{F^{\mathrm{en}}_2(x)}{F^{\mathrm{ep}}_2(x)}=\frac{4d_{V}(x)+u_{V}(x)+10S(x)}{4u_{V}(x)+d_{V}(x)+10S(x)} $$

It then states, without proof, that in the low-x limit this becomes:

$$ \frac{F^{\mathrm{en}}_2(x)}{F^{\mathrm{ep}}_2(x)}\rightarrow{1}\,\,{\mathrm{as}}\,x\rightarrow{0} $$

Now, I'm hoping any particle physics experts can help me understand why this is the case...?

I believe $S(x)$ is the sea PDF (what exactly this is elludes me!). These expressions might also be useful:

$$ \int^1_{0}u_{V}(x)\,dx=2 $$

$$ \int^1_{0}d_{V}(x)\,dx=1 $$

So, my question is, can anyone help me understand this and also show me how this result is come to in Thomson. It's very frustrating when these important results are not explained in greater detail and show mathematically to be true.

Context/background reading from Thomson's Cambridge lectures around slides 194/195.

  • $\begingroup$ I don't follow. Could you show me? $\endgroup$ – DarthPlagueis Nov 22 '15 at 23:34

He's relying on isospin symmetry.

The integrals you exhibit are for the proton, but the the form factors in the ratio are proton in the denominator and neutron in the numerator.

The claim is that the up-distribution of the proton is a good proxy for the down-distribution of the neutron and vice versa, and that the sea distributions are identical.

That is \begin{align*} u^{en}_V(x) &= d^{ep}_V(x) \\ u^{ep}_V(x) &= d^{en}_V(x) \\ S^{en}_V(x) &= S^{ep}_V(x) \end{align*}

None of those things is exactly true, but they're pretty good approximations at low momentum fraction.

With those substitutions, the identity should be obvious.

  • $\begingroup$ I still don't follow. So how exactly is the expression evaluated for values of x? I'm just seeing a lot of functions of x... :/ $\endgroup$ – DarthPlagueis Nov 22 '15 at 23:38
  • $\begingroup$ They're not. The limit in $x$ is simply necessary for isospin symmetry to be a reasonable approximation. $\endgroup$ – dmckee --- ex-moderator kitten Nov 22 '15 at 23:39
  • $\begingroup$ It's not obvious? Apologies. I can't see how substitution can allow me to evaluate for different x. I really don't follow this at all. Sorry $\endgroup$ – DarthPlagueis Nov 22 '15 at 23:40
  • $\begingroup$ Could you show the substitutions for x tends to 0 and also to 1? $\endgroup$ – DarthPlagueis Nov 22 '15 at 23:42
  • 1
    $\begingroup$ So what approximations can be made in the high x limit i.e., $x\rightarrow{1}$? $\endgroup$ – DarthPlagueis Nov 22 '15 at 23:56

The parton model assumes that nucleons are composed of three valence quarks that share the momentum of the nucleon in an approximately equitative way. That means that the valence quark pdf's have a peak around $x = 1/3$, and have much lower values at $x \to 0$ and $x \to 1$.

In addition to that, you have interactions between quarks (nowadays we know that these interactions are described by QCD, but the details are not too important). The interactions produce loops of virtual quark/antiquark pairs, which constitute the quark sea. Virtual particles with low energy, and therefore low momentum, are easier to produce than energetic virtual particles (in fact, if the quarks are massless, there is an infrared divergence). In conclusion, the quark sea is dominated by quarks with $x=0$, and in the limit $x\to 0$, $S(x) \gg u_V(x), d_V(x)$. The ratio of the proton and neutron form-factors is $$\lim_{x\to 0}\frac{F_2^{en}(x)}{F_2^{ep}(x)} = \lim_{x\to 0} \frac{4 d_V(x) + u_V(x) + 10 S(x)}{4 u_V(x) + d_V(x) + 10S(x)} = \frac{10S(0)}{10S(0)} = 1$$

On the other hand, at $x\to 1$ is very difficult to produce energetic virtual particles, so $S(x) \ll u_V(x), d_V(x)$. Here, the difference of masses (which breaks the isospin symmetry) becomes important, since light particles gain more momentum in scattering processes than the heavier ones. The lightest quark is the up quark, so we expect that $u_V(x) \gg d_V(x)$, and the ratio of form factor is now $$\lim_{x\to 1}\frac{F_2^{en}(x)}{F_2^{ep}(x)} = \lim_{x\to 1} \frac{4 d_V(x) + u_V(x) + 10 S(x)}{4 u_V(x) + d_V(x) + 10S(x)} = \frac{u_V(1)}{4u_V(1)} = \frac{1}{4} $$

You can compare this approximations with the predictions of some models (lines, triangles) and empirical results (squares, diamonds): enter image description here

Plot taken from: A. Heidari and M. Ghorbani, "An Analytical and Numerical Approach to the Self-Consistent Method for Computing the Proportion of $F_2^n/F_2^p$ Using the 3He and 3H Nucleuses’ Structure Function and EMC Ratio," Journal of Modern Physics, Vol. 3 No. 1, 2012, pp. 124-128. doi: 10.4236/jmp.2012.31017. (Open Access link here)


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