In his blog at http://motls.blogspot.com/2010/09/can-antimatters-gravity-be-repulsive.html Lubos Motl writes
"...neutrons contain a slightly different mass contribution from the antiquarks (antimatter!) than the protons ..... where the difference is something of order 1%, ..."
I would like some reliable references to this derivation, the simpler the better, assuming the usual standard model, of course. I assume it is somewhere in the computation of the masses of the baryons, but where, and why?
There is no way to decompose the proton into "matter" and "antimatter" because gluons are their own antiparticle, and the glue field contributes most of the mass. The anti-quark contributions to the nucleon mass are what Lubos is talking about, and these are about 1% only because quark loops in low-lying hadrons are considered to be 1/N corrections, and so a few percent, and you lop off a little just to be on the safe side.
Since he is trying to refute a manifestly idiotic theory that antimatter falls up, there is no harm in this underestimate--- it is adequate for the argument he is using it for.
I started writing an answer to this weeks ago, as a followup to my answer on your related question, but I completely forgot about it until Ron bumped the question :-P Anyway, it's worth checking what happens when you take the calculation I did in that answer and "translate" it to the neutron. Spoiler alert: this doesn't amount to a 1% difference, I'm just posting it to show how one might go about calculating this sort of thing from a phenomenological perspective.
Assuming that the proton and neutron are identical under isospin reflection, we can make the following correspondences:
$$\begin{align*}f_{\mathrm{u}/\mathrm{n}} &= f_{\mathrm{d}/\mathrm{p}} \\ f_{\mathrm{d}/\mathrm{n}} &= f_{\mathrm{u}/\mathrm{p}} \\ f_{\mathrm{\bar{u}}/\mathrm{n}} &= f_{\mathrm{\bar{d}}/\mathrm{p}} \\ f_{\mathrm{\bar{d}}/\mathrm{n}} &= f_{\mathrm{\bar{u}}/\mathrm{p}}\end{align*}$$
Recall that in the other answer, I calculated the energy distribution of the proton (approximately) like this:
$$E_m \approx p\sum_{i\in\{\mathrm{uds}\}}\int_{0}^{1}x\,f_{i\,/\mathrm{p}}(x)\mathrm{d}x$$
The thing is, if you just plug in the correspondences above, the only difference is that the contributions from the up and down quarks are switched (and similarly for antiquarks). So you get exactly the same result of 41% matter and 9% antimatter. This reflects the isospin symmetry of the proton and neutron, which is unbroken if you consider the quarks to be massless.
In order to find a difference between the energy distributions of the proton and neutron, we therefore need to break the isospin symmetry by incorporating the fact that the up and down quarks have different masses. So let's restart from this equation and reexamine the simplifications I made:
$$E_m = \sum_{i\in\{\mathrm{udcsbt}\}}\int_{Q_1^2}^{Q_2^2}\int_{0}^{1}\sqrt{x^2p^2 + p_T^2 + m_i^2}\,f_{i\,/\mathrm{n}}(x,Q^2)\mathrm{d}x\frac{\mathrm{d}Q^2}{Q^2}$$
As in the other answer, I'm still going to ignore the heavy quarks and the incidental transverse momentum $p_T^2$. I won't ignore the masses, but I will still assume they are small relative to the momenta involved, which means I can expand the square root as
$$\sqrt{x^2p^2 + m_i^2} = xp\biggl(1 + \frac{m^2}{2x^2p^2} + \dotso\biggr)$$
so in addition to the same terms I calculated in the other answer, we have corrections of the form
$$E_m \approx p\sum_{i\in\{\mathrm{uds}\}}\frac{m_i^2}{2p^2}\int_{0}^{1}\frac{x\,f_{i\,/\mathrm{p}}(x)}{x^2}\mathrm{d}x$$
As it turns out, using the same data I used for the other calculation, if you take $p \approx 1\text{ TeV}$ as is roughly the case at the LHC, these corrections come out on the order of $10^{-5}$ (for the strange quark) to $10^{-8}$ (for up and anti-up).
Dropping $p$ to about $100\text{ GeV}$ does increase them enough to make a difference of about 1%, i.e. 42% quark energy and 10.2% antiquark energy in total. But I wouldn't trust that calculation too much, even less than I'd trust the other one I did, mostly because the integrand in the correction term is singular in $x$ by an extra factor of $x^{-2}$ compared to what I was calculating before, and the contributions of small-$x$ regions in which the PDFs have not been measured (and the assumption $m_i \ll xp$ is not valid) is accordingly enhanced.
