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?

  • 3
    $\begingroup$ Hi, the keyword you may want to search is "parton distribution functions" (acronym PDF; too many references). They are functions telling you how many quarks, antiquarks, and gluons with various momenta are included in the proton or neutron. Because $p$ and $n$ differ by one quark, $u$ and $d$, and because the mass of these two quarks differs - 4 or 7 MeV and equally for the antiquarks, it follows that the heavier antiquark (anti d) will start to disappear faster etc. $\endgroup$ Commented Aug 21, 2011 at 6:14
  • $\begingroup$ Off the top of my head I would direct you to E866 (NuSea) at Fermilab as one fairly recent experiment, though most of the PDFs published are from combining decades of data from many different experiments. $\endgroup$ Commented Aug 21, 2011 at 15:08
  • $\begingroup$ durpdg.dur.ac.uk/HEPDATA/PDF is one of the recommended sources for global PDF fits. I'll expand this into an answer later. $\endgroup$
    – David Z
    Commented Aug 21, 2011 at 22:14
  • $\begingroup$ There is also the point that gluons are their own antiparticle, and glue contributes most of the mass. $\endgroup$
    – Ron Maimon
    Commented Sep 3, 2011 at 23:51

2 Answers 2


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.

  • $\begingroup$ I also realized now that Lubos wasn't talking about the antimatter composition of the proton and neutron, but about their difference. This difference would be zero if isospin were exact, as Zaslavski said, and is therefore much smaller than just the pure antiquark contribution to the proton mass. But all this is silliness, there is certainly a difference of 1 part in 10^{-6} and this is enough for his argument to work. $\endgroup$
    – Ron Maimon
    Commented Sep 10, 2011 at 3:46
  • $\begingroup$ I assume that as well as gluons, a considerable amount is also carried as kinetic energy of the various particles. I would accept these two answers, but would like to first ask some "ab initio" people for their results, which may be better because of all the nonlinearities at the low quark energies involved. By the way, how low is that in nuclear matter at rest? $\endgroup$ Commented Sep 10, 2011 at 18:32
  • $\begingroup$ @sigoldberg1: How do you split the quark part of the energy from the gluon part? The quarks carry glue fields. The easiest way is to ask what is the dependence of the proton mass on the quark mass, and that is small, and can be understood from changes in the pion condensate. But I don't know if this is good enough, because it doesn't get the antiquark contribution, but the quarks are relativistic, so they are antiquarks about 1/2 the time. The question itself has no relativistically invariant meaning, as is good, because "antimatter falls up" violates relativity (and thermodynamics). $\endgroup$
    – Ron Maimon
    Commented Sep 10, 2011 at 20:31

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.


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