# Which arguments for $m_u \approx 0$ are still in the market?

The RPP note on quarks masses has traditionally carried, and it is still there, the comment that

It is particularly important to determine the quark mass ratio mu/md, since there is no strong CP problem if $m_u$ = 0.

But in recent versions they have also added some remarks about how the calculations from the MILC and RBC collaborations show that this mass is no zero.

So, does it still survive some argument for $m_u=0$? And, by the way, does the strong CP problem request exactly 0, or is it enough if it is approximately zero?

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Here is an obscure paper in which the up quark mass is zero but it gets an effective mass: sciencedirect.com/science/article/pii/037026939190435S –  Mitchell Porter Nov 19 '11 at 5:42
There is a very good review by Michael Dine arxiv.org/abs/hep-ph/0011376. If you already did not, I urge you to read it. Anyhow, as you can check from eq.(91) in that review, the measurement of the electric dipole of the neutron grants the smallness of the $\theta$ parameter, making CP violation very near to zero in strong interactions. This, in turn, should grant that $m_u\approx 0$. It is an experimental evidence so far, while a better theoretical understanding would be needed. –  Jon Nov 21 '11 at 9:53
@Jon, you should make that an answer. Seems that the paper captures the essence of the question at hand. –  Larian LeQuella Nov 25 '11 at 15:19
arxiv.org/abs/hep-ph/9403203 is the deepest theoretical discussion I have found. –  Mitchell Porter Oct 26 '12 at 5:57

The question OP is proposing is linked to the question of the mass formulas. Here, what really matters is if the mass of the u quark is indeed very near zero and if one has some compelling theoretical reason to believe this.

The strong CP problem could not be of much help here as pointed out in the Dine's review. The reason is quite simple: If one should have a $\theta$ term into QCD Lagrangian, the neutron would have a measurable electric dipole. From experiments we know that is not the case and a lower bound is fixed. But the electric dipole of the neutron does not depend only from the mass of the quark u and so, having $m_u\approx 0$ is a sufficient condition but not necessarily the right one.

From a theoretical stand point, from QCD sum rules a lower bound for the masses of u and d quarks can be estimated. The main reference is S. Narison, QCD as a Theory of Hadrons (Cambridge University Press, 2007). I report here the estimation given in this book for the sake of completeness (chapter 53 in the book):

$$(m_u+m_d)(2\ GeV)>7\ MeV.$$

This grants a small but yet finite mass and whatever mass formula should satisfy this bound. Of course, this is consistent with $m_u\approx 0$. But a more recent review (see here) gives $m_u\approx 3\ MeV$ that is not so small but it is on the strong interaction scale. Smallness of $m_u$ and $m_d$ masses makes chiral symmetry a very good yet approximate symmetry.

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The block equation there is dimensionally inconsistent. Should there be a radical on the LHS or $\text{MeV}^2$ on the RHS? –  dmckee Jan 18 '13 at 20:29
@dmckee It is perfectly consistent with the quark masses evaluated at 2 GeV. Reference book is the one cited. Please, check it. –  Jon Jan 18 '13 at 20:46
Ah...am I to understand that "(2 \text{ GeV})" indicates the energy at which the masses are evaluated rather than a quantity multiplied by the mass? In that case I retract the objection: I misunderstood the notation. –  dmckee Jan 18 '13 at 20:49