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

+ 6 like - 0 dislike
3906 views

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?

This post has been migrated from (A51.SE) asked Nov 19, 2011
retagged Apr 19, 2014
Here is an obscure paper in which the up quark mass is zero but it gets an effective mass: http://www.sciencedirect.com/science/article/pii/037026939190435S

This post has been migrated from (A51.SE)
There is a very good review by Michael Dine http://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.

This post has been migrated from (A51.SE)
@Jon, you should make that an answer. Seems that the paper captures the essence of the question at hand.

This post has been migrated from (A51.SE)

+ 3 like - 0 dislike

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.

This post has been migrated from (A51.SE)
answered Nov 25, 2011 by (345 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.