It is well known that the most of the proton (or any other hadron with light quarks) mass is not made up from quark masses, but it is dynamically generated by QCD mess inside. I've also heard that, even if quarks would be massless, protons (and other hadrons) would still have a nonzero mass.

However, if proton mass does not (in most part) arise from the quark masses, from which dimensionful constants does it arise?

I've heard that proton mass arises from spontaneous symmetry breaking of scale invariance. However, this is a troublesome explanation, or non-explanation at best, because it opens more questions:

If a theory is scale invariant, how can it pick a scale when breaking this symmetry? The proton mass is a constant, so how can the scale invariance be broken across entire universe the same way? Is there a field, very resistant to change, that permeates entire space to ensure the constancy of proton mass?

  • 2
    $\begingroup$ There's a nice overview here (400KB PDF). $\endgroup$ – John Rennie Dec 6 '14 at 8:00
  • $\begingroup$ @JohnRennie, I guess that equation (11) is the closest to answering my question. However, this differential equation still requires a dimensionful integration constant to be solved. It is not explained from which dimensionful constants does this integration constant arise, so my question remains unanswered. I guess the integration constant is $\Lambda_{QCD}$, but from which dimensionful constants does it arise? Or is it itself a fundamental constant? $\endgroup$ – Varin Esan Dec 6 '14 at 8:31
  • $\begingroup$ Also, if $\Lambda_{QCD}$ is a constant incorporated in QCD, then how does it make sense to say that pure QCD (QCD without quark masses) has a scale invariance which gets spontaneously broken? $\endgroup$ – Varin Esan Dec 6 '14 at 8:38
  • $\begingroup$ The point is that once you measure the coupling constant at some scale you can use RG running to find where the coupling becomes nonperturbative. This scale is $\Lambda_{QCD}$. In some sense the boundary condition is what "gives the equations a scale". Much above this scale the physics is insensitive to this value and so it is scale invariant. Below the scale invariance is spontaneously broken. $\endgroup$ – JeffDror Dec 6 '14 at 12:08
  • 2
    $\begingroup$ @JeffDror Is $\Lambda_{QCD}$ a fundamental constant? If so, how can a theory containing it be scale invariant? If not, how does it arise from the fundamental constants? $\endgroup$ – Varin Esan Dec 6 '14 at 16:24

I think its easiest to understand this if one has a minimal understanding of QFT. I'm not sure about your background knowledge but hopefully this isn't gibberish to you.

The QCD Lagrangian for massless quarks is given by, \begin{equation} {\cal L} = - g \sum_i \bar{\psi} _i A _\mu \gamma ^\mu \psi _i - \frac{1}{4} F _{ \mu \nu } F ^{ \mu \nu } \end{equation} where the fields are, $ A _\mu $ and $ \psi _i $. The only constant in the equation is the coupling constant, $g$. Therefore, we see that there is no single scale in the Lagrangian. Naively one would say that the theory is scale invariant.

However, there is a subtlety. We haven't fully specified the theory. We have yet to say what the value of the coupling constant is. The problem is that QFT causes the strength of an interaction to depend on the scale which its measured. Luckily, we know how to calculate how a coupling changes with scale (this is done is every full year QFT course), \begin{equation} \frac{ d \alpha }{ d \log \mu } = - \frac{ b }{ 2\pi } \alpha ^2 \end{equation} where, $ \alpha \equiv g ^2/4 \pi $ and $ b $ are calculable numbers. For QCD with the SM fermions we have, \begin{equation} b = 7 \end{equation} From here its easy to solve the differential equation above and get the coupling as a function of the scale, $ \mu $, \begin{align} \frac{1}{ \alpha ( \mu ) } &= \frac{1}{ \alpha ( \mu _0 ) } - \frac{ b }{ 2\pi } \log \frac{ \mu }{ \mu _0 } \\ \alpha_s (\mu) &= \frac{ \alpha _s ( \mu _0 ) }{ 1 + \alpha _s ( \mu _0 ) \frac{ b }{ 2\pi } \log \frac{ \mu }{ \mu _0 } } \end{align} Therefore, we can measure the coupling at some scale and then know what it is at every scale. As pointed out by the OP, we can already see breaking of scale invariance since the couplings depend on scale.

Now we move on to the relation to $ \Lambda_{QCD} $. This is conventionally defined as the scale where the coupling becomes infinite. From the running above we see this occurs when, \begin{equation} \mu \equiv \Lambda_{QCD} = \mu _0\exp \left[ - \frac{ 2\pi }{ b \alpha _s ( \mu _0 ) } \right] \end{equation}

Here we see that the scale only depends the field content (through $b$) and Natures choice for the coupling.

  • $\begingroup$ "We now fully specified the theory and the theory still appears scale invariant." It doesn't, because the formula for $\alpha(\mu)$ contains a preferred scale. No need to analyze whether the theory becomes nonperturbative to see that. And it still remains unclear whether $\Lambda_{QCD}$ is a fundamental constant or derived from some fundamental constants. $\endgroup$ – Varin Esan Dec 7 '14 at 17:03
  • 1
    $\begingroup$ You are right. I updated answer with hopefully an answer to your second question. $\endgroup$ – JeffDror Dec 7 '14 at 17:31
  • $\begingroup$ Thanks, now it's much clearer. I guess that "Nature's choice for the coupling" should be (more precisely) "Nature's choice for the coupling at a given scale" or equivalently ""Nature's choice for the scale at which the coupling becomes infinite". From that I conclude that in quantized theory this scale becomes fundamental constant instead of the coupling (which isn't even a constant but depends on scale). Is it correct to make that conclusion? $\endgroup$ – Varin Esan Dec 7 '14 at 17:50
  • $\begingroup$ I think its a philosophical question about what Nature chooses. I like to think that Nature chooses a coupling since its in line with how we think of all Lagrangian parameters, however I don't know if there is a right answer. More precisely, I like to think that "the coupling Nature chooses" is the unscreened coupling, i.e., at $\mu \rightarrow \infty $, while the running coupling is just an effective coupling. $\endgroup$ – JeffDror Dec 7 '14 at 18:38
  • $\begingroup$ But your formula for $\alpha(\mu)$ seems to suggest that $\alpha(\mu\rightarrow\infty)=0$. $\endgroup$ – Varin Esan Dec 7 '14 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.