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  Manifest covariant color superconductivity?

+ 1 like - 0 dislike
827 views

@JiaYiyang wrote:

I think color-superconductivity would count. Although the formalism treating it is often not manifestly covariant (which is not surprising since e.g. it's often dealt with in thermal QFT context),  it's still a QCD phenomenon after all.

How can it be manifest? Doesn't the permanent presence of matter always break Lorentz covariance?

asked Jan 6, 2016 in Theoretical Physics by Arnold Neumaier (15,787 points) [ no revision ]

I have not seen a completely manifest covariant treatment. The question is why bother?

@JiaYiyang Well, since you only wrote ''often'' I concluded that you seem to have treatments in mind where covariance is manifest.  

Why bother? Since in the context of your remark that I cited covariance is essential (though not necessarily manifest covariance). But it seems to me that there is no covariance since matter always defines a preferred frame. 

 Well, since you only wrote ''often'' I concluded that you seem to have treatments in mind where covariance is manifest.  

Sorry for the confusion, I wrote "often" simply to be safe, just in case there is actually a covariant treatment out there I haven't seen. 

The difference between color-superconductivity and ordinary electron-superconductivity is that the dynamics is treated relativistically (because it's often high-density that is discussed), so in expressions such as $\langle q q\rangle$, the $q$ is still relativistically quantized. But the thermal state certainly breaks manifest covariance. (Aside: maybe I should also say, another difference is that in color-superductivity there's an obvious channel for pairing, i.e. color-Coulomb interaction, which is a plain QCD effect; while in normal electron superconductivity, the pairing mechanism is much subtler, i.e. phonon exchange. In this sense, color-superconductivity theory is much closer to QCD than electron-superconductivity to QED.)

In any case, we know physically when evaluating (thermal-sense) color-condensate such as $\langle qq \rangle$ there is a contribution from the true vacuum, so the true vacuum vev has to be defined.

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