What does the "UV" in "UV completion" stand for?
Also, I'm not sure which tags I should tag this question with.
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9$\begingroup$ The Wikipedia article answers this in its first sentence. $\endgroup$– ACuriousMind ♦Apr 9, 2015 at 15:32
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6$\begingroup$ I'm voting to close this question because it displays insufficient prior research $\endgroup$– John RennieApr 9, 2015 at 15:38
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2 Answers
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UV stands for Ultraviolet and it is referring to a special kind of divergences in quantum field theory. In NLO loop diagrams, we often encounter divergences (infinite integrals) when we investigate what happens at $k \to \infty$, where $k$ is the internal momentum of a virtual particle in Feynman diagrams. These are precisely the divergences we call "UV divergences". A UV divergence is often due to the fact that we assume the theory is valid up to infinite energies, where in reality this is not true (there are new physics at high energy scales which will change the theory).
A "UV complete" theory is a theory which UV divergences do not occur because this theory will be valid at all scales. An example of a UV complete theory is String Theory.
answered Apr 9, 2015 at 15:39
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UV = Ultra-Violet = High energy = Small length-scale.
answered Apr 9, 2015 at 15:30
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1$\begingroup$ Well seeing how there are two (named) energy bands above UV, I thoroughly disagree with assessment that UV is "high energy." That would be X-rays and $\gamma$-rays. $\endgroup$ Apr 9, 2015 at 15:50
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2$\begingroup$ @KyleKanos: It is QFT speak for "high energy", though, just like IR is QFT speak for "low energy". $\endgroup$– ACuriousMind ♦Apr 9, 2015 at 16:28