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comment What is the difference between leptons and baryons?
When cosmologists talk about "baryons", what they really mean (usually) is all standard model particles i.e. everything that isn't dark matter. For part of that source, they use that terminology, and the other part uses the more precise particle physics terminology in which a baryon is a composite state of 3 quarks. It's understandable that this would lead to confusion.
Dec
17
answered How far can light go?
Oct
17
comment Hopf Algebras in Quantum Groups
The keyword to look for here is Tannaka-Krein duality (e.g. on nLab), a natural extension of Pontryagin duality. Perhaps someone else here is capable of giving some simple explanation of it, but I don't think I can do better than what's already on nLab or Wikipedia.
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Jul
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comment Is speed of light and sound rational or irrational in nature?
I'm not sure in what sense a system with $c=\pi$ would have "very complicated and perhaps even inconsistent behavior under Lorentz transformations". A Lorentz transformation is simply a linear map on $\mathbb R^4$ preserving the quadratic form $(ct)^2-x^2-y^2-z^2$. Even leaving $c$ as a formal parameter, the theory is exactly what we teach in introductory courses. One can just as easily rescale $t$ such that $c=1$ or $c=\pi$ or any other positive real number; the same theorems in dimensional analysis ensure these are all equiconsistent.
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Feb
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comment How can stars make up 0.5% of whole universe?
Minor terminology quibble: neutrinos aren't baryonic matter. They are, however, ordinary matter.
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Dec
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comment Operator-state correspondence in QFT
I'm not saying that we never do things like this in physics, but that it isn't exactly what is meant by "local" (at least to me). For example, in the standard Penrose diagram for Minkowski space, past timelike infinity is mapped to a single point, but if you want to talk about local processes occurring in a neighborhood of that point you really have to blow it up to resolve that. There's also a technical issue as to in what sense the limits converge. On AdS this would be a whole different story of course, but in flat space it doesn't make much sense to regard past infinity as a single point.
Dec
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comment Operator-state correspondence in QFT
@Axion I agree with Prahar's comment. Perhaps another way to say this is that in a CFT, 0 is literally a point, in that we can compute correlation functions between fields at 0 and other points. In an ordinary QFT, past infinity isn't such a thing. The idea of contracting past infinity to a single point seems inherently nonlocal, in that if I send two wave packets back in time in opposite directions in flat space, I expect them to be getting farther away from each other, not converging to the same point...