# Tag Info

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To be honest, much of this feels like very irresponsible journalism, partly on the part of the BBC and very much so on the part of Science alert. If you're looking for an accessible resource to what the paper does, the cover piece on APS Physics and the phys.org piece are much more sedate and, I think, much more commensurate with what's actually reported ...

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Well, suppose that you're in the office and you go to the coffee machine. You notice that there is an incredibly tiny puddle of coffee left in the pot. The pot is almost completely empty, but it's not totally empty either. Why is this? One theory is that it could be a complete coincidence -- the pot was going to have some coffee level or other in it, and ...

33

The articles are a little on the hysterical side, but I think they are just saying that violation of CP-symmetry means there must be violation of T-symmetry. T-symmetry means that physical laws are unchanged if we reverse the direction time flows. Classical theories obey T-symmetry, and it seems intuitively obvious that quantum mechanics would as well. But ...

18

If we write the field strength in terms of "electric" and "magnetic" fields $\vec{E}$ and $\vec{B}$, the relevant expression can be written as $$\text{Tr}F_{\mu\nu}\tilde{F}^{\mu\nu}=4\,\text{Tr}\vec{E}\cdot\vec{B}.$$. Under parity transformations, $\vec{E}\rightarrow-\vec{E}$ and $\vec{B}\rightarrow\vec{B}$, while under charge conjugation, $\vec{E}\... 14 The usual action for Yang-Mills theory is, using differential forms $$S = \int \operatorname{tr} (F \wedge \star F)$$ where$\star$is the Hodge dual. Now note that the integral of a differential form is always defined with respect to an orientation, and the Hodge dual is also defined with respect to an orientation. Parity is reversing the orientation, which ... 11 Parity involves a transformation that changes the algebraic sign of the coordinate system. Parity is an important idea in quantum mechanics because the wavefunctions which represent particles can behave in different ways upon transformation of the coordinate system which describes them. Under the parity transformation: The parity transformation changes a ... 11 I'll give a slightly more mathematical answer. First let me expand upon what chirality is all about. Quantum fields transform on specific representations of the Lorentz group. The irreducible representations are known as the$(A,B)$representations and they are labelled by two integers or half-integers$A$and$B$. If you have never seem this, please see ... 9 Short answer: nothing has been seen. Long answer: Questions like this on the experimental limits in particle physics can usually be answered by looking things up in the Particle Data Group's annual Review of Particle Physics. There is a summary online version and an extensive (but free!) print version. EDIT: Here (pdf) is the full section on conservation ... 7 Here is an answer that assumes the question did not require a great deal of care and precision in the response. 1) Topology'' is used because$F \tilde F$integrated over a manifold is a topological quantity. (One of my preferred discussions of this integral is in the second volume of Weinberg's series on QFT.) 2) A susceptibility$\chi$tells us the ... 7 We have the standard model of particle physics, which has been well validated by experiments during the last decades or so, to the point that one can say that it is a convenient encapsulation of all the data. The table of elementary particles which is axiomatic in the theory, has again axiomatically a table with the antiparticles, the theory is symmetric in ... 6 I know this is a year old question, but I am going to attempt an answer. As far as I can tell, this is not really a caveat. The reason for this is that I can always set the overall phase of the quark mass determinant to be zero with a chiral U(1) transformation. For a discussion of this see for example the chapter on theta vacua in Weinberg's QFT book. The ... 6 There is not, because the combined transformation$CPT$is a symmetry of all Lorentz-invariant systems. The$P$-violating decay distribution observed by Wu et al. is also a$C$-violating distribution, because polarized anti-cobalt would have had the opposite sign of asymmetry. (However no one has ever made, or probably will ever make, polarized anti-cobalt,... 6 The phase of a single constant$y$is unphysical and may be completely eliminated by the redefinition $$y \to t\cdot \exp(i\alpha), \quad \phi\to \phi\cdot \exp(-i\alpha)$$ which doesn't change the Lagrangian and for a given$\alpha$,$y$will be real positive. One could also get rid of the phase by transforming the fermions. So there can't be a CP-... 6 You say that$K_0$and$\bar{K}_0$oscillate because while you create a state with definite strangeness (either$K_0$or$\bar{K}_0$from a strong decay) what you observe decaying are mass eigenstates ($K_S$and$K_L$) that are a linear combination of$K_0$and$\bar{K}_0$. What this means is that you start with a state of definite strangeness, you write ... 5 Depending on the way you look at them, there are several kind of kaons:$K^0$and$\overline{K^0}$are the kaons produced by strong interaction. They have a definite isospin and strangeness quantum numbers made respectively of$d\bar{s}$and$\bar{d}s$. However they can oscillate meaning that they transform spontaneously in each other:$K^0 \leftrightarrow \...

5

You understand that $\mathcal C$ violation is required, as if it weren't, processes related by $\mathcal C$ that violated baryon number conservation would balance, i.e. $$P \to Q B \qquad \mathcal C:\qquad \overline P \to \overline Q \, \overline B$$ would result in no net baryon number violation. In these expressions, $B$ is a fermion carrying baryon ...

5

My apologies if this answer is more simplistic that you were looking for, but you did ask why phases are related to CP-violations. Most QFTs are CPT invariant, so a CP-violation is also a T-violation. If you consider an S-matrix element to be the amplitude for a transition from $a$ to $b$, i.e. $\langle a|b\rangle$, then the T-version should be $\langle b|... 5 We have a pretty good idea of the thermal history of the universe. Combining this with the Sakharov criteria for baryogenesis allows one to calculate the necessary CP violation in terms of the strength of the bayon-number violating interaction and how far out of equilibrium the universe was. Taking a purely SM approach and having only sphalerons as baryon-... 5 There's no reason for supposing that matter and antimatter should have been produced in equal quantities in the Big Bang. It's simply what is the most "natural" (inverted commas because this isn't well-defined and different people might find it different things more natural). Compare the related question: what is the total net electric charge of the ... 5 What matter are not the behavior of the dipole moments$\vec{\mu}$and$\vec{d}$themselves, but the Hamiltonian terms$-\vec{\mu}\cdot\vec{B}$and$-\vec{d}\cdot\vec{E}$,* showing the interactions of the moments with external fields. Under parity, the electric field (being a polar vector) changes sign, but under time reversal, it does not. (This can be ... 4 Your question is riddled with ^'s in equations making it hard for me to understand the body of your question. If I understand your question "why is there no weak isospin vacuum angle in analogy with the one in QCD?," then I can answer it easily: Suppose we write that CP-odd term in the Lagrangian. Then, to remove it, all you need to do is to look for a U(... 4 The sentence in Peskin's and Schroeder's book that "the weak interactions preserve CP and T" is a bit misleading but there is a sense in which it is right. Experimentally, CP and T is known to be violated and CPT is always a symmetry. Theoretically, CPT is always a symmetry, too – it's proven by the CPT theorem. The CPT transformation is effectively a ... 4 1) Note that$p^\alpha p^\beta \Pi_{\alpha\beta}$is gauge invariant, but$\Pi_{\alpha\beta}$is not. This implies that$a$and$b$are not separately gauge invariant, only their sum (or difference, in your convention) is. For example, Diakonov and Eides [Sov. Phys. JETP 54 (2), 232-240] write down a spectral representation corresponding to$a=0$, but many ... 4 In the basis for$K^0$there is an oscillation between$K_L$and$K_S$. However, this is not the basis we make measurements in. We measure the decay products, which are the pions, and what is found is there is a probability for the production of$2$or$3$pions that match the lifetime of$K_S$and$K_L$respectively. 4 Short answer (we're hung up on terminology): If you use the word "Leptogenesis" exclusively for theories with lepton violating vertices, then "Dirac Leptogenesis" is a misnomer. If you use the word "Leptogenesis" more broadly, to describe theories where leptonic interactions are important for Baryogenesis, then calling it "Dirac Leptogenesis" is OK. Long ... 4 Anti-neutrinos are produce in copious quantities by nuclear reactors, in the atmospheric interactions of cosmic rays and can be produced on-demand at accelerators. Indeed, given the nature of the interaction of normal and anti-neutrinos it is arguable that anti-neutrinos are easier to deal with experimentally at low energies. It is worth noting that the ... 4 So, now that I've sounded off, T2K has a preliminary finding that contradicts what I said at moderately good confidence. Figures. The free-space oscillation formulae depend on the masses of the neutrino flavors (well, on the differences of the squared masses of the mass states), and anti-particles have the same mass as their normal counterparts, so the ... 4 The operation that maps particles to antiparticles is just$C$. (This is somewhat of a simplification. A better thing to say is that in theories with$C$symmetry, you can pair particle states with the same spacetime quantum numbers but the opposite internal quantum numbers. When$C$is violated, there may exist no pairing that gets the quantum numbers right.... 4 Let$G$be a nonabelian gauge group and$G'$be a global, classical symmetry. Suppose there exists a$G^2 G'$anomaly, which is to say that the triangle diagram with two$G$currents and one$G'$current does not vanish. We say the$G'$symmetry is anomalous. This implies the instantons associated with the gauge group$G$may change the$G'\$ charge. Also, ...

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