# 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 ...

32

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 ...

17

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}\... 13 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 ... 13 The text by Lumo may have been a bit confusing but it's the other way around: the possibility to redefine the phases of the vectors leads to a reduction of independent angles and phases in the CKM matrix, but there's still one complex phase that can't be rotated away. Imagine that you change the phases of the kets$u,c,t;d,s,b$by six multiplicative ... 10 Good question! Regarding (2) baryon number is certainly violated at Planckian energies. If you can make a black hole, you can eat up baryons. Luboš Motl's argument that you linked to is correct in this regard. Whether you can make a believable scenario of quantum gravity driven baryogenesis at the Planck time is up in the air as far as I know. It's the old ... 8 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 Remember that the theta term appears in an exponential$e^{i\theta n}$inside the path integral. If$\theta n$shifts by$2\pi N$, for any integer$N$, the exponential is unchanged, and all path integrals have the same value. The integral$n = \frac{1}{32\pi^2} \int F \wedge F$is not arbitrary either. It's a topological invariant, and it's normalized so ... 6 Surely someone has mulled over why the universe might exhibit such a non-intuitive and thus interesting asymmetry? As always, when it comes to valid and important physical theories, the reason why the Universe has non-intuitive features is simply that the intuition is wrong. Arguments based on wrong intuition are irrational and unscientific. Rationally ... 6 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 ... 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 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 The spin-statistics thing isn't a problem, it is a theorem (a demonstrably valid proposition), and it shouldn't be addressed, it should be understood and celebrated. The Higgs field gives us interactions between chiral fermions and the Higgs,$yh\cdot \chi_\alpha\eta^\alpha$which produces mass terms$m \chi_\alpha\eta^\alpha$if the Higgs field has a ... 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 ... 4 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 ... 4 Cecilia Jarlskog proposed this invariant already in 1973 and it was mentioned in the original Kobayashi-Maskawa paper. For three families, it's easy to see why it is nonzero iff the unitary matrix in$U(3)$can't be brought to the real, orthogonal i.e.$O(3)$form. It's because after the 5 phase redefinitions of the up-type-quark and down-type-quark ... 4 Noether's theorem does not apply to discrete symmetries like C, P, and T. Only continuous symmetries generate local conservation laws. For discrete symmetries you get multiplicate rather than additive conservation laws so they are somewhat less useful. Also note that T is an anti-unitary transformation so it is a little more subtle than the others. On the ... 4 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|...

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.

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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 ...

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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....

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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 ...

3

I realise I'm responding to an old question, but perhaps you're still in need of an answer. I'm going to use $B^0_s \to \phi \phi$ as an example. It is a transition of a pseudoscalar meson to two vector mesons. There are 3 possible configurations for angular momentum, $\ell$: aligned, anti-aligned or perpendicular. In the LHCb angular analysis of this decay,...

3

surely someone has mulled over why the universe might exhibit such a non-intuitive and thus interesting asymmetry? Oh yes, definitely. I have for one (though I haven't made a significant contribution to the question)! :) There are a number of "left-right symmetric" models out there which usually involve a group like $SU(2)_L \times SU(2)_R$ where the \$SU(2)...

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