Tag Info

Hot answers tagged

14

Dear Chad, you misinterpret the statement that "the known sources of CP-violation are not enough to explain the matter-antimatter asymmetry in the Universe." You seem to think that the statement means that the known CP-violating parameter (namely the CP-violating phase in the CKM matrix) and the processes based on it are qualitatively insufficient to ...


11

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, ...


8

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 ...


8

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 ...


7

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 ...


7

Presumably you are asking about the communication ambiguity in physics: can we unambiguously specify what we mean by "a right handed coordinate system" to a correspondent far away without a pre-arrnage communications channel (i.e. using SETI)? For a long time the answer seemed to be "no", but the discovery of parity violation in 1957 changed the answer to ...


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 ...


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

Good question! There actually isn't a term for this that I know of. The most common use of such a term would be to classify a particle, for example "the 'polarity' of the electron is matter-polarity," but in that case most physicists would just say "the electron is a matter particle." There is a mathematical operator called the charge conjugation operator, ...


5

1) Does antimatter-matter symmetry exist? Yes there is a CP violation and the whole Nobel prize thing. On the other hand there is CPT symmetry which is very protected. So call it what you want. As for the popsci articles... I would express my thoughts, but this is a family site. 2)Does CP violation explain matter-antimatter imbalance? It's certainly ...


4

A clear recent review of flavor physics, including CP violation, is in the TASI lectures by Gedalia & Perez. The parametrization-independent measure of how much CP violation is present in the Standard Model is called the "Jarlskog invariant"; it's explained in those lectures, but might be a useful keyword if you're searching for other resources. If you ...


4

I haven't read that book, but I did read Feynman's discussion of (sounds like) exactly the same thing. Easy: Tell the aliens how to build a telescope, then describe the configuration of some galaxies near them. OK OK, but suppose we rule that out: We can't see any objects in common. Easy: Send them circularly-polarized radio waves (thanks @Anonymous Coward). ...


4

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 ...


3

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

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 ...


3

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 ...


3

To do this, the man needs to build a particle accelerator and measure Kaon decays, or some other process involving higher quark flavors. Everything else is CP invariant, so he wouldn't know for sure.


3

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 ...


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 ...


3

No, it's not true. Suppose I'm floating in outer space (presumably in a space suit or something else to keep me alive). I'm still me, and I still know that, for example, my left hand is the one on the left, and my right hand is the one I can write with. Even on Earth, we don't need environmental clues to distinguish left from right; it's more a matter of ...


3

As dmckee wrote, the term "symmetry" has a fully uniform meaning. It is not used ambiguously in any way and for the same reason, it is not overused. Symmetries are really important in physics and that's why they're used so often. (We also use "symmetries" with various well-defined adjectives such as "global", "local/gauge", "approximate", "broken", ...


3

The question OP is proposing is linked to the question of the mass formulas. Here, what really matters is if the mass of the u quark is indeed very near zero and if one has some compelling theoretical reason to believe this. The strong CP problem could not be of much help here as pointed out in the Dine's review. The reason is quite simple: If one should ...


3

I'd like to point out that there is a small probability that the assumption on which the question is based: "As I hope is obvious to everyone reading this, the universe contains more matter than antimatter," may not be true, depending on the result of the Aegis experiment at CERN. That's because, as Professor Orzel stated in his answer to this ...


2

In reply to the second partenthetical question, I wrote that matter created from energy in particle physics experiments is "generally" in the form of particle-antiparticle pairs . This is too restrictive. Quantum numbers have to be conserved, and they are conserved in pair production, but there can also be associated production of mesons etc: For example ...


2

Time-reversal operator is anti-unitary, meaning, basically, that for any c-number $a$: $$T\,a\,T^{-1} = a^*$$ Now, If you have a T-invariant Lagrangian term ${\cal L}_{term}$: $$T{\cal L}_{term}T^{-1}={\cal L}_{term}$$ Then if multiply in by $a$: $$ Ta{\cal L}_{term}T^{-1}=TaT^{-1}T{\cal L}_{term}T^{-1}=a^*{\cal L}_{term}$$ So you need $a$ to be real if you ...


2

CPT is a general theorem of quantum field theories: Specifically, the CPT theorem states that any Lorentz invariant local quantum field theory with a Hermitian Hamiltonian must have CPT symmetry. Questioning CPT invariance is questioning the foundations of modern physics theory, which is probably the reason you cannot find anything on this. The label ...


2

In the Standard Model, the lepton sector does not have CP violating couplings (at tree level). The quark sector however has CP-violating couplings (through the CKM matrix). The PMNS matrix (describing neutrino mixing), may have a complex phase (implying CP violation). Whether it has a nonzero phase or not remains to be tested experimentally. This is ...


2

No. Conservation of energy is generated by the continuous time translation symmetry $t \rightarrow t + \epsilon$. This is a differenty symmetry than the discrete time reversal symmetry $t \rightarrow -t$. Violating the latter symmetry does not mean that you violate the former symmetry.


2

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible