Tag Info

Hot answers tagged

181

Imagine you are an electron. You have decided you have lived long enough, and wish to decay. What are your options, here? Gell-Mann said that in particle physics, "whatever is not forbidden is mandatory," so if we can identify something you can decay to, you should do that. We'll go to your own rest frame--any decay you can do has to occur in all reference ...


34

The statement is true for decays, where lifetimes can be measured. It is not true for interactions though. A suicidal electron meeting a positron has a good probability to disappear, together with the positron, into two gamma rays, at low energies. Electron-positron annihilation It is intriguing that this is not true for neutrinos. If an electron ...


12

This is not exactly true. It is believed that net charge is conserved, but there is a weak process called electron capture, where an electron is captured by a nucleus, (usually from an inner "orbital" so there is a spectroscopic signature), a neutrino is emitted and a proton changes to a neutron. So therefore your textbook is wrong!


8

Dark matter candidates are all "beyond the standard model" physics, which means that they represent an extension of the model into something more comprehensive. Some of these extensions are pretty minimal (I think both axions and massive, sterile neutrinos are in that category) others are quite comprehensive. Many theorists have favorite models and ...


5

There is no non-trivial one-dimensional representation of $\mathrm{U}(1)$ on a scalar field $\mathbb{R}^4\to\mathbb{R}$, but on complex fields $\mathbb{R}^4\to\mathbb{C}$, we have the one-dimensional "phase" representations by $$\phi\mapsto\mathrm{e}^{e\mathrm{i}\chi}\phi$$ for $e\in\mathbb{Z},\chi\in\mathfrak{u}(1)\cong\mathbb{R}$ for $\mathrm{U}(1)$ ...


5

Yes, that is the Higgs potential of the Standard Model. Note that a $\phi^3$ term is forbidden by symmetry (it would not be an $\mathrm{SU}(2)$ scalar), and $\mathcal{O}(\phi^5)$ terms would be non-renormalizable, so this is really the only potential we can write down that does not need other new physics.


4

This can be explained by thinking about the coupling of fermions to the $SU(2)$ weak gauge field. Let's recap what we know Weyl fermions necessarily appear in two complex representations of the Lorentz group $L$ and $R$. Only fermions in the $L$ representation of the Lorentz group couple to the $SU(2)$ gauge field. CPT is a symmetry of the theory. Now ...


3

You're right that the reaction fails to conserve baryon number. The change in strangeness is a strike against the reaction, but not a fatal one; after all, the strange $K$ mesons decay into various mixtures of zero-strangeness mesons, charged leptons, and neutrinos. The thing to notice is that only the charged weak current, mediated by the $W$ boson, ...


3

Answer to the main question: It is a well regarded fact that the terminology unified electroweak interaction is a bit of an abuse of terminology. What the term means is that both Quantum Field Theories, the Hypercharge ($U(1)_Y$) and Weak ($SU(2)_L$), are unified in a common framework, which predicts the low energy electromagnetism ($U(1)_{em}$) through the ...


2

The string-net condensation is a general construction to obtain gauge fields and fermions. The chiral fermion problem refers to the fact that in the Standard Model (SM), the SU(2) gauge field only couples to the left-handed fermions but not the right-handed fermions. However in the (early version of) string-net condensation, the emergent gauge field will ...


1

All these links are accessible at a non-mathematical level, and they are by recognized scientists (with the exception of the first link). (1) To start, see the "Simple English Wikipedia", which explains what the Higgs effect is, and the reason for the Higgs effect: http://simple.wikipedia.org/wiki/Higgs_field. (2) The difference between the Higgs boson and ...


1

A spin-spin interaction is really a magnetic moment - magnetic moment interaction, where the magnetic moment of each particle is proportional to spin. [Of course, it might be a chromomagnetic moment - chromomagnetic moment interaction if two quarks are interacting, as they are here.] In any case, the interaction term goes like $\vec S_1 \cdot \vec S_2$. ...


1

HyperLuminal asked: "Does that mean that electrons are infinitely stable?" Think about Dirac's model of an electron, which includes left and right handed contributions. Now add the (Nobel-worthy) Brout-Englert-Higgs idea, that the left-handed bit interacts with a condensate of weak hypercharge, while the right-handed bit does not. This suggests a ...


1

What is meant by fractional scaling dimension is exactly what is says: Given a dilatation $x\mapsto\lambda x$, the field/operator $\mathcal{O}(x)$ behaves as $$ \mathcal{O}(\lambda x) = \lambda^h\mathcal{O}(x)$$ with $h\in\mathbb{R}$ a possibly fractional or even irrational number. The prime example of quantum field theories in which a fractional scaling ...


1

Yes and no. Electrons - and all other elementary particles - may be viewed as microstates of very tiny black holes. As one considers increasingly heavy elementary particles (e.g. those in the Hagedorn spectrum of string theory), they increasingly morph into black hole microstates. When the elementary particle masses sufficiently surpass the Planck scale, ...



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