# Tag Info

## New answers tagged standard-model

0

I totally agree with anna v's answer. Here I will just give some reasoning that you may use to determine the nature. Forbidden If the conservation of $Q$, Lepton number or Baryon number is not obeyed. Weak interaction If their are neutrino's involved, or the quantum numbers $S$, $C$ or $\tilde B$ are not conserved then the interaction weak (given that it ...

7

Life is not so simple, as in all high energy interactions there is a probability of a large number of particles appearing at the main interaction which will subsequently have decays through the weak or electromagnetic interaction. If one sees jets of hadrons in the detectors the strong interaction is involved, but the main vertex may be electromagnetic, ...

1

$$\mathcal{L}_{SM}=\mathcal{L}_{EW(after ~symmetry~breaking)}+\mathcal{L}_{QCD}+\mathcal{L}_{loc}+\mathcal{L}_{gf}$$ $$\mathcal{L}_{EW}=\mathcal{L}_K+\mathcal{L}_N+\mathcal{L}_C+\mathcal{L}_H+\mathcal{L}_{HV}+\mathcal{L}_{WWV}+\mathcal{L}_{WWVV}+\mathcal{L}_Y$$ $$\mathcal{L}_{loc}=\mathcal{L}_{gol}+\mathcal{L}_{int}$$

0

I cannot find the comment you cite in the link you posted. I think the limit on the higgs mass can be easily obtained by just using the fact that $M_h^2 = \lambda v^2, \lambda < \sqrt{4\pi}$ where $\lambda$ is the Higgs self coupling. Thus one gets $M_h^2 < \sqrt{4\pi}v^2 \sim (870 GeV)^2$ The violation of the Ward Identity just requires you to add ...

2

An explicit mass term violates Gauge invariance, because left and right particles belong to different representations 2.At one loop, the lepton mass is given by $m_{1L} = M_{bare} + \Delta M_{1L}(\mu = m_{1L})$ this condition uniquely defines the bare mass. The correction $\Delta M(\mu)$ is anyway proportional to some power of the yukawa, thus is very ...

0

In any theory with chiral fermions where Left and Right fermions don't have the same charges such fermions cannot have an explicit mass therm and can acquire mass only with spontaneous symmetry breaking. In the standard Model all fermions except Right handed neutrinos are charged under the SM gauge group and all L/R counterparts live in different ...

1

Short answer is we don't know, there may be right handed neutrinos and left handed antineutrinos that either don't interact through the weak force or do so extremely less than neutrinos. Sterile neutrinos a theoretical candidate for these particles.

5

It is not possible to fairly explain the origin of masses – and most other things – without "any mathematics" at all. Charged fermions get masses through their cubic "Yukawa" interactions with the Higgs $$L_{Yuk} = y\cdot h \psi_L \psi_R$$ When the Higgs field is $h=v$ in the vacuum, the simple part of this cubic term generates the quadratic term $$m \... 0 Yes, you can in a vector theory. But in the SM you cannot do that because of the chiral structure of the model: l_L \sim \mathbf{2} but e_R\sim (e^c)_L\sim \mathbf{1}. So you need a scalar which transforms as \mathbf{2}. 2 In fact, the Yukawa Lagrangian is (more or less) only the term \mathcal{L}_Y = -g \bar{\psi}\psi \phi. The (massless) Dirac Lagrangian for fermions and Klein-Gordon Lagrangian (plus potential) for the Higgs are not shown in your formula. The main difference between the Yukawa Lagrangian and the simpler -g \bar{\psi}\psi \phi is that the Standard Model ... 1 To answer first to your last question, it is an experimental fact that only left-handed fermions are affected by weak interactions. The couplings of the fermions to the W bosons is proportional to their weak isospin T_3. Therefore, right-handed fermions must have T_3=0. Left-handed fermions are observed to have the same isospin (charged currents ... 0 What you explained is correct, given a chosen language and a formal system there is always a minimal program that generates the rest of the string and you can use that as a measure of complexity. But there is a practical problem to implement this a as form of comparing two theories. It has been shown that it is not computable, that is, there is no program ... 3 There is a recent survey of canonical quantum gravity and its confrontation with exciting experimental data: R.P. Woodard, Perturbative Quantum Gravity Comes of Age, Int. J. Modern Physics D 23 (2014), 1430020. http://arxiv.org/abs/1407.4748. Woodard writes in the introduction: All of the problems that had to be solved for flat space scattering ... 0 Generating CP asymmetry from scattering process is indeed possible. Actually, already in leptogenesis people have considered such processes because they dominate at high energy (T>M_N) but it turns out that they have negligible effects on the final asymmetry (e.g., hep-ph/0309342). A field \psi, which has CP violating interactions with fermions f, ... 2 Part 1: The branch of string theory which actually tries to match experiment is called string phenomenology. The state of the art in string phenomenology is that, starting from different forms of string theory (heterotic string theory, M-theory, F-theory...), it is possible to define space-time geometries, arrangements of branes, background fluxes... such ... 1 I will address the title question: does string theory explain the existence of 3 generations of quarks leptons because of the word "explain". Physics is about measurements and observations and mathematical models which not only fit the measurements and observations but also have predictive power. Otherwise the model is just a map, not a physics theory.... 0 The official string theory website says this: Theoretical physics has not explained why there are three generations of particles that make up matter. Maybe string theory will come up with an answer for this. That's really where it stands. In fact, there's another question on physics SE here, where one of the answers says The question as to why ... 1 It seems like you're fixing the representation of \mathrm{SU}(2) to be T^i=\frac{1}{2}\sigma^{i} (i.e., the fundamental representation). This makes sense if you're talking about the Higgs mechanism. Now you want to find the generator Y for the \mathrm{U}(1) part of \mathrm{SU}(2)\times\mathrm{U}(1). Let's call that generator Y. Now, any ... 2 The de Broglie-Bohm interpretation is a way of putting quantum mechanics in a classical-like format. It is somewhat contrived in a way. It starts with the wave function put in polar form \psi~=~Re^{-iS/\hbar} and the Schrodinger equation for a particle moving in a potential$$ i\hbar\frac{\partial\psi}{\partial t}~=~-\frac{\hbar^2}{2m}\nabla^2\psi~+~V\psi $... 4 The de Broglie-Bohm (dBB) interpretation really only applies to nonrelativistic quantum mechanics. It can handle spin just fine, but isn't well adapted to special relativity, quantum field theory, particle creation or destruction and so on. For example, if you have a situation that is well described by the Schrödinger equation or the Schrödinger-Pauli ... 4 The phase of a single constant$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-... 0 Processes of the type$u\bar u \to s \bar s$are allowed, but suppressed. See their eq.7. However, the processes of the type$uu\to ss$are forbidden, indeed, by the flavor symmetry ($u$and$s\$ have different charges and you can see that they don't get balanced).

Top 50 recent answers are included