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4

Yes, massive particles such as W-bosons, Z-bosons, quarks, and leptons couple to the Higgs field via the cubic (Yukawa) interaction, so they may also exchange the virtual Higgs. Yes, because the virtual particle is massive, one gets the Yukawa potential that includes the exponential dumping with distance. This "Higgs force" is much less fundamental and ...


3

In the Standard Model, electric charge $Q$ is actually part weak hypercharge $Y_W$ and part weak isospin $T_3$ $$Q = T_3 + \frac{Y_W}{2}$$ which can be either positive, zero (electrically neutral), or negative. In this framework, that's it. If, in fact, there is another type of electric charge (and its associated anti-charge), I believe it would be the ...


17

No, there are only positive and negative charges. Or, more carefully stated, if there is a another type of charge, then electromagnetism is not what we are currently thinking it is.1 Electromagnetism is a $\mathrm{U}(1)$-gauge theory, which relies on introducing the covariant derivative $$ D_\mu = \partial_\mu - \frac{e}{\hbar}A_\mu$$ acting upon matter ...


2

Mathematically, electric charge current 4-vector conservation refers to the invariance of theory under U(1) transformations, so there aren't different types of electric charge (like in SU(n) theories) excepting the usual plus-minus. Moreover, the fact of conservation of physical quantity means that corresponding operator commutes with hamiltonian which is ...


2

Look at each vertex independently. The vertex including the photon also involves a hadron, so the exchange is going to be strong mediated. The eta carries certain quark flavors. These can-not have come from the photon, so they came from the exchange particle. You have to conserve angular momentum between the initial and final states, whic may involve a ...


1

For me, the most clear reference on the optical theorem and on Cutting rules is chapter 7.3 of Peskin & Schroeder.


1

Have a look here, and here. cutting is essentially a shortcut for calculating complicated diagrams.


1

You should take it completely literally. (Quibbles about the Higgs field vs the Higgs boson are misguided. Particles don't acquire masses until the point at which the Higgs boson appears, so attributing the particle masses to the Higgs boson is just as correct.) However, there is a simple way to picture this. The concept of a Higgs boson is completely ...


0

By no means not a complete answer, more a criticism of @luksen’s one. It is posted here because the text is too long to fit in the comment field. First of all, the spin is not a well-defined concept for composite particles. More precisely, whether the spin of a particle is defined depends on how the “particle” is defined. Look at an atom: it has the nucleus ...


9

"Binding a massless particle into a small space" is a good phrase for a popular discussion, but it is not the only way to picture the Higgs mechanism. Another perspective comes from the fact that every particle inside some interaction field behaves exactly like its energy or momentum has changed. This concept is called canonical momentum, in contrast to the ...


1

The plots are "expected from background" (thin line) vs "observed" (thick line); the horizontal axis is energy (in GeV), with a peak at 125 GeV. On the left is the raw data - the frequency with which certain energies were observed (note it's a log axis); the plot on the right is the "statistical significance" in standard deviations. The peak is at 5 sigma - ...


1

Actually, no. The original Dirac's concept was the annihilation in the literal meaning of that word. According to that concept, anti-particle was a particle which has anti-energy, that is, negative energy. But, after discovery of positron, and when collision of it and electron was performed, and showed that the result are two gamma-photons (when slow ...


3

A standard simple answer (for the standard Higgs boson field) is that a particle acquires mass by passing through this field, which changes the particle's inertia (thus appearing as acquiring mass which is a measure of inertia among others) Of course the standard Higgs boson is still investigated (if it is the standard one and not some variation of other ...


15

Short answer: do not take it literally, without further context. In order to understand the Higgs boson's role in the Standard model, it is necessary to take a closer look at the framework in which we describe elementary particles: quantum field theory. In this approach, particles are described as excitations of fields that spans all spacetime. The ground ...


96

The Higgs field (note it is the field that is important here, not the Higgs boson itself, which is just a ripple in the Higgs field) gives particles mass in the same sense that the strong force gives the proton mass (context: $99\%$ of the mass of the proton comes not from the mass of its constituent quarks, but from the fact that roughly speaking the quarks ...


0

If you have two terms then you would have two vertices which contribute to a certain graph. For your first term, as far as I can tell, you would have a vertex for $\phi_i+\chi_j\to\phi_k+\chi_l$ given by $\propto g_3^2(\lambda^a)_{ik}(\lambda^a)_{jl}$. The general recipe to derive the Feynman rules is to feed your Lagrangian into the path integral and just ...


2

One cannot ignore quantum mechanics when one is talking of photons. Photons, in addition to giving rise to transitions of bound states or being emitted by changes in energy levels of bound states ( atom, molecule,lattice) can also interact elastically with electrons and the fields that collectively build up around matter. Now a liquid which has a solute and ...


1

Photons do not get reflected from molecules in a solvent. Typically a photon will be absorbed and will excite one of the electrons in the molecule into a higher energy state. In a liquid this energy is usually transferred to other molecules by collisisons, so no photon is re-emitted. The result is that the absorbed photon just disappears. There are ...


0

For massive particles the intuition of thinking of spin as a rotation is correct. In the rest frame, a massive ($M^2>0$) particle has momentum $$p_0^\mu=(M,0,0,0).$$ Remember that the quantity $p^2=p_\mu p^\mu$, for arbitrary $p_\mu$ is invariant under Lorentz transformations. In the case above the subgroup of the Lorentz group leaving $p_0^\mu$ ...


1

The best way to understand spin is actually to consider the Dirac Equation $$ i\hbar \frac{\partial }{\partial t}\Psi=\left[c\sum_i{\alpha_i p_i}+mc^2\beta\right]\Psi $$ or more compactly: $$(i\gamma^\mu\partial_{\mu}-m)\psi=0$$ The solutions to the Dirac equation are collections of complex valued fields called spinors. The spinor solution actually ...


0

Sorry, but I don't understand what you mean when you say that quarks have a structure. What we know is that quarks are point-like particle, like was confirmed from SLAC years ago. Quarks aren't so much different from Leptons, unless they are "colored".


1

The Feynman diagram is completely analogous to the Feynman diagram of the decay of neutron (beta-decay): The muon $\mu^-$ splits to a muon neutrino (there is a mistake on top, it should say $\nu\mu$, not $\nu_e$, sorry) and the $W^-$-boson, and the latter splits to $e^-$ and $\bar \nu_e$. The word "decay" doesn't mean that the original decaying particle ...


6

Elementary particles have quantum mechanical spin. This induces a spin magnetic moment, independent of the presence (or, indeed, absence) of a (net) electric charge. This is how the neutron attains its magnetic moment (as you already mentioned). The case of the neutrino magnetic moment is slightly confusing, as they are not completely understood yet. ...


-1

Imagine this sheet of paper is the known universe: The paper is the known universe. The two dots on the sheet are atoms or particles. The empty space on the paper is space/time. Space/time is continuously expanding from every point on the paper (indicated by the arrows), except for those points where a particle or atom exists. Because space/time is ...


0

For what it's worth, I've always had the same feeling that the spin should have some sort of reason behind it. It seems so unsatisfying to be told more or less that "it just came that way." Is there any deeper sort of explanation at all? I recall a paper in AJP from years ago called "What is spin?" by Ohanian, but I didn't put in the effort to follow it. I ...


0

A function can be interpretable as a scattering amplitude if that function satisfies the axioms of relativistic S-matrix theory [1]: Lorentz invariance Unitarity (Not realized by the beta function, but may be dropped if the function is interpreted as a Born approximation to the exact amplitude) T, C, P invariance (only for strong nucl. interactions) ...


0

I've come to the conclusion that the language used in the paper is probably not completely accurate. The paper mentions determining displacement based on "reflection time." I believe that the device actually uses trigonometry and the angle formed by the bounced laser beam to determine the displacement of the torsion pendulum. The angle is measured using a ...


0

Can you cite the paper, please? Assuming that it's a modern version of an old experiment, my first guess would be, that the observation uses the fact that the light will induce a constant moment on the torsion pendulum. The response of the pendulum will be an oscillatory motion, for short amounts of time (seconds to minutes), that motion around the original ...


1

I am very ignorant here, but.. Surely the question is whether you want to do statistical mechanics or not. Finite temperatures will allow you to ignore the energy sources - but still enable you to calculate the production of particles. I don't see how you would ever use it for scattering problems involving small numbers of particles. On Noldig's answer: ...


2

This answer is mostly rephrasing Trimok's correct answer in other words. The super-Poincare group is supposed to be an extension of the Poincare group, which contains the Lorentz group and translations. We will complexify the Lorentz group. The Lie group $G:=SL(2,\mathbb{C})\times SL(2,\mathbb{C})$ is (isomorhic to the double cover of) the complexified ...


2

It is maybe simpler to consider all the generators as representations of $SL(2,C)$, so, using spinor indices, you will have : $M^{\alpha \dot \alpha \beta \dot \beta}, P^{\beta \dot \beta}, Q_\alpha, \bar Q^\dot\beta$ Indices are raised and lowered with the Levi-Civita symbols $\epsilon_{\alpha \beta}, \epsilon^{\alpha \beta},\epsilon_{\dot \alpha \dot ...


4

You have to check if the temperature is small compared to the chemical potential. In heavy ion collisions, the chemical potential is rather low but the temperature is very high, so one have to stick with thermal field theory. In compact stars the densities and therefore the chemical potential is at the MeV scale whereas the temperature is at the keV scale so ...


2

The algorithms used are as many as the experimental setups times the detectors used in the setups. They are built to fit the detectors and not the other way around. The common aspects are a few 1)charged particles interact with matter ionizing it and one builds detectors where the passage of an ionizing particle can be recorded. It can be a bubble ...


0

The actual effective collision energy of the LHC is less than the beam energy, because the machine is not really colliding the protons, but only their constituents, the quarks and gluons. Imagine you are shooting two shotguns at the same point. Instead of the two shells hitting each other, you will, at most, get occasional collisions of the pellets that are ...


1

Well, if you have the time... CERN has all the technical design reports for its detectors online at http://cds.cern.ch/. They are excellent reading material. Start with a search for "ATLAS technical design report" and "CMS technical design report" and work your way trough the references in those documents. Once you understand the geometry of the detectors ...


-1

The new constant U 2.30x10 -28 J-m proves that the only diference between hydrogen atom and neutronc lays only in the distance r between the elctron and proton . Dividing constant U by the energy of hydrogen atom gives exactly the radius of hydrogen atom in any of its states , while dividing the same constant U by the nuclear energy one Mev = ...


4

The Higgs gives mass to other fields because it couples to them, producing terms in the QFT Lagrangian that look like what you ordinarily would call a mass term, without us having ever to write down an explicit mass term. Thus, the Higgs "gives rise" to the field/particle species property "mass", since it allows us to get massive theories without having to ...


1

As far as we know, there is no particle (or field) that gives rise to energy in a way analogous to the Higgs mechanism. Energy is a conserved quantity - there is a symmetry associated with it. Now mass is equivalent to energy, as per Einstein's equation $E=mc^2$, but not all particles have (rest) mass. On the other hand, every particle has some energy ...


1

John Rennie's answer is good, but I'll try to explain intuitively why the symmetry breaking leaves some symmetry unbroken. Start with a sphere. You can rotate a sphere in three independent ways—around the x axis, around the y axis, and around the z axis, if you like. All of these are symmetries of the sphere, i.e., they leave the sphere unchanged. These ...


2

Elementary particles are conveniently divided into the fermions and the gauge bosons. The fermions are what we think of as matter, e.g. protons (i.e. quarks) and electrons, while the gauge bosons provide the forces that act between the particles of matter. Fermions get their mass from an interaction with the Higgs field called a Yukawa coupling, and the ...


2

The Particle Data Book, published by the Particle Data Group is probably the nearest to what you're asking for. The actual book is a massive tome costing a fortune. I'm not sure how much of and in what form it's online, but some intrepid Googling should find you most of what you want. As fqq points out the book is available from Phys. Rev. D vol. 86 Issue 1 ...


4

Scientists calculated this knowing that in the time between the Big Bang and the Era of Recombination there was a large 'soup' of superheated particles, which cooled down as the universe was expanding. Atoms could not form because every time an electron tried to 'orbit' a proton it was knocked out of orbit by a high energy photon. These high energy photons ...


10

The calculation is described in detail in the Wikipedia article on recombination. If you consider the ionisation of hydrogen as a reaction: $$ p + e \rightarrow H + \gamma $$ Then you can write down an expression for the equilibrium constant as a function of temperature using the Saha equation: $$ \frac{n_pn_e}{n_H} = \left( \frac{m_ek_BT}{2\pi\hbar^2} ...


3

It is a non-perturbative effect because it is 1-loop exact. The triangle diagram is actually the least insightful method to think about this, in my opinion. The core of the matter is the anomaly of the chiral symmetry, which you can also, for example, calculate by the Fujikawa method examining the change of the path integral measure under the chiral ...


0

An additional means of distinction between the two is known as the Glashow resonance. The process occurs when a high energy ($E\sim6.3$ PeV) antielectron neutrino hits an electron at rest in the experiment and creates an on-shell $W^-$ boson. Since the earth (and pretty much everything we know) is only made up of matter and not antimatter, this process ...


4

You may decompose (irreductible) representation of groups as a sum of (irreductible)representations of subgroups. Starting from a traceless symmetric irreductible representation of $SO(D-1)$: $$R_{ij} = \frac{1}{2} (v^iv^j+v^jv^i) - \frac{1}{D-1} \delta_{ij} ( \sum\limits_{i=1}^{D-1} v_k v_k),\, \text{with} \, (i,j) \, \text{in} \,[1, D-1] \tag{1}$$ ...


1

A picture is worth a thousand words. Three-jet event. Electronic display of an electron-positron collision in the ALEPH detector at CERN, the European particle physics laboratory near Geneva. The display shows a cross-section of the detector, with the beam tube in the centre (blue) surrounded by various detector components (blue and red). The electron ...


3

Usually the word "annihilation" is reserved for interactions whose inputs are two particles that are each other's antiparticle, and your example is not annihilation by that definition. But you can have an interaction where two different quarks go in and something else comes out, as ACuriousMind already mentioned. We happen to have a special word for ...


3

Well, you can just take the ordinary quark/anti-quark/gluon vertex for that and get a red/anti-green gluon (which will then decay/hadronize further, since it is color-charged and thus confined).


1

A particle interpretation of QFT answers most intuitively what happens in particle scattering experiments and why we seem to detect particle trajectories. Moreover, it would explain most naturally why particle talk appears almost unavoidable. [My italics: the answer to your question.] http://plato.stanford.edu/entries/quantum-field-theory/#TakSto The ...



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