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19

Massless photon Photons interact with the "Higgs doublet" but they don't interact with the "ordinary" component of the Higgs field whose excitations are the Higgs bosons. The reason is that the Higgs vacuum expectation value is only nonzero for the component of the Higgs field whose total electric charge, $Q=Y+T_3$ where $Y$ is the hypercharge and $T_3$ is ...


18

The show you watched seems to get two concepts mixed up: Supersymmetry and Dark Matter. The existence of Dark Matter is strongly hinted at by comsological and astrophysical considerations. It is the easiest explanation for several observations we make in the universe. Supersymmetry on the other hand provides a candidate particle. The lightest ...


9

Just conserve angular momentum. If I have two photons on a collision course, their spin can either be aligned or anti-aligned, since photons must have spins lying on the same plane as their motion by virtue of their masslessness. Then, you can either add one to one to get two, or you can subtract one from one to get zero. If you have a decay to two ...


8

Yes there are "virtual" Higgs bosons. A virtual particle isn't really a particle but a ripple / disturbance in a field. So a virtual electron is a ripple in the electron field. A virtual higgs is a ripple in the higgs field. Virtual particles are just a convenient conceptual model for describing field disturbances in terms of particles. Matt Strassler ...


7

The difficulty with Higgs boson is it's high mass, so in order to create it, you need lots of energy (125GeV, using $E=mc^2$). What is important to give particles mass is s the Higgs field, not the Higgs boson (which is an excitation of the field). The problem is that you have mixed the concept of real particles and "virtual" or "force carrier" ...


7

In few words: All the data gathered on particle physics can be beautifully classified in what is called the Standard Model. It is based on group symmetries in the behavior of particles, three groups two of them special unitary groups and one a simple unitary group. SU(3) x SU(2) x U(1) strong weak electromagnetic each group representing ...


7

It's certainly possible for a particle's mass to come partially from kinetic energy of massless particles; for example, about half of a proton's mass is the kinetic energy of its gluons. But the kind of mass that fundamental particles have, the kind that comes from the Higgs mechanism, doesn't appear to be of that kind. Maybe someday we will discover that it ...


7

Courtesy of the question Bound State of Only Massless Particles? Follows a Time-Like Trajectory? the answer to your question is no. If you take the example of a glueball formed from two gluons, although the gluons are massless the glueball has a rest mass. In his answer to the above question Ben Crowell argues that the glueball must move on a timelike ...


7

Linear terms can be thought as source terms. They are important to define the effective potential (which is the Legendre transform of the (log of) the partition function with respect to the source). I'm not sure why one would say that one can forget about them, since, for instance, they imply a non zero value of $\langle \phi\rangle$ even in the symmetric ...


6

It is important to distinguish between the Higgs boson and the Higgs field. The Higgs field is present in all of space and it has a nonzero value everywhere, because that's the lowest-energy configuration, whereas the Higgs boson is an excitation of this field which takes quite a bit of energy to get going. I'm definitely no expert, but here goes a very ...


5

There are two considerations: a) if you follow a bit the current theories, the manifestation of the Higgs decays can vary, depending on the parameters of the models, and there are many models. b) Hadron colliders, in contrast to e+e- ones are "dirty", there are enormous backgrounds that have to be understood, and the few clear decay channels proposed by ...


5

The full answer is unknown, i.e. nobody can tell you why the electron is only 0.00484... times as heavy as the muon. In fact, all interactions of massive particles (leptons) with the Higgs are of the same form (called a 'Yukawa interaction'), but for every particle, there is a different constant of proportionality ('Yukawa coupling constant'). For ...


5

That really depends on what you call necessary. If you completely forget all about $SU(2)_L$ (say, in an alternate universe with no Weak Interactions). Then mass terms in the Lagrangian for quarks and leptons are not forbidden by any symmetry and you would not need the Higgs field to generate the mass of the quarks or of the electron. Now, in OUR ...


5

It's a scenario that has heavy scalars and relatively light gauginos, so it's one example of a class of "split SUSY" or "mini-split SUSY" scenarios that have survived most of the constraints. In this kind of scenario, collider bounds put the lightest superpartners, namely the winos, above about 270 GeV. Gluinos are constrained to be somewhere north of a TeV, ...


5

This question is similar to "where is the electromagnetic field?" And the answer is: the electromagnetic field is everywhere; it exists at every point in space-time, but it simply happens that its average value is zero (or close to zero) at points far away from charges, currents, and waves. The Higgs field, like the electromagnetic field, is a quantum ...


5

An obvious difference between the two ways of thinking about it you mention is that in the case of the Higgs mechanism, there is an observable particle excitation of the field associated with it, which was found recently. Furthermore it should be noted that the Higgs mechanism only concerns the mass generation of some elementary particles. The mass of ...


4

The Standard Model predicts that Higgs Bosons could be formed in a number of ways but the probability of producing a Higgs boson in any collision is always expected to be very small. If you make some assumptions you can estimate the rate. So based on the production rate at the LHC operating at 7 TeV. The total cross-section for producing a Higgs boson at ...


4

The massless photon: The zero mass is not due to a special value of the Weinberg angle, the angle which determines the mass of the other three bosons $W^+$, $W^-$ and $Z$ The mass is zero because the vacuum expectation value of the Higgs field doublet is single valued rather than two valued. This means it can in principle always be expressed by. $\langle ...


4

The decay channels were discussed here: Shape of the Higgs branching ratio to ZZ Approximately, the Higgs decays to the heaviest pair of particles that is still consistent with energy conservation. The precise rates are calculable by the Standard Model - if you substitute the so-far unknown Higgs boson mass as input. Of course, if the world isn't ...


4

The model – so far compatible with everything you know – isn't just an empirical model and it won't be found in 2015. It was already found in the 1960s, with the QCD completion in the early 1970s, and it's called the Standard Model; it is the most complete realistic example of a quantum field theory (or, using a more specific classification, a gauge theory ...


4

You are right, in this case, scalar means Lorentz invariant field. But it is not invariant under the transformations of SU(2)xU(1) of the electroweak model. And it is a scalar under the SU(3) of QCD. So the four real components of the Higgs are indeed invariant under space-time transformations. Physicists are usually not very clear in these distinctions, ...


4

Gauge Bosons Mass terms for any gauge bosons are forbidden since they are not invariant under gauge transformations. Suppose you have some symmetry $ SU(N ) $ with generators $ T ^a $. To be a symmetry there must be a set of gauge bosons which I denote $B _ \mu ^a $. The mass terms for these bosons are \begin{equation} - m ^2 B _ \mu ^a B ^{a, \mu} ...


4

They can't be the same thing. As Wikipedia says, it's possible to calculate certain properties of glueballs from QCD, including their masses, and the masses don't come close to what we've observed for the Higgs boson. Also, the Higgs doesn't have color charge, so it doesn't interact with gluons, whereas a glueball would. That would make a large difference in ...


4

Linear terms are important. But in a Poincare covariant QFT, one can always remove them by shifting the field by a constant computed as a stationary point of the Lagrangian. If there is only one stationary point, it must be a minimizer (to have the energy bounded below), then this gives a unique normal form without linear terms. If there are multiple ...


4

First, to be clear on what the graph is showing: as a function of the possible mass of the Higgs, it plots the fraction of Higgs bosons that will decay via each individual channel. Before we knew the mass of the Higgs boson, a plot like this one was useful for identifying the best channels to look at to detect the Higgs in various mass ranges. For example, ...


3

We have the functional of the external source $J$, which gives us v.e.v.s of field operators, by functional differentiation: $$e^{-iE[J]} = \int {\cal{D}}\phi\, e^{iS[\phi]+iJ\phi} $$ $$\phi_{cl}=\langle\phi\rangle_J = -\frac{\delta E}{\delta J}$$ Where $\langle\phi\rangle_J$ is the v.e.v of $\phi$ in presence of external source $J$. That could be ...


3

Yes, it can. But the probability is so small that it's barely worth considering, and that's why you won't see it in most plots of Higgs branching ratios (the relative probabilities of decaying into different kinds of particles), nor mentioned in lists of what the Higgs can decay into. Roughly, the probability of the Higgs boson decaying into a particular ...


3

There is an aspect to this question that nobody seems to have addressed and that is, although the higgs (the 'radial' component of the field) is neutral, and therefore doesn't interact with the photon at 'tree level' we still see the decay $h \rightarrow \gamma \gamma$. This is because, roughly, by quantum effects a higgs will fluctuate into a particle/ ...


3

This is a very interesting question. It turns out that in the standard model before symmetry breaking the only particle with a mass term is the Higgs boson itself (and possibly the neutrinos) The fermions and weak gauge bosons acquire mass through the Higgs mechanism and more mass in nature is generated by the confinement of quarks inside the nucleons, but ...


3

The first part of your question is difficult to answer because it is unclear what you mean by "potential energy functions." Depending on what you mean at the moment there could be one for all the particles, or one for each particle, or really any partition of the particles into subsets. "Potential energy function" is really just a way of talking about the ...



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