# Tag Info

## Hot answers tagged beyond-the-standard-model

62

This seems rather incredible that these two seemingly conflicting announcements come on the same day. The pre-print for the Nature paper by the BMW group was placed on arXiv in 2020 around the same time as the muon g-2 Theory Initiative paper (submitted on 8 Jun 2020 and last revised 13 Nov 2020, published here) that the Fermilab collaboration referenced in ...

56

There are some standing anomalies that could be explained by non-gravitational dark matter interactions. For example, Fermi-LAT is an indirect detection experiment (i.e. an experiment that looks for the debris of a dark matter decay that occurred far from Earth), and it currently reports an excess of gamma rays. There are occasional claims that nontrivial ...

42

It's not detecting the particles that is hard, it's making them in the collisions. Although the LHC collision energy is 14TeV, collisions aren't between the protons but rather between individual quarks inside the protons. Since the energy is shared between the three quarks in a proton the actual quark-quark collision energy is a lot less than 14TeV. Even ...

36

One of the searches performed at the LHC consists in selecting events in which two high energy photons are produced ($\gamma\gamma$ channel) and in computing their invariant mass - the energy of the photon pair in its rest frame $m_{\gamma\gamma}$ - distribution. The standard-model predicts these events to be fairly common (mainly produced by direct QCD ...

30

Measuring $w$ is actually what I do for a living. The current best measurements put $w$ at $-1$ but with an uncertainty of $5\%$, so there's a little room for $w \ne -1$ models, but it's not big and getting smaller all the time. Indeed, we'd all be thrilled if, as measurements got more precise, $w \ne -1$ turns out to be the case because the $\Lambda$CDM ...

30

One thing that stops us from having a theory of everything is actually quite simple. Gravity as we understand it, thanks to the strong equivalence principle, is not a force. It is entirely geometrizable because there is actually no coupling constant between a physical object and the "gravitational field". This means that there is no a priori way to ...

30

Gell-Mann's totalitarian principle provides one possible answer. If a physical system is invariant under a symmetry group $G$ then everything not forbidden by $G$-symmetry is compulsory! This means that interaction terms that treat irreducible parts of a reducible field representation differently are allowed and generically expected. This in turn means that ...

25

One can pinpoint the technical error in LQG explicitly: To recall, the starting point of LQG is to encode the Riemannian metric in terms of the parallel transport of the affine connection that it induces. This parallel transport is an assignment to each smooth curve in the manifold between points $x$ and $y$ of a linear isomorphism $T_x X \to T_y Y$ ...

25

I think this question contains a misconception unfortunately caused by popular science descriptions of the Standard Model. The question seems to assume there needs to be some concrete source that particles "get" mass from, as if mass is a resource like money and the Higgs field is giving it out. But that's not right. In a generic field theory there is no ...

23

Charge conjugation is extremely slippery because there are two different versions of it; there have been many questions on this site mixing them up (1, 2, 3, 4, 5, 6, 7, 8, 9), several asked by myself a few years ago. In particular there are a couple arguments in comments above where people are talking past each other for precisely this reason. I believe ...

19

This is only semantics. A reducible representation $\mathbf R$ of the symmetry group can be decomposed into a direct sum $\mathbf R_1 \oplus \cdots \oplus \mathbf R_N$ of irreducible representations. A field that transforms as $\mathbf R$ is the same thing as $N$ fields, which transform as $\mathbf R_1, \dots, \mathbf R_N$. When talking about fundamental ...

15

The $g$ factor describes the magnetic moment of a spinning particle. The $g$ factor for a classically spinning particle is equal to 1, but in the "basic" (ie, non-interacting) quantum field theory of spin-1/2 particles you would expect this number to come out to be 2, or in other words you would expect $g-2$ to be zero. It turns out that in "...

13

If you make a gauge theory about $\mathrm{SU}(N)$, you also have one about $\mathrm{SO}(N)$ since $\mathrm{SO}(N) \subset \mathrm{SU}(N)$. The representation theory of $\mathrm{SO}(N)$ is more complicated since it has not that many nice properties compared to $\mathrm{SU}(N)$ - the latter preserves orthogonal, complex and symplectic structures while the ...

13

My understanding of this question is really two different questions. Let me answer each of these in turn. 1) What is the relation between the CKM and PMNS matrices? To see how this works consider the relevant quark interaction terms without any choice of basis, - m _d \bar{d} d - m _u \bar{u} u - i W _\mu \bar{d} \gamma ^\mu P _L \bar{...

12

Yes, physics has learned things on both concepts, but only gradually. The value of the mass 125 GeV is in the sub-130-GeV region that favors supersymmetry, or makes it necessary according to some, because the pure Standard Model predicts a catastrophically unstable vacuum for such low Higgs masses. It is also below 135 GeV which is where it should be ...

12

This is a slight abuse of terminology, related to talking about 'second quantization.' The word 'wave function' in this case really refers to the 'one particle wave function,' which happens to correspond to the solutions of the (linear) classical equations of motion. It does not refer to the 'wave functional,' ie the Schrodigner representation of the full ...

12

Strings are not quanta. They are not excitations of something, they are the fundamental objects from which standard string theory starts building its model. In quantum field theory, particles only appear in the theory once it is quantized. The classical field theory corresponding to a QFT doesn't know anything about particles. In string theory, the classical ...

12

As a consequence of $E=mc^2$, to create a heavy particle (i.e. large $m$) requires a large amount of energy ($E$). Since the LHC only generates a finite amount of energy in the collisions, there may be particles that are too heavy to be produced. This looks likely to be the case for superpartners (if they exist).

12

Irreducible representations are always determined by some numbers, labeling the representation, which correspond to the eigenvalues of some observables which are invariant under the (unitary) action of the Lie group. If the group represents physical transformations connecting different reference frames (Lorentz, Poincare',...), these numbers are therefore ...

11

The thing to understand is how we tag neutrinos for flavor in the first place. Neutrinos are created and destroyed in reactions that also involve a charged lepton (electron, muon or tau). At vertex level these are $$W^\pm \to l^\pm + \nu_l$$ and various rotations. The flavor of a neutrino is defined as coincident with that of the charged lepton produced. (...

11

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 let'...

11

In the modern effective field theory point of view, there's nothing wrong with non-renormalizable theories. In fact, one may prefer a non-renormalizable theory inasmuch they tell you the point at which they fail (the energy cut-off). To be concrete, consider an effective lagrangian expanded in inverse powers of the energy cut-off $\Lambda$: \begin{...

11

No, the discovery of the Higgs boson has little to do with validating the Higgs mechanism, or the breathtaking role of the Higgs field Yukawa couplings giving masses to fermions. There are several "Higgsless models" accommodating the Higgs mechanism just fine; and a very heavy and broad (and hence barely visible, σ-like) Higgs would have little to ...

10

Mainly because you need complex representations for the fermions such that anomalies cancel. Real representations don't work though these also cancel anomalies since these give large radiative masses. $SO(n)$ has both the tensorial representations(single valued) that are always real and spinorial representations (double valued). For $SO(n)$(or more ...

9

We should distinguish between the different types of anomalies that can arise in quantum field theory. An anomaly is a symmetry of the classical action that is not preserved in the quantum theory. A gauge anomaly causes a gauge symmetry to be broken leading to a violation of a Ward identity which is needed to ensure that unphysical polarization states and ...

9

I have not seen the film. But this was not "supersymmetry versus multiverse". It was "supersymmetry without multiverse" versus "supersymmetry with multiverse". According to quantum field theory, a light Higgs boson (light compared to "grand unification" energies) should still look heavy because of virtual particle effects, unless these effects mostly ...

9

At present for physics research, a phenomenologist is a theoretical physicist who is well grounded in the current physical theories and at the same time understands the data and can create detailed theoretical models that can predict the behavior of future experiments. In this context, phenomenology is the study of the way current theories fit the data and ...

9

The $\nu_e$ is a mixture of three mass eigenstates $\nu_1,\nu_2,\nu_3$, at least two of which are massive. The mixing coefficients form the PMNS matrix. For neutrinos, mass and flavor are not simultaneous observables, so the $\nu_e$ does not have a well-defined mass of its own.

9

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 ...

9

There is a limit on the number of flavors in Quantum Chromodynamics, behind that limit Color Confinement can no longer exist. The Beta-function that describes the interaction strength at different scales (at one loop) is: $$\beta(g) = \frac{g^3}{16 \pi^2} \left( - \frac{11}{3} N_c + \frac{2}{3} N_f \right)$$ This is negative for $N_f = 6$ quark flavors ...

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