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1

$R$ Parity is a discrete $Z_2 $ symmetry while an $R$ symmetry is a global continuous symmetry. If you use a $Z_2$ symmetry to build your model then each field can just be either odd and even, that's it. If you impose a continuous symmetry then there are an infinite number of possible choices of $R$ charges. From a model building perspective, a continuous ...


1

Since John is not addressing positrons one should know that positrons are easily created once a photon has more energy than twice the mass of the electron, in electron positron pairs. This can be seen clearly in this bubble chamber picture: where the positron is shown in purple on the right. One knows they are electrons (positrons) because of the ...


1

It really depends on the meaning of "new". If "new" means a particle which exists in nature, but was so far unknown (think like the gas helium, it did exist before his discovery), then yes. If "new" means something which is theoretically possible, but not found under normal conditions in nature, then yes. If "new" means something engineered which could ...


1

Classical physics describes the movement of the center of gravity of extended bodies, which, when poorly taught, in the mind of the student becomes equivalent with "classical physics being a theory of point particles". That, of course, is utterly false, even on the level of the classical description. A center of gravity is a vector, not a point. ...


12

Creating anti-protons is straightforward in principle because any high energy collision produces a shower of protons, antiprotons and various types of pions. The pions decay in a few nanoseconds, so you just have to wait for the pions to decay then separate the antiprotons from the protons. At Fermilab a 120GeV proton beam was collided with a nickel target ...


0

An electron is not a point particle. Point particles don't exist. The world is governed by quantum mechanics, which describes physical systems in terms of quantum mechanical observables, which are represented by Hermitian operators. Different observables represent different ways in which you can interact with a given system and copy information from it. For ...


9

Gluons and photons are similar in the sense that they are both massless gauge bosons. They do, however, correspond to different gauge symmetries: photons arise due to $\mathrm{U}(1)$ symmetry, while gluons follow from $\mathrm{SU}(3)$. This leads to a different number of particles: there is only one photon, while there are eight different gluons, ...


5

From the perspective of fundamental quantum field theory, gluons and photons are quite similar. Both of them are gauge bosons, meaning that their existence is required by a mathematical mechanism called local gauge invariance. However, as particles, there isn't any particular connection between them. For instance, there's no reason they both have to be ...


1

There are two things that define a particle physics model (at low energies). The first one is the gauge group G we want the model to be symmetric under. For the Standard Model (SM) we set this to $G=SU(3)\times SU(2)\times U(1)$ (for good experimental reasons!). This will uniquely determine the number of gauge bosons needed to make the model consistent. The ...


4

The four quantum field theories (QCD, QED, QFD, and EWT) unite quantum mechanics and special relativity. They are all fully understood, complete, and proven. In your quote for the standard model, there are not four distinct field theories, the electroweak has united the electromagnetic and the weak in one field theory, the electroweak theory. The ...


7

The total angular momentum of a meson is the sum of the spins of the two quarks and their orbital angular momentum. Excited states can have $L>0$ and therefore $J>1$.


2

A theory with N=2 supersymmetry, where particles have two superpartners, has mirror symmetry built in. Nir Polonsky wrote some papers about an N=2 extension of the standard model (e.g.). The main problem for such a theory are the chiral Yukawa interactions between fermions and the Higgs field, which give fermions their mass in the SM. The mirror symmetry of ...


2

The number of gauge bosons is restricted by symmetry: a given theory with a certain gauge invariance admits as many gauge bosons as there are generators of the corresponding gauge group. For example, there is one generator for $\mathrm{U}(1)$, resulting in the existence of a photon. $\mathrm{SU}(3)$ admits eight generators, which yield eight gluons. This is ...


0

I discussed this with a colleague yesterday and I think that I get it. It has to do with the normalisation of the group generators (momentum operators). Within each representation of the translation symmetry we are free to normalise the momentum operator as we like. Then it is natural to choose this normalisation such that all representations have the same ...


10

Spin is best understood as an intrinsic angular momentum. It is probably easier to understand the concept for a charged particle. A classical charged particle moving along a circle has an angular momentum and the "circuit" has a magnetic moment. Further, the two are proportional to each other. It is experimentally found that a charged particle like an ...


3

Spin arises from the need to represent the rotation group $\mathrm{SO}(3)$ upon our Hilbert space of states. We need such a representation because the rotations (together with space translations) correspond to the non-relativistic changes of reference frames. Since states are only determined up to rays in the Hilbert space, the true space of states on which ...


2

There are different layers of reconstruction, at each step the amount of data is reduced with the goal of inferring the momenta, type and direction of the particles produced first in the collision: pulse shape reconstruction: the electronic signals caused by particles interacting with the detector cells are digitized at a rate of 40 MHz at LHC. Some ...


2

There experiment which has measured the most stringent limit on neutron to anti-neutron oscillations (i.e. produce neutrons, let them fly for some time and then look if you find anti-neutrons) has used a 130 micrometer thick and 110 cm diameter carbon foil. This target had a probability greater than 99% for anti-neutrons to interact (and thus produce ...


9

Detectors at particle colliders are layered like onions around the collision vertex. The CMS detector at CERN First there are charged particle sensitive detectors where charged particles leave tracks because of ionisation, but mass density is low so strong interactions do not happen often; their momentum can be measured by the curvature in the ...


-1

As CuriousOne mentions in the comment, if the new fermion has no charge under any of the gauge symmetries then it will not interact with anything (except gravitationally) so we barely have any constrains on such particles. On the other hand, if you are thinking more about the lines of a fermion that is "similar" to $e, \mu$ or $\tau$, but just heavier, then ...


0

For these kind of questions I think the pdg booklet is your best friend. You can find the data for $B^0$ online here: http://pdg8.lbl.gov/rpp2014v0/pdgLive/Particle.action?node=S042 and follow the link for $K$ or $K^∗$ modes. In a first look, I couldn't find any decays including muons, which means that the first process is more probable. See the link for the ...


1

For the massless case, one needs to show that $W^\mu = \lambda P^\mu$. Equation (10.53) provides a basis for an arbitrary four-vector and then expands $W^\mu$ in that basis. Imposing the two conditions $W\cdot P=0$ and $W \cdot W=0$ completes the proof by showing that all other "components" in that basis vanish.


1

Strictly speaking the prefix semi means half, but it's often used in the sense of partial. A good example of this would be semiconductor. So semileptonic just means partially leptonic.


0

For massive force carriers, one finds (in natural units) an exponential dependence $\mathrm{e}^{-mr}$ that prevents long-range forces with massive mediators. Restoring SI units, one sees that the compton wavelength is the length at which the damping is exactly $\mathrm{e}^{-1}$.


4

The criterion for gravitational radiation is (conjectured to be, pending direct evidence) a changing quadrupole moment in the mass distribution, so an accelerating mass distribution does not always radiate, but can do so if the acceleration changes the quadrupole moment. This is in contrast to electromagnetic radiation, which occurs when the charge ...


2

The expression refers to the situation where the particle, such as an atom, containing charged constituants, is coupled to the (quantised) electromagnetic field. Then, if the atom is in an excited electronic state, it can decay to a lower state by emitting a photon (a quantum of the electromagnetic field). This process is known as radiative decay.


1

From a decoherence point of view, fields are more fundamental as they give rise to particle-like behavior from the wave behavior if interactions with the environment are strong. In the end though, quantum mechanics only describes correlations between macroscopic changes in detectors (or other materials), so whatever kind of ontology you want to take in the ...


1

Simply because it is usually taught from historical, heuristic and pragmatic point of view, rarely from axiomatic point of view (e.g. Wightman axioms, as mentioned in a comment by ACuriousMind). This is because it is taught to be useful, as most QFT calculations boil down to scattering and decay amplitudes, and as Sean Carroll said: Heuristic QFT, on ...


7

An elementary particle is defined as an irreducible representation of the Poincar\'e group. These were classified by Wigner in 1939. This was done via the little group construction. The important representations are (metric signature $(-,+,+,+)$ $p^2 = 0$, $p^0 < 0$ - The little group is ISO(2). All finite dimensional representations of this group are ...


0

I think that the answer is that there is a flavor symmetric octet representation and a flavor antisymmetric octet representatio, while the decuplet is totally symmetric. Therefore, when you consider the spin and flavor wavefunction of a baryon for an octet baryon you have: $\chi(spin)\cdot\phi(flavor)=\frac{1}{\sqrt{2}}(\chi^{1/2}_s\cdot ...


1

I'm trying to give a less technical answer. It's not rigorous but should give you the idea how spin and the regular rotation related. Maxwell's equations say in order to have magnetic field, you need a ring current. This can be achieved by giving angular momentum to charged particles. This can be orbital or simply because the particle is spinning. This was ...


4

It's not necessarily true that most of the photons that strike a wall will be absorbed and turned into heat. The whitest white paints can have a light reflectance value of up to about 85%. There isn't a "wavelength corresponding to white color". An ideal white surface reflects as much as possible of all wavelengths in the visible spectrum. That sounds ...


0

1) No, substances almost never completely absorb photons. Otherwise you could not see them. In case a substance would absorb all photons (which is quite hard to achive intentionally) it would be pitch black even if you shine arbitrarily strong light on it (-> black-body). 2) It will be reflected back and forth, but only a finite amount of time. This is ...


1

Almost always, when photons hit matter or interact with it, they are not reflected in the way a billiard ball bounces off a billiard table edge. Rather, they are absorbed, the absorber rises into a metastable state, and then a new photon is emitted on the decay of the metastable state. Sometimes, though, when photons undergo an interaction with a lone ...


2

Usually phase diagrams, e.g. of water, are shown as pressure vs temperature. However, we could just as well write a phase diagram as temperature vs density. This is because the equation of state of matter is a relation between pressure, density and temperature. The above phase diagram should be interpreted as showing: what is the state of matter at a certain ...


2

1) Why don't we consider finite dimensional representations of this group? As you said, we ask (anti)unitarity, so it is impossible to find finite-dimensional representation. 2) Why associate the Lorentz group to fields? The essence of the answer is what Trimok already said in his comment: the "translational part" of the Poincarè group is already ...


1

The SIS accelerators are heavy ion accelerators, and the German for heavy ion accelerator is SchwerIonenSynchrotron (my capitalisation), hence the abbreviation SIS. There is more info in this article.


1

SIS-100/ SIS-300 is an accelerator under construction for FAIR (Facility for Antiproton and Ion Research) in Darmstadt, Germany. see - http://cern.ch/AccelConf/e08/papers/mopc100.pdf I believe, but am not sure, that the -100 and -300 refers to the magnetic rigidity (i.e. Magnetic field * bending radius) of the accelerators, which determines the maximum ...


-1

You asked HOW it is that the bonded neutron is stable, but the free neutron is not: 'What happens inside the nucleus that makes neutrons stable?' This is an ontological question and these are the hardest to answer. The best answer you can get in terms of conventional physics is differences in binding energy, as Lagerbaer explained. The Table of Nuclides ...


1

Yes! Neutrons are electrically neutral, but they have a magnetic moment. You can accelerate a bar magnet with a magnetic field, so you can also accelerate a neutron with a magnetic field. For most beams, the change in energy is pretty negligible, but there's a major exception for ultra-cold neutrons (UCN), which just so happen to be my specialty. UCN have ...


2

Because muons don't carry a color charge, and hence don't participate in the strong interaction...


4

For a scalar field $\phi$, the most widely used convention, based on my experience, is to write the Lagrangian with kinetic and potential terms, followed by interactions like so, $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2 - \frac{1}{2}m^2 \phi^2 - \sum_{n \geq 3} \frac{\lambda^n}{n!}\phi^n$$ where $\lambda_n$ are coupling constants. (We could not have a ...


0

A perhaps not completely rigorous, but easy to understand derivation: It can be shown that the magnitude of angular momentum of circularly polarized classical radiation of frequency $\omega$ and energy $E$ is given by $E/\omega$. If we now assume that radiation is quantized in packets of energy $E = \hbar \omega$, we arrive at the conclusion that the ...


4

The sun gets its energy from the pp-chain. The first step is the two protons forming the diproton (Helium-2): $$ \,^1_1H+\,^1_1H\to\,^2_2He+\gamma $$ where the $\gamma$ is the photon (of energy about half an MeV). This quickly $\beta^+$-decays into a deuterium by converting a proton into a neutron: $$ \,^2_2He\to\,^2_1D+e^++\nu_e $$ where $e^+$ is the ...


-2

Perhaps this helps: https://de.wikipedia.org/wiki/Benutzer:HolgerFiedler/Strukturen/Summary It explains how the photons electric field changes to the magnetic and back and there are no points of zero energy. But what you have to keep in mind is that radio waves are cliffs rent waves, they are omposed from a lot of photons and the radio wave is zero at all ...


8

the only difference between matter and anti-matter is simply charge Nope. All "quantum numbers" reverse for them. In particular, for the example you consider, there's Baryon number, which is different for baryons and anti-baryons, and in particular, for a neutron and its anti-particle, an antineutron. It is in fact possible to differentiate between the ...


1

It's all about spin. The conservation you mentioned is the key to it - you could conserve energy and/or momentum by tweaking the resulting KE, but you're still left with unconserved spin. Historically, observations showed an upper limit to the electrons which was not high enough to include the missing energy that is found if you don't include neutrinos.


4

The momentum is harder to deal with than the energy. If a stationary neutron decays into an electron with momentum $(a,b)$ and a proton with momentum $(a,-b)$, then there is no way to conserve 3-momentum without the creation of a third particle with momentum $(-2a, 0)$ And while real numbers wouldn't work out this nicely, it would be obvious that the ...


3

An electron is a charged particle, charge conservation would not work as the neutron has zero charge. In addition it would have been detected with its interaction as its energy would be similar to the energy of the other electron seen. The neutrino was posited as a weakly interacting particle exactly because it was not caught by the detectors, and because ...



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