# Tag Info

## New answers tagged particle-physics

0

Classical electrodynamics has a lagrangian for the classical fields, see discussion here . The photon is an elementary particle and does not have a classical existence. Here is on page 5 the Lagrangian for a photon

2

Here is a (partial?) list of new hadrons discovered at LHC experiments $\chi_b(3P)$: a $b\overline{b}$ bound state, discovered by ATLAS in 2011 $\Xi_b(5945)^0$: a $bsu$ bound state, discovered by CMS in 2012 $\Xi_b^\prime(5935)^-$ and $\Xi_b^\star(5955)^-$: $bsd$ bound states, discovered by LHCb in 2014 $P_c(4380)$ and $P_c(4450)$: $c\overline{c}uud$ ...

2

The common hypothesis supposes that there is a slight asymetry in a transformation (i.e., slightly more likely in one direction than the other), which is called the "violating CP symmetry" (occuring in the weak interaction). See wikipedia "baryon asymmetry" and "CP violation".

1

Let me start by examining your last line about kinetic and potential energies. Potential energy is NOT a property of matter. But kinetic energy IS! Here is a short derivation of why kinetic energy is a property of matter: $E=mc^2$ in rest frame of the particle. But if you switch to a different inertial frame in which the particle is moving at speed v, $E' ... 2 Recent particles, which were confirmed by CERN, are pentaquarks and also it has been observed that the Bs0 meson decay in 2 muons. Both of these though had been theoretically predicted long ago. 0 When right handed neutrinos are introduced, they imply$L$violation through their Majorana nature. Their decay into lepton and Higgs doublets $$N\,\longrightarrow l\,\Phi^\dagger$$ violates the lepton number (since$N$has zero lepton number being Majorana). These decays take place out of thermal equilibrium, therefore a net lepton asymmetry is produced. ... 0 The history of attempting to solve, overcome the conundrum that wave particle duality presents is fraught with inventive thought experiments. But in every case a fault has been found in the reasoning. In your proposed experiment you define a phosphorus 'particle' that's able to illuminate itself, and thus no interference involved in observing the particle as ... 4 The Large Hadron collider was closed for a year and more for an upgrade in energy from 7 to 14 TeV. They have started runs in the summer but there is nothing solidly announced, though there are some exciting hints , which need more statistics. 2 1) You can of course write down these amplitudes in any basis you choose, as long as you take into account matrix elements of the CKM and PMNS matrix. 2) There is indeed a difference here, neutrinos are produced exclusively by the weak interaction, whereas quarks can be pair produced by the strong (or electric) force, or produced by weak decays. 3) in ... 0 Only charged particles can be deflected: electrons, protons, and other charged hadrons and mesons. Light and X-ray can't. 1 To make it short, you need "proper" Mass/Energy Eigenstates and "proper" flavour Eigenstates. The mixing theory means that you have real coherent states. This works very well since the mass differences of Neutrinos are extremely small and the Energy is high. For the charged leptons this is not the case. For an electron to turn into a Tau it needs a lot of ... -3 Identification of particles and anti-particles Good question. It's nice to see somebody thinking about physics. Shame it's an old question, but hey ho, it's never too late for physics. The identification of an electron as a particle and the positron as an antiparticle is a matter of convention. We see lots of electrons around us so they become ... 1 Here is one possible way to evaluate this expression using FeynCalc: << FeynCalc` SUNF[a1, a2, a3] SUNF[a4, a2, a7] SUNF[a7, a8, a1] SUNF[a5, a6, a3] * SUNF[a9, a4, a5] SUNF[a8, a9, a6] // SUNSimplify[#, Explicit -> True, SUNNToCACF -> False] & // Simplify and the answer is: 1/4 SUNN^3 (-1 + SUNN^2) i.e. what the OP got from doing the ... 7 Free neutrons are unstable, and decay to a proton, electron, and electron antineutrino with a half life of about 10 minutes. https://en.wikipedia.org/wiki/Neutron In most cases, the electron escapes but the proton captures another electron from its environment, making a hydrogen atom composed of one proton, one electron (and no neutrons.) In some rare cases ... 1 Depending on the way you look at them, there are several kind of kaons:$K^0$and$\overline{K^0}$are the kaons produced by strong interaction. They have a definite isospin and strangeness quantum numbers made respectively of$d\bar{s}$and$\bar{d}s$. However they can oscillate meaning that they transform spontaneously in each other:$K^0 \leftrightarrow ...

16

There is a misconception in your question, specifically: how does a neutron weigh more than itself plus 2 extra particles It doesn't. A hydrogen atom is composed only of a proton and an electron, but no neutron. Hydrogen is shown diagrammatically shown below: Image Source If a neutron were included, then it becomes the isotope of hydrogen, ...

0

The way to approach the problem initially is to consider what you know about the reaction $$n \longrightarrow \mathrm{p}^+ + \mathrm{e}^- + \bar{\nu}_e \,.$$ Because of the relativity principle we can consider the reaction in the rest-frame of the neutron without loss of generality and we know that both (three-)momentum and energy are conserved. ...

0

In Richard Beth's Mechanical Detection and Measurement of the Angular Momentum of Light, bright light from a mercury arc lamp was circularly polarized and passed through a half-wave plate (which reverses the sense of circular polarization) attached to a torsion pendulum. A bit of clever experimental design sent the light through the half-wave plate twice so ...

2

Yes and it already happened. http://physicsworld.com/cws/article/news/2012/mar/19/neutrino-based-communication-is-a-first From the arXiv:1203.2847 Beams of neutrinos have been proposed as a vehicle for communications under unusual circumstances, such as direct point-to-point global communication, communication with submarines, secure communications ...

-2

If you look at the standard model you will only find gluons. This is very clear and should settle any doubts. (Pions are a historical relic of the middle of the twentieth century which only provide an approximation.)

2

What happens to a particle and antiparticle that collide? The 511keV/c² electron is typically converted into a 511keV photon, and the 511keV/c² positron is converted into another 511keV photon. However it needn't be a 1:1 conversion. Check out positronium where you can read that the triplet state's leading decay is to three gammas. That's three photons, ...

2

"Matter can never be destroyed, so what happens to those particles? Do they just disappear? Where does the mass go?" It's not true that "matter can never be destroyed". According to classical understanding, yes, mass was always conserved and was never destroyed. But that's not entirely correct. The meaning of the well known equation $E=mc^2$ is that energy ...

3

The following diagram and explanation from Cornell University's page A Brief Introduction to Particle Physics may be of help: (Note, as correctly mentioned by @HDE in the comments, the term 'mini Big Bang' is a bit misleading, but the main point remains as @Jon Custer mentioned in the comments: The mass gets converted into energy. And energy can be ...

2

So let's start from the relations you gave and transform one of them from ket to bra. $$\left|i\right> = \mathcal{ CPT}\left | \bar{i}\right>$$ $$\left<f\right| = \left< \bar{f}\right| (\mathcal{ CPT})^{\dagger}$$ Using the CPT invariance condition, $\left(\mathcal{ CPT} \right)T \left(\mathcal{ CPT}\right)^{-1}= T^{\dagger}$, It is easy ...

1

I'll give you some points here. If you just treat $K_\alpha$-rays as a transition from principal number 2 to 1, $\Delta E \approx \frac{3}{4}Z^2$ Hartree, which has unit of keV when $Z>10$. Use characteristic Moseley's law, check http://en.wikipedia.org/wiki/Moseley%27s_law setting $k_1, k_2$ appropriately and using $\Delta E = h \Delta f=hf_{\text ... 1 To start with what are identical to each other are the elementary particles of the standard model. of particle physics. When complex composites of these particles are built this complete identity starts differentiating. In interacting with each other quantum numbers enter and energy states. One proton may be indistinguishable from another proton , but a ... 1 Particle antiparticle potential/hypothetical pairs exist in vacuum as a mathematical description, necessary for calculations of interactions between elementary particles. These mathematically annihilate and reappear within the heisenberg uncertainty principle.In the Hawking radiation case the virtual pairs at the event horizon have a probability one of ... 3 The quote is correct but a bit misleading. The statement "In doing so it also liberates particles known as neutrinos" includes electrons also which are the other particle that is released in neutron decay, and is the way that beta decays were discovered. The neutrino was discovered because neutron decay showed a three body momentum spectrum for the ... 1 Setting $$\theta := \theta_1 + \theta_2$$, the momentum of the Higgs boson (candidate) with respect to the lab $$\| \textbf p_{lab}[~H~] \| = \| \textbf p_{lab}[~\gamma_1~] \| ~\text{Cos}[~\theta_1~] + \| \textbf p_{lab}[~\gamma_2~] \| ~\text{Cos}[~\theta_2~] = (E_{lab}[~\gamma_1~] ~ \text{Cos}[~\theta_1~] + E_{lab}[~\gamma_2~] ~ ... 0 How does a molecule form? At the most general level the idea is that there exist lower energy states with the atoms in the molecule closer to each other, and the original joint state of the atoms had a nonzero ability to transition into that lower energy state and give up some energy. The rest is really some thermodynamics. If everything is hot and dense ... 2 You're question is interesting because it is connected to the notion of elementary particle. As mentioned by anna v, the elementary particles (fermions) of the standard model have very specific properties under the symmetry of the standard model (SU(2)_L\times U(1)_Y \times SU(3)_c): they lie in the fundamental representation of the group, which in ... 11 How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles? Here I will only consider elementary, non composite particles. All the hadronic resonances are composite particles of quark antiquark combinations as well as the neutron . The standard model of particle ... 4 Inmediately is not really true, there is some proportionality. Sorry I am not answering directly about "the standard model", this is quarks and leptons. But they will fit the general pattern, you will see. Let me first consider all the "particles" listed in the particle data group file. Most of the particles decaying via photons have a half-live about ... 2 How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles? Nobody has an issue calling the electron a particle. Ditto for a neutron. It's stable in a nucleus, and the fact that a free neutron decays in circa 15 minutes doesn't much matter. It's similar for a muon, ... 4 I think the most direct answer to this would be the fact that a heavier particle can decay into many different lighter particles for different reactions. The abundance of occurence of these relations are const. Again the same heavy particle can be created in multiple types of collisions of various different lighter particles. Thus we cannot say that the ... 24 Take for example an electron and a muon. The muon is unstable because it decays into an electron and two neutrinos in about 2\mus. But a muon is not in some sense an excited electron. Both particles are excitations in a quantum field and they are both as fundamental as each other. The electron is stable only because there is no combination of lighter ... 2 Your question seems quite general, but perhaps you're confused about what "decay" is. When we say something "decays" we don't always mean that it's somehow "breaking up" into it's constituent parts. In fact, we hardly ever do. The heavier particles aren't really "transient interplay of the stable forms", unless I misunderstand, and that isn't something that ... 3 The intrinsic parity of a pair of particles is the product of the intrinsic parities of the particles. The convention is that that matter particles have positive parity and antiparticles have negative parity, so a pair of matter particles should have positive intrinsic parity. However that's not quite the entire story, because electrons must obey the ... 4 I expect you are familiar with the Big Bang model, seen here . It is a mathematical model using mathematical solutions from General Relativity and the Standard Model of particle physics . The BB developed to describe astronomical observations and the SM developed to describe particle physics observations. The SM describes how particles/nature behaves as ... 10 You say: Now, when we talk about energetically favourably bound systems, they have a total mass-energy less than the sum of the mass-energies of the constituent entities. and this is perfectly true. For example if we consider a hydrogen atom then its mass is 13.6ev less than the mass of a proton and electron separated to infinity - 13.6eV is the ... 4 This happens because of a property of the strong force, called Asymptotic Freedom. This causes the interaction between quarks to get asymptotically weaker as the distance between them decreases. This is the reason why quarks are always found in a bound state and are not freely available in nature. The strong force confines quarks to a region where they ... 2 Your mistake is coming from your treatment of the orbital angular momentum in the case of a 3 body-decay. You have to take into account the orbital angular momentum between 2 pions L_1 and the orbital angular momentum L_2 between the third pion and the barycenter of the first 2 pions. The conservation of the total angular momentum imposes that$$\vec{1} ... 3 The best solutions of the challenge are available in these papers: http://jmlr.org/proceedings/papers/v42/ 0 Since you asked this question there have been a couple of confirmation of exotic particles which consist of four or more quarks. For example, Z(4430) recently observed in LHCb, was already discovered by Belle long time ago. This exotic particle has a$c\bar{c}d\bar{u}\$ quark structure. This would lead us to think how color confinement would be satisfied? As ...

1

You are right saying that the only advantage of the higher lumi accelerator will be to operate for a shorter amount of time. Indeed you can build up the same statistics just running longer at a lower lumi. But if you contextualize this, you find out important consequences. With physics programmes that already extend over decades, a factor 10 less luminosity ...

1

The textbook An Introductory Course of Particle Physics by Palash B. Pal offers a (slightly unsatisfactory) answer. The author writes that Such symmetris are often called horizontal symmetries. The name is a pictorial reminder to the list of quark fields in Eq. 17.1: transformations in this group act horizontally in this arrangement. I don't have the ...

0

I'm not actually a high-energy guy, and my knowledge of jets is all second hand, but I did do a graduate summer school on the topic one year. If I recall correctly... Jet energy is the total energy of particles making up the jet. Jet energy resolution is the experimental limit on how well that quantity can be known. You'll note that both of these ...

1

Is there any size of photon if so what is it? The photon is an elementary particle among the others which form a basis for the standard model of particle physics. The Standard Model of elementary particles (more schematic depiction), with the three generations of matter, gauge bosons in the fourth column, and the Higgs boson in the fifth. The ...

5

A photon is a unit ("quantum") of excitation of the quantum electromagnetic field. Thinking roughly of the quantum field as a vast collection of quantum harmonic oscillators, each oscillator corresponding to a mode of vibration of the field, we specify the quantum field's state by stating how many quantums above the QHO ground state each mode oscillator is ...

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