# Tag Info

## Hot answers tagged particle-physics

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$\mu$s. 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 ...

15

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, ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

3

The best solutions of the challenge are available in these papers: http://jmlr.org/proceedings/papers/v42/

3

I presume you are talking about spontaneous radiative transitions. If you allow stimulated emission or collisional de-excitation then obviously this depends on the external environment. The basic answer is that you have to do the calculation quantum mechanically. Some transitions are quick, but others (known as forbidden transitions) can be very long ...

2

If there is a relative absence of electrons on the nuclei then this absence acts as if it were a positive charge. In fact, the absences can even sometimes behave like bona-fide particles; they are called "holes" in semiconductor physics. It would make a lot more sense if electric charge were the negative of what it is. Sadly, that is not likely to happen ...

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} ... 2 Charge is a conserved quantity. If the incoming photon on a molecule is of the appropriate energy , an electron can be kicked from a low energy level to a higher energy level .Since the charge of the photon is zero , the molecule remains neutral. If the energy of the photon is high enough the electron gets kicked out, the molecule becomes positively ... 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 ... 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 ... 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 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 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 ... 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, ... 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

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 ...

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 ...

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 ...

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

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 ...

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