New answers tagged

2

The strong nuclear decay process emits alpha particles that are just helium-4 nuclei. The weak nuclear decay process emits beta radiation that is an electron (usually) with the internal to the nucleus the process $n~\rightarrow$ $p~+~e~+~\nu_e$, or in some cases if energetically possible $p~\rightarrow$ $n~+~e^+~+~\bar\nu_e$. The electromagnetic process is a ...


2

The closest thing to an answer would be, that if a neutron in a nucleus decays into a proton and an electron (beta radiation), the ligher electron is the one that can escape the nucleus. The proton is heavier, and also strongly bound by the strong nuclear force, so it stays there. The same goes for neutrons - they aren't emitted in simple nuclear decays, ...


1

No, a weak decay doesn't imply a change of $S$. For example, the decay of the neutron – the basic part of the beta-decay – has $S=0$ both in the initial and final state. So the first proposition is false and only the second one is true.


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A somewhat meaningful and useful concept for massive particles is the Compton wavelength: https://en.wikipedia.org/wiki/Compton_wavelength#Limitation_on_measurement The Compton wavelength of a particle is determined by its rest mass. The position of a particle cannot be measured with a precision less than half its reduced Compton wavelength. In this sense, ...


1

A point particle is an idealization of a particle. It simplifies calculations by using a 0 dimensional object instead of a normal particle in calculations where size, shape, and structure are irrelevant. For example, in the theory of, say, electromagnetism, scientists will talk about a point charge - a particle represented by a point that has a non-zero ...


2

A heavy Higgs is easier to detect because it can decay in ways that lead to an easy detection, e.g. in two W or Z bosons. The background processes that lead to the same signature as what you would get if you have a Higgs decaying into two W's are very low, so you don't need to produce all that many Higgs particles to detect it. But at lower masses, there are ...


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The CKM matrix describes oscillations but that doesn't mean that it doesn't matter in any other process. The CKM matrix matters pretty much in every process involving quarks. In particular, $d\bar c$ primarily contains two mass eigenstates of quarks. And they are eigenstates from different families. $d$ is the first generation of quarks, $\bar c$ is the ...


1

If I have an object A, will a second object exists that is an exact copy of A and cannot be distinguished from the original? If A is a very simple object such as an electron, a nucleon, or even an atom or a small molecule, yes. It is actually an important notion in quantum mechanics that indistinguishable objects (fermions, bosons) collectively behave ...


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The answer to your question is we do not know, and will never know. As you stated it, the uniqueness property depends on our ability to perceive a difference. But that ability is always limited to some level of detail, beyond which differences, if any, are no longer perceived. For instance some people has the idea that each elementary particle contains ...


1

You are correct that quantum mechanics is the basic framework of nature, but not everything in this basic level is quantized, in the sense of coming in a discrete spectrum. Even the spectrum of the hydrogen atom at very high n has such close spacing where it can be called a continuum. The first quantum level is bound atoms. The second level is bound ...


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The continuum radiation from a material is the conversion of thermal energy into electromagnetic energy. As per Wikipedia All matter with a temperature by definition is composed of particles which have kinetic energy, and which interact with each other. These atoms and molecules are composed of charged particles, i.e., protons and electrons, and kinetic ...


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To add to the other answer: Your intuition is right in a way. The fact that the $W$ and $Z$ bosons are so heavy is the reason for the weakness of the interaction. For example, $\pi^+$ mesons can decay over the weak interaction, the process is described by the following Feynman diagram: According to the Feynman rules, the probability amplitude of such a ...


7

When e.g. a neutron decays, there is no "real" W-boson inside, in the sense that it could be detected at every point. Instead, the decay of the neutron involves a "virtual" W-boson, a W-boson that only exists for a very short time. Quantum mechanics allows the energy conservation law to be violated by $\Delta E$ for a very short time $\Delta t$ as long as $\...


2

There is not, because the combined transformation $CPT$ is a symmetry of all Lorentz-invariant systems. The $P$-violating decay distribution observed by Wu et al. is also a $C$-violating distribution, because polarized anti-cobalt would have had the opposite sign of asymmetry. (However no one has ever made, or probably will ever make, polarized anti-cobalt,...


4

It is not a postulate that the Hilbert space for QFT is a Fock space. In fact, for interacting theories is often almost surely not a Fock space. The requirements for a Hilbert space to be the space of a QFT is that the Wightman axioms are satisfied. For free theories, a suitable Fock representation of the canonical commutation relations satisfies the ...


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What is the relation of the particle's resonance with the cross-section? And how it is actually measured? The most striking feature of resonance particles is that they are extremely short lived particles. The lifetime of the order of 10-23 s. can travel about 10-15 meters before decaying . How can we know anything about particles we cannot detect? For ...


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If a particle changes flavor, it's a charged-current weak decay. Example: $n\to pe\bar\nu$. If there's a neutrino in the final state, it's a weak interaction. Decay example: $\pi^+\to\mu^+\nu$. See also neutrino scattering. If parity isn't conserved, it's a weak interaction. Examples: $K^0 \to 2\pi$ and $K^0 \to 3\pi$. Note that kaon decays and $K\...


3

During proton therapy, most of the damage is actually done in the last few mm before the beam stops - at the point called the Bragg Peak Yes, the penetration distance is largely determined by the energy above a few MeV; as the particle slows down, it starts to dump more energy per unit length. Quoting from "The physics of protons for patient treatment" (...


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See, the thing is that spin is actually a vector — it has also a direction. When considering such vector in quantum mechanics, 2 observables describe it completely: its norm ($S$) and projection on one of the axis (usually, $S_z$). For a spin-$\frac12$ particle the norm is $S=\frac12$ and $S_z = \pm \frac12$. Then, for a system of 3 spin-$\frac12$ particles,...


2

We don't need to separate electrons out in order to observe them. The structure of an atom, as revealed in electron transitions (atomic spectroscopy) is clearly based on orbitals at specific energy levels, with a two-electrons-per-orbital limit. And, the collective behavior of unpaired electrons that gives rise to ferromagnetism, and subtle spectroscopic ...


3

Spin was assigned to elementary particles so that conservation of angular momentum would hold in the quantum mechanical framework of elementary particles and nuclei. The Stern–Gerlach experiment involves sending a beam of particles through an inhomogeneous magnetic field and observing their deflection. The results show that particles possess an ...


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Electrons And Spin From Scientific American Unfortunately, the analogy breaks down, and we have come to realize that it is misleading to conjure up an image of the electron as a small spinning object. Instead we have learned simply to accept the observed fact that the electron is deflected by magnetic fields. If one insists on the image of a spinning ...


0

You don't write which context this is in, so this answer may be inappropriate for you, but for an ionized gas of pure hydrogen (i.e. no heavier elements) in the interstellar medium (which contains also electrons, or else the protons would quickly repel each other), the main cooling mechanism above $T\sim10^6$ K is Bremsstrahlung, where the electron lose ...


0

You got it almost right. The symmetry of the wavefunction describing a system made of 2 identical boson must be even by the interchange of the 2 bosons because of the Pauli principle. Interchanging the 2 bosons position introduces a factor $(-1)^l$ with $l$ the angular momentum quantum number. Thus among the 2 values allowed by the conservation of angular ...


0

Those definitions are either for the mean velocity and the mean acceleration respectively, or for the instantaneous speed for non-accelerated motion (the first). To find the instantaneous velocity, acceleration or actual position over time in the general case you need to solve the equations of motion, which will result in different "formulas" than those ...


0

If you already know the path to be taken, along with the time stamp for each position; or having measured the actual path, and the corresponding times, then in either case you have all of the information for the kinematics. The more usual problem is that you start with less information, but you know the dynamics: then Newton's Laws of Motion can be used ...


1

The force is given by: $$ F = -\frac{dp}{dt} $$ where $p$ is the momentum of the particles being ejected. The momentum is given by: $$ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} $$ And the momentum is related to the energy by the relativistic equation for the total energy: $$ E^2 = p^2c^2 + m^2c^4 $$ so: $$ p = \frac{\sqrt{E^2 - m^2c^4}}{c} $$ The ...


3

In fact it turns out to be very useful to generate cold, or slow, neutrons. A neutron which has gotten close to thermal equilibrium with some material at room temperature (kinetic energy $E=kT=p^2/2m\approx\frac1{40}\rm\,eV$) has a de Broglie wavelength $\lambda = h/p \approx 2\,\text{Å}$ which is comparable to the interatomic spacing in most materials. ...


4

Pseudoscalar and vector are terms that indicate the total spin and parity of the resonance. Pseudoscalar particles have spin 0 and parity -1, while vector particles have spin 1 and parity -1. The Particle Data Review lists your particles as: $D^\pm\qquad\qquad\qquad\quad 0^-$ $D^0\qquad\qquad\qquad\quad 0^-$ $D^*(2007)^0\qquad \qquad 1^-$ $D^*(2010)^\pm\...


0

Q^2 is the Mandelstam variable t, i.e. the four momentum transfer squared where the s channel is the x axis in the feynman diagram. where p1 p2 are incoming. In the same link it is seen that at the relativistic limit : The dot product for t (Q^2) is p transverse to the incoming beam direction of p1 , p2 are incoming. etc. Is this Q the same if ...


0

The electric force acting on a particle can be expressed as: $$\vec{F} = q \vec{E} $$ And also as: $$\vec{F} = m \frac{d^2x}{dt^2}$$ The charge is a scalar, so $\vec{F}$ behaves as $\vec{E}$ because of the first equation and it doesn't change sign under time reversal due to the second one. So $\vec{E}$ doesn't change sign under time reversal. Let's ...


0

There are flavor changing weak interactions mediated by the charged $W^\pm$. In the cases you cite I would agree that flavor symmetry would not permit these interactions $uu\rightarrow cc$ etc.


1

All mesons are unstable and most have quite short half-lives even by particle physics standards, so they are only to be found (as on-shell particles1) in the presence of energetic events. And if you look closely enough the process that creates them is essentially2 always one of banging two particles together pretty hard. 1 That is, I am excluding virtual ...


4

The 'rule' that "matter can not be [created or] destroyed" simply isn't true. Matter can be created and destroyed under the right circumstances, it's just that those circumstances are not met for macroscopic quantities of matter in places where humans live. This means that the conservation of matter can be taken as true in almost all of chemistry and large ...


0

According to the CPT thoerem, the choice of positive parity for particles and negative parity for antiparticles is just as arbitrary as the choice of positive charge for protons and negative charge for electrons.


-2

Kinetic energy and photon absorption Particles - when moving into a magnetic field - have gained before kinetic energy. As GRB wrote: Every time a charged particle has to be accelerated, a photon has to be involved. If you want to linearly accelerate a charged particle, you have to shoot photons at its back. To be precise this one of the possibilities ...


4

Quarks are charged, so the only possible answer is "yes". Nor do you have to isolate quarks to test this. By choosing Baryons whose valence content consists entirely of a single type of quark, we can get those in motion and look to see if they are a current. So we choose a $\Delta^{++}$ (valence content $uuu$), a $\Delta^-$ (valence content $ddd$), or a $\...


3

Fast neutrons are produced as a product of most nuclear reactions. In fission reactions, the mean energy of the released neutrons is approximately 2 MeV. Since the mass of the neutron is 939 Mev/c2, their velocity is 0.065 c. In fusion reactions, like the deuterium-tritium reaction, the neutron gets 14 MeV, and thus its velocity is 0.17 c. Larger ...


20

There isn't a simple answer to that. Colour arises when the light absorption or emission of a system is dependent on the wavelength. For example chlorophyll (i.e. plants) is green because it absorbs red and blue light so only the green light is reflected and reaches our eyes. So the question would be how does the light absorption and emission of trillions ...


10

The 'color' would be an ultra-bright burst of gamma rays as the trillions of electrons rush apart, frying both you and your eyes to a crisp. More seriously, if you confined the cloud of electrons, it wouldn't emit any particular color on its own -- for instance, there could be no optical transitions since there are no nuclei. If you shined light on it, it ...


0

The particle is not larger nor more massive. We normally describe particles using quantum field theory, and in QFT particles do not have a size. This is discussed in the answers to Why do physicists believe that particles are pointlike? and it would be worth reading through them. To say that a particle is a point is a bit of an over simplification. We ...


4

Online data analysis is cursory analysis done as the data is collected. It is often used for the purpose of selecting which events to save to disk or tape to be analyzed later (an event "filter"). Given that the current CERN experiments will be taking, in the next run, data at rates exceeding a terabyte per second, this notion is essential. In fact, the ...


2

Because interactions conserve the third component of weak isospin, $T_3$. So the incoming $T_3$ in a vertex must be equal to the outgoing $T_3$. For example:


1

Lets look at the weak isospin doublet of for the first generation of quraks \begin{equation} \begin{pmatrix} u\\d' \end{pmatrix}_{L} \end{equation} Where $d'$ is represents the quantum mixing of $d$ and $s$ quark. From gauge theory, we know that, the role of gauge bosons is to transform a particle to another one, which living in the same mutiplet. So, if a ...


0

String theory postulates that of the elementary particles we currently know about, each relates directly to low-energy string vibrations, the presence of multiple holes causes the string patterns to fall into families. Each hole in the Calabi-Yau space is a group of low-energy string vibrational patterns. If the C-Y has three holes, then three families of ...


1

Tl,dr: Entropy is the right definition, because it's incredibly useful in the description of statistical and thermodynamic systems. Whether or not it quantifies "disorder" in whatever sense of the word is completely irrelevant - it just so happens that it can be interpreted that way. Entropy is not a measure of disorder. At least not really. Then again, ...


0

Typically, we solve a partial differential equation (analytically if we can, but often we can only do numerical calculations), where in the input of the problem we put the known quantities Physics started with observations of nature, and mathematics developed which could model the observations and measurements and predict future set ups. Beginning with ...


0

That is a good question but hard to answer. So far experimental particle physics is concerned, no signature of quark substructure observed till date. Nevertheless imagination of a theorist goes far beyond. There are few proposal already in the literature notably the idea of Preons. Which was considered as the constituent particles of quarks. But its a ...


-1

Of course they can. Some particles arent elementary, imagine Glueballs Neutrons, Protons, they all conclude other particles. But an elementary particle (electron, quark, so on..) are described by one single string. It is also important, what particles you are talking about, because there exist open and closed strings, which all have other mathematical ...


0

Since the electron moves in spacetime and has mass, it produces gravitational waves. that can be derived from General Relativity.



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