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In answer to the main question, matter does, in fact, "pass" through other matter. Starting from the macro scale (stars , galaxies), down to the micro scale (atoms), it happens all the time. The "free" movement of matter starts to get impeded, as the atoms start making latices (solids, crystals). But even at this scale, as Rutherford demonstrated, matter ...


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I think that you have to neglect the s-quark mass not its momenta in your expression. Anyway, if you perform the standard procedure for calculating this integral you will find an expression which is a function of all the possible external momenta and the metric tensor as well. Then you can carry the limit $p_s \to 0$ you will find the formula which are you ...


0

The spatial wavefunction is $Y_L^m(\theta,\phi)$. When exchanging the two particles, the spatial wavefunction becomes to $Y_L^m(\pi-\theta,\pi+\phi)$. Mathematically, we have $Y_L^m(\pi-\theta,\pi+\phi)=(-1)^L Y_L^m(\theta,\phi)$. If $L$ is odd, the spatial part is antisymmetric, otherwise symmetric. BTW, you may post this question as a comment of the ...


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Most materials rubbed against another (different) material will result in a transfer of charge from one to the other. This is known as the triboelectric effect. The amount of charge transferred can vary considerably and depends on a number of factors including how much energy went into rubbing as well as humidity of the atmosphere. The triboelectric series ...


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Solar neutrinos have energies under 10 MeV. There isn't enough energy to make heavy leptons. That's really the whole story.


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1) There exists the classical electromagnetic wave as described by Maxwell's equations. 2) The photoelectric effect showed that these electromagnetic waves are composed of photons, with energy E=h*nu , where nu is the frequency of the classical wave. Single photon experiments have been performed by limiting the intensity of the beam to one photon at the ...


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The interaction between two charged partices occurs through a change of momentum. All very well, but the next question is how we calculate the momentum exchange, and this is where quantum field theory comes in. The interaction between two electrons is described as a disturbance in the quantum fields involved. Quantum field theory gives us an expression for ...


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The deuterium nucleus is a boson, with spin $\hbar$ (and positive parity). Unlike other stable nuclei, deuterium doesn't have any bound excited states; however if it did they would also have integer spin. The deuterium atom is a fermion, which may have spin $\frac12\hbar$ or $\frac32\hbar$, to be combined with the orbital angular momentum (which is zero in ...


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Protons, electrons, and neutrons too. Hydrogen is a boson, deuterium is a fermion.


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Briefly, no. Charged particle interaction is fundamentally a non-instantaneous photon exchange. The interaction can be written or expanded as the ''naive'' instantaneous interaction ($e^2/r$) plus a photon exchange portion which contains exactly a $-e^2/r$ term and a non-instantaneous (retarded in time) term. The author calls the first term the Coulomb ...


1

The definition of $s$ is the following: $$ s=(p_1+p_2)^2,$$ where $p_1$ and $p_2$ are the 4-momenta of each colliding proton. For head on collision of particles with same energy and momentum (like in LHC) these explicitly read $p_1=(E_p,\vec{p})$ and $p_2=(E_p,-\vec{p})$. Plug this in the definition: \begin{eqnarray} s &=& p_1^2 + p_2^2 +2p_1\cdot ...


1

All you need to do is conserve energy and momentum in the lab frame. Firstly you conserve energy in lab frame: \begin{equation} E_{\gamma 1} + E_{\gamma2} = E_{\pi} = 1.3GeV \end{equation} Then you work out what the pion's momentum was (still in the lab frame) using the mass-energy-momentum relation where the $E_\pi$ is the total kinetic and mass energy: ...


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Looks like the Laplace transform of a Gaussian function, which is well known, e.g. done here, and here. You will have to expand the square to make use of the identities, of course.


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For anyone with similar problems: The following observation has helped me immensly: We have in fact four particles directly related to an electron: A left-chiral electron $\chi_L$, with isospin $-\frac{1}{2}$ and electric charge $-e$, A right-chiral anti-left-chiral-electron $(\chi_L)^c=\chi_R$ with isospin $\frac{1}{2}$, electric charge $+e$ A ...


5

Yes, it is correct to say that the Higgs boson, just like other elementary particles, get its mass from the interaction with the Higgs boson – which means "with itself" in the case of this particle. More concretely, the mass may be derived from the Higgs potential (energy density) $$ V(h) = \frac a4 h^4 - \frac b2 h^2 + c$$ where the additive shift $c$ ...


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There are several ways to measure the frequency of a photon. The most direct way is to "mix" the photon beam with another of a precisely known, slightly different frequency. This produces signals whose frequencies are the sum and difference of the 2 beams. If your mixing frequency is well chose, the difference will be low enough to be directly countable in ...


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There are a few ways to think about, the worst being trying to imagine a particle as a solid little ball behaving in some oscillatory way. Particle are described by quantum field theory. So for a particle, say an electron, there is a corresponding quantum field (electron matter field in this case). The particle of the field is an excited mode of the field, ...


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Two ways to think about this: instead of thinking of one photon, think of the Quantized Electromagnetic Field. This quantum field is spread throughout all space and time and is everything electromagnetic. An elementary way to think about it is as a collection of quantum harmonic oscillators, one for each of the classical modes of Maxwell's equations. If you ...


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a particle can be described by a wave-function. The frequency of the wave is given by the de Broglie relationship


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Roughly sketched, for the quantized Dirac field one has: \begin{equation} \hat\psi(x)\sim \int d\mathbf{p}\, \sum_r \bigg[ u^{r}(p)\, \hat a^r_\mathbf{p}\,e^{-ipx}+v^{r}(p)\, {\hat b^r_\mathbf{p}}^\dagger e^{ipx}\bigg], \end{equation} where $r=\pm1$ denotes helicity. The ${\hat a^r_\mathbf{p}}^\dagger$ operator creates a helicity-$r$ particle state when it ...


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I think the short answer is that they mix, because nothing prevents them from not doing it. The fact is that states with equal quantum numbers in general do mix. The experiment will tell you how much they mix. If you find a very or almost zero mixing you can start thinking to add a symmetry that prevents some mixing to happen, which was the case of a lot ...


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All the experimentally observed particles that are believed to be fundamental particles, but for neutrinos, in the Standard Model get a mass from the interaction with the Higgs boson. Please note that this aspect of the dynamics of the Standard Model is still under experimental investigation, despite there is data that clearly points towards this direction. ...


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It is mainly measurement and detector errors that make up the width in the plots you show. The Monte Carlo simulates the detector resolution and folds in the theoretical values when it says that the width agrees. The real width is expected to be much smaller. In this we see that the real width is only given as a bound by the experiments the CMS ...


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The decay width of a particle is antiproportional to its lifetime. Looking at the partial width of the $H \rightarrow \mu \mu$ decay, one could expect that the lifetime of the Higgs is large. This would be correct, if the Higgs could only decay to muons. In other words: The Higgs decaying to muons has a low probability (a low branching ratio). This comes ...


2

We then talk about a left-chiral electron we do it in an informal way, you are correct that a massive particle cannot be inherently chiral. To see this, let us remember that he handedness of an elementary particle depends on the correlation between its spin and its momentum (helicity). If the spin and momentum are parallel, the particle can be said to be ...


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Acceptance is some measure of the fraction of events or the spacial or momentum distributions of events that will be registered in the detector. The exact meaning varies from experiment to experiment and sometimes from analysis to analysis within a single experimental data set. Understanding the acceptance of a experiment for a particular signal can be a ...


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It is instructive to look at chart of isotopes , number of neutrons on the x axis and protons on the y. The stable (black) isotopes diverge from the diagonal, more neutrons are needed to neutralize the coulomb repulsion of the protons, for stability. The main forces are the coulomb force (repulsive) and the strong force (attractive) , but the specific ...


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Even if the the strong nuclear force is the most powerful force over subatomic distances. The electrostatic force is almost always significant, and, in the case of beta decay (which also involves a transformation of a proton into a neutron), the weak nuclear force is also involved. But the basic answer is yes, you cannot two protons together (Helium 2 ...


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I think there can be confusion around what 'binding energy' and 'mass excess' mean. The Wikipedia entry on Deuterium has links to explanation them, which may clear it up if the following doesn't. IF you could start with isolated protons and neutrons and assemble your own nucleus, the mass (energy) balance of the result would be the sum of the isolated ...


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For a real particle to be off-shell, it is sufficient to be in an external field of some sort. It is not necessary to be "absorbed". "Absorbed" are virtual particles, for example, "virtual photons" who describe non propagating fields like a Coulomb one. For a Compton scattering, the real electron is "off-shell" during interaction with the real photon, i.e., ...


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Electron positron annihilations can give mu and tau neutrinos as well as electron neutrinos. For a calculation of the probabilities see for example Mu and Tau Neutrino Thermalization and Production in Supernovae: Processes and Timescales. You might also be interested to read DavidZ's answer to Why does electron-positron annihilation prefer to emit photons?. ...


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Skobeltzyn recounts his research on particle physics in his text "The early stages of cosmic ray particle research", which has been mentioned in another answer. I found it in the book The Birth of Particle Physics edited by Laurie M. Brown and Lillian Hoddeson in 1985. It looks like he was the first one to try using Wilson chamber to detect tracks of ...


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You might want to look at D. SKOBELTZYN's paper The Angular Distribution of Compton Recoil Electrons Nature 123, 411-412 (16 March 1929) | doi:10.1038/123411a0; but it is behind a 'paywall' and only a short abstract is available for free. (FWIW, Nature, Lond., 1929, v. 123, No. 3098, p.411) EDIT (11/27/2014): See also: Dimitry V. SKOBELTZYN, THE EARLY ...


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According to Bazilevskaya's Skobeltsyn and the early years of cosmic particle physics in the Soviet Union paper (emphasis mine), Skobeltsyn demonstrated a series of photographs with the cosmic ray tracks at the Cambridge conference presided by Ernest Rutherford on 23–27 July 1928, where they made a strong impression on the audience. The comprehensive ...


1

The proton is not fundamental. It is made up of quarks and gluons. It is these constituents that are colliding in the LHC to produce, in your example, a Higgs boson. The quarks and gluons only carry a fraction of the energy of the proton. In addition, the colliding gluons or quarks in general do not have the same momentum. Therefore some of the energy will ...


1

If you assume that whatever generates the mixing patterns of quarks and leptons (beyond the SM) has no underlying symmetry and that nature chose $V^{CKM}$ and $V^{PMNS}$ randomly within the set of $3\times3$ unitary matrices, then it is natural to expect mixing between families because the probability of randomly selecting $V^{CKM}=V^{PMNS}={\mathbb 1}$ is ...


0

Exactly as you mention in the second part of your post. Leptogenesis takes place at temperatures higher than the electroweak scale $T_{EW}\simeq100\,\mbox{GeV}$. Therefore, heavy neutrinos decay into lepton and higgs doublets, whose fields are still all physical.


1

This is misconception that light is some kind of 'mix' of waves and particles. Instead, It actually IS both waves and particles at the same time, you can't separate them from each other. So probably, the answer could be: you see particles as well as waves.


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A field and a particle are two different concepts and it is well that one should separate them. A field can be classified as a scalar field, a vector field, a spinor field or a tensor field according to whether the value of the field at each point is a scalar, a vector, a spinor or a tensor, respectively. For example, the Newtonian gravitational field is ...


3

The Higgs Field is believed to permeate the universe, and the Higgs Boson is just an excitation of one of the four components the Higgs Field! The Higgs field needs high amounts of energy to be excited, so when Higgs Boson is "created", its energy level is usually many orders of magnitude higher than the ground energy level of its surroundings, and hence the ...


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It would be physically impossible to be able to "see" light as anything other than a particle (photon). The only time photons, or any other subatomic particle for that matter, can be described as a wave is when we are NOT looking at them.


3

The uncertainty in any particular measurement is $\sigma_E$. Resolution for these devices is almost always stated in relative terms as here, but take it like this because it depends on the energy measured. So just multiply by the energy. That is, express your signal in $\mathrm{GeV}$ and then find $$ \begin{align} \sigma_E = \left(\frac{0.1}{\sqrt{E}} ...


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You are seeing particles. However there's more to this than meets the eye so I need to explain exactly what I mean by this. Light is neither a particle nor a wave. Instead it is a quantum field. As a general rule while light is travelling it appears as a wave, but when the light quantum field is exchanging energy with anything it does so in quanta that ...


2

A) What is the piece of theory which dictates that electrons interact via the weak force with other electrons and protons, and how can this force be understood in terms of what I am more familiar with i.e a coulombic interaction and dipole moments ... Electrons interact via the electromagnetic force dominantly with other electrons and protons, not the ...


0

I think that we may add that in the confining phase, the QCD-string descritpion of quarks (say, mesons, which are bound states of quark/anti quarks) is that these particles sit at endpoints of "QCD-strings" (I use "" to distinguish this from the normal superstring which is a well defined object, though it failed for the moment at describing exactly ...


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Quarks as we know them are fundamental particles, which means that they do not have smaller constituents. This however does not imply that they cannot decay. A particle in quantum field theory does not need to have constituents to decay into, it can in principle decay into any particle its corresponding field couples to (interacts with), as long as it obeys ...


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The $u$ & $d$ quarks decay into $d$ & $u$ quarks and bosons (e.g., W bosons)--this is effectively what happens to the hadrons in weak interactions. This (incomplete) chart shows, for instance, $$ u\to d+W^+\\ d\to u+W^- $$ There isn't anything sub-quark, as far as the standard model goes.


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The current understanding of quarks is, that they are a fundamental particle. This means for the energy scales currently available in particle accelerators all quarks have behaved like point-like particles. Due to the strange nature of the color-field (the energy stored in it increases with distance instead of decreasing) if you break a proton apart (which ...



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