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1

The short answer would be not really (at least the way you've described it). The problem with any sort of physical piston is that high energy particles vibrate and will collide with the piston. This would have several undesirable effects. Firstly, it could damage the piston depending on how its made. Secondly, the collision would heat up the piston causing ...


0

That depends on what you call a particle. And what you mean by breaking down, exactly, In the context of this definition of particle, of course. And, most importantly, what you mean by guarantee. It you have that, it's either trivial - just looking up current state of physics. Or it can not be answered in a meaningful way. It makes no sense to begin ...


2

Degenerate matter is essentially squezed atoms. It's not man-made, But it is perfectly squeezed atoms, all of them. And real squezed in the sense of static pressure. In cores of stars, the pressure get's so high - only after the end of hydrogen burning - that the electrons can no longer stay withe the nuclei. See Degenerate matter and Electron degeneracy ...


2

Once you get up to relativistic speeds, there's no such thing as a "hard" material. If you could get a pair of macroscopic pistons to collide at the location of a test atom, the dynamics of the collision would be dominated by collisions between the atoms in the pistons. You can collide a light particle with a heavy nucleus — this is what's done at RHIC and ...


4

The experiments at LHC hit protons on each other at total energy of 7 TeV. In comparison, a flying moscquito has a kinetic energy of 1 TeV: a trillion electronvolts, or 1.602×10−7 J, about the kinetic energy of a flying mosquito[12] Could one collide two flying bees of 3.5TeV kinetic energy and get a proton proton collision? i.e. give the total ...


4

Can we squeeze atoms? Yes. High pressure changes the wavefunctions of electrons in atoms. See for example Accurate Wavefuctions for the Confined Hydrogen Atom at High Pressures. One effect of this is to increase the rate of electron capture by the nucleus, since s electrons will spend more time at/in the nucleus at high pressure. See the lecture: ...


0

See the comments for a correction on this. There is one more conservation law: energy. You don't have the velocities of the particles, but looking at A: $$\nu_e + p \to e^- + \pi^+ + p$$ There is a proton before and after, simplify: $$ \nu_e \to e^- + \pi^+ $$ Something that has almost no mass turning into two massive particles? That should raise any ...


0

Like dmckee said in the comment, electrons' charge is an intrinsic part of what it is and it cannot be lost. That being said, if a hydrogen atom (which consists of one proton and one electron) loses its electron, it will simply have a positive charge while the whatever gained the electron will both weigh more (due to the added mass of the electron) and have ...


4

PDG is discussing charged current interactions, $\nu N \to \mu^- X$ and $\bar\nu N \to \mu^+ X$. These are not charge conjugate processes. The neutron is $udd$. With the neutrino, $\nu$, we need a $W^+$ for the charged current interaction, $$ \nu \to W^+ \mu^- $$ We then need $$ W^+ d \to u $$ Note that the neutron contains two down quarks. The case of the ...


4

The only dimension-five operators allowed by the SM are neutrino masses, $(HL_i)(HL_j)$. So we mostly talk about dimension-six operators because for almost any question they're the first higher-dimension operators that can appear.


1

The standard model of particle physics is a theoretical framework that encapsulates almost all elementary particle data to date. The full Lagrangian takes pages. In your comment: @Danu I understand the 6/7 Wightman axioms but fail to capture how does the concepts of fundamnetal particles, quarks-leptons, or bosons mediating forces etc. come from those. ...


1

There is an official convention for a positive muon being a nucleus, that a positive muon with one electron is muonium (Mu) and a positive muon with two electrons is muonide (Mu-). See Names for Muonic and Hydrogen Atoms and Their Ions. For a negative muon replacing an electron in helium, I see both $He\mu$ and $^{4.1}H$ in the same paper: Kinetic Isotope ...


0

Your question covers numerous topics, but I will try to answer several of them here. For a proton (I am most familiar with this case), the proton asymptotically becomes a black disk Ref. The disk part is because it is Lorentz contracted at the collision point. The same process happens for nuclei as well (see the bottom left of here for example). The "black" ...


2

Yes, e.g. all three Mandelstam variables $$ s~:=~(p_1+p_2)^2 ~\approx~ (m_1+m_2)^2 + \frac{m_1m_2}{2} ({\bf v}_1-{\bf v}_2)^2 ~>~0,$$ $$ t~:=~(p_1-p_3)^2~\approx~ (m_1-m_3)^2 - \frac{m_1m_3}{2} ({\bf v}_1-{\bf v}_3)^2 ~>~0,$$ $$ u~:=~(p_1-p_4)^2~\approx~ (m_1-m_4)^2 - \frac{m_1m_4}{2} ({\bf v}_1-{\bf v}_4)^2 ~>~0,$$ are strictly positive in ...


2

The actual masses are accessible in theory, but not from mixing measurements. Cosmological measurements could give us a useable handle on the sum of the masses (though until we settle the hierarchy questions this may not provide a unique answer), or the combination of a much better model of supernovae plus a precision measurement of the differences in ...


0

Since the 4-momentum must be conserved, there is no angular dependence in a 2-body decay in the CM frame. But If you are asking yourself what is the phase-space of such process, you should take a look at equation 4.72 of the book I mentioned in the comments: $$\Gamma(M\to 1 + 2)=\dfrac{|\vec{p_1}|}{32 \pi^2 M^2} \int \mathrm{d}\Omega ...


4

First, to be clear on what the graph is showing: as a function of the possible mass of the Higgs, it plots the fraction of Higgs bosons that will decay via each individual channel. Before we knew the mass of the Higgs boson, a plot like this one was useful for identifying the best channels to look at to detect the Higgs in various mass ranges. For example, ...


1

http://www.quantumdiaries.org/2014/04/04/moriond-2014-new-results-new-explorations-but-no-new-physics/ Top quark mass has been revised upward.


2

I am not exactly sure which low energy cutoff you refer to; however, there is a low-energy cutoff for photons that I am aware of. Photons with energies on the order of $H_0\sim10^{-33}\text{eV}$ would be super-horizon modes. That is, their wavelengths would be on the order of the Hubble radius, $H_0^{-1}=14.6~Gly$. Larger than this would mean that the ...


2

Top Quark, with a rest mass of approximately $\ 173.07 \space GeV/c^2$ .


2

Recall $|E,l,m \rangle$ is the joint eigenvector of the Hamiltonian $H$, the total spin $L^2$, and a spin component (typically z) $L_z$, and the $E$, $l$ and $m$ label their respective eigenvalues. Notice all three of these operators act on the single particle that we're considering. Also recall $\langle k |$ is an element in the momentum basis, also in the ...


4

In simple terms QCD as a "background" usually refers to LHC research where hadronic jets create a lot of particles that clutter up the results you're trying to see. I think it has become a slang term and the use is discouraged. ABCD method is a tool used to separate the particles of interest (signal) from the "other stuff" (background) made by the jets. ...


1

I would guess, and it can only be a guess, that Stewart is referring to weak measurement. There is a rather vague description of this in New scientist. Annoyingly I can't track down the original paper, but if Stewart's book was written in 2013 the timing fits.


4

Since $\pi^0$ is a pseudoscalar particle, we have $$\langle 0|J^\mu_{em}|\pi^0 \rangle =0,$$ and the pion cannot decay into two leptons with a simple photon exchange. In the Standard Model, the leading-order contributions for this process are a box diagram and a $Z^0$ exchange, as you can see in fig. 1 of arXiv:0806.4782 (replacing a $c$ quark by a light ...


0

Let me address your questions one by one. Why is it said that lepton number is conserved in Standard Model (SM)? How do I know that lepton number is an Abelian charge? The SM Lagrangian is invariant under the fermion transformations, $$ \psi \to e^{iL\theta}\psi $$ where $L$ is assigned such that $e^-$, $\mu^-$ and $\tau^-$ leptons and ...


1

Electron doesn't 'get into' the shape of the orbital. An atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom. This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom's nucleus. The term may also refer ...


-1

Why is it said that in standard model lepton number is conserved? Because the standard model is a mathematical model specifically fit to what the data tells us, and lepton number is conserved according to the data. How do I know that Lepton number is an abelian charge? It is additive in the number of leptons and antileptons reduce the lepton ...


1

I'll only address your first question here. To start off with a sidenote, I think the idea that mass is a fundamental property of a particle has been on shaky ground ever since Einstein showed the equivalence of mass and energy. I can hardly imagine it took very long for people to come to the conclusion that mass cannot be a fundamental property of ...


0

This was answered well in the comments but I'll write it up as a proper answer. There are two main questions here: Why is there an asymmetry between matter and antimatter? given that there's an asymmetry, why did matter win out? The second question is answered rather easily. The Big Bang produced more of one of the two types and everything, from ...


3

It really goes deeper than just a theoretical demand on a particular domain. The Hamiltonian for any system must be unitary, because that preserves the total probability at one. This is important because if I start with some state and let it evolve for a while the system must afterwards exist in some state which means that the sum of the probabilities taken ...


1

I will offer two reasons. First, unitarity of mixing matrices insures that probabilities sum to one. The probability of an oscillating neutrino having electron, muon or tau flavour should equal one. Second, because the neutrino mass matrix is Hermitian it is diagonalised by a unitary matrix.


4

It's a theoretical demand : $$ \begin{pmatrix} \nu_{e}\\ \nu_{\mu}\\ \nu_{\tau} \end{pmatrix} = \begin{pmatrix} U_{e1} & U_{e2} & U_{e3} \\ U_{\mu1} & U_{\mu2} & U_{\mu3} \\ U_{\tau1} & U_{\tau2} & U_{\tau3} \end{pmatrix} \begin{pmatrix} \nu_{1}\\ \nu_{2}\\ \nu_{3} \end{pmatrix} $$ You know that all states ...


1

In the Standard Model, fermion number is not conserved. Lepton number is conserved, because of an accidental symmetry. One cannot write down a renormalizable, gauge and Lorentz invariant operator that violates lepton number conservation in the Standard Model. A Majorana neutrino would violate lepton number conservation by two units. To see this, consider, ...


0

Just summing together all the comments and providing some more explicit calculations, we have conservation of four-momentum (which is the amalgamation of the conservation of energy and conservation of momentum), we have: $$p^{\mu}_{1}+p_{2}^{\mu}=p_{1}'^{\mu}+p_{2}'^{\mu}+p_{\pi}^{\mu}$$ Taking the inner product of each side with itself, we get: ...


1

Yes, it's a misconception, or not - or both. What do you call "matter"? Let's call matter particles with a rest mass. So, everything that's made up of elementary particles is matter. Now here's the catch: To the best of our knowledge, elementary particles are pointlike, i.e. they really don't have any extend in space, they don't really "occupy" any space. ...


2

There are two $SU(3)$ symmetries that are often discussed with regards to QCD. There is a gauge symmetry which corresponds to color charge which is mediated by the gluon and there is an approximate global flavor symmetry which acts on the flavors of the quarks (turns an up into down quark for example). All stable hadrons are color singlets and thus don't ...


0

When a speedy particle collides with another particle, it loses some or all of its enormous Kinetic Energy. As energy is conserved, this Kinetic Energy is converted into particle-antiparticle pair (mass is a form of energy). They come in pair for the sake of other conservation laws. For example, if the produced pair is electron-positron, here you can see the ...


2

I'm not entirely sure what the answer to this question is. It's probably not friction heat alone as Parth Vader claims, since you can ignite coarse steel wool with the flame of a match, and yet the significantly finer "atomized" 100-mesh aluminum will not ignite under the same conditions ("atomized" refers to the manufacturing method of molten metal ...


2

For over forty years accelerator technology has been giving antiproton and positron beams, the easiest to create anti particles. Positrons are the simplest because once the energy of electrons is accelerated to the values over pair creation of e+e-, the the brehmstrahlung photons, gamma ray energies, will create electron positron pairs when interacting with ...


0

Morsley's experiment was based on Bohr's atom model, it showed that the value of the nuclear charge increased by one, as the atom's place increases across the periodic table. The method was to knock out an electron, and look for the emmission-spectrum as another electron fell from the outer shell to the inner one. This would produce a hydrogen-like ...


5

Anti-matter is created from interactions. They are created as a way to conserve energy and other quantum numbers such as charge. One such interaction would be the $Z^0$ decay: $Z^0 → ν_e + \bar{ν_e}$ In this case lepton number is conserved. They do annihilate producing energy (usually photons, but they can produce other particle-antiparticle pairs) when ...


0

While grinding metal, sparks are produced because of Friction. The rotating grinder cuts through the metal molecules, rubbing against them and producing heat. Some particles get loose in this process and burn because of this heat. From Wikipedia, "Steel is an alloy of iron, with carbon being the primary alloying element, up to 2.1% by weight. Carbon, other ...


0

This is a speculative question and can only have the answer that mathematically it is conceivable that the theory of everything ( TOE) may contain higher mass gravitons. After all in the electroweak interactions the photon, the Z and the W are all to start with zero mass carriers of the electroweak force, and then symmetry breaking makes the photon massless ...


1

I don't know what a "pile" of fuel is. I assume you mean a container full of it. Gasoline needs oxygen to burn, and it needs the correct mixture. Too little oxygen and burning is impossible. Too much oxygen causes the same problem. To achieve ignition with Gasoline, you need between 1.4 and 7.6% petrol vapour (by volume) in the air. Ouside this range ...


0

Since electrons don't interact through the strong interaction, an electron-quark "atom" is on the face of it the same as an electron-proton atom. (Except maybe weak interaction decays, I'm not entirely sure.) However: a free quark has never been observed in experiments, and it is widely believed - but not proved - that the theory of strong interactions does ...


2

Both statements are correct. Only left-handed electrons and left-handed neutrinos participate in weak interactions. The projection operators $$ P_L = \frac{1}{2}(1-\gamma^5)\\ P_R = \frac{1}{2}(1+\gamma^5)\\ $$ satisfy the relations $$ P_L \gamma_\mu = \gamma_\mu P_R\\ P_LP_R=0\\ P_LP_L=P_L\\ P_L + P_R =1 $$ From this it follows that $$ j^\mu=\bar u_e ...


1

I think K-K is still of much research interest. The K-K theory, which unifies YM fields and gravity are there since a long time. However, these theories have some consistency problems. In the original K-K theory, it was assumed that the 5D metric functions do not depend on the 5-th coordinate. This was the main reason of inconsistency. If the 5th ...


0

Regarding neutrons in neutron stars the answer is a direct extension of the argument use by madR. In a neutron star there are mostly "free" neutrons and the question then is why they don't all beta decay into electrons and protons? Well, some of them do, but the point is that when the electron/proton (there are equal numbers of each) numbers build up then ...


0

I think there are two parts to this answer. The first is to do with an ensemble of neutrons in a dense fermion gas and the second is to do with the strong nuclear force between two neutrons (in a many-body nucleon system). Neutrons in a dense gas will be degenerate. That is to say that the Pauli exclusion principle prevents more than two neutrons (spin up ...


0

The standard model of particle physics is the encapsulation of an enormous amount of data gathered in elementary particle experiments. The group structures of the model fit the symmetries observed in the data and the model predicts the behavior for future experiments, for example the existence of the top and the Higgs The model starts before symmetry ...



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