New answers tagged

0

Consider a Dirac particle say a electron. In the lab frame if the electron is travels in the z direction with a z component of the spin is -1/2 ,then it is corresponds to left handed electron. Since the electron has a definite mass,so it can travels only less than the speed of light.Now consider case where the observer is travels with a velocity higher than ...


3

The entropy of a single atom does not make sense per se, unless you specify the preparation. The entropy of a single isolated atom, fixed at a point, is indeed not defined – the entropy is, after all, a property of an ensemble not of a system. The entropy of an ensemble of isolated atoms prepared at a specific energy, on the other hand, is well defined (this ...


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In one sense, this should not be a coincidence, because the mass and the charge radius are actually determined by more fundamental quantities (quark masses, strong force coupling constant). So given that those quantities have their particular values, in theory it inexorably determines what "mass times charge radius" is going to be. What should be a ...


0

"Perhaps a sagging potential is one of ratios rather than sizes." Your conclusion is correct, but the idea is not restricted to the sagging well. This is a neat exercise in scaling, and it all comes from the choice of parametrization of the potential for given length $L$. To see this better in the present case, just rewrite the hamiltonian in terms of ...


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If we did have decay of massless particles then there'd be no reason a photon wouldn't decay from an energy of $h\nu$ into two photons of energy $h\nu/2$. Eventually all our photons would be red-shifted into oblivion and we'd have no light left in the universe.


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EM Waves are basically spinnings of photons. To create a EM Wave, you basically move some electrons in a directional way(think of an antenna). Electron is a charged particle as well as protons. Charged particles emit photons, and if you emit photons in an ordered way such as this: You are seeing a dipol antenna. When you apply negative voltage(intense ...


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The energy is carried in individual photons. A photon with twice the frequency has twice the energy. x-rays are made of photons with higher frequencies. The energy is transferred as kinetic energy as with the photoelectric effect. The energy of a photon is calculated as E=hf (Energy=Plank's constant x the frequency).


1

No. Feynman diagram calculations are much more complicated; the $\alpha^n$ scaling is just one piece. For example, consider the amplitude for $N \gamma \to (N+1) \gamma$, where $N$ photons interact to turn into $N+1$ photons. The lowest-order Feynman diagrams contain one electron loop connecting all the photons together, so the cross section, according to ...


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Somewhat reluctantly, I decided to add to the otherwise excellent technical answers above, since none confronted the fundamental false premise of your question, "why do charged leptons not mix?". Of course they do. Let me review its antecedents as you seem to be aware of the phenomenon, when you have all quarks, ups and downs, mix, and not just the downs, ...


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An electron being a ball of uniform mass and charge is not consistent with its observed gyromagnetic ratio. The charge must be pushed out and the mass must be pushed comparatively inwards to satisfy the existing ratio of about 2. See Classical proof of the gyromagnetic ratio $g=2$


1

At lowest order, color screening comes from virtual quark/antiquark pairs, just like charge screening comes from electron/positron pairs in QCD. The diagram/effect is also referred to as 'vacuum polarization' and is shown below. Charge antiscreening comes from virtual gluon pairs; its diagram is below.


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In an imaginary world where the leptons and the neutrinos are all massless, they would exist only in the left-handed helicity version. In this imaginary world the decay of the pion in lepton plus neutrino would be prohibited: you could not conserve both the total spin (zero) and the total momentum, in that this would imply emitting two particles both with ...


2

First, let's clear up some terminology: the usual statement "Majorana fermions are their own antiparticles" is correct, but confusing because the words we usually use to describe neutrinos are made for Dirac fermions. If neutrinos had no mass at all, there would be two independent types of neutrino: a left-handed and a right-handed neutrino. These particles ...


0

The PDGlive webpage lists their fit as $$ \text{BR}(D^0 \to \pi^- e^+ \nu_e) = (28.9\pm0.8)\times10^{−2} $$ Looking at the experimental analyses from which this combination was obtained, e.g. this CLEO analysis, we find that inclusion of charge-conjugate states is implied throughout this report In other words, I believe that PDG lists the branching ...


2

The color of the photon is related to its frequency $f$, which can be related to the energy of the photon by the expression $E = hf$, where $h$ is Planck's constant. Thus the different colors of the emitted photons describes their different energies. The next step is to determine why specific elements emit certain colors. This has to do with the different ...


0

One way to think about it is similar to a chemical equilibrium. Electron-hole pairs are spontaneously generated every now and then from random thermal fluctuations, and when an electron collides with a hole they annihilate with each other (some fraction of the time). The frequency with which electrons and holes collide is $np$. In steady-state, this ...


0

"Why is $np$ always equal to $n_i^2$ ?" Well, first of all, the easy way to answer your question "Why is $np$ always equal to $n_i^2$ ?" would be simply to notice that $np$ is independent of the Fermi level $E_F$, and thus independent on the fact that the semiconductor is doped or not. In the case of a non-degenerate semiconductor (i.e. when $E_F$ is ...


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The process is discussed at the parton level – both in the initial form and the desired form – so the conversion cannot depend on PDFs. Now, the Mandelstam variable $t$ is equal to $$ t = (p^\mu_1 - p^\mu_3)^2 = m_1^2+m_3^2 -2 E_1 E_3 +2 \vec p_1\cdot \vec p_3 $$ in the "mostly minus" metric convention. The masses of particles are fixed and the total energy ...


2

Electrons and holes occupy their states according to the Fermi-Dirac distribution, which has a single parameter $E_f$, the Fermi level (assume a fixed temperature). Provided $E_f$ is in within the band gap and far from the band edges, the (energy integral of) Fermi-Dirac takes an exponential form $\propto e^{E_f}$ for electrons and $\propto e^{-E_f}$ for ...


1

There are two mistakes. As AccidentalFourierTransform pointed out, the coefficient $7.181\times 10^{-16}$, when converted from MeV to eV, should give $7.181\times 10^{-46}$. Mega means a million, and it to the fifth power gives $10^{30}$, not just $10^{15}$. In this way, the OP has to add a $10^{-15}$ factor to his result. That makes his result $10^{-3}$ ...


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I want to offer a different perspective from the already existing answers, which all seem to somehow refer to the Standard Model or other specific physical theories to say that mass is not an integral multiple of some fundamental mass unit, hence not discretized. The reason why mass is not like that - and can indeed conceivably have continuous values in a ...


1

I consider the scattering process $A+B \to 1 + 2$. The differential cross-section is always given by \begin{equation} \begin{split}\label{eq1} d\sigma &= \frac{1}{(2E_A)(2E_B)|v_A - v_B|} \frac{d^3p_1}{(2\pi)^3} \frac{1}{2E_1} \frac{d^3p_2}{(2\pi)^3} \frac{1}{2E_2} \left| {\cal M} \right|^2(2\pi)^4 \delta^4( p_A + p_B - p_1 - p_2) \end{split} ...


0

Particles are simply high momentum wave states interacting weakly with matter, i.e. their "existence" is observer dependent. Trying to derive them from some form of free field equation is therefor useless and so is the assumption that they are a general phenomenon. They are a highly likely phenomenon for energies that are much higher than the typical em ...


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Yes it can transfer momentum. But alpha or beta particles are very few compared to photons in star emission


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Hadronic jets deposit a significant fraction of their energy in the electromagnetic calorimeter, for example because they can contain neutral pions that decay as $\pi^0\to\gamma\gamma$, bottom/charm mesons with semi-leptonic decays... Therefore the jet reconstruction algorithm uses energy deposits from both electromagnetic and hadronic calorimeters, so that ...


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Edit: I am leaving this in because some effort has been made to present how decisions are made on complicated channels.The simple answer by @atlas-insider clarifies the general point that the OP is asking. From the sample ATLAS paper given in the comments Search for supersymmetry at $\sqrt{s}=13\ \rm TeV$ in final states with jets and two same-sign ...


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Those of us who've worked at JLAB (and those who worked at SLAC) know that energetic electrons create a lot of hadronic junk when incident on significant amounts of matter. Think about Deep Inelastic Scattering. Once you have an electron with energy in the few GeV range or higher there is a significant chance of creating pions or other light mesons in the ...


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Although the situation is quite simple, and equations of motion can be written without much difficulty, these equations cannot in general be solved in terms of simple functions, except in a few special cases. Numerical solution is necessary in most cases, and 'orbits' will not be stable. Motion along the perpendicular bisector will be an oscillation but not ...


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Anna V gave the correct explanation. Only when weak bosons are created on mass shell, e.g. at collisions @ Ecm = M, can you apply total angular momentum conservation at a single vertex (production and decay). On the other hand, even for off mass shell bosons chirality (read helicity for ultra relativistic particles) imposses costraints at each vertex.


1

In the second step, when you take the hermitian conjugate, you should have $$u^\dagger(\gamma^{\mu}p_{\mu}-m)^{\dagger}=0\tag{1}$$ instead of $$(\gamma^{\mu}p_{\mu}-m)^{\dagger}u^{\dagger}=0\tag{2}$$ because $(AB)^\dagger=B^\dagger A^\dagger$. Also, you wrote $$ (\gamma^\mu p_\mu-m)^\dagger=(-\gamma^{\mu}p_{\mu}-m)\tag{3} $$ which is wrong. The correct ...


0

I guess the smallest mass would be the neutrino's, though their mass hasn't technically been determined, but it is thought to be of the order 0.05 eV.


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Fundamental values are quantised so I believe there is a 'minimum' or 'smallest' mass. If you look at the similarities between fundamental electromagnetic and gravitational interactions then its clear to see the two are very similar, with gravitational interactions being related to Mass and EM related to Charge. There is a minimum quantised Charge (e, or ...


0

If we work on the theory that over decades we have been able to with the assistance of more unique scopes been able to break down these sub atomic particles then in theory, as we have in discovering quarks, these particles must be made up of even smaller particles. Our perception is only confronted by our current knowledge and capabilities, therefore in say ...


1

Though the question is off topic, its a rare opportunity for a young person to connect with more senior physicists, so I'll share my thoughts: There is a vast, beautiful mathematical world waiting for you to discover - you haven't even seen a deep treatment of mechanics yet. It takes a lot of work and a lot of dedication, to grasp, but we're all here ...


0

The Heisenberg uncertainty principle is a basic foundation stone of quantum mechanics, and is derivable from the commutator relations of the quantum mechanical operators describing the pair of variables participating in the HUP. You are discussing the energy time uncertainty, . For an individual particle, it describes a locus in the time versus energy ...


0

It is possible that the wavelike behavior in a double slit experiment is just the outcome of particle distribution. One example is on my link at the top of my page. No one has ever offered any justification as to why particles can be proven wrong and waves can be proven right. No one can even explain how a wave without a medium can work. On the other hand ...


3

This reaction can proceed through multiple charged weak interactions (i.e. instances of a virtual $W$ boson being emitted or absorbed, the flavor-changing weak processes of the standard model); provided sufficient center-of-mass energy, of course. Schematically: $$ \small{ \begin{array}{cc} \pi^- \!\! \equiv \begin{Bmatrix} \overline u \cr d \end{Bmatrix} ...


4

Finally, I am now able to provide an answer to my question. The weak charged current interaction is described by the gauge field $W_\mu^\pm$ through the interaction Lagrangian term: \begin{equation} \mathcal{L}_I = -\frac{g}{\sqrt{2}} (\overline{u}_L, \overline{c}_L, \overline{t}_L) \gamma^\mu {W_\mu}^- V_\text{CKM} \begin{pmatrix} d_L \\ s_L \\ b_L ...


1

You are free to choose which axis you make a measurement on and doing so will always yield and eigenvalue of the spin operator in that direction: you will always measure $\pm\hbar/2$ in whichever direction you choose to measure. The reason is that the state vector of the particle exists in a superposition of states with various probabilites $$ |\psi\rangle ...


0

The nearest possible analogy to a Galton board will be a quantised electric field, interacting between the electric fields of surface electrons from the edges of a slit and the electric field of the particle (an electron or a photon), one direct to the slit. A common field would explain even the longtime distribution pattern from single shoted particles and ...


0

Beta decay occurs, approximately, in nuclei where the Fermi energy of one species of nucleon is higher than the first unoccupied orbital for the other species. In these nuclei energy can be liberated by turning one type of nucleon into another --- the new nucleon moves into the available, lower-energy orbital. By definition, a nucleus with neutron excess ...


1

Defining precisely what are all the quantum numbers is a difficult question because it depends highly on the model under consideration, even for the standard model. In particular any U(1) symmetry leads to a quantum number, and similarly some U(1) subgroup of non-abelian groups that commute with all other interactions can also be associated to quantum ...


25

Mesons are not elementary, they are composed of quarks. So take a look at their quark content. The charmed eta meson consists of a charm and an anti-charm quark, denoted $c\overline{c}$. An anti charmed eta meson would therefore be an anti-charm and an anti-anti-charm (which is just a charm) quark, i.e. $\overline{c}c$, which is obviously the same as ...


1

I would assume that to mean the zeroth component of the energy-momentum four-vector, for which $p^\mu p_\mu=-m^2$ is an expression of the full version of the famous mass-energy equivalence formula (plus or minus, depending on a sign convention you can choose). $p^\mu p_\mu$ is Einstein notation for, in this case, $p^\mu p_\mu=\frac{1}{c^2}E^2 ...


0

The real world is messy and annoying, so maybe. But it shouldn't. Magnetic fields are created by currents, and the direction of a current depends on both the direction in which charge carriers move and the sign of the charge that is moving. When you spin a whole capacitor, in any orientation, you move both positive and negative charge carriers in the same ...


3

[The figure shown in the OP question above ...] is experimental data for the ratio $R = $ [...] as a function of the centre of mass energy $\sqrt{s}$ The so called cross section ratio $$R[~\sqrt{s}~] = \frac{\sigma^{(0)}[~e^+~e^- \rightarrow \text{hadrons}, \sqrt{s}~]}{\sigma^{(0)}[~e^+~e^- \rightarrow \mu^+~\mu^-, \sqrt{s}~]}$$ ?? Actually, ...


18

Interstellar space is an excellent vacuum, but it's not a perfect vacuum. For example Earth is constantly bombarded with protons from the solar wind, which stream outward uninterrupted until the heliopause when matter from other stars becomes more dominant. If there were, say, an antimatter star nearby, the place where its stellar wind of antiparticles met ...


1

This is because $dn_i$ can be arbitrary. You get an infinite number of equations by choosing different $dn_i$. For these equations to be statisfied simultaneously, you need the coefficient to be zero. \begin{equation} \ln n_i + \alpha + \beta \epsilon_i = 0 \end{equation} Note because you have included Lagrange multipliers, $dn_i$ can be treated as ...


2

Let $$ (\gamma^\mu p_\mu-m)u=0 $$ Using the property $\overline{AB}=\bar{B}\bar {A}$, we have the following: $$ 0=\overline{(\gamma^\mu p_\mu-m)u}=\bar u \overline{(\gamma^\mu p_\mu-m)} $$ Now, use $\overline{A+B}=\bar A+\bar B$: $$ 0=\bar u (\overline{\gamma^\mu p_\mu}-\bar m) $$ Next, as both $m$ and $p_\mu$ are real numbers, we have $\bar m=m$ and ...


3

The way you combine quantum systems is not by summing their wavefunctions, but by taking the tensor product, see this question. In particular, if the systems you are composing are just single electrons described by square-integrable wavefunctions in $L^2(\mathbb{R}^3,\mathrm{d}x)$, then a system of $N$ electrons is described by a square-integrable function ...



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