New answers tagged

1

Trackpy is a Python package for particle tracking in 2D, 3D, and higher dimensions. http://soft-matter.github.io/trackpy/stable/ https://github.com/soft-matter/trackpy The Matlab Particle Tracking Code Repository Daniel Blair and Eric Dufresne http://site.physics.georgetown.edu/matlab/ Particle tracking using IDL John C. Crocker and Eric R. Weeks ...


1

Intensity in this context often refers to only the number of alpha particles incident on a unit area per unit time. If you assume that the alpha particles only slow down and none of them are stopped completely, then the number passing through any area does not change with depth and the intensity in this sense is unchanged. Naturally the intensity in terms of ...


2

Actually, i would think you are rigth, too. Usually, such a Problem is described via the Bethe-Bloch Formula. If you do an alpha-particle experiment, you can even measure the attenuation of alpha-particles in air, for example. So maybe he meant something else and just put it queery?


4

There are lots of ways to make antimatter "naturally". One of the most common is pair production. A high energy photon is converted into a particle / anti-particle pair. For example, a photon with energy greater than about 1 MeV ($E > 2 \, m_\mathrm{electron}c^2$) can turn into an electron positron pair (some more considerations are needed to conserve ...


0

Antimatter, although only in the form of positrons, is produced by many nuclides during the β⁺ decay. I can not get any reliable source, but vast majority of such β⁺ nuclides seem to be artificially prepared in a reactor, so this is perhaps not a truly natural source. Other article, named "Antimatter from bananas" states otherwise. The concentration of ...


3

You understand that $\mathcal C$ violation is required, as if it weren't, processes related by $\mathcal C$ that violated baryon number conservation would balance, i.e. $$ P \to Q B \qquad \mathcal C:\qquad \overline P \to \overline Q \, \overline B $$ would result in no net baryon number violation. In these expressions, $B$ is a fermion carrying baryon ...


4

velocities don't add up like A + B = C. This is only approximately true in the limit of every day perception. You have to consider your coordinate frames relative to each other if you approach relativistic (meaning a fraction of the speed of light) velocities. $s = \frac{u + v}{1 + \frac{u v}{c^2}}$ would be the added velocity s of two objects moving ...


1

What I assert is that instead of the line being formed because it 'follows the path of an electron or muon', it is a chain of interactions. That is, there IS a disturbance, but the path begins with an electron-atom (or even an atom-atom, doesn't matter) interaction, and instead follows like a row of billiard balls (atoms) hitting each other and revealing ...


0

I built a small cloud chamber at home years ago, powered with the Americium taken from a smoke detector. Dry ice, alcohol, a container, and an electric field ... didn't work very well, but you could see little ion trails. I should make a better one ... And yes, very little energy is lost in forming the ion trails. For the cloud chamber to work the ...


1

Atoms are electrically neutral. Because of this they shouldn't attract or repel each other - but atoms do show a slight attraction, which is the reason most molecules form. This is called the residual electromagnetic interaction. In short, the positive parts of one atom attract the negative parts of the other, and vice versa. There is a good little diagram ...


2

This could be probably the closest one that I know: Data Analysis in High Energy Physics: A Practical Guide to Statistical Methods Olaf Behnke (Editor), Kevin Kroninger (Editor), Gregory Schott (Editor), Thomas Schorner-Sadenius (Editor) ISBN: 978-3-527-41058-3 http://eu.wiley.com/WileyCDA/WileyTitle/productCd-3527410589.html Especially Chapter 11 is ...


3

No, atoms have the same number of protons and electrons so they have no net charge. On the other hand ions (cations and anions) would be repelled or attracted depending on their net charge. Atoms are bound together in a molecules by different means like covalent bonding, ionic bonding (which can be easily explained in terms of electrostatic forces) or ...


1

Do any physical real entities exist? All the particles we consider elementary (the most well-known being the electron) do not occupy any finite amount of space, i.e. are pointlike, at least according to our present understanding of the standard model. However, that understanding could change as new theories emerge or as experimental evidence is found. As ...


1

I'm going to say no, because I interpret a "physical entity" as something which we can observe, and therefore confirm it's existence. For instance: Particles (0 dimensional): Mathematically are points, but when we observe them we observe them to have sizes because of the observation process (bouncing photons off atoms, electrons off other electrons, even at ...


0

Working in dimensional regularization, in the MS bar scheme consider the renormalization group equation for the strong coupling $$\mu\frac{d\alpha}{d\mu}=-\beta_0\alpha(\mu)^2-\beta_1\alpha(\mu)^3-\beta_2\alpha(\mu)^4+\ldots$$ Reordering terms we get an expression we can integrate ...


1

The expected discoveries (resonances) in the energy range up to $100~$TeV depends on the theory that you decide to believe in. None of the theories beyond the Standard Model has been experimentally validated yet. If you only believe in the Standard Model, then there are no new resonances expected as the theory was completed with the discovery of the Higgs ...


3

The main goal is to measure accurately the parameters of the Higgs boson and its interactions. It is a proposal to repeat what was done with the SPS and its discovery of W and Z and electron positron collider LEP built to solidify accurately the standard model parameters. New physics will come up with deviations from the standard model, but these ...


0

While this may not be an objective answer, I personally like G4Beamline. They have plenty of built in materials and have all the particles in the PDG. Further, it's particularly easy to use and gives a nice graphical output (as well as numerical).


1

See if this argument works - I am making this up on the spot so there is definitely space for argument... Most of the interactions with the electrons will not be "head-on collisions" but rather electrostatic interactions. If we get to a certain distance of an electron, it will feel the force and undergo acceleration. If the time of the interaction is short, ...


5

Photons mediate the electromagnetic force. Atoms are not necessary for photons to exist. You just need charged particles (electrons, protons, etc) to interact with each other from a distance. There are many ways for a photon to be created and destroyed. Depending upon its wavelength, as it propagates in free space , it could "disappear" and a pair ...


2

Gluons could form glueballs (theoretically) because they interact in the QCD model that is part of the standard model (there are three gluon vertices, for example). QED lacks such entities because there are no three (or four) photon vertices. Since there is no theory of quantum gravity, there is no answer to your question. @JohnRennie (see above comment) ...


0

The amplitude for low-energy muon scattering mediated by $Z$ bosons is given the the muon's "weak charge," in the same way that the amplitude for scattering mediated by photons is given by the electric charge. In units where the weak charge of the neutrino is 1, the weak charge of the charged leptons is suppressed; I think at first order it's ...


4

You have no way to know if the virtual particle is a gamma or a Z. Actually, it does not really make sense, since no measurement can tell you what has been exchanged. You always have to envisage all the possibilities resulting from the perturbative series used to describe the theory. It's a bit as asking in the double slits young experiment, which slit has ...


2

The Higgs field has a nonzero vacuum expectation value ("vev") $\langle h\rangle =v$. What matters is that it's nonzero; the sign is a matter of conventions. In fact, the Higgs field is a complex one so the phase of $v$ could be chosen to be arbitrary complex, too. This nonzero vev gives masses to other particle species (whose "vevs" are zero) because of ...


2

pfnuesel's answer is absolutely correct and very satisfying to understand, and you should read it before this one. There is, however, a route which would permit $\pi\to e+\nu$ even if the electron were massless. The conserved quantity which suppresses that decay is not spin, but total angular momentum. There exists an electron+neutrino wavefunction with ...


3

The electron-positron pair has a center-of-mass reference frame where the momentum is 0. Obviously, there exists no one-photon system with positive energy which has 0 momentum, as the energy-momentum relation for a photon is $E = p c$.


0

No. A pair of neutrinos is pulled from the vacuum. One of them interacts with one of the quarks via the weak force, and they both change identity: the quark to another kind, thus changing the neucleon; the neutrino to an electron, which escapes. (The negative charge unit also moved from the quark to the lepton.) The electron escapes as the beta ...


21

Since the spin of the charged $\pi$ is $0$, the spins of the daughter particles need to add up to $0$ as well, i.e., their spins need to be anti-parallel. That's nothing else than the conservation of angular momentum. Assuming the anti-neutrino to be massless, it is always right-handed. Right-handed means that the momentum vector and the spin vector are ...


2

The underlying problem of your trap design is that two dipoles can't make a quadrupole! You are probably aware of the multipole expansion of electric (or magnetic) fields. This is where the terms "dipole" and "quadrupole" come from. As a physics student, one often encounters the exterior multipole expansion: When there are charges inside some volume, and ...


0

Not at all. The split of the 6 component electromagnetic field into a 3 component electric field and a three component magnetic field is an artifact of choosing an inertial frame. A different frame will break the same electromagnetic field into different parts. Just like if you have combinations of jumping to the future and teleporting in space then when ...


0

I do not think many people attempt to find the absolute total internal energy of a system. Rather they will pick and choose what needs to be counted to suit their purposes. If interested in statistical mechanics then treated atoms as billiard balls which can rotate and vibrate, have translational kinetic energy and potential energy due to the inter atomic ...


2

How do we know that elementary particles possess definite parity? From the fitting of experimental data. Here is a review from 1965 , when we were still discovering the plethora of particles and started classifying them according to their quantum numbers. Since spin and parity are closely related quantities, there is usually some advantage in ...


4

Why, yes, they do! Probably the most decisive way to show this involves "handedness" of neutrinos. I won't get too much into the details here, but neutrinos are nearly massless and travel at nearly the speed of light; and when something travels at the speed of light you can describe its spin as either "right-handed" (momentum and spin are in-line) or ...


0

Seeing how the other answers mention things like Lagrangian and quite a few details of the Standard Model, I feel that a more popular level explanation is needed. This comes with the pitfall of not being accurate, but for that you have other sources. I will pull a rabbit out of my hat and answer without a single (real) formula. What does "mass" mean to a ...


9

The mass always means the same thing – but in different theories, one uses different equations and other tools to express the mass. Inertial mass $m$ is the quantity expressing "resistance of the object with respect to acceleration", i.e. the coefficient that enters Newton's $F=ma$. The gravitational mass is what enters the formula for the gravitational ...


7

Particle physics has a set of elementary particles , some of which have zero mass. In this table the general public has heard of the electron and maybe the photon. The mass of the electron has been measured consistent with classical definition of mass. The mathematical theory of elementary particles is called the standard model and the table has the ...


3

Those who write than the Standard Model (SM) is natural interpret it as a fundamental theory, rather than an effective theory or effective field theory, with an unphysical cut-off taken to infinity, $\Lambda\to\infty$. The bare parameters and loop corrections diverge, but are in any case considered unphysical. Only renormalized Lagrangian parameters ...


0

Ok, I thought about it again and with the help of your comments I would explain it this way (please let me know what you think about it): The top quark mass can be determined via the formula $m_t^2 = p_t^2 = (p_W + p_b)^2 = p_W^2 + p_b^2 + 2p_Wp_b = m_W^2 + m_b^2 + 2 E_W E_b(1-\beta_W\beta_bcos(\gamma))$ with $p$ denoting the four-momenta and $\gamma$ ...


3

If we indeed suppose that such an exotic material did indeed exist, then we just have to consider the flux of solar neutrinos from proton-proton fusion. This is about $10^{11} cm^{-2}s^{-1}$ at about 0.4 Mev each. So a one-metre square sheet of annixxite, would generate, in Watts: $$ {1\over2} \times 10^{11} \times 10^4 \times (0.4 \times 10^6) \times ...


0

Almost. You have to write the proper symmetrized wavefunction and use it to take the expectation value of $\hat\mu_{1}+\hat\mu_{2}+\hat\mu_{3}$. Each operator acts on one slot, like this: $$\hat\mu_1\left|a,b,c\right> =\mu_a\left|a,b,c\right>\\ \hat\mu_2\left|a,b,c\right> =\mu_b\left|a,b,c\right>\\ \hat\mu_3\left|a,b,c\right> ...


0

In addition to the arguments already given (units), the 4-momentum $p^\mu$ is defined from the 4-velocity $u^\mu$ via: $$ p^\mu = m u^\mu $$ with $u^\mu = \frac{d x^\mu}{d\tau} = \gamma \frac{dx^\mu}{dt} =\gamma (c, \vec{v})$. Hence: $$ p^\mu = (\gamma mc, \gamma m\vec{v}) = (E/c, \vec{p})$$


-3

Another factor here: We build these huge atom smashers because we want a look at things that are normally sealed inside larger particles. The energy is to shatter the box they're in.


0

Perhaps you could fill us in on some more details about what your application is. There are a whole bunch of ways to produce plasmas and the techniques vary largely by the application. For example DC or RF fields applied across a very small gap are used for plasmas that need to be produced at atmospheric pressures (DBDs). At the other end of a scale, a ...


87

Isn't the universe full of Higgs bosons, making up the Higgs field? No. In particle physics, it is understood that the underlying (more fundamental) object is the field, not the particles. Particles are excitations of the fields that can be measured, and always carry certain properties like charge, mass, spin etc. The field that you are most familiar ...


3

Ad 1) Yes and no. There are certainly Higgsbosons in the Universe. Everything we can create at the LHC is created in other events too. Cosmic rays can have more energy than the LHC beam, so there will be Higgs bosons for sure, just as an example. The problem is to bring such a sophisticated detector in place, in addition you don't really have the experiment ...


5

Let me answer your second question first. A free neutron has a larger mass than a free proton (on the order of 1 MeV). An electron's mass is about half an MeV so it is energetically possible (meaning total energy is conserved) for a neutron to decay into a proton and an electron and still have enough energy left over to form an anti-neutrino (moving almost ...


3

Yes! However, it very much depends on what you mean by atoms. For example - as ACuriousMind above mentioned, one of the more common ones is muonium. This is a muon bound to a proton (instead of an electron bound to a proton, as in hydrogen). You can solve this particular model quite easily using the same methods used for solving the hydrogen atom (i.e. ...


0

If quantum mechanics is probabilistic, there is no reason for a particle to be in one place and not the other, but particles do make up their minds... but how? This topic is obscured by common confusions about what quantum mechanics entails. There are several different explanations of what is happening in reality, if anything, to bring about the ...


1

I think it's impossible, because you'd break electromagnetism if the spectrum had a single neutral Higgs in a 2HDM. I presume both doublets obtain VEVs such that 8 real dof are supposed to be reduced to 5 dof after EWSB, with 3 dof eaten by massive gauge bosons. But if the Higgs doublets have a single neutral component, one of the charged ones must be ...


2

In fact, for Quantum Mechanics Nature is intrinsically probabilistic: there is no law describing with probability $1$ a single event. In accordance with Copenhagen interpretation of the quantum formalism this must not be considered as a limitation to our knowledge, but it is simply the manner as things exist in Nature, it is an ontological claim. In this ...



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