# Tag Info

## Hot answers tagged particle-physics

34

There is no established quantum theory of gravity. Hence, at the microscopic level of particle, we don't know what is going on gravitationally between particle, but it isn't going to be the "inverse square law" we know, just like electromagnetism between two charged particles is, on quantum scales, not just an "inverse square law", but a rich variety of ...

7

Short answer: you don't. Slightly longer answer: You're using beams of particles, and you focus each of them as much as you (practically1) can so that the particles in each beam are reasonably close together. The result is a wide variety of interaction distances from far apart through near misses to closer interactions still. You mentioned electrons ...

6

Ignoring the quantum effects that make such a situation both improbable and overtaken by stronger but shorter acting forces - if we take Newton's gravitational equation with the inverse square law: $$F = \frac{G \cdot M_1 \cdot M_2}{R^2}$$ and if, in theory, you get 2 objects to occupy the same exact space - (ignoring quantum and other difficulties there ...

5

Probably you are interested in the magnetic force between two moving charges which is $$\vec{F}=\frac{\mu_0}{4\pi}\frac{q_1 q_2}{r^2}\vec{v}_1\times (\vec{v}_2\times\hat{r})$$

4

What you want is essentially the Biot-Savart Law. For a point charge that is moving slowly compared to the speed of light (which is also a condition for the Couloumb law that you give to be true, by the way), Biot-Savart says that a point charge makes a magnetic field like: $\vec{B}=\frac{\mu_0}{4\pi}\vec{v_1}\times\frac{\hat{r}}{r^2}$, where $\vec{v_1}$ ...

4

The spin of a vector boson in any dimension is spin 1. What changes with the number of dimensions is the number of degrees of freedom associated with a given spin. A massless vector in four dimensions has two independent degrees of freedom, which can be seen from the rank of what's called the "little group" in the literature. It is the subgroup of the ...

3

If dark energy would consist of particles, it would dilute with the growing radius of the universe to the third power, since the total number of particles would stay the same while the volume increases. What observations found was that dark energy rather behaves like a constant which does not thin out, that's why it is also known as the cosmological ...

3

In the Ising anyon model, there are three topological charges, $1,\sigma,\psi$. $\sigma$ can be thought as carrying a Majorana zero mode, and $\psi$ is an ordinary fermionic excitation. This can also be understood in the context of $p_x+ip_y$ superconductor, where $\sigma$ is the non-Abelian vortex and $\psi$ is the Bogoliubov quasiparticle. Braiding $\psi$ ...

3

You state your age as 13, and it is not clear how much you know about elementary particles and interactions. This is the table of elementary particles in the standard model of physics And these are the forces with which the elementary particles interact and finally create matter as we see it everyday. The quarks within the proton and neutron interact ...

3

The 3 pions can be considered as 3 states of the same particle, the isospin being used to label the 3 states. Since pions are bosons, the total wave function must be symmetric (Pauli principle). The total wave function is the (tensorial) product of space-wave function, spin wave-function and isospin wave-function. Spin wave-function is symmetric since pions ...

3

The paradox you describe is even worse than the effect of gravity alone: Electrostatics works the same, attracting most matter at everyday distances very strongly by comparison. That is obvious for opposite charges, but even neutral matter attracts as electric interaction induces dipoles. Even where you have equal charges, which do repel each other, you do ...

3

It is by lepton number and charge, but you can't get energy/momentum to balance. In the $\nu_e$ rest frame there isn't enough energy to make the products. If there is a nucleus around, you can imagine the $\nu_e$ emitting a virtual $W^+$ making the $e^-$, the $W^+$ scattering electromagnetically off a nucleus to deal with the momentum, then decaying into ...

3

Beliefs have no real place in physics, at least should not have. The proposal of particles not seen comes from the theoretical models in order to complete them. For example the omega minus was predicted more or less exactly from the symmetries of the weak SU(3), and was later found .A more exact prediction than the Higgs. The Higgs as a particle might not ...

3

Such information can be inferred from the differential cross section. Different spins and parities lead to different angular distributions of decay products / scattering partners. You find these correlations by using one decay product to define an axis and measuring the distribution of the other decay products with respect to that. This requires a lot more ...

3

If there were a $t$-channel or $u$-channel diagram for this process, it would have to involve a vertex where an electron changes into a muon and some other particle. There is no such vertex in the standard model.

2

It is not an exponential, it is the result of the available phase space. The higher the mass the smaller the probability from the phase space of n particles sharing the available energy that two particles will have a large invariant mass. The background depends on the energies and experiments. One uses for the phase space Monte Carlo simulations that in ...

2

The transformation of a quark field under a group require you have to choose a representation of that group. It happen that the fundamental representation and the anti fundamental (bar) of $SU(N)$ with $N>2$ are inequivalent in the sense that there no non singular matrix independent of the representation chosen that allow us to make a change of basis and ...

2

The relevant formula is Einstein's: $E^2 = p^2 c^2 + m^2 c^4$, where $E$ is energy, $p$ is momentum, $m$ is mass, $c$ is speed of light. If $p=0$ then the particle is at rest and we get the famous equation $E=mc^2$. For photons, $m=0$, and we get $E = pc = \hbar \omega$. Here $\hbar$ is Planck's (reduced) constant, and $\omega$ is the angular frequency. ...

2

You may wish to look at http://arxiv.org/abs/nucl-th/0104037 (Nucl.Phys. A694 (2001) 295-311) - they give figures for Th-224. For low deformation, the moment of enertia is about 5000 m fm^2, where m is the nucleon mass and fm is, I guess, Fermi (femtometer). I would think the moment of inertia of Th-232 should not be much different.

1

In your second diagram, there is implicitly a gauge boson source of your $q\bar{q}$ pair production. It could be a gluon, a photon or a $Z^0$. This gauge boson has to be attached to something, reasonably your single quark leg as in the first diagram.

1

Take a look at Figure 2(a) of the 1fb$^{-1}$ $B^0 \to K^*\gamma$ branching fraction measurement paper: http://arxiv.org/abs/1209.0313v1 The inclusive $B^0 \to K^\pm\pi^\mp\pi^0$ background is modelled as a black dashed line. From the looks of this plot, the background contribution from $B^0 \to K^*\pi^0$ with a missing $\gamma$ is negligible. You have ...

1

One can write the magnetic momentum operator in the following form: $$\mu=g_p\bf{s_p} +g_n\bf{s_n}+\frac{l}{2}$$ where $\bf l$ is a angular momentum, $g_p$ and $g_n$ are gyromagnetic ratio for proton and neutron respectively, $\bf{s}_p$ and $\bf{s}_n$ are sin operators.The coefficient $\frac{1}{2}$ in front of $\bf l$ is appear because of the contribution in ...

1

There are several experiments on $\mu$ decay planned, I know of an upgrade of MEG at PSI, who are looking for $\mu \to e \gamma$ and I saw a poster on a new experiment looking for $\mu \to eee$ a while back at a workshop. Furthermore, there is a so-called $B$-factory collider in Japan, Belle, that is dedicated to electroweak precision and flavor physics by ...

1

Would the human start emitting photons and die? EDIT: This answer is an answer to the original question regarding "a billion" positions. The question was subsequently edited to now read "2.3*10^28" positrons. That is not cool. The human would start emitting photons. This is exactly what happens during a PET (Positron Emission Tomography) scan at your ...

1

The rest mass of an electron is 0.511 MeV. When an electron and a positron annihilate their mass turns to energy (two 0.511 MeV photons) so for each annihilation an energy of 1.022 MeV is released. One electron volt is $1.602 \times 10^{-19}$ joules, so in joules the energy released is $1.637 \times 10^{-13}$ J. You ask what happens if $2.3 \times 10^{28}$ ...

1

If there where magnetic monopoles, the force between them in static conditions would be exactly the same as that described by Coulomb. You'd need to replace the electric charges with the magnetic charges, and possibly the universal constant as well. All of this is just a consequence of the symmetry of Maxwell's equations with a magnetic source.

1

The $p$-value statistic is used frequently in physics, most notably in the search for the Higgs boson at the LHC. The $p$-value (which I denote with $\lambda$) is the probability of obtaining such a large $\chi^2$ statistic by chance if the null hypothesis is true, $$\lambda = p(\chi^2 \ge \chi^2\text{ observed} | H_0)$$ i.e. the area in the right-hand ...

1

I think I actually have an answer to this question even though I put a bounty on it. The idea is that the amplitude is maximal when the $W$ bosons are produced on-shell (since the propagators are the largest in this limit). This allows us to treat the creation of the $W$'s separately from the rest of the diagram. The momenta of the particles in the lab ...

1

Boy, that term gets thrown around in a number of ways. In a MC generator context it sometimes means "everything but radiative corrections", but I don't know if that is the way the authors of Pythia mean it. On that assumption a "second hard process" would be a final state interaction that is modeled separately of the corrections; re-scattering in a ...

1

In order for a W boson to be exchanged, there has to be a vertex where a neutrino has turns into the corresponding charged lepton or vice-versa. $\nu_e \to W^+ + e^-$, for example. Hopefully you can see why the equivalent for a muon neutrino ($\nu_\mu \to W^+ + e^-$) couldn't happen.

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