# Tag Info

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 ...

0

It's Fermi repulsion between the quarks in the nucleons, not between protons and neutrons, that prevents the overlap. You wouldn't claim that a hydrogen and helium atom should overlap because they are distinguishable - obviously you need to consider the repulsion between the electrons they are made of. The same argument applies to nucleons.

0

Atoms/particles are accelerated with an appropriate force, according to the second Newton "law".

1

On a windy day, sound travels faster "downwind" than "up wind". The "particle accelerator" is the sun - through the medium of heating the atmosphere and causing wind. If you travel with the same speed as the air, you will see sound travel at its usual speed; but if you are standing still, sound "from upwind" will arrive a little bit faster than sound "from ...

1

(Moving this from my comment to an answer) Yes, the electric field simply penetrates the glass wall and charges (the electrons) placed in that field will feel a force and move. The glass does not really interact with the charges on either side, so you might as well remove it completely (theoretically).

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

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 ...

1

There is a very nice demonstration that you can treat holes as positively charged carriers in the Hall effect. As you may know, to observe the Hall effect we place a semiconductor in a magnetic field and pass a current through it. We observe a polarization (voltage) at right angles to both the current and the magnetic field as a consequence of the Lorentz ...

0

Holes are technically just vacancies, but when an electron jumps into this vacancy, one more hole is created, but this constant filling up of a hole and creation of a hole appears as if the hole is moving, that is it appears as if there are these +ve carriers which are moving in the direction of the current (opposite to that of the direction of electron) ...

1

I think it helps to look at why things get stuck at all. If you have two things that could together get to a lower energy level than they are at now, and they can give that excess energy up to something (like sending photons into deep space) that isn't going to give it back (at least for a long time), then they can get stuck (at least for a while). So why ...

1

The decay $$D_s^{**+} \rightarrow D_s^+ \pi^0$$ is a strong decay (flavors charm and strangeness being conserved). Thus, parity must be conserved. The final state particles are both pseudo-scalars, and hence the product of their parities is positive. Thus, the final state parity equals $(-1)^{L_f}$, where $L_f$ is the final state orbital angular momentum. ...

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

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

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 ...

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 ...

0

To clarify upon the other answers: There is no magnetic force between non-moving charged particles. Other answers posted here have shown that if there is motion between two charged particles, there will be a magnetic force between them given by the Biot-Savart law. Actually, the Biot-Savart law covers both moving and non-moving cases. If the velocities are ...

0

Since the two pions are in an L=0 state, they cannot have isospin 1. Therefore $\pi^+\pi^0$ must be an on isospin 2 state.

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 ...

0

First of all, infinitesimal small distances are not allowed in the quantum world simply due to the Heisenberg uncertainty principle - the Newtonian force law doesn't hold at these distances. Apart from that, new forces arise that repel at short distances, a simple $H_2$ molecule is an example for that. When you manage to collide particles instead, the whole ...

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 $\vec{v_1}$ ...

5

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

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 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

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 ...

0

I'm not sure to interpret properly what you have in mind with this question but the u-channel of the reaction $A_1 + A_2 \to A_1 + A_2$ would correspond to the reaction: $A_1 + \bar{A_2} \to A_1 + \bar{A_2}$. Except if $A_2$ is its own anti-particle, the u-channel is different from the s-channel and thus you can't add both amplitudes. I'm using the ...

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

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

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 ...

0

The second one is a "quark diagram" or "quark model diagram". These are not 'true' Feynman diagrams, as they only represent the quark fields. The gluon field is implied, and it's an exercise for the reader to fill in the gaps. The only pedagogical source I can find is this: http://ned.ipac.caltech.edu/level5/Cottingham/Cott1_5.html (See the caption of ...

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 ...

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.

0

Well, quark anti-quark pair can be created by a photon in a process like $$\mu^- + \mu^+ \to \gamma \to q + \bar{q} \,,$$ which is just the time reversal of a typical Drell-Yan process, with the intermediate photon shown explicitly. Not sure if this really answers your question since you seem to be assuming a single quark initial state.

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.

-4

It is my understanding that antiparticles are the opposites of each other such as electron positron pair -- which means the positron is an electron flipped 180 degrees on its axis --same angular velocity, same size but PHOTONS are not particles, they are electromagnetic waves which rotate 360 deg for every wavelength so in effect are "positive" 50% of time ...

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

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

If I understand your question correctly, the answer is no. I myself am not a particle physicist but as far as I know, in contrast to massive particles, photons cannot deposit their energy partially. Either they interact and get destroyed or they do not. Of course this way, you can make a grid and detect, where photons are coming in, however, as they can not ...

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 ...

0

You can compute the feynman rule for the $\phi$-$\phi$-$\chi$ vertex by taking $$e^{-i \int \mathrm d^4x L_\mathrm{full} }\frac{\delta}{\delta \phi^a} \frac{\delta}{\delta \phi^b} \frac{\delta}{\delta \chi^c} e^{i\int \mathrm d^4x L_\mathrm{full}}$$ where $L_\mathrm{full}$ is the sum of the free and interaction Lagrangeans and afterwards remove any ...

0

Not necessarily simple. I don't really know what energy is. Energy is conserved. This makes us think of it as some sort of "stuff" that can be changed from one type to another, but never created or destroyed. This is sometimes a good way to think of it, but not always. You can talk about the velocity of an electron, but it isn't really right. You are ...

0

The answer by Surgical Commander covers the title question. I will address what gives photons different energy levels is my question. The creation of light in Classical Electrodynamics, no photons, is continuous. Macroscopically it was observed and the theory fitted the data that the acceleration of charges, i.e. giving increasing energy to a charged ...

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

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 ...

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 ...

1

The symbol $s_{NN}$ is in OP's context of RHIC the Mandelstam $s$-variable in a Nucleus+Nucleus collision. The $s$-variable is also known as the square of the center-of-mass energy.

0

In quantum mechanics, the magnetic moment operator is related to the spin operator by: $\vec\mu = -\left(\frac{e}{mc}\right)\vec{S}$ In other words, they are directly proportional up to some known physical constants. This means that measuring the spin of an electron is exactly equivalent to measuring its magnetic moment: if you obtain either quantity, you ...

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.

0

Until today, there is little review literature about AdS/CFT given how young the subject is. This might change soon, when the book of Erdmenger and Ammon apprears. Until then, this review of Sean Hartnoll is a great reference with much focus on the applications of charged black hole solutions for boundary condensed matter systems. In this review, one can ...

0

This choice just depends on the experiment conducted. Some plots expect the trendline to give an implied value $mc^2$, while others track $(mc^2)^2$. My personal favorites on these plots are Curie plots which purportedly show that the square of the mass of the neutrino is negative. Good luck determining what that means. :)

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 ...

Top 50 recent answers are included