# Tag Info

0

The way I like to understand particles is with an analogy to a our solar system. Particles like the stars and planets make up the whole, the whole being something we perceive as a complete entity or system. The whole is the "wave function" and the particles are the units that make up the wave function. Also, a massless object still contains particles. To ...

0

The idea that the particle is "spread out" over a region is closer to true. As you say, in QM the position of a particle is described with a probability distribution. Only that's not entirely a correct description. It would be more correct to say that particles don't have position, they have superposition. The superposition is described using a complex ...

6

Events in high energy physics detectors that can't produce useful data, mostly because they are the result of soft scattering events, are discarded by multiple layers of trigger circuits. What these circuits do is prescribed by so called trigger menus, which are based on theoretical predictions about a large number of known and hypothetical physics event ...

0

From the machine side, a symmetric $p\bar{p}$ collider can have only one beampipe, so it is much simpler. On the other hand if you fill it with many bunches they will start to collide all around the machine. You may manage to separate their orbits, but they will still feel the fields reciprocally generated (long-range beam-beam interaction) that will limit ...

2

The experiments, both CMS and ATLAS report 2.5 and 3 sigma candidates, but not at the same spot/channel. The place to look is at Cern's document server , asking for "supersymmetric" for example in conjunction with CMS or Atlas. This general talk is about limits . Lubos Motl in his blog discusses an Atlas 3 sigma possible excess and there are links there. ...

0

So the naïve answer is that the charge density of nucleon is roughly $1\,e/\rm fm^3$, while an atom has (neglecting the spike at the nucleus) charge density roughly $-1\,e/\rm Å^3$. Each of these is pretty huge in SI units! The conversion factors are \begin{align} 1\,e &= 1.60\times10^{-19}\,\rm C \\ 1\,\rm fm &= 10^{-15}\,\rm m \\ 1\,Å &= ...

0

Let's be clear on what we call a particle. It is an object of which you can measure its physical properties like energy, momentum, charge or spin. None of those is space and for a good reason : space (or time) is not an intrinsic property of a particle. Space is a useful mean to describe the universe and the particles in it, nobody could deny that. But as ...

3

There are no such things as particles in the physical world. The correct description of "small things" in classical mechanics is that the dynamics of the motion of the center of mass of an extended object is the only relevant physical quantity while internal degrees of freedom like rotation, vibration, magnetization, temperature etc.. can be ignored. That ...

1

Electric charge is not special. Every charge is replaced by its opposite under the C conjugation. For instance, electric charge goes from positive to negative and vice versa. Color charge goes from blue to antiblue and vice versa. Mutatis mutandis with green/red color charge. Every charge is sent to its anti-charge.

0

As far as I know, there used to be a Kilpatrick limit on sparking at a given RF frequency. At the typical frequencies around 2 or 3 GHz one can get up to 50 MV/m, which is the max field I've heard of in the context of the ILC and other linear colliders. According to "New Techniques for Future Accelerators II: RF and Microwave Systems" edited by Mario ...

2

There are already antiproton proton pairs experimentally created at LEP , Proton–antiproton pair production in two-photon collisions at LEP The reaction is studied with the L3 detector at LEP. The analysis is based on data collected at e+e− center-of-mass energies from 183 GeV to 209 GeV, corresponding to an integrated luminosity of 667 pb^−1. The ...

1

The reaction $$\gamma\to e^+ e^-$$ is actually forbidden on kinematic grounds - one can show that momentum and energy cannot be simultaneously conserved if there is not something else to take "excess" momentum, e.g. a heavy nucleus nearby - pair production from gamma rays happens in metals, for example, but not in vacuum. Of course, given enough energy, it ...

0

If the particles are small you can assume laminar flow and use Stoke's Law to estimate the drag force $\|F_\text{drag}\| \propto \mu R w$ as a function of viscosity $\mu$, radius $R$, and fall speed $w$. The gravitational force in terms of $R$, gravity $g$, and the particle density $\rho$ is given by $\|F_\text{grav}\|\propto g\rho R^3$ setting the two equal ...

1

This depends entirely on the kinematics. Imagine I'm throwing an electron at a positron. If the two interact, then the resulting photons (two or more, although the more you add, the less likely the process is of occurring) will have energies entirely dependent on how fast you threw them together. If you throw them together hard enough, you can create heavier ...

2

All of charge, lepton number and lepton-flavor number should be conserved in the reaction. Because the reaction is charged-current, the neutrino will be converted into a charged lepton and because it is a neutrino (and not an anti-neutrino) it must be a negatively charged lepton (to conserve lepton number). That means that the quark involved must be ...

1

However it makes sense that gravity can't travel faster than light because of the force-carrying photons Whilst it makes sense that gravity can't travel faster than light, we don't actually know this for sure. What we do however know is that the force of gravity is not conveyed by photons. Even electromagnetic force is not conveyed by photons - hydrogen ...

3

It is the way one reads/writes Feynman diagrams, a particle going backwards in time is the antiparticle. The electron radiates a gamma, and continues to meet the positron , annihilating charge with another photon. Two real particles are needed for momentum conservation in the center of mass, and two photon vertices are the simplest case within the standard ...

3

Leptons and quarks are fermions. ( Fermions are particles with half integer spins.) You can, like the author has, divide them into three generations on basis of their masses. The Higgs boson is a boson. (Bosons are particles with integer spins.) The Higgs boson (which happens to be electrically neutral) is part of a completely different category of ...

-2

Since photons are massless (just Google photon mass), the answer has to be absolutely not! That said, there will be diffraction at the edges of the shadow and/or the shadow may be imperfect or the shadow of a gauze shade or infrared radiation passing through an otherwise dark window - in such cases and probably more besides, there will be radiation pressure ...

7

A shadow is a lack of light. Therefore, a shadow has no mass, for a shadow is not an object or energy. Shadows can go faster than light in certain cases because they are not objects. In the same way, a vacuum has no mass.

3

This decay (occurring via the strong interaction) violates the charge conjugation since $J^{PC}(\pi^0) = 0^{-+}, J^{PC}(\rho^0) = 1^{--}, J^{PC}(\eta'^0) = 0^{-+}$. The charge conjugation transforms a particle in its anti-particle. In the case of the 3 particles involved in this decay, they are all their own anti-particle, and the effect of the charge ...

2

This is a bubble chamber antineutron event This picture, taken in the Berkeley 30-inch propane bubble chamber in 1958 Antiprotons enter from the top with momentum 684MeV/c . At the arrow one of the antiprotons in the beam disappears, shown with purple on the right. Then a vertex appears out of nothing where an annihilation event is obseved into ...

1

A beam of anti-protons are fired into a charge-exchanger. If an anti-neutron is produced, S1 and S2 will not detect anything, but the final detector will. ([Picture courtesy of John Rennie])2 [Here is a book on the matter.]3

6

Your question asks why the "current quark masses" [see http://pdg.lbl.gov/2011/download/rpp-2010-booklet.pdf at page 21] of the quarks that make up a proton don't add up to the mass of the proton. The problem is that, for the light quarks, the "current quark masses" are very different from the "constituent quark masses" [see wikipedia]. "Constituent quark ...

2

The equivalence principle tells us that energy and mass are really just two sides of the same coin, and are related by $E = m c^2$. Rearranging, we get that $m = E/c^2$, so instead of asking where all that mass comes from, let's ask where all that energy comes from. In the case of the proton, there are some quarks and gluons that make it up, and those ...

1

The three quarks you talk about are usually called the valence quarks of the proton, and their contribution to the mass of the proton is not it. In particle accelerators, when we hit protons with high energy beams, we discover that protons are made of a cluster of smaller constituents (like quarks and gluons, which constantly are created and destroyed in ...

0

If everything were at zero temperature you probably* would not be able to distinguish between the past and future. Mathematically time would still exist, in the same way that spatial directions still exist on a completely featureless plane, but since it would not be measurable (even if there were something around that could do a measurement!), it is ...

0

No. In fact the official definition of the second is: The duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom at absolute zero. So time is still alive and kicking at absolute zero.

2

Your idea does not seem to work if you have two particles at different temperatures. Assume you "stop" one of them but not the other. Then does the time slows down for only one particle and not the other? or how would you explain that?

1

Although there are different types of "radiation," their common effect is to transfer some/most of their energy to the material they "hit," resulting in the breaking of the atomic bonds and or structures of the material. When "enough" bonds and/or structures are broken, the material will fail. Since the electrical characteristics of electronic components are ...

1

Presumably you are referring to semiconductors. A hole physically exists in that it is the absence of an electron. Just like a hole in a piece of paper physically exists. However, if you are asking whether the hole is a particle, then no, it does not physically exist. In a semiconductor, we deal with electron hole pair (EHP) generation. When this happens, ...

2

Ok, the previous answer by Alchemist is totally reasonable, but I think we could add a bit of "what is real?" into this discussion without getting metaphysical. A hole is a perfectly well-defined mathematical concept, in the same way that an electron is a perfectly well-defined mathematical concept. The mathematical concept of an electron in the theory of ...

3

That is true indeed. A hole has no physical existence. It is just the absence of an electron that creates the illusion of a positive charge at that point. You can find it in Boylested Electronic Devices and Circuit Theory that it's a theoretical thing.

0

In the standard model, there is no elementary spin 0 boson being electrically charged (but there are many charged spin 0 composite particles). However, in many extensions such as supersymmetry, there are such particles: the scalar partner of the electron, the selectron carries the same charge as the electron. The anti-selectron is the spin 0 partner of the ...

2

The main force keeping the electrons around the nucleus is the electromagnetic one. Electrons do interact gravitationally and weakly but those are very much weaker forces. In principle if the masses were exactly inverted, it would just change the definition of positive and negative charge, which is arbitrary. Generally changes in mass affect the orbitals. ...

2

Major differences no, because what really collides are constituents of the protons (called partons): the valence quarks, the quarks and anti-quark due to quantum fluctuation and the gluons. The same thing happen with neutron. However, since in proton $u$ quarks carry more momentum than $d$ quarks (because proton contains 2 $u$ valence-quarks and 1 $d$ ...

1

Short-short version: We can measure the mass of electrons. They have some. Surely that is enough. Details: To perform the measurement obtain a beam of electrons of known energy. You can know the energy either by knowing the accelerating potential or by measuring the energy with a velocity selector. Pass the beam through a region of known and constant ...

5

It's risky to think about subatomic particles in a classical way, but maybe we can get something from it if we're careful. Electrons orbiting an atom in a state with well-defined angular momentum quantum number $\ell$ have wavefunctions described by the spherical harmonics. The $s$ shell, with $\ell=0$, has spherical symmetry; this state really is ...

3

If we think as electrons around atoms classically, then electrons would irradiate electromagnetic energy, losing momentum and thus collapsing into the nucleus, and atoms couldn't exist. Therefore your question makes no sense. The correct description of an atom is using quantum mechanics, which means there is no orbits on atoms. There is only the solution of ...

1

No. Gauge invariance is not a real physical symmetry but a mathematical property of the formalism while renormalization is more deep property related to the scaling of the coupling constant. One can think interaction that breaks gauge but is still renormalizable. It is fact that QED and QCD are gauge theories but gravity is a gauge theory and not ...

3

You should always apply energy conservation, and it ought to hold in all reference frames, including the frame in which the sigma is at rest. In the sigma's rest-frame, $$E_{\text{initial}} = E_\Sigma = m_\Sigma$$ and $$E_{\text{final}} = E_\Lambda + E_\pi \ge m_\Lambda + m_\pi$$ Thus we have that, $$E_{\text{initial}} < E_{\text{final}}$$ The ...

1

Energy conservation always applies. Your mistake is in thinking that adding masses will solve the problem instead of clarifying some aspects. In the case of the sigma-zero the decay at rest allows to see that the sum of the constituent masses is larger than the mass of the sigma-zero. For a decay to happen there should be energy left over to go to the ...

2

Your assessment of the transitions which can occur, and hence the photons which can be emitted, is correct. However, the colliding electron does not go to one of the energy levels in the atom (as Sebastian already correctly pointed out). What happens is that the colliding electron can deposit its energy in the bound electron, 'promoting' it from the ground ...

1

Now I assume that the question is asking the following: When an electron of energy 12.1eV collides with this atom, photons of three different energies could be emitted. Show on the diagram (with arrows) the transitions responsible for these three photons. Because from one single collision the emission of three photons doesn't make much sense to me. The ...

2

Photons can be emitted when electrons change energy levels. You say that you have worked out where a 12.1 eV difference is. In an ordinary hydrogen atom, the electron will be in the $n_1$ state. Now, what energy state will the electron be in if an ordinary hydrogen atom absorbs 12.1 eV of energy? After absorbing that energy, the electron can lose energy ...

0

... is this what is responsible for Radioactive Decay? Sort of ... radioactive decay occurs when the final state (result of the decay) has lower energy than the initial nucleus and the barrier is occasionally surmounted. This barrier penetration and the virtual pair at the black hole horizon are both quantum processes, so they have a little physics in ...

1

A spin-spin interaction is really a magnetic moment - magnetic moment interaction, where the magnetic moment of each particle is proportional to spin. [Of course, it might be a chromomagnetic moment - chromomagnetic moment interaction if two quarks are interacting, as they are here.] In any case, the interaction term goes like $\vec S_1 \cdot \vec S_2$. ...

0

I have to take issue with Demosthene's textual analysis in a comment on the question. To examine the physics rather than the way chemists talk, we should look at a tree level Feynman diagram for the reactions. Any reaction. Both the reactants and the products touch the participate in a vertex with the force carrying boson, so both have an equal claim to be ...

2

You are correct, it is redundant. You could also say they have the same spin and parity.

2

If certain reaction is allowed, the inverse reaction (if energy and momentum are conserved) will be allowed too. So neutrinos (not antineutrinos) can interact with neutrons to yield electrons and protons. For your second question: leptons can interact via electromagnetism, too. Electrons and positrons are leptons, and they interact mediated by photons: ...

Top 50 recent answers are included