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54

Most electromagnetic radiation is of very high frequency - the magnetic field changes many times per second. This means that the compass just doesn't have time to "follow" the magnetic field changes. The only thing that does affect a compass is a DC magnetic field - usually this is a large piece of iron etc. that gets magnetized (e.g. by the earth's ...


26

The frequency is one very good argument (and I guess the most important factor) but it might be worth also looking at the magnitudes of the fields. The earth's magnetic field has a strength of roughly $31┬ÁT$. The intensity of the sunlight hitting the earth is about $1300W/m^2$. Since the intensity is related to the electric field $E$ of an electromagnetic ...


19

Basically the same reason as what Floris said, but this also has another important aspect: Visible light has a far too small wavelength to affect a compass. Not only does the field oscillate too quickly around an average of zero – even at any single “snapshot” in time of the electromagnetic wave, there would nowhere be a large region where ...


7

The $\nu_e$ is a mixture of three mass eigenstates $\nu_1,\nu_2,\nu_3$, at least two of which are massive. The mixing coefficients form the PMNS matrix. For neutrinos, mass and flavor are not simultaneous observables, so the $\nu_e$ does not have a well-defined mass of its own.


5

It's probably easier to measure gravitational masses for anti-hydrogen or anti-ions, because those particles can be cooled and trapped using only electromagnetic fields. The problem is that antiparticles are generally "hot" when they are created. The processes that create antiparticles have an energy threshold beneath which they don't work at all (this is ...


5

An experimentalist's answer, Our observations tell us that baryon and lepton number are conserved, within the accuracies of our experiments and observations. This means we have chosen as a standard model SU(3)xSU(2)xU(1) because in the group structure of the possible representations of all the quantum numbers assigned to the particles and resonances we ...


5

They are both allowed, but they are flavor-changing neutral currents that are loop-supressed. The $B^0\to\gamma\gamma$ decay occurs by the Feynman diagrams from arXiv:1010.2229 and I imagine that kaon decay proceeds similarly. They are so suppressed that the $B^0$ decay hasn't been measured, but PDG report an upper bound on the branching ratio of ...


4

We (physicists) believe the reason is this: known symmetries and conservation laws. For example, the mutual annihilation of a proton and positron would remove $2\,e$ charge units from the Universe. This violates the conservation of charge principle, which can be seen to arise from the application of Noether's theorem to the global gauge symmetry of the ...


4

If the accuracy of the experimental measurements is smaller than the width of the gaussian then the shape describes the probability distribution one should get for any decay in particle physics. If the experimental accuracy is larger than the width of the decay then one gets a gaussian from the randomness of the experimental error . Example J/psi in ...


4

An atom can be positively charged as well as negatively charged. The charged form of an atom (or molecule), which has either more or fewer electrons than protons, is called an ion. It cannot be both at the same time, though, and that is because you are viewing positive and negative charge as relative things ("atom A is negatively charged in atom C's view"), ...


4

you see the beam of light because it lightens the molecules in the air. The photons you see have been diffused by the molecules of air. In the vacuum photons won't change of direction (they go at the speed of light) and you wouldn't see the beam passing in front of you.


4

Is there anyone knowledgable enough in this area who would be able to comment on some of the possible theoretical/ hypothetical implications of the existence of spin 3 particles? Is there any thought that their existence could imply additional fundamental forces? If you look at the presentation linked in the link you gave , in page five, you will see ...


4

From scholarpedia: The Unruh effect is a surprising prediction of quantum field theory: From the point of view of an accelerating observer or detector, empty space contains a gas of particles at a temperature proportional to the acceleration. Direct experimental confirmation is difficult because the linear acceleration needed to reach a temperature 1 K ...


3

All quarks have baryon number 1/3, so that the nucleons can be built up. Baryon number is a conserved quantity in the standard model . In models where the proton can decay, it is not conserved, but no proton decays have been detected up to now. One has to realize that the quantum numbers S,C,B,T are attributes for all quarks and have a value, even if that ...


3

As irish guessed in the comment, this is about beam alignment and tuning. A high intensity beam can damage the beam pipe (as in cut right through to the helium jacket) or cause a unusually hard superconducting magnet quench. The former is disastrous and the latter has some potential outrun the quench protection with similarly unhappy consequences. Having a ...


3

The refractive index of air is about 1.0003, so the speed of light in air is about 0.9997$c$. You can work out the energy of the proton at this speed using: $$ E^2 = p^2c^2 + m^2c^4 $$ where the momentum is: $$ p = \frac{m_p v}{\sqrt{1 - v^2/c^2}} $$ I get the energy to be a shade over 38GeV.


2

The short answer to "are there anti-photons" is "yes", but the disappointment here is that anti-photons and photons are the same particles. Some particles are their own antiparticles, notably the force carriers like photons, the Z boson, and gluons, which mediate the electromagnetic force, the weak nuclear force, and the strong force, respectively. Particles ...


2

The answer to the first question is no. It is a basic but subtle point... Science is about adopting the simplest explanation of phenomena. This is usually referred to as Ocam's razor. With very exotic, high-energy theories like string theory or here a gravitino (I think you have a typo!) there is no single clear signature that can not be explained by ...


2

We normally assume the well is due to an electrostatic potential i.e. the potential energy of the electron is lowered because it interacts with some system of electrical charges. However photons do not feel any force from an electrical charge because they are themselves not charged. So if a photon is emitted by the electron at the bottom of the potential ...


2

There is such a thing as a neutrino initiated reaction and we can confidently calculate the rates for them. By that I mean that we can compute---accurately---the rate for stable particle to be transformed in a neutrino flux, and we can test those calculation with controllable sources (reactors and artificial beams). Then we can take that---well ...


2

Answer transposed from a comment: the $K$, $D$, $B$ have nonzero "flavor quantum number" (strangeness, charm, and beauty, to be specific). The analogy you should pursue is the $J/\psi$ or "charmonium," made of a charm quark and charm anti-quark.


2

They do. It's the third component of isospin, which came about before Murray Gell-Mann's quark model. \begin{equation} I_3 = \frac{(N_u-N_\bar{u})-(N_d-N_\bar{d})}{2} \end{equation}


2

Yes, wikipedia has a table which lists the 19 free parameters that need to be tuned by experiments. These include, as you already said, the masses of the elementary particles including the Higgs Boson, and some other notable ones are: CKM Mixing angles and CP-violation phase. Gauge coupling of he three symmetries (U(1), SU(2), SU(3)). Higgs VEV


2

The curve you show is not a Gaussian. It is a binomial distribution with $n=100$ and $p=0.5$ (if it is an unweighted coin). This arises from processes where there are two outcomes. It approximates a Gaussian/Normal distribution when $n$ is large. This distribution has little to do with particle detection other than perhaps as a means of explaining what is ...


2

It turns into kinetic energy a la $E = mc^2$. When a particle decays into a lighter particle, that lighter particle is usually traveling much faster than its parent particle was. So, for instance, at the LHC they'll catch particles that come out of a decay, add up their total energy (counting rest mass and kinetic energy), and use that to determine the mass ...


2

In a comment engineer has suggested using radio frequency induction, and this seems an excellent idea. The technique is widely used, for example in plasma etching, and it's not especially high tech. There is a Wikipedia article that describes the technique, though this is a bit short on practical advice. A Google search returns lots of likely looking hits.


2

The difference, I think, is that $\overline B{}^0 \to B^0$ and $B^0 \to \overline B{}^0$ are inverse processes, and so a difference between them can be ascribed to broken $CP$ symmetry, as you say, or to broken time-reversal symmetry. Only with the additional (strongly motivated) assumption that $CPT$ is an exact symmetry of nature can you definitely say ...


2

Yes. Any interaction! Entenglement is only a quantum version of correlation. Let supose that you have two $\frac{1}{2}$ spins that interact one to another via magnetic interaction. This interaction can produce an entenglement. But this is not the only one. Some Amount of electrons are entangled in a metal due electrodynamic interaction. You can produce ...


2

Though the entangled states are non-local in some sense, they are not non-local in the sense of allowing superluminal transfer of information, by the no-communication theorem and others. There is no force "responsible" for entanglement. In fact, in quantum mechanics, there is no clear notion of such a thing as "force", although you might derive classical ...


1

Jezstarski is mostly correct, The para-positronium (p-PS) state ends up being the main mode of annihilation of positronium (PS). Positrons can annihilate in at least eight different ways but once ortho-positronium (o-PS) forms in a void/vacuum, it has additional time to undergo another mode of annihilation. P-PS annihilates in under 125 picoseconds. O-PS ...



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