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0

You can look at data from the ACE and Wind spacecraft at NASA's SPDF CDAWeb. Both of these spacecraft have solid state telescopes that go up to tens to thousands of keV for protons. The data is freely available to anyone.


1

I suggest the following link Most Particles Decay — Yet Some Don’t! as a complement to the response above.


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

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 ...


0

In order to create a top quark pair you need at least an energy of $2 m_t$. At Tevatron, the energy available is about 1 TeV per beam, where beam1 is a proton beam and beam2 an anti-proton beam. For simplicity, let's pick-up a parton with an energy $m_t = 173$ GeV from a proton and another parton with $E=m_t$ from the anti-proton. The ratio of the energy ...


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 ...


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 ...


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

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 ...


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.


0

Your reasoning is essentially correct. Assuming that $\alpha_\text{weak}^4 \ll \alpha_\text{EM}\alpha_\text{weak}^2$, we can conclude that (a) is less probable than (b). As ACuriousMind notes, to be certain one should actually compute the full diagram, but the integrations from the loops usually give quite reasonable numbers that don't change the picture ...


1

In physics, fundamental particles are typically treated as point particles. In this approximation, they have no size or shape whatsoever. They sort of have a location, but we can never exactly pinpoint this location in space, because quantum mechanics tells us that a particle never has an exact location. The classical model of the electron does yield a ...


0

First, you have to realise that in the nucleus there are two main forces that you have to consider (relevant to this question, at least): the electromagnetic force and the nuclear force. The former interests particles with an electric charge (protons, electrons), the latter holds protons and neutrons together - remember that protons would repel, so you need ...


1

A diagram which is first order in $\alpha_\text{EM}$ would have to have one vertex, because $\alpha_\text{EM}\propto g^2$ where $g$ is the factor associated with each vertex (and the amplitude corresponding to the diagram gets squared). There's only one possible vertex in QED, namely the photon-electron-positron vertex, and it's impossible to arrange this in ...


0

I hear that analogy too. Spin 0: any rotation left the "object" invariant, like a circle who rotates. Spin 1/2: half rotation to het the initial state of the object, and here we are: any figure of the playing card "has spin 1/2". Spin 1: any non figure card, like who knows, the ace of clubs. One integer rotation to get is as it was initially. Spin 2: no ...


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 ...


0

You're right that the book's answer seems to have the wrong units; most likely the author meant to say $p_{max}^2$ in place of $p_{max}$. Also, both instances of $2m$ in the book's formula should really be $2m_p$. Assuming these two typos are resolved as just mentioned, it's clear what error the book is probably making. The book seems to be assuming that ...


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 ...


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

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.


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 ...


1

"Spin parity" isn't a thing. It's saying the xi baryon has spin $\frac{1}{2}$ and positive parity; they're separate properties whose names tend to be run together for some reason. As for why we use the word spin even though some of the angular momentum may be orbital: it allows you to imagine the $\Xi^-$ as an elementary particle which has the same amount ...


1

Is it possible that you're misinterpreting the energy diagram for the finite and infinite wells? The vertical scale is the energy, not the height. The spatial degree of freedom is the horizontal axis. Note that the potential is unchanging with distance except at the boundaries where the potential is discontinuous; a particle is free within the well.


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

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 ...


1

A short answer to questions 2 and 3: In Mermin-Wagner's paper the short-range condition is stated as $\sum_{\bf R} {\bf R}^2 |J_{\bf R}|<+\infty$. For interactions with (or more precisely majorized by a) power law decay $|J_{\bf R}| \sim R^{-\alpha}$, this requires $\alpha > D+2$, where $D$ is the space dimensionality (i.e., $\alpha >4$ for $D=2$ ...


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 ...


1

Relativistic mass is a weird concept that creates a lot of problems. I describe what mass really is in this post of mine. With that in mind, the mass is constant, even when a particle is accelerating. Its on this invariant mass that gravity acts on really, so in your example gravity will become weaker with separation, because the mass in reality stays ...


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 ...


0

As far as I understand it, the curves indicate the fraction of energy released at each point by each type of (ionising I suppose) particle thrown at the material. It can be interpreted as the percentage of energy lost in the neighbourhood of a point $M$ located at depth $d$ in the material. If you were to integrate the curve over the whole depth, you would ...


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.


0

My "answer" to this question is, for the moment at least, do more reseach and update this current note with more details as I discover them. I am reluctant to withdraw this question for the moment, I asked it ahead of myself and my knowledge level but I will return to it, even if only for my own personal notes. The answer to my question may lie in a ...


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 ...


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 ...


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 ...


0

I can think some speculative or unorthodox answers, and sure others can do, so please allow me to mark this answer as Community Wiki: three generations make a nice number of degrees of freedom for a GUT model. Assuming that the neutrinos have companions of the other chirality, one generation has 36 degrees of freedom. With this, the MSSM happens to have ...


1

The interference pattern appears when the two slits are at a distance of the order of the wavelength of the incoming waves. This is classical wave dynamics, nothing quantum to it. The quantum part is that particles are actually waves, and have an associated de Broglie wavelength, which depends inversely on their mass. Presumably, making your objects heavier ...


0

Gauge symmetry imposes local conservation laws, which are called Ward Identities in QED and Slavnov-Taylor identities for non-Abelian gauge theories. Those identities relate amplitudes or limit them. An example of those constraints imposed by gauge symmetry is the transversality of the vacuum polarization. To be more precise, gauge symmetry does not allow ...


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 ...


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 ...


0

My knowledge is okay about these things, so I will do my best. Gravity is a property of space-time that is a vector quantity and is accumulative with matter as it increases in mass and density relative to the volume occupied by that matter and energy. It becomes more significant as matter, in a geometrically defined region of space curves that space. The ...


-2

No, there are just spin-0, spin-1 and spin-1/2 particles, even the Higgs-Boson is 0.


-1

Well, photons behave both as a wave and as particles, and the photons in coming from the beam of light "enlighten" the particles in front of you.


1

For gravity, it doesn't matter whether or not mass is infinite, since the gravitational acceleration $\vec{g}$ is independent of the accelerated mass: $$ \lim_{m \rightarrow \infty}\frac{\vec{F}_G}{m} = \vec{g} \ne 0 $$ if the attracting mass is finte. This is what one calls equivalence principle and which makes gravity quite special, since every object ...


0

Well first of all, physically this can never happen, but as a hypothetical the answer is no. Because mass acts as inertia and if an object has an infinite amount of it, it will never accelerate. \begin{equation} \lim_{m \rightarrow \infty} \frac{\vec{F}}{m} = \vec{a} = 0 \end{equation} This is of course if you literally mean infinity, if you mean a lot of ...


-1

In physics infinity does not means that actual infinity.It only means that the parameter that we taken is something very much bigger than other parameters. the answer of your question is "yes".Mathematicaly F get the value of zero only indicate it is infinitesimally small


1

It seems to be an error (due the the units mismatch). You should doublecheck the errata page and perhaps contact the publishers. http://www.umsl.edu/~chengt/gaugebooks.html The errata is for the 1997 edition. It doesn't seem to mention this page.


0

To my knowledge, measuring any kind of gravitational interaction between sub-atomic particles is impossible with current technology. This is due to the fact that gravity is very weak, and the gravitational interaction between particles is negligible even at the high energy scales of the LHC. Furthermore, we do not have quantum theory of gravity (yet) hence, ...


0

Since you've posited full containment the conceptually "easy" approach is to try to pick out the sign of the beta in the decay state. Difficulties: The decay may happen a frame or two down the data stream. Drat you, oh 2.2 ms lifetime! Both options are likely to shower, and while the positron ought to form a double or triple shower when it annihilates, ...



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