# Tag Info

4

I will try to illustrate the sizes for you in my answer. However, with the examples you gave from the article and TV show, that is nonsense...they may be right with the size comparisons, but that is not an illustrative point. You kind of have to take that type of pop-science with a (large) grain of salt. First, let's familiarize ourselves with units. We'll ...

2

The mass of the neutrinos are estimated to some tenths of an $\mathrm{eV}$. The masses of atoms are mostly between 1 and 300 $\mathrm{GeV}$. Thus, considering the masses, this golf ball comparison isn't okay in my opinion. In my mind, comparing the mass of the Moon to the mass of the Solar System would be more realistic. Their sizes can't be easily compared,...

3

All objects and fields that have a nonzero mass, energy, or momentum interact gravitationally, and so do neutrinos – although they're very light and hard to produce so the gravitational force from any neutrinos we know is undetectable at this time. Neutrinos also have negligible but nonzero interactions with the electromagnetic field. They're uncharged and ...

1

A neutrino is thought to interact only through the weak force and gravity. They interact primarily, though, through the weak force (perhaps explaining the Martin/Shaw comment). Interestingly, since the neutrino has a minuscule mass (as opposed to none at all), it could have tiny neutrino magnetic movements, therefore allowing the possibility that it could ...

0

I totally agree with anna v's answer. Here I will just give some reasoning that you may use to determine the nature. Forbidden If the conservation of $Q$, Lepton number or Baryon number is not obeyed. Weak interaction If their are neutrino's involved, or the quantum numbers $S$, $C$ or $\tilde B$ are not conserved then the interaction weak (given that it ...

3

In a nutshell, no. Part of the problem seems to be that you misunderstand the fundamentals of string theory. The strings do vibrate. The frequency of these vibrations determines the type of particle and the energy of the string determines the energy of the particle. Second, your understanding of the uncertainty principle isn't quite right. Yes, we cannot ...

7

Life is not so simple, as in all high energy interactions there is a probability of a large number of particles appearing at the main interaction which will subsequently have decays through the weak or electromagnetic interaction. If one sees jets of hadrons in the detectors the strong interaction is involved, but the main vertex may be electromagnetic, ...

1

As far as I am concerned, I think you shouldn't take the question particle/anti-particle too literally. Here is why: Take you first picture. The bubble on the right hand side does represents a process which does not have a well defined stop or starting point. You could interpret it as (an this is how it is typically done) a particle anti-particle pair being ...

0

Let us consider what would happen if CP was actually a symmetry. Then, arguing non-perturbatively, the overlap between a $K^0$ state and a $\bar K^0$ after time $t$ would be $$\langle \bar K^0| \exp(-i H t) K^0 \rangle.$$ Let $U$ denote the operator implementing CP symmetry, such that $U^{-1}=U$. Then, since $[U,H]$ by assumption,...

1

Producing ultra-heavy elements in nature is not easy. So their absence "in nature" does not mean they cannot exist or cannot be created given the right conditions. Some details: The valley of stability becomes increasingly n-rich, so neutron capture reactions are essential. To get beyond lead requires rapid neutron capture in the r-process. The ...

2

Because no rational process can make them. I've been over the tables, and there are only a couple of possible reactions to get there for any nuclei, and they require two rare ones. Alpha particle capture just isn't going to cut it. Look at the curve; you need more neutrons. We remember that all elements heavier than iron have primary sources as neutron star ...

1

First of all, physical descriptions usually cannot be proven in a mathematically sense. Of course they should make sense and therefore mathematically sound. In particular when it comes to particle and antiparticles which is a concept of relativistic quantum field theory (QFT), actually a wavefunction as it exists in quantum mechanics no longer makes sense. ...

0

I cannot find the comment you cite in the link you posted. I think the limit on the higgs mass can be easily obtained by just using the fact that $M_h^2 = \lambda v^2, \lambda < \sqrt{4\pi}$ where $\lambda$ is the Higgs self coupling. Thus one gets $M_h^2 < \sqrt{4\pi}v^2 \sim (870 GeV)^2$ The violation of the Ward Identity just requires you to add ...

2

An explicit mass term violates Gauge invariance, because left and right particles belong to different representations 2.At one loop, the lepton mass is given by $m_{1L} = M_{bare} + \Delta M_{1L}(\mu = m_{1L})$ this condition uniquely defines the bare mass. The correction $\Delta M(\mu)$ is anyway proportional to some power of the yukawa, thus is very ...

0

In any theory with chiral fermions where Left and Right fermions don't have the same charges such fermions cannot have an explicit mass therm and can acquire mass only with spontaneous symmetry breaking. In the standard Model all fermions except Right handed neutrinos are charged under the SM gauge group and all L/R counterparts live in different ...

0

The magnetic force on a particle is given by the Lorentz Force: $F=q(\vec{v}\times \vec{B})$, where $\vec{v}$ is the velocity of the particle and $\vec{B}$ is the magnetic field Using Newton's second law, $m\vec{a}=q(\vec{v}\times \vec{B})$ where $m$ is the mass of the particle This is equivalent to: $ma=q(vBsin\theta)$ where $\theta$ is the angle ...

-1

The least you may require of something to be able to call it a particle is that it can be established where it is when, that it interacts, however weakly, with the objects in its environment. Since according to relativity theory a massless particle must move at the speed of light and at that speed there passes no time at all, the particle -its state- is ...

2

When the atoms get close to each other, the electrons of the first begin to feel the nuclei of the second, and vice versa. So, for example, when two hydrogen atoms approach each other, we have to consider the the potential due to all four particles (and the Pauli principle, but that issue doesn't have a bearing on the kernel of your question). The ...

3

As explained in this paper, the dominant effect is due to gravitational interactions, which can yield overdensities up to a factor of $10^3$ for neutrino masses of the order of 1 eV. The clustering of relic neutrinos can be modeled well using the collision free Boltzmann equation (Vlasov equation) where the densities evolve under the influence of the ...

8

I thought about this a little more since I prompted your question and there are a couple of other complications worth noting. This may be more like a comment than an answer, but it's too long for a comment. First, and directly addressing your question: the cosmic neutrino background is already present before the gravitational well of the star forms. If ...

8

It requires that they lose energy somehow (to drop hyperbolic orbits into periodic ones). There are two basic mechanisms available: gravitational scattering and weak scattering. In both cases we expect the interaction to be elastic, but that doesn't mean the neutrino has as much kinetic energy in the star's frame afterward the interaction as before: it ...

-1

There is also the confining strong interactions (quarks, gluons, color charge) which give hadrons mass independent of the Higgs (interaction, mechanism). There is also the dominant decay mode of the first detection of the Higgs boson which turned out to produce a pair of gluons, not to mention the masses of other fundamental particles / EM bosons which ...

2

Give it a try. The four momentum of a massive particle is $P~=~(E,~\vec p)$ and the invariant momentum interval is $$P^2~=~E^2~-~|\vec p|^2c^2.$$ Since E = $\gamma mc^2$ and $\vec p~=~\gamma m\vec v$ then $P^2~=~m^2c^4$ using $\gamma^2~=~1/(1 - v^2/c^2)$. Now suppose that this particle is converted into a photons with zero mass. This means $P~=~(h\nu,~h\nu/... 0 Consider it in condensed matter physics framework: we all know that around Dirac point, an electron can be considered as massless; however, if we turn on some kind of interactions, say spin orbital interaction, there would be no more linear dispersion, which give the particle somehow a mass. 0 The mass of a quantum particle is derived from the degree to which it is "linked" to the Higgs Field (a job done by the exchange of Higgs Bosons). For instance, a photon has no rest mass as it has no amount of link to the field. In order to "assign" mass to a particle, you would have to increase how much linkage there is between it and the Higgs Field ... 4 According to the Particle Data Group [1], Mesons are [strongly interacting particles that] have baryon number$\mathcal{B}=0$. In the quark model, they are$q\bar{q}'$bound states of quark$q$and antiquark$\bar{q}'$. This definition has two sentences, which in fact are somewhat different. The second one defines the quark-model mesons, that you're ... 2 From the last email of the particle data group : The 2016 edition will be published in summer. PDG Books containing Summary Tables and review articles as well as Booklets will be mailed in fall. Starting this year, the Data Listings will only be published online. 1 What we intuitively think of as "solid objects" are actually electromagnetic force-fields repelling each other. So you are correct; atoms are 'empty' in that they contain no solid objects or things. On the other hand, they are 'full' of basic force field which, in the aggregate, on a macro-scale, creates the illusion of 'solidity' that is what we perceive to ... 1 Your question is very broad, and actually popular science books about string theory, particle physics or cosmology are pretty much obliged to write about the subject in much greater depth than here. Anyway here's my short attempt starting with the quantum scale and using QM rather than QFT. We have plenty of equations, all of which incorporate the Planck ... 12 Because of the Pauli exclusion principle, it's extremely difficult to compress atomic matter beyond a certain density. It's not impossible, because there are always higher-energy electron states available, but there's a very strong force opposing it (called electron degeneracy pressure). This is what it means for space to be full. If you define "empty space"... 3 A charged particle like electron maybe is point-like (of radius zero), but it is "long-handed" as it is "felt" far away. In this sense it is not so "point-like". 55 Although it's commonly said that fundamental particles are point particles you need to be clear what this means. To measure the size of the particle to within some experimental error$d$requires the use of a probe with a wavelength of$\lambda=d$or less i.e. with an energy of greater than around$hc/\lambda$. When we say particles are pointlike we mean ... 10 Yes, elementary particles such as electrons and quarks (inside protons) are point-like or at least, their internal structure is incomparably smaller than the size of the atom. So the atom is mostly empty space. However, that doesn't mean that atoms may penetrate each other. Matter is impenetrable because of a combination of the uncertainty principle that ... 0 Yes, you can in a vector theory. But in the SM you cannot do that because of the chiral structure of the model:$l_L \sim \mathbf{2}$but$e_R\sim (e^c)_L\sim \mathbf{1}$. So you need a scalar which transforms as$\mathbf{2}$. 3 The upper limit you mention reflects the hypothesis that photons in vacuum could have some tiny rest mass. But it seems to be more important that c is the velocity of massless particles such as photons in vacuum. However, there is no real vacuum in the universe: Not only that even in outer space you will always find some interstellar atoms. But also, the ... 15 We can't measure to infinite precision; so even if a particle had in fact zero mass we couldn't experimentally measure it to the infinite precision needed to justify this; which is why certain amount of judgement is called for, and that judgement is made in the context of a theoretical framework. The second point to make is that all particles with zero rest ... 2 Because glueballs have energy, and$E = m c^2$says that energy is equivalent to mass. (Or another way to say it is that if you "zoom out" far enough that you can't see the constituent gluons that form the glueball, than you just lump all their energy into an effective glueball mass.) The energy can be thought of as just being the kinetic energy of the ... 5 If you had a gas of photons in a perfect cavity and these photons had energy$E~=~h\nu$, then for$N$photons the cavity would have a mass$m~=~Nh\nu/c^2$of photons. Glueballs as similar. The gluon carries two color charges (really color plus anti-color) and they can interact with each other. This forms a self-bound system that confines the massless gauge ... 4 Because in relativity the mass of a collection of particles is not necessarily the sum of the masses. Even two photons (treated as a unit) can have mass. Consider the total four-vector of a system with component four-vectors$(E,\hat{z}E/c)$and$(E,-\hat{z}E/c)$. It has mass$(mc^2)^2 = (2E)^2$. 8 Consider the average beta decay, which at nucleon level looks like $$n \longrightarrow p + e^- + \bar{\nu} \,. \tag{1}$$ The distribution of electron energies (as measured in proton's frame) is controlled by the phase space of the products. We observe an electron energy spectrum consistent with these physics. What you propose is essentially that this ... 4 This argument might conceivably work for weak decays, though I believe there is evidence to the contrary. This comes to mind, but I won't 100% swear it's quite what you're after. Peterh mentions in a comment that weak decay rates (e.g. in supernova afterglows) appear to be independent of the local density of dark matter. There is no reason to believe the ... 10 There are indeed massless particles. As of 2015 there were two known massless particles (both gauge bosons): the photon (carrier of electromagnetism) and the gluon (carrier of the strong force). It should be noted, however, that gluons are never observed as free particles, since they are confined within hadrons. Gravitons (if discovered) would be another ... 48 Here is a quick & simple answer until professionals arrive. On the Standard Model, it is zero. This$< 1.10^{-18} \frac{\mathrm{eV}}{c^2}$is an experimental upper limit (i.e. if it has a rest mass, because of physics beyond the Standard Model, it must be smaller as this value). This value is very small, compare to the estimated rest mass of the ... 24 The experimental detection of slow neutrinos is indeed a big problem, but one that is very important. The cosmic neutrino background is at a temperature of around 2K and likely to consist of non-relativistic neutrinos for plausible neutrino rest masses - with a density of around 340 cm$^{-3}\$ (all flavours). It is at this low temperature for precisely the ...

55

Strictly speaking, it is indeed incorrect that neutrinos travel at "close to the speed of light". As you said, since they have mass they can be treated just like any other massive object, like billiard balls. And as such they are only traveling at nearly the speed of light relative to something. Relative to another co-moving neutrino it would be at rest. ...

1

If I'm understanding correctly, you're asking whether an arbitrary QFT admits an asymptotic Fock basis. If a QFT does admit a Fock basis, you can talk about particles and do scattering theory in momentum space and construct position operators for single particles. But it's not guaranteed that a given QFT admits asympotic Fock bases. Quantum field theory ...

-1

No, no, no. Helicity and chirality are not the same thing in the massless limit. They are unrelated. Helicity is an extrinsic physical property related to the alignment of spin and momentum; chirality is related to weak interactions. Chirality is more akin to electric charge or strong color charge than it is to momentum.

3

In QFT, a single particle does not scatter, hence its (renomalized) wave function in an interacting theory is the same as the corresponding asymptotic wave function in the asymptotic Fock space. However, the multiparticle picture breaks down as the interacting Hilbert space cannot be identified with the asymptotic Fock space, by Haag's theorem. Thus ...

2

The conventional way to handle bound states in relativistic quantum field theory is the Bethe-Salpeter equation. The hydrogen atom is in QFT usually treated in an approximation where the proton is treated as an external Coulomb field (and some recoil effects are handled perturbatively). The basics are given in Weinbergs QFT book Vol. 1 (p.560 for the Bethe-...

1

In a simple word, particles are thought to be excitations in fields. Particles are not interactions but we notice particles when they interact because we are capable of noticing the change in interaction. (How I see is it's not the EM field that creates photon but photon itself is an excitation of the quantum electromagnetic field.)

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