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27

Basically, the answer is no, it's not possible. When we produce neutrons for research purposes, we have to produce them using nuclear reactions. They come out of the nuclear reactions with energies that are determined by the reaction, are not otherwise under our control, and that are on the MeV energy scale of nuclear physics. Examples of a neutron source ...

14

Although a neutron is electrically neutral, it has a non-zero magnetic dipole moment. It interacts with a magnetic field to give a potential $$U = \vec{\mu} \cdot \vec{B}$$ A gradient of magnetic field strength will give a force $$\vec{F} = \nabla|\vec{\mu} \cdot \vec{B} |$$ It's not possible to produce large, sustained field gradients, nor is it ...

14

It would violate the law of conservation of baryons. Baryons (half-integer-spin particles, i.e. s=1/2, 3/2, 5/2,... interacting through the strong force) cannot be created at will, but must conserve the total baryonic number: protons and neutrons both have $+1$ baryon number, while their antiparticles, antiproton and antineutron, have baryon number $-1$ ...

11

The photon does couple directly to charged stuff, e.g. via Compton scattering. This is indirectly related to the spin, as direct interactions between fermions are hard to construct. The neutrino on the other hand does not couple immediately to any other matter particle. It requires a force-carrier. Now as it turns out the only force carriers that care about ...

10

A neutron contains (on average) 1 up quark and 2 down quarks. The decay to a proton occurs when a down quark emits a W$^-$ particle and changes to an up quark. This gives a proton with two up quarks and 1 down quark. The W$^-$ particle decays to an electron and anti-neutrino. However an antiproton contains 2 up antiquarks and 1 down antiquark, which is ...

9

The standard model is very successful in its group structure in ordering all observed particles. To introduce a particle with charge and zero spin, you will need a different model that would also accommodate the symmetries observed experimentally and fitted by the standard model. So the answer to "why" is "because" we have not seen any and can model well ...

8

Neutrinos having no charge means they don't participate in electromagnetic interactions, which are the strongest (at least long range). Them being leptons means they don't interact with the strong force (which is, as the name says, strong), hence they can only interact via the weak force, which is, as the name says, weak. Photons on the other hand do not ...

8

the only difference between matter and anti-matter is simply charge Nope. All "quantum numbers" reverse for them. In particular, for the example you consider, there's Baryon number, which is different for baryons and anti-baryons, and in particular, for a neutron and its anti-particle, an antineutron. It is in fact possible to differentiate between the ...

8

Neutron decay is not a electromagnetic phenomena at all. It is governed by the weak nuclear force. This is well supported by fact that the lifetime of the neutron fits neatly into weak universality. Secondly the neutron is not it's own antiparticle. The anti-neutron is a distnct particle.

5

The reason is that the $SU(2)$ invariant in $\mathbf{2}\otimes\mathbf{2}$ (or in their complex conjugate $\mathbf{2}^*\otimes \mathbf{2}^*$) is given by contracting the two $\mathbf{2}$ with the anti-symmetric $2\times 2$ matrix $\epsilon_{ab}$, as $i\tau_2$ is. In the case at hand the two $\mathbf{2}^*$ are $\bar{Q}$ and the $\Phi^*$. You could form another ...

5

It's a matter of history. When George Stoney developed Stoney units in 1881, or when Robert Millikan performed the oil drop experiment in 1909, it wasn't yet known that it was possible for anything to have a charge smaller in magnitude than the charge of an electron. By the time the quark model was proposed, in 1964, the use of the "elementary charge" being ...

5

Briefly, but perhaps less briefly than the footnotes to the tables, and reflecting my own personal biases: The strings like $I^G(J^P) = 1^-(0^-)$ represent the quantum numbers of the particle. $I$ is isospin: the pion, with isospin 1, is a member of an isospin triplet ($\pi^+,\pi^0,\pi^-$). $J$ is spin: the pion is spinless. $P$ is spatial parity: the pion ...

4

I addressed this a little at your other question but this one is more like physics. Yes, they're decay product lists. Beware that if all the modes are only "seen" or "not seen," you are sort of looking at the hairy edge of what's experimentally accessible. The indented lines are subtypes of the same reaction. Using your example, a neutral particle can ...

4

The sun gets its energy from the pp-chain. The first step is the two protons forming the diproton (Helium-2): $$\,^1_1H+\,^1_1H\to\,^2_2He+\gamma$$ where the $\gamma$ is the photon (of energy about half an MeV). This quickly $\beta^+$-decays into a deuterium by converting a proton into a neutron: $$\,^2_2He\to\,^2_1D+e^++\nu_e$$ where $e^+$ is the ...

4

For a scalar field $\phi$, the most widely used convention, based on my experience, is to write the Lagrangian with kinetic and potential terms, followed by interactions like so, $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2 - \frac{1}{2}m^2 \phi^2 - \sum_{n \geq 3} \frac{\lambda^n}{n!}\phi^n$$ where $\lambda_n$ are coupling constants. (We could not have a ...

4

The Standard Model Yukawa interactions must be $SU(3)\times SU(2) \times U(1)_Y$ gauge invariant. The down-type Yukawa interaction is $$\mathcal{L} \supset -y_d \bar Q \phi d_R + \text{h.c.}.$$ This is indeed gauge invariant. The $\bar Q d_R$ form a colour singlet ($3^* \times 3$), the $\bar Q \phi$ form an $SU(2)$ singlet ($2^*\times2)$, and the whole ...

4

The difference between the Higgs boson and the bosons of the three/four fundamental (depending whether you include gravity as a quantized theory or not) actions is that the latter are associated with gauge symmetries, while the Higgs plays a role in spontaneous symmetry breaking. Photons, W- and Z-bosons, gluons and gravitons arise from the requirement that ...

4

The momentum is harder to deal with than the energy. If a stationary neutron decays into an electron with momentum $(a,b)$ and a proton with momentum $(a,-b)$, then there is no way to conserve 3-momentum without the creation of a third particle with momentum $(-2a, 0)$ And while real numbers wouldn't work out this nicely, it would be obvious that the ...

4

It's not necessarily true that most of the photons that strike a wall will be absorbed and turned into heat. The whitest white paints can have a light reflectance value of up to about 85%. There isn't a "wavelength corresponding to white color". An ideal white surface reflects as much as possible of all wavelengths in the visible spectrum. That sounds ...

4

A particle is defined as an irreducible representation of the Poincar\'e group. These were classified by Wigner in 1939. This was done via the little group construction. The important representations are (metric signature $(-,+,+,+)$ $p^2 = 0$, $p^0 < 0$ - The little group is ISO(2). All finite dimensional representations of this group are ...

3

Lets analyse the Majorana condition and the Majorana mass term. A massive Majorana neutrino $\chi_j$ (a Majorana spin $1/2$ fermion) having mass $m_j>0$ can be described in a local quantum field theory (eg. the standard model) by a four component spin $1/2$ field $\chi_j(x)$ which satisfies the Dirac equation and the Majorana condition which reads: $$... 3 The Lagrangian$$\mathcal L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \mathcal L_\textrm{free} + eA_\mu J^\mu \tag{1}$$where A_\mu is the 4-potential, F_{\mu\nu} = \partial_{[\nu}A_{\mu]} is the field tensor, \mathcal L_\textrm{free} describes fields other than A_\mu, and J^\mu is the 4-current density expressed in these other fields, describes a ... 3 Interesting question! This may not directly address your question, but I think it might be relevant, and it's too long to fit in a comment. Some questions that arise here are: What do we mean by a negative-energy object? Do we have any reason to think that they're worth searching for? If so, is there any method that we know should, at least in principle, ... 2 Within the limits of accuracy of the experiments, no negative energy particles, with negative invariant mass, have been leaving a signature. Particle accelerators deliver beams at fixed energies. All the event ordering algorithms depend crucially on energy and momentum conservation, that is what missing energy and missing transverse momentum are all ... 2 How would you define a negative energy particle? Is that one that, when it hits your detector, takes a fixed amount of energy out of it? That's trivially forbidden by the third law of thermodynamics, otherwise you could construct a fridge that can make negative temperatures. The only way to escape that would be by requiring, that the particle can measure the ... 2 Because muons don't carry a color charge, and hence don't participate in the strong interaction... 2 How is it possible to accelerate a neutron? Neutrons have a dipole moment, so they may be 'accelerated' insofar as they will turn in a magnetic field – that is their primary interaction with the electromagnetic field. It is possible to accelerate a charged particle in an electric field, how is it possible to accelerate a neutron? Neutrons also ... 2 An electron is a charged particle, charge conservation would not work as the neutron has zero charge. In addition it would have been detected with its interaction as its energy would be similar to the energy of the other electron seen. The neutrino was posited as a weakly interacting particle exactly because it was not caught by the detectors, and because ... 2 The important point is the fact that such a mass term breaks the gauge symmetry (Edit: I am assming that you want to build the Majorana mass term using SM available fields -- no extension considered -- of which there is only \nu_L). Namely, the desired term is (one generation suffices):$$\frac{1}{2}\,M\, \nu_L^T \,\mathcal{C}^\dagger\,\nu_L\, +\, ...

2

If you are considering massless neutrinos there is no such a diagram since all interactions would preserve flavor. If you take instead massive neutrinos, you are probing lepton flavor violation within the SM since the new interaction with $\varphi$ respects flavor. It is thus very very small, being controlled by the neutrino masses. In turn, it is therefore ...

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