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1

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

It depends on what theory of everything nature follows. There are some theories that predict that protons decay, so if one of the protons decays the particle with disintegrate as one proton and two neutrons are not a stable system.


0

Spin is the total angular momentum, or intrinsic angular momentum, of a body. The spins of elementary particles are analogous to the spins of macroscopic bodies. you can look at http://en.wikipedia.org/wiki/Spin_%28physics%29


2

Good question! I can maybe guide you in the right direction, although I only found this post because I wanted clarity myself. The reduced width idea comes from the R-matrix formalism (a good paper is by Descouvemont and Baye here). The most basic understanding of it is that the most general cross-section for an interaction of two nuclei (in which a ...


0

When two atoms approach each other, there comes a point where it is no longer possible to describe the system as "two atoms". Rather, it becomes "one molecule". This is a gradual transition. When the atoms are far away from each other, they behave like separate atoms: one wave function for the atom on the left and an identical atom on the right having ...


0

This is an attempt at hand-waving over an effect that is not really explained in the statement you gave. In a periodic (lattice) arrangement of identical atoms, the wavefunctions of electrons can also be described as having the same periodicity. In fact, it is the best (in the sense of most practically useful) description for this sort of system. Then your ...


0

To justify giving mass to a would-be massless particle, scientists were forced to do something out of the ordinary. They assumed that vacuums (empty space) actually had energy, and that way, if a particle that we think of as massless were to enter it, the energy from the vacuum would be transferred into that particle, giving it mass. A mathematician named ...


1

Comparing the radius of the H atom to positronium is misleading because (in the classical sense), the H atom has the electron on the circumference of a circle (with the Bohr radius) whereas positronium's constituent particles' equal masses produce two similar elliptical orbits orbiting a common center of mass. If those ellipses happen to be circles, then ...


0

The Higgs field is actually two complex scalar (spin 0) fields so there are two particle and two anti-particle excitations (quanta). The pair of fields transform as an electroweak doublet which essentially means that the Higgs field quanta interact with the electroweak gauge field quanta (W and B bosons). In addition, the Higgs field has a peculiar ...


2

The energy is: $$ E = \frac{n^2h^2}{8mL^2} $$ Your mistake is that you have $L$ not $L^2$ in the denominator so your answer is a factor of $L$ too small.


0

Yes, an electron is just some wave, as you say, in the electron field, as it is for any particle. You can also interpret in a broad sense that a field needs to be perturbed at a particular point in spacetime for you to have a non-zero odd of measuring it a that point, although this simple picture is complicated by quantum phenomenas. The energy of a ...


2

Short answer: We can measure their energy and momentum distribution functions in the nucleus. We do this by interacting with them individually, either knocking them out of a nucleus left otherwise undisturbed (quasi-elastic scattering) or by exciting them to higher energy states inside the nucleus (many inelastic scattering reaction backed up by data from ...


5

Evidence that there are distinct protons and neutrons in nuclei starts with the Pauli term (pairing term) in the semiempirical mass formula of the liquid drop model. Furthermore, all nuclei with even numbers of protons and neutrons have nuclear spin of zero. This is consisent with shells being filled with spin up and spin down pairs of nucleons, each pair ...


3

ATLAS has no experiment-wide definition of "fiducial", it basically means sensitive to signal. The definition is confusing because, unlike most experiments, ATLAS (and CMS, D0, CDF, etc) doesn't just define the physical area where the experiment is sensitive, they also define collision properties. This means the definition of fiducial isn't limited to the ...


3

This presentation (NB: PDF) has a "jargon" page that states, Fiducial (Webster's): Taken as a standard of reference Founded on faith or trust Having a nature to be trusted Fiducial Volume (Particle Physics): The volume used to make physics measurements The volume where the detector is assumed to be well understood With the ...


2

In some detector experiments, The response at the periphery of the detector is poorly understood. The majority of background events interact in the periphery of the detector. The periphery of the detector is the final shielding. Some parts of the detector may be broken. In such cases, results from such parts of the detector are ignored. The results are ...


1

The Dirac operator $H=-\alpha\cdot i\nabla+\beta m$ is self-adjoint on $L^2(\mathbb{R}^3,\mathbb{C}^4)$. Therefore you may write a general solution as $\psi(t,x)=e^{-itH}\psi_0$, $\psi_0\in L^2(\mathbb{R}^3,\mathbb{C}^4)$. The problem with explicit solutions of the "eigenvector type" (for an energy $E$) is that they cannot belong to the $L^2$ space, for the ...


1

There are two methods for making predictions for LHC searches: Data-driven. If you are searching for a signal, extrapolate the background prediction from a "sideband" region to a "signal" region of phase space. Monte Carlo (MC). Generate events with in pseudo-experiments (event generation from matrix elements, followed showering, hadronization, and ...


-2

Also : Lepton number conservation law, respectively Barion number conservation law, are laws which explain if any reaction can occur or not. Every particle has its own leptonic (barionic) number, and for the reaction to occur, the sum of those numbers in the right side of the equation MUST be equal to the sum of leptonic numbers in the left side of the ...


0

Although there is already an accepted answer, I want to give some further ideas: While the Bragg-peak is the "perfect" solution to get a high percentage of effective dose to the desired tissue, the exact positioning of this peak in the tumor tissue can be challenging due to misalignments of CT/MRI data and the application gantry and furthermore due to ...


0

proton collisions from a cern blog (LHC) Anything is possible provided the conservation of energy holds. the collision results in a shower of all types of particles,.. in a proton-proton collision “anything” can happen, provided some important principles are respected, such as energy and momentum conservation. How many quarks? The graphic ...


1

Your guess is essentially correct. If you want to think about the nucleus in a shell-model sort of way, you can say that the nuclear spin $J$ is the vector sum of the spins $S_i$ and orbital angular momenta $L_i$ of all the nucleons in the nucleus. In the deuteron, for instance, the nucleus must be antisymmetric under exchange of the "identical" proton ...


1

Presumably you are talking about mesons. First of all, parity is not (-1)^L for a meson, it is (-1)^(L+1). This is because you have to take into account the fermion-antifermion relative parity for the quark-antiquark pair, which is negative. Secondly, yes, it is L that matters and not J (which includes S). Thus, the low-lying mesons (pi, rho, D, D*, ...


2

Some practical information first - the charged particle beam passing through the matter suffers from energy (and also angular) straggling. That means that even if an ideally monoenergetic beam is used, there will be always a finite volume with Bragg peak losses. The bigger initial energy, the bigger is the volume. Protons stopping at 40 mm have straggling ...


1

There is a thing called energy, and it is conserved in the sense that if it leaves a region, you now have less of it in that region unless or until some more comes back into the region. If you have some fundamental particles or composite objects they can have different energy based both on their state and how far apart they are from each other. But if they ...


0

Well, I know some but not all of the answer numerically. But the main answer is "electromagnetic repulsion" and "Pauli exclusion principle." The nucleus is very, very small compared to the entire atom, so nuclear effects are not much of it. Off the top of my head, I've seen the calculation for how much of the pressure in an ordinary metal (which is basically ...


0

It's made up of many particles. I would assume some of them "orbit" the center. I know many electron orbitals have angular momentum. The orbitals of nuclei are less well-understood, but I would expect the same is true.


0

The angular momentum of the nucleus is the combined contribution of the spin-orbit angular momenta of the constituent particles. In order for an entity to have orbital angular momentum of its own it must some conceptual orbit: electrons in the atom, protons and neutrons in the nucleus, atoms in a molecule. That's why the angular momentum of a nucleus is ...


0

Influence of internal EM forces on the sphere cannot be calculated based on the field momentum outside the sphere - it is more complicated since the field is not simply moving with the sphere when the sphere moves in a general way. Also, the electromagnetic defect is present, but it is explainable purely with mutual EM forces. No effect of zero-point field ...


1

this electromagnetic mass $m_{elec}$ has to be added to the standard "mechanical mass" of the sphere to give the total observed mass of the object.  Would this view be accepted by most physicists today? That is a wrong idea. Why? There is no reason (in this case) to count quantity $m_{elec}$, defined based on EM momentum distributed in the whole ...


2

It is indeed the case that the Casimir force for a three dimensional sphere is repulsive. http://arxiv.org/abs/hep-th/9406048 Neither myself nor the authors of this paper know an intuitive explanation of why this is the case. It is important to remark, however, that this result actually disproves the naive and very common "mode counting" argument that ...


-4

There is an important alternative to the Higgs mechanism. This will be explained by first stating that photons are only massless particles when they are freely propagating. If a photon is confined to a specific volume, for example in a hypothetical reflecting box, then the photon is forced to adopt the frame of reference of the box and the confined photon ...


0

I don't think that would be possible since there is vacuum outside of the electron as well, everything between the electron and the nucleus is vacuum.


1

You mixed up eq43 with eq44, you should replace $\left|B\right\rangle$ with $\mathcal{M}(\uparrow\uparrow\rightarrow B)$ But in any case, in the derivation of this expression an important assumption has been made, one which does not hold for pions! The bound state must be non relativistic to a good approximation. This is true for heavy quark mesons, not ...


2

Since the collider is a big circle, even if you mounted it vertically it would make no difference, because half of the time, the particle would have to oppose the gravity. But the other answers are also correct in that the magnitude of any effect due to gravity is too small to make a difference anyway.


4

Your reasoning is correct in theory, but flawed in magnitude. Placing the Large Hadron Collider vertically would have little to no effect on the final velocity of the particle. Since the particles fired through the collider are nearly massless, the potential energy of the particles is very, very small. The corresponding increase in final velocity will be ...


0

It really won't help much. It's the weakest of the fundamental forces, and it can't be used to overcome fundamental issues regarding the speed of light without some exotic features (wormholes, negative masses) which we don't have good access to on Earth.


-5

Uhole Model This model is based on John A. Macken´s proposal,[2] that universe is only spacetime and can be seen as a sea of energetic waves (Dipole Waves), traveling at light speed. From Macken´s Dipole Waves (DW model), Policarpo Y. Ulianov [5] defines a fundamental particle, named Ulianov Hole (uhole),[3][4] that can seems as an elastic tube that ...


1

There is no well established theory predicting the neutrino mass (even if of Majorana type). A particle is its own anti-particle if the field describing the particle is a real field. That means obviously that the electric charge is 0 but not necessarily that other charges are zero. For instance, if the neutrinos are Majorana neutrinos, they would still have ...


3

If there were a $t$-channel or $u$-channel diagram for this process, it would have to involve a vertex where an electron changes into a muon and some other particle. There is no such vertex in the standard model.


1

In QED there are vertex only with the same two lepton line and one photon line. I mean that there is not vertex when $\mu\to e\gamma$ in QED.


0

When someone refers to the quadrupole moment of a nucleus, they usually assume cylindrical symmetry and are referring to the $zz$ component of the tensor $Q_{ij}$: $$<Q_{zz}> = 3z^{2}-r^2$$ For a spherical nucleus, $<Q_{zz}>=0$, but this doesn't mean it has no size. So no, in general you can't get an estimate of the size of the nucleus from the ...


4

Such information can be inferred from the differential cross section. Different spins and parities lead to different angular distributions of decay products / scattering partners. You find these correlations by using one decay product to define an axis and measuring the distribution of the other decay products with respect to that. This requires a lot more ...


0

It's Fermi repulsion between the quarks in the nucleons, not between protons and neutrons, that prevents the overlap. You wouldn't claim that a hydrogen and helium atom should overlap because they are distinguishable - obviously you need to consider the repulsion between the electrons they are made of. The same argument applies to nucleons.


0

Atoms/particles are accelerated with an appropriate force, according to the second Newton "law".


1

On a windy day, sound travels faster "downwind" than "up wind". The "particle accelerator" is the sun - through the medium of heating the atmosphere and causing wind. If you travel with the same speed as the air, you will see sound travel at its usual speed; but if you are standing still, sound "from upwind" will arrive a little bit faster than sound "from ...


1

(Moving this from my comment to an answer) Yes, the electric field simply penetrates the glass wall and charges (the electrons) placed in that field will feel a force and move. The glass does not really interact with the charges on either side, so you might as well remove it completely (theoretically).


3

Beliefs have no real place in physics, at least should not have. The proposal of particles not seen comes from the theoretical models in order to complete them. For example the omega minus was predicted more or less exactly from the symmetries of the weak SU(3), and was later found .A more exact prediction than the Higgs. The Higgs as a particle might not ...



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