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0

Gluons are never found in isolation. If you had a way of directly detecting the mass, charge, and spin of particles, and detector finds something with no mass and no charge and spin 1, it's a photon, definitely not a gluon. Besides, photons interact electromagnetically, whereas gluons don't. The way modern detectors are constructed, they typically have a ...


1

Indeed you cannot measure gluons, since the fact that they are color charged implies they are always confined. Thus, while photon may freely escape from the colission, gluons, like quarks, may not. That is the reason why you measure in your detectors hadronized matter, whose color content is "white".


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Such operators are ill-defined in an interacting theory because whatever counterterms we try to subtract, their expectation value in any finite-energy state will diverge. The closest operators that are well-defined are densities of charge – number operators with signs labeling antiparticles – because the divergent contributions naturally cancel for them. ...


0

In fact, your question is not so clear. I try my best. The Yukawa potential is an exchange potential, so it is based on the particle which is exchanged between 2 interacting particles. So if that particle (a boson) is coupled to the Higgs boson, it will get mass and the potential between the 2 interacting particles should change from U(r)~1/r to U(r)~ ...


1

In your previous question, you say that these two images (a)  (b)  have been taken of the same area of your sample but with a different lens and lighting configuration. I think that I believe this: in this first image, there are two dark areas (the left shaped sort of like New York State, and the right shaped sort of like a heart), and in the second ...


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I'm not sure what you're asking, but it's not really true to say that particles interact with the Higgs field. A quantum field like the electron field interacts with the Higgs field and the result is that the electron field is massive i.e. its excitations (electrons) have a mass. If you consider the Higgs boson, rather than the Higgs field, then the ...


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This is a bit pedantic, but when you say catch free neutrons -- I assume you mean those that exist from non-discrete sources undergoing reactions (transmutation, decay processes, etc.) rather than neutrons that are produced in the unbound state from a specific source, be it "fusors" or a radioactive source coupled with a target material or a source that has ...


1

If you want to stick to a n-particle-interpretation, for the electron, your answer is the Dirac Equation. However, it breaks down as soon as you reach sufficiently large energies, a unified description is only possible within quantum field theory. For the photon, the situation is more difficult as it only appears as excited state of the electromagnetic ...


0

The implication of the question is that gravity is already at the "limit" of our perception, and therefore any force that might exist, but weaker than gravity, we might not be able to perceive it. With the passage of time, we have improved our perception abilities tremendously. If we continue to improve, and there are other fundamental forces weaker than ...


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A state like $ \frac{1}{\sqrt{2}}(a^\dagger_+(\vec p)a^\dagger_+(-\vec p) + a^\dagger_-(\vec p)a^\dagger_-(-\vec p))|0\rangle$ would be an example. It is both entangled in spin and entangled in momenta.


2

Let $f(x,y)\in L^2(\mathbb{R}^{2d})$ and $\Omega$ the vacuum of the symmetric Fock space $\Gamma_s(L^2(\mathbb{R}^d))$. Suppose there is no $f_1,f_2\in L^2(\mathbb{R}^d)$ such that $f(x,y)=f_1(x)f_2(y)$: then $f_s$ (the symmetrized of $f$) is an "entangled" two particle state of $\Gamma_s(L^2(\mathbb{R}^d))$. This is created by $$\frac{1}{\sqrt{2}}\int ...


4

You can estimate the neutron lifetime using dimensional analysis. Beta decay is correctly described by the well known four-fermion Fermi theory, so the amplitude must be proportional to the coupling $G_F\approx10^{-5}\text{GeV}^{-2}$ (the Fermi constant). The decay rate is proportional to the squared amplitude: $$\Gamma\propto G_F^2\thinspace.$$ $\Gamma$ ...


0

Physicists collide particles to study their behavior under extreme conditions. New unknown particles can be created in high energy collisions or new unknown processes may be observed. Our equations describing the particles are predicting certain behavior and physicists are testing if the particles really behave like that. And they are hoping to find a ...


1

The high energies of the LHC helps in a sense "reproduce" energetic eras of the universe right after the big bang, or even reproduce collisions of high energy particles in space. During those periods of the universe, the energies were high enough for more massive particles to be produced (Recall the more mass a particle has, the more energy it needs to be ...


1

In space there are phenomena leading to gigantic energies or temperatures. This can create such particles. In fact, there are collisions of high energy particles in space. We here on earth reproduce these effects using a collider like the LHC. But still the maximum energies per particle we meassure comming to us from space are far from reached with any ...


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The link posted by akhmetelli suggests it's because of parity nonconservation: in the center-of-mass frame for the scattering, at high energy, only left-handed particles and right-handed antiparticles participate in the weak interaction. The two particles may always scatter through an angle of 0º; the technical term for that is "they missed." So the ...


0

The simplest way to see that all products of your kind vanish is to notice that one of the objects (bilinear invariant) $T_{\mu\nu}$ is a self-dual 2-form while the other $T^{\mu\nu}$ is anti-self-dual, and their contraction without a complex conjugation has to vanish. A self-dual (anti-self-dual) antisymmetric tensor obeys $$T_{\mu\nu} = \pm \frac i2 ...


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Looks like the answer is quite messy - I cannot reproduce it here - ...


0

Z boson can decay to a neutrino-antineutrino pair. Your this process is the reverse of the decay. Direct combination to photon(s) is impossible, because photons don't interact weakly, and neutrinos don't interact electromagnetic. Direct combination to Higgs were (IMHO) possible, because it doesn't contradict to a preservation law, but I think it has an ...


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The neutrino and the anti-neutrino can annihilate to create a $Z$ boson. But the mass of the $Z$ boson is around $90$ GeV, so in order to create such a boson, the neutrinos need to be high energetic. Theoretically, a Higgs boson could be created as well, but for that an even larger amount of energy is needed, since the Higgs boson is heavier than the $Z$ ...


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Yes, the Z boson can decay into a neutrino and anti-neutrino, and the process you describe is just the time reverse of this.


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The charge conjugation symmetry is conserved in the strong interaction. In the above decay mode, neutral rho meson decay into two neutral Pion, neutral rho meson has -1 and neutral Pion has +1 for C such that decay violates C symmetry. Therefore it is forbidden.


-4

The weak force is like why water stays together in a system. The strong force is why when the atoms in the bottoms of your feet don't go through the ground when you walk. Objections?


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This is a nice idea that has been studied thoroughly, and continues to being studied. You are talking about a type of sub-critical reactors referred to as Accelerator Driven Systems (ADS). Here is a nice review Review:Basics of accelerator driven subcritical reactors, where references to previous studies are given. For modern advances you might like to check ...


6

Anomalies (not anamolies) are a whole subject whose basics are covered by one or several chapters of almost any good enough quantum field theory textbook so it's counterproductive to retype this whole chapter here. But generally, in quantum field theory, anomalies are quantum mechanical effects breaking symmetries that exist in the classical theory – ...


2

There is some debate about whether it exists or not, but there has been some research into what is called a quark star. This article (should be a free link, but here is the arXiv version in case) suggests that A recent calculation for cold and dense QCD strange quark matter including corrections to order $O\left(\alpha_s^2\right)$ indicates that ...


2

A good place to look to get an idea of the experimental bounds is at Eötvös experiments, which attempt to measure deviations in the inertial mass from the gravitational mass. Another way to think about this, is if there was a force we can not yet model, it would show up as an effective change in the inertial mass in these experiments. So far, these ...


1

We may ask, what holds a negatively charged electron together (since it has no nuclear forces), if an electron is all made of one kind of substance, each part should repel the other parts. Why, then, doesn't it fly apart? But does the electron have "parts"? Perhaps we should say that electron is just a point and that electron forces only act between ...


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From an experimentalist's point of view in order to measure something or set a limit, a specific theoretical framework has to exist. Take as an example the limits measured for the electric dipole moment of the electron. The experiment is specifically designed on the supposition that the signal will come from a T asymmetric theory proposed. As you can see ...


4

There can certainly be extra forces. For example the Higgs particle can be regarded as the carrier of a fifth force, though as you'll see from the answers to that question whether or not this is really a fundamental force is debatable. More generally there have been discussions of possible extra forces for decades, for example Brans-Dicke theory postulates a ...


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With a sufficient "tolerance", one may of course envision forces that are weaker or much weaker than gravity. Experimentally, one may only improve upper bounds on the strength of the new forces. On the other hand, there exist rather strong theoretical arguments that gravity actually has to be the weakest long-range (power-law) force for the consistency of ...


1

As the previous post mentioned, forget about the concept that the electron is actually spinning. Spin, like rest mass and electric charge, is an intrinsic property of subatomic particles. Yes, it's angular momentum. No, nothing is spinning. Although many physicists today do not like this explanation, special relativity introduces a useful analogy with mass. ...


3

First, the electron isn't actually spinning. Physical objects made up of collections of electrons and protons (and neutrons) can have angular momentum because they rotate; the electron does not get its angular momentum for the same reason. Second, the magnetic moment of an object with angular momentum L is proportional to $$ \mu \propto \frac{qL}{M} $$ ...


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[...] $\Delta^+ \rightarrow p + \pi^0$, [...] $\Delta^+ \rightarrow n + \pi^+$, which process is favored: the proton and neutral pion or neutron and charged pion [?] Since the kinematics (and corresponding "phase space" factors) for the two final states are presumably as good as equal, the evaluation of the branching ratio $$\text{BR} := ...


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Don't think about annihilation as something exceptional. Annihilation is just a type of interaction and there are many other possible interactions. Now I don't use the word "interaction" in the sense of "4 fundamental interactions", but in the sense of possible process in the quantum world where particles are destroyed and created. So the question should ...


7

If it were possible for one object to pass through another object, then it would be possible for one part of an object to pass through a different part of the same object. Therefore the question asked here is equivalent to the question of why matter is stable. See this question on mathoverflow. That question was more about the stability of individual atoms, ...


5

Yet there are all kinds of reactions that cause light, seemingly without providing the infinite amount of energy needed to accelerate particles to light speed. If you're imagining that there are photons at rest within the flashlight when it's off and the flashlight accelerates photons to light speed when it's turned on, then I can see why you're ...


2

Photons have no mass, they always travel at the speed of light. Not so for massive particles. You could try building a collider with massless particles, but you would fail for several reasons: The only freely propagating massless particle is the photon, but it does not have a charge, so you cannot bend it to follow a circular collider. You could still build ...


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It takes an infinite amount of energy to accelerate a particle with mass to the speed of light. A photon does not have mass, thus can move at the speed of light. Note that a photon does not accelerate; the moment it is created it moves at the speed of light.


3

The Higgs field is not giving mass to all other particles. There are particles that acquire mass differently - for example neutrinos. Or the yet undiscovered particles of dark matter probably don't get mass from the Higgs field. Also please notice the difference between the Higgs field and the Higgs boson. The Higgs field is giving mass to some particles, ...


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Things are not empty space. Our classical intuition fails at the quantum level. Matter does not pass through other matter mainly due to the Pauli exclusion principle and due to the electromagnetic repulsion of the electrons. The closer you bring two atoms, i.e. the more the areas of non-zero expectation for their electrons overlap, the stronger will the ...


6

Gluons are bosons, they have spin one and are mediators of the strong force. They have mass 0 and there is no weak interaction vertex with a gluon. The extract is talking of the Higgs field, not the Higgs boson. A field in physics is A field is a physical quantity that has a value for each point in space and time at that point. The Higgs field ...


1

Luminosity is necessary in order to turn number of interactions to crossections, because theories provide crossections to compare with experiments. Experiments measure number of interactions. A well known crossection, as is Bhabha scattering, substituted on the right will give the luminosity to be used in the other observed interactions in the experiment. ...


2

If I'm understanding, you're asking: How does the Higgs field "give" particles mass? This is an explanation given by Henry of MinutePhysics (To be clear, we're talking about the Higgs field and NOT the Higgs Boson, which is merely an excitation leftover after the process we're about to explain. But I digress…back to mass!) To begin, we need to know what ...


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Does a particle enter/interact with the Higgs Field when created, or at some other time? After reading your question a couple of times as well as your comments, it occurs to me that you're picturing something like this: a massless particle is created, interacts once with the Higgs field to acquire a permanent classical like mass which it then ...


7

I don't think you understand QFT. To be fair, I'm no expert myself, but I can certainly point out where you're going wrong here. A particle does not enter the Higgs field. However, the particle field that gets mass from the Higgs field does interact with the Higgs field. What this means is that in the Lagrangian of your model, there exists a term that will ...


1

There are 5 standard model (SM) multiplets per generation of fermions. The SM gauge group is $\mathcal{G}_\text{SM} = SU(3)_C \times SU(2)_L \times U(1)_Y$. Various multiplets can then be written as $\mathcal{G}_\text{SM} \ni x = (C,T)_{(Y)}$, where $C$ denotes colour multiplet, $T$ weak isospin multiplet and $Y$ hypercharge value. Multiplets (1st ...


4

First, note that we are quite sure what the overall nuclear spin is; we are not sure how to obtain it mathematically from available models. Due to the phenomenon of color confinement, there are no gluons at low energies in QCD (the theory underlying nuclear physics). Importantly, you can't say there are this or that many gluons in any proton or neutron. ...


6

At first, consider two particles decay: $A\rightarrow B + e^-$ Where A is initially rest. So $\vec{p_B}+\vec{p_{e^-}}=0$ now \begin{align} \frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released} \\ \frac{p_{e^-}^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released} \tag{1} \end{align} see here you have uncoupled equation (equ.1) for $p_{e^-}$ ...


2

You're correct: the unique thing about beta decay is that there's a three-body final state. In the reference frame where the decay takes place at rest, the daughter nucleus, beta particle, and neutrino share the momentum roughly equally, and because of the mass scales the beta and the neutrino take the bulk of the energy. It's pretty straightforward to show ...



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