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## Hot answers tagged baryons

109

Free neutrons in flight are not deflected by electric fields. Objects which are not deflected by electric fields are electrically neutral. The energy of the strong proton-neutron interaction varies with distance in a different way than the energy in an electrical interaction. In an interaction between two electrical charges, the potential energy varies ...

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Suppose that the strong nuclear force were instead caused by Coulomb interactions. Since we know how strong the binding energies are (of the order of $\sim 1\ \text{MeV}$, as can be gleaned by say, looking at a table of alpha particle energies) and how far apart the nucleons are (about a proton radius, or $a_p\sim1\ \text{fm}$) we know how charged the ...

21

No, a three-quark baryon can not be be made out of two quarks and one anti-quark (and vice versa) as this would necessarily give the particle color. Each quark carries one of three colors (red, blue, green) and each anti-quark respectively carries anti-color. Color is an additive quantity when constructing particle and the result must be color-neutral, i.e. ...

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The up quark has a charge of $+2/3$, the down has a charge of $-1/3$. If you have a bound state of charged particles, the total charge is just the charge of the elementary constituents. The neutron consists of one up quark and two down quarks, so the total charge $Q$ is: $$Q = 2/3 + 2 \times (-1/3) = 0$$

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Pentaquarks contain three quarks and a quark-antiquark pair, and they are baryons, since baryons are defined as having an odd number of valence quarks.

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Rob's answer is the simplest and probably best, but let me add another approach. We know that nuclei are made out of protons and neutrons. Protons repulse each other, but somehow, if you get them close enough, they stick together extremely strongly. This already suggests that there is another force in play! So even if you completely ignored neutrons, you ...

14

Short answer: yes. You should look in to the history behind the dark matter hypothesis. It started not from the examination of cosmology and the CMB, but from the motion of galaxies in clusters and stars orbiting around galaxies. See, the vast majority of ordinary matter in every galaxy is contained in the gas between the stars, not the stars themselves. ...

12

Briefly, a hadron has to be color singlet ${\bf 1}$ under the $SU(3)_C$ color gauge group, due to color confinement. Examples: A single quark $q$ transforms in the fundamental representation ${\bf 3}$ of $SU(3)_C$, and is hence not allowed. See also related Phys.SE post here. A diquark $qq$ belongs to the tensor representation ${\bf 3}^{\otimes 2}:={\bf 3}\... 11 The isospin is different.$I=0$for the$\Lambda^0$and$I=1$for the$\Sigma^{0}$. This makes the$\Lambda^0$an isospin singlet state but the$\Sigma^0$is part of an isospin triplet. There are quite few other examples e.g. compare a proton (uud with$I=1/2$) with a$\Delta^{+}$(uud with$I=3/2$). 11 It's a good question, and the answer is surprisingly simple and physical. There is indeed no fundamental objection to having a superposition of particles of different charge. But it turns out this is not stable. This is basically due to wavefunction collapse, or more sophisticatedly, due to decoherence. Imagine having a single particle in a superposition ... 11 The charges of the quarks must be simple fractions of the electron charge$e$, because otherwise there would be a breakdown of charge conservation in quantum corrections. The fractions do not need to be$-\frac{2}{3}$and$\frac{1}{3}$specifically. In simple models with$2n+1$quarks making up the proton (the number of quarks must be odd, so that the ... 10 Symmetry and statistics. The quarks being fermions dictate a fully antisymmetric wavefunction of the three constituents of the baryon. The color wavefunction is antisymmetric, so the combined spin&flavor wavefunction must be symmetric. The spin 1/2 combination (SU(3) octet) is of mixed symmetry and so is the flavor symmetry of the baryon octet you ... 10 Both SU(3) flavor and SU(2) isospin are approximate symmetries of the Standard Model at low energies. Consider physics below the proton mass, where we can talk about the pions and kaons that are the avatars of these symmetries. At energies this low, it doesn't make sense to talk about the heavy quarks (charm, bottom, top), so we're left with the light quarks:... 10 But the Sun in our solar system also is in the state of plasma and yet doesn't act like a blackbody Wrong, the sun radiation is approximately fitted as a black body. The word "black body" does not describe the frequencies, but the assumption that it absorbs all radiation falling on it and re-emits it. Here is the sun, and it fits the black body formula ... 9 You are correct to point out that there's no symmetry that forbids a state with isospin 3/2 and spin 1/2; in the nomenclature, this is also called a$\Delta$resonance. The Particle Data Group lists two such particles, with mass 1620 MeV and 1910 MeV. They exist, but they are heavier than the spin-3/2$\Delta$at 1232 MeV. The reason why is isospin, ... 9 The names up and down don't refer to electric charge$Q$but are rather references to isospin charge$I_3$. 8 Short answer: nothing has been seen. Long answer: Questions like this on the experimental limits in particle physics can usually be answered by looking things up in the Particle Data Group's annual Review of Particle Physics. There is a summary online version and an extensive (but free!) print version. EDIT: Here (pdf) is the full section on conservation ... 8 I think I have worked out my confusion so I thought I should post it as an answer. The original question was not well-posed; hopefully this will help anyone else who has similar misunderstandings. Keeping spin in the picture, the space of states for an individual quark is the tensor product of the three-dimensional flavour space with the two-dimensional ... 8 That would be an anti-Xi baryon. Production of Xi / anti-Xi pairs in discussed in Branching Fraction Measurements of psi(2S) Decay to Baryon-Antibaryon Final States. So yes it does form something and it is a fault of the applet. 8 What do you think a$\Delta^+$or$\Delta^0$is, if not an excited nucleon? (To be sure the$^{++}$and$^{-}$states do not correspond directly to a nucleon, because there is no allowed lower spin state with that valence content.) The thing I am not sure about is how closely these excitation match the ones you are envisioning. They match nuclear excitation ... 8 First, the neutron$n$and the proton$p$don't have "excited counterparts" in the decuplet, either, do they? Now, the two multiplets are completely different. One has eight$SU(3)_{\rm flavor}$components, the other has ten. So it's clearly invalid to call one group "excitations" of the other. Because the quark content (charge and strangeness) is the same,... 8 One is talking quantum mechanics and attributed quantum numbers to elementary particles. A simple quantum number is charge and it it assigned to quarks ( and antiquarks) as +/-1/3 or +/-2/3 as in the table Charge is connected with the electromagnetic force. Flavor is assigned as a quantum number to each quark, and it is connected with the weak ... 8 I asked a question very much like this several years ago (in person, not online, but someone else asked it here): "what's the baryon asymmetry of the proton?" Thinking, of course, about the three valence quarks and the so-called "sea" of quark-antiquark pairs. After talking with several people in my department and at conferences and at other places I ... 7 A neutron consists of three quarks$u d d$(up down down quarks). The up quark(u) carries charge$2e/3$and the down quark(d) carries a charge$ -e/3 $. Thus$2e/3-e/3-e/3=0$7 To achieve a nonzero baryon asymmetry, one needs to satisfy the so-called Sakharov conditions: Baryon$B$violation C-symmetry violation and CP-symmetry violation Interactions out of thermal equilibrium If at least one of these "asymmetries" or "imbalances" is missing, the total$B$of the Universe will remain zero. The Standard Model preserves$B$... 7 This answer is basically equivalent to OP's own answer using slightly different words. Firstly, we will need a formula $$\dim (V^{\odot n})~=~\begin{pmatrix} n-1+\dim V \cr n \end{pmatrix} \tag{1}$$ for the dimension of the symmetrized tensor product$V^{\odot n}$of a finite-dimensional vector space$V$. The proof of formula (1) is left as an exercise. ... 6 A similar question is the following. How can$\pi^0$and$\eta$in the$SU(3)_F$meson octet both have the same$SU(3)_F$flavor content? One could answer that this is because$\pi^0$is part of an isospin triplet of pions with$I=1$, while$\eta$is an isospin singlet with$I=0$. Or one may point out that their explicit ket linear combinations of quark ... 6 There exists a standard model of cosmology, i.e. accepted as the current status of research, called the Big Bang. This makes extensive use of the known interactions of particle physics encapsulated in the standard model., and assumes a singularity at the beginning of the universe where the energy seen in the universe now was originally generated. I like ... 5 Good question! There actually isn't a term for this that I know of. The most common use of such a term would be to classify a particle, for example "the 'polarity' of the electron is matter-polarity," but in that case most physicists would just say "the electron is a matter particle." There is a mathematical operator called the charge conjugation operator, ... 5 U-spin is very similar to isospin which interchanges u and d quarks, except, now, U-spin interchanges d and s quarks, so, in the Eightfold-way weight multiplets, it transitions to lower right from upper left---unlike for left to right for isospin. So, behold!, in the baryon octet, the$\Sigma^-$is part of a U-doublet, the$\Sigma^0$mixing with the$\...

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