# Tag Info

1

No- because electric charge can be measured (in principle) up to an infinite distance from the source, i.e. it's range (and influence) is infinite. Colour charge can only be measured up to a range close of approximately $r_p$, the radius of the proton, i.e. its range (and influence) is confined to below a fixed distance. As for whether the quark's electric ...

2

I don't see the analogy. Isolated color charged particles (quarks) have not been observed, this is not the case electric charge particles (like electrons) because we observe total net electric charge. You might mean that fractional color has not been observed, neither fractional electric charge. In that case, it's true, but this is not what confinement ...

2

If you want an even more everyday example than Emilio Pisanty's example: "no scattering" would mean that the would be scattering object in question (modelled by the short range potential in Emilio's answer) would beget no change the the forward travelling wave. Otherwise put, an observer sensing the incoming plane wave could not tell whether or not the ...

1

The only way we know that quarks exist is by a series of deep inelastic scatterings with leptons and with protons. This is a reconstructed event at LEP quark antiquark at 12:00 o'clock and 4:00 o'clock gluon the third one. The lepton colliders have the advantage that most of the energy taking part in the collision can be detected in four pi ...

4

Forward scattering need not be equivalent to "no scattering" - and, indeed, will only rarely be indistinguishable from it. In the usual scattering-theory setup, you have an electron coming in in a plane wave $$\psi(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}=e^{ikz}$$ and impinging on some short-range potential. This will add to the wavefunction a scattered ...

0

Usually, Cronin effect is given in terms of the central-to-peripheral nuclear modification factor for $dAu$ collisions at midrapidity $$R^h_{CP}(p_t)= > \frac{(1/N^C_{coll})dN^h/p_tdp_t(C)}{(1/N^P_{coll})dN^h/p_tdp_t(P)}$$ where $C$ central, $P$ reipheral, $N_{coll}$ the average number of inelastic $NN$ collisions.If hadronization is ...

2

Axial charge'' refers to the (isovector) axial coupling constant $g_A$ of the nucleon $$\langle p|A_\mu^a|p\rangle = g_A \bar{u}(p)\gamma_\mu\gamma_5\tau^a u(p)$$ where $A_\mu^a=\bar{\psi}\gamma_\mu\gamma_5\tau^a\psi$ is the QCD axial current, $|p\rangle$ is a nucleon state with momentum $p$, $u(p)$ is a free nucleon spinor, and $\tau^a$ is an isospin ...

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It looks like the following article is relevant: http://arxiv.org/abs/hep-ph/0402256 (published in Nucl. Phys. A). (The phrase you quote is probably from lectures http://www.physik.uni-bielefeld.de/~borghini/Teaching/HIC-Seminar/SoSe2013/Francois_SPhT2006-1.pdf by one of the authors of the article): "The Cronin effect was discovered in proton-nucleus ...

1

If, by "first principles" you mean without any observation or experimental input, I don't think we have the answer yet. We don't know why we have the particles/fields we do. But if we're given those particular particles/fields, then we know that consistency tells us what the Lagrangian must look like and leaves us little room for fixing coefficients of some ...

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Is there a still a chance to Regge theory nowadays? Regge trajectories are having a revival in string theories. String theories are candidates for the Theory of Everything (TOE) , i.e unification at the quantum level of all four forces, including a quantized gravity. If you google "regge trajectories and strings 2013" you will get a large number of ...

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First, to check the decomposition of a product of representations, you may use, as noticed by user26143, the tool Form Interfact to Lie. Choose Tensor product decomposition, then choose $A_1$ for $SU(2)$, or $A_2$ for $SU(3)$,click sur "Proceed", type your representation, and click on "Start" to have the decomposition. The name of the representations in this ...

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Because $SU(2)$ is not the same as $SU(3)$ ? The closest analog to your $SU(3)$ case would be two doublets: $\mathbf{2} \otimes \mathbf{2} = \mathbf{1} \oplus \mathbf{3}$, as you already know :) Afaik, $SU(3)$ has two independent $SU(2)$ subgroups, i.e., it has two "$L^2$" operators. You can still do Clebch-Gordan-style coefficients calculations but it ...

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