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1

My QCD knowledge is very limited - I haven't learnt it anywhere else other than in some qualitative discussion in Thomson's particle physics textbook. So I was only seeking a qualitative answer to aid my understanding in PDFs. Thanks to @anna_v 's answer I realised I hadn't gone through Thomson's own slides of the chapter and the answer lies there all along. ...


0

I copy these definitions of the scattering variables" I am not able to see what this has to do with the value of x. Does this mean when x decreases, the four momentum q of the gluon also decreases? If so, why is it the case? The definition of x is dependent on the four momentum carried by the propagator, in the diagram small $q$ vector , which is ...


1

So the Parton distribution functions are "essentially" the probability distributions (very suspicious how both are called pdf) of a probe particle to interact with a component inside your hadron. They depend both on the momentum at which we probe $Q^2$ and the fraction of momentum carried by said parton, the Bjorken x. They are mostly known through ...


7

Your quote already contains the answer: Symmetric, antisymmetric and mixed.


2

Interesting analogy between the magnetic dipole moment of a neutron and the strong force. However, the two are very different. Anna v has already discussed the issue of the radiation of photons from neutron, so I'll just say a bit more about the strong force. Fundamentally, the force among the quarks is described by quantum chromodynamics (QCD), which is a ...


2

In fact, a neutron has a finite magnetic dipole moment despite it is neutral ... Therefore, a neutron could be accelerated by electromagnetic fields even if its electric monopole is zero. Dynamics of motion When light falls on a particle moving in the same direction, this particle is accelerated. If photons are not completely absorbed, they are re-emitted ...


5

The radiation for a neutron in a magnetic field has bravely been calculated. They conclude: The calculations in this paper are mainly of theoretical interest, as good pedagogical examples in the classical and quantum theories of radiation. Physically the process is not observable,because the radiation rate of the neutron is very small, They give an ...


0

Radiation from a nucleus is discussed. For example x-rays are emitted from electrons and Y rays are emitted from the nucleus. https://www.sciencedirect.com/topics/physics-and-astronomy/gamma-radiation There are even interesting articles about the field that is formed by photons emitted from the nucleus and how it affects the arrangement or separation of the ...


1

You may well be misunderstanding several things. The $N_f^2-1$ flavor charges do no count flavors. It is easiest for you to see that by imagining a world of just two flavors, u and d, namely isospin, for $N_f=2$. Then the three generators/charges effect infinitesimal flavor rotations and are just the three Pauli matrices. So you see that $\sigma_1$ just ...


1

Quarks can never be observed isolated, since they only exist in confinement. What you are asking about is basically the conversion of a proton into a neutron. Even then, the proton cannot decay in isolation (except if there is a incident antineutrino with sufficient energy), and there are basically two main types of cases where the proton can do this ...


3

The most common example of this is beta plus decay. In this process one of the up quarks in a proton decays into a down quark and a $W^+$, and the $W^+$ then decays into a positron and electron neutrino. As a result of the decay the proton converts to a neutron. As you say, the process violates conservation of energy and that means it cannot occur unless ...


-1

A bare single up-quark could not decay into a down-quark, as this would indeed voilate energy-mass conservation. A transition to a down-quark could happen within some interaction with another particle.


4

The deuteron is an isospin singlet (if it were not, one would expect $|nn\rangle$ and maybe $|pp\rangle$ to be bound states) with isospin wave function: $$ |I=0, I_3=0\rangle = \frac 1 {\sqrt 2}[|pn\rangle-|np\rangle]$$ so the nucleons aren't in isospin eigenstates (that is: they're each already a mixture of neutron and proton, a la Schrodinger's cat). ...


6

First of all, regarding quarks and gluons: Of course, we should be able to understand the quadrupole moment of the deuteron starting from QCD, but this is complicated (or has the be done numerically, using lattice QCD). However, there is a large separation of scales, given by the fact that the neutron and proton are heavy, and the deuteron binding energy is ...


2

Indeed, this must be a doubly weak decay, as you can see from the BR in PDG, and must involve two virtual Ws to eliminate two s quarks. The basic reaction is $$s\rightarrow u W^{-} \to u d \overline{u}, $$ and you need two of those. So, overall $$ dss \to d ~~ u d \overline{u} ~~ u d \overline{u}\leadsto udd+ d\overline{u}, $$ with a $\overline{u}...


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