# Tag Info

1

This is an interesting question. I think the real difficulty is understanding the difference between Noether charges and what we normally call the charge of a particle. Noether charges are operators that when acting on states give the values which we normally call "charges" of different particles. In other words the charges of different particles are the ...

0

It is small wonder there is no definitive answer to this question here. But I can pass along the best answer I got. I asked the same question, in a slightly different format, to a blog "Of Particular Significance" that is popular with Physics Stack Exchange. My question was whether the famous Goldstone Mexican Hat potential had a value of 245 GeV at the ...

5

They can't be the same thing. As Wikipedia says, it's possible to calculate certain properties of glueballs from QCD, including their masses, and the masses don't come close to what we've observed for the Higgs boson. Also, the Higgs doesn't have color charge, so it doesn't interact with gluons, whereas a glueball would. That would make a large difference in ...

1

According to the following paper and commentaries, general relativity can be derived from a standard model matter field equation combined with some other consistency criteria. If new matter such as dark matter is found then the given procedure could give a new theory of gravity or it might just lead back to general relativity. How quantizable matter ...

0

Your question is riddled with ^'s in equations making it hard for me to understand the body of your question. If I understand your question "why is there no weak isospin vacuum angle in analogy with the one in QCD?," then I can answer it easily: Suppose we write that CP-odd term in the Lagrangian. Then, to remove it, all you need to do is to look for a ...

7

Special relativity is used in the SM formulation. It is kinematics, so somehow more basic than interactions between bodies. A QFT derivation of General Relativity has been the Holy Grail of the field for many years. In the early times, Feynman, Dirac, and the others tackled this problem, but after decades of failures it was more or less considered ...

1

An experimentalist's view: I do not see the need to search further for why the three quarks add up to the electron charge than that given by the group structure of the Standard Model. The SM is very successful in organizing into beautiful symmetries the particle and resonances data gathered the last sixty years or so. There is no experimental reason to ...

1

Now when we operate parity operator, does that mean we are taking any physical entity at x to −x. Or we are just reverting axes of the co-ordinate system? Well, either operation should adhere to the same rules, and you mention the correct term: it depends on whether we see the operation as active or passive. Either view has the same end result: we move ...

1

Yes it fluctuates but it is a very small fluctuation. Note that unstable particles have a decay rate or width $\Gamma$ that is related to its lifetime $\tau$ by $$\Gamma=\frac{\hbar}{\tau}$$ when you measure the mass/energy of such particles in experiments you always get a Lorentzian or Breit-Wigner distribution like this from which you can measure the ...

0

Take a simple quantum mechanical potential that describes an atom. The mass of the atom is fixed. Take hydrogen. The electron is in an orbital around the proton which is a probability distribution of its location in time and space: if you measure it, i.e. interact with it, where you may find it. Correspondingly there exists an energy width to the energy ...

0

Benford's law is pretty cool. It states that, for many sets of data, a leading digit of n has a probability of $Pr(n) = log_{10}(1+1(n))$ Plugging in our n values we find that we can expect low values of n to have a higher probability of being our leading digit. The most (initially) boggling thing is that our $Pr(1) = .301$ stays independent of units. If ...

2

There is no such tree-level interaction in the conventional theory. But then free-propagating neutrinos are not in pure flavor states except by chance anyway, so any pair of neutrino and anti-neutrino1 could participate in a vertex $$\nu_l + \bar\nu_l \to Z^0 \,,$$ which is roughly equivalent to Drell-Yan in the charged lepton sector with a projection into ...

7

When it comes to fundamental charges, the (left-handed) up-type quarks actually have either the same values of the charge as the down-type quarks, or exactly the opposite ones. It just happens that the electric charge isn't a fundamental charge in this sense. Let me be more specific. All the quarks carry a color – red, green, or blue – the charge of the ...

0

If you are looking for symmetry, I think one should point out that there IS a particle with a -2e/3 charge and a particle with a +e/3 charge. They are the up antiquark and down antiquark respectively. Now, following that, you would very reasonably ask the question why we observe more up quarks than up antiquarks, and other follow-up questions like ...

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The names up and down don't refer to electric charge $Q$ but are rather references to isospin charge $I_3$.

6

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 $2e/3-e/3-e/3=0$

18

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: $$Q = 2/3 + 2 \times (-1/3) = 0$$

3

Because $2/3-1/3-1/3=0$.

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