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11

In principle, yes. You can reverse any decay process and the corresponding synthesis will be valid - in this case, since $H_0\to\gamma\gamma$ happens, then $\gamma\gamma\to H_0$ will also happen, assuming the kinematics work out. However, the corresponding probability is very small. Out of all the possible things that could happen when two photons cross ...


5

The mass of a fundamental particle turns out to be quite an elusive concept, because massless particles act as a source of gravity and they carry momentum. What then is special about mass? Where mass comes in is in explaining the relationship between the total energy of a particle and its momentum. For any particle we have the expression for the total ...


4

Neutrino flavor is defined as agreeing with the flavor of the charged lepton participating in the interaction, so that the neutrino in the reaction $$ \nu + A \to \mu + X \,, $$ is defined to be a muon neutrino and the one in $$ \nu + n \to e + p $$ is a electron neutrino by definition. We have no way of knowing the alleged flavor of a neutrino ...


3

If we know the classical physics theory of the electromagnetic force (and we do), we can guess what the quantum mechanics theory for it should be (and then test with experiment, and as far as we can tell we've guessed correctly). We can do likewise with any classical force. (Although the strong and weak theories were not found by starting from any classical ...


3

For a real answer, each particle would have to be discussed individually and that might get long. Dark matter possibly being Neutrinos has certainly been proposed and in many ways, Neutrino's lack of interaction makes them a good candidate, as they are essentially "dark" - though "invisible" is perhaps a more accurate term and Neutrinos fly through stuff ...


2

A classical "shell" of charge of radius $R$ will appear to have a field, outside the shell, corresponding to the charge all being at the center of the shell. If the charge on the shell is $q$ then a charge $dq$ being pulled in from infinity will cost an energy $k_e~q~dq / R$; without loss of generality this is the same if $dq$ is spread over an infinitely ...


2

Mesons and Baryons are the most common objects quarks make. However, they are not the only ones. There are tetraquarks (=4 quarks, in the form $qq\bar{q}{\bar{q}}$, due to color confinement), such as the Z(4430) confirmed by LHCb with a significance of 13.9$\sigma$, see this for relevant references. The treatment of these objects theoretically is more ...


1

Within the standard model: It cannot! Even though there is CP violation in the SM (as you state), the amount is not enough to give the ratio of $ \frac{n_B}{n_\gamma} \sim 10^{-10} \ .$ Furthermore, the electroweak phase transition (EWPT) in the SM is not even a second order transition, but merely a crossover. In order to render the transition ...


1

Taking the trace of an operator over all the states/particles effectively means taking a sum over all the eigenvalues of the operator (the charges) over these states/particles. So what matters is how many states you have and with what charges. The numbers that you are referring to are just the appropriate multiplicities of the states. For example the first ...


1

Indeed you find the Feynman rule by differentiating the term in the Lagrangean w.r.t. the fields attached to the vertex. In the current-current case, you need to be more careful with spinor indices than in the general gauge coupling case. There are four independent spinor indices on such a vertex. You get something proportional to $$ V \propto (\gamma^\mu ...


1

I am not sure whether the standard model has a Landau pole at $10^{34}$ GeV but there are two obstacles to providing a definite answer to the question: (1) perturbation theory is no longer valid when the coupling constants get large, and (2) $10^{34}$ GeV is well beyond the Planck scale, so that ignoring the effects of (quantum) gravity is not valid. If ...



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