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First a word of caution: The information in public science videos like this is often like a game of telephone: A researcher tried to explain his/her results to a science communicator in lay terms, and then the communicator tries to further improve on the analogies to make the video more intuitive or appealing. The final product often has a tenuous connection ...


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Basically all of QFT is done in natural units in which the speed of light and the reduced Planck constant are set to unity, $$c=\hbar=1.$$ There is rarely any need to work in any different system $-$ you need to be doing something very specific to do so, and if that is the case you always note it explicitly. Otherwise, the default is to assume natural units. ...


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The reason why calculating amplitudes using the Feynman calculus formalism is tedious is that they stem from a perturbative treatment that is formulated by upholding, above all, manifest Lorentz-invariance. While this is extremely useful for developing and formalising the theory (e.g. to detect anomalies easily), it tends to obstruct practical calculations - ...


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Another way of saying the same thing, is that if/when a meson is in a $b \overline b$ state it can annihilate through gluons and form a $r \overline r$ state with the same quark flavours, and likewise a $g \overline g$ state. The 3 states all mix into each other: you can't have a $b \overline b$ meson because it won't stay a $b \overline b$ meson. The ...


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Due to colour confinement, observed free particles (hadrons) must be "colourless" or "white", i.e. a colour singlet. A necessary (but not sufficient) condition for a colour singlet is that it is invariant under the $\text{SU}(3)$ colour gauge symmetry, which automatically rules out "pure" $r\bar{r}$, $b\bar{b}$ and $g\bar{g}$ ...


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Evgeniy, I don't agree with your answer. If you want to put in the explicit $\hbar$ and $c$ and thus work in Gaussian units rather than natural units, I believe that you should be dividing the dimensionless strong coupling constant $g$ by $\sqrt{\hbar c}$, not $\hbar c$, everywhere that it appears, so that the equations are $$F_{\mu \nu}=\partial_{\mu} A_{\...


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My understanding is that this is due to momentum conservation. A gamma ray cannot create an e-/e+ pair without interacting with a nucleus in order to conserve momentum. This is why the virtual gamma ray here can only create a virtual (off-shell) e-/e+ pair. The change that occurs in the electron after the photon emission is just a change in energy/momentum. ...


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It would be double counting, since total phase rotations of the quark wave function are already part of the model and the photon that makes them into a gauge symmetry already exists. The total gauge group is SU(3) × SU(2) × U(1), so the question "where has the U(1) gone" has as its answer that it already was included. In a gauge theory you can only ...


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