Luboš Motl
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 9h answered Why is $\sqrt{v_1+v_2+\ldots} =246$ GeV in multi-Higgs models? 17h awarded units 1d answered Invertibility of Dirac matrices 1d comment When can the constant of proportionality in an eq be set equal to 1 and when not? If the coefficient is one, then the coefficient is one. If the coefficient is not one, the coefficient is not one. There obviously can't exist any other "general rule" that would say something about the value of a coefficient in all situations in physics or mathematics or any science. Some of the coefficients are totally physical, others may be changed by various changes of the conventions (and people usually try to make as many coefficients in their basic laws equal to one as possible), and many coefficients in sciences etc. depend on measurements, different levels of evidence, and so on. 1d comment When can the constant of proportionality in an eq be set equal to 1 and when not? 1d revised Why does Special Relativity apply to more than just light? added 95 characters in body 1d answered Why does Special Relativity apply to more than just light? 1d revised Are asymptotic states in scattering experiments really momentum eigenstates? added 162 characters in body 1d revised Are asymptotic states in scattering experiments really momentum eigenstates? added 162 characters in body 1d revised Are asymptotic states in scattering experiments really momentum eigenstates? added 479 characters in body 1d answered Are asymptotic states in scattering experiments really momentum eigenstates? 1d comment Difference between 1 unit of a physical quantity and the unit of that physical quanity and how is the unit of derived quanities obtained? It's a convention that simplifies things. In similar laws, especially in electromagnetism there are sometimes extra coefficients such as $4\pi$ in different conventions. When it's possible to set things to one, we can. The force is a synonym to "mass times acceleration". If you used $F=9ma$, all numerical values would be different by a factor of 9, but it would be easy to translate everything back to the normal language where $F=ma$ because $F_{your} = 9 F_{our}$. 1d comment Difference between 1 unit of a physical quantity and the unit of that physical quanity and how is the unit of derived quanities obtained? The point is that if we have a unit, like "1 kilogram", we treat other masses, like "5 kilograms", basically in the same way as if they were the corresponding numbers such as 5 in my example. So the unit itself, 1 kilogram, plays the role of the number 1. But this is cheating because we could have chosen 1 to represent "1 gram", too. That's why the convention has to be specified by adding the name of the unit. But the number in front of the unit is 1 because the word "unit" means "one unit". Similarly, if we say "3 units", it may mean "3 kilograms". The letter -s in "units" says it's $>1$. 1d comment Difference between 1 unit of a physical quantity and the unit of that physical quanity and how is the unit of derived quanities obtained? Why is what set to one? The numerical value in front of the unit is $1$ because the word "unit" linguistically means the number "one". So your question is equivalent to the question why an individual is one person. It's because the word "individual" means one, among other things. If we measured the mass in "pairs", the elementary piece could perhaps be "2 kilograms" instead. But we wouldn't gain anything out of it, except for confusion analogous to yours. The special feature of the number one is that it doesn't hurt when you add another "one times" in front of anything. 1d answered Difference between 1 unit of a physical quantity and the unit of that physical quanity and how is the unit of derived quanities obtained? 1d awarded Enlightened 1d awarded Nice Answer 2d comment Are quadrilinear weak boson couplings possible? They're not just possible. They're unavoidable by gauge invariance in the presence of the kinetic terms $F_{\mu\nu}F^{\mu\nu}$ for the gauge fields. 2d comment Heisenberg's uncertainty principle derivation in a ring No, Mikael, you don't run into this problem if $x$ is not periodic. The delta-functions have the infinite uncertainty of the other quantity, the product is an indeterminate form, and may be said to obey the inequality when properly regulated. 2d comment Quantum entanglement definition Dear @user115519 - the logic is exactly the other way around than you suggest. The "noise" (decoherence) that ruins (or makes harder, and requiring fixes or repetitions) a quantum calculation isn't due to "too much entanglement". It's due to the quantum computer's being measured - and therefore disturbed - by the environment which reduces the entanglement. These reductions of the entanglement is something we need to prevent.