Hot answers tagged

2

The hierarchy problem has to be framed in the context of beyond standard model physics. You have to distinguish between 5 mass scales, namely $m$: the mass of the particle in concern, e.g. Higgs mass $m_H$. $\Lambda$: the UV cutoff scale of the regularization scheme (in dimensional regularization (DR), $\frac{1}{\epsilon}$ plays the role of $\Lambda$, ...


1

The answer to your question is stress-energy, not mass. Mass is just a certain form, a manifestation of something deeper, being energy. A proton and a black hole, both have stress-energy, and both cause spacetime curvature. It's a commonly made mistake that gravity, and therefore a black hole, is caused by matter. In fact the spacetime curvature is related ...


1

General relativity does not distinguish between matter and energy. As far as gravitation goes they are the same and are related by Einstein's famous equation $E = mc^2$. The gravitational field of a proton is the same whether we treat it as a mass of $1.673 \times 10^{-27}$ kg or an energy of $938.2$ MeV. So whether or not the Higgs field confers a rest mass ...


1

A consequence of the Heisenberg uncertainty principle is that nothing can ever standstill or not oscillate. Everything has to oscillate. I do not think this is a correct deduction. If one measure momentum there is a limit to the accuracy of measuring position , is all that the HUP says. It is an envelope where measurerements of momentum and position at the ...


1

Yes you are right that for electrons or muons, their Yukawa coupling is certainly NOT $O(1)$. You can have two interpretations of Mark Thomson's remark: The ratio of electron mass vs top quark mass is around $10^{-5}$. It is certainly NOT $O(1)$, but it is not as bad as $10^{-20}$ when comparing fermion mass with Planck mass. So Mark Thomson means that the ...


1

Yes, the values of the masses are real positive numbers. Recall how you find them out of the complex matrix Y. Note first that $Y Y^\dagger$ is hermitian, and has positive-eigenvalues, and so can be written as $$ Y Y^\dagger = U D ^2 U^\dagger $$ for some unitary U and diagonal real D with no zero entries, for simplicity. Take the positive square roots. ...


1

First let's consider this in the context of a very simple toy model, which however is not the Standard Model. In particular, we will just imagine one fermion and one scalar field with a Yukawa coupling \begin{equation} S = \int {\rm d}^4 x \left( - \frac{1}{2}(\partial \phi)^2 -V(\phi )+ i\bar{\psi} \gamma^\mu \partial_\mu \psi - m \bar\psi \psi - g \bar\psi ...


Only top voted, non community-wiki answers of a minimum length are eligible