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Hints: Then potential term $\frac{1}{2}(\nabla\phi)^2$ is semipositive definite and is only zero for a $x$-independent configuration $\phi$. If one completes the square of the potential $$V(\phi)~=~\frac{\lambda}{4}\phi^4-\frac{\mu^2}{2}\phi^2~=~ \frac{\lambda}{4} \left(\phi^2-\frac{\mu^2}{\lambda}\right)^2-\frac{\mu^4}{4\lambda},$$ then it becomes clear ...


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With a Lagrangian like: $\mathcal{L} = \partial_\mu \phi^\dagger \, \partial^\mu \phi - V(\phi) = \mathring{\phi^\dagger} \mathring{\phi} + \partial_i \phi^\dagger \, \partial^i\phi - V(\phi) $, the Hamiltonian is: \begin{equation*} \mathcal{H} = \frac{\partial \mathcal{L}}{\partial \mathring{\phi}} \mathring{\phi} + \mathring{\phi^\dagger} ...


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Let me begin with David Miller's illustration of the Higgs mechanism. Miller depicts the Higgs field as a throng of journalists and politicians at a cocktail party. A famous politician, Margaret Thatcher (the only bad thing about this analogy), enters the room, playing the role of a particle. As she passes, the crowd gather around her, resisting her ...


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It seems to me that there is nothing fundamentally wrong about your statements, thus I also don't see any contradictions. The electroweak unification states that, as you said, the $Z$ and the $\gamma$ are different linear combinations of $B^0$ and $W^0$. This all works very nicely, there is just the problem of the masses which is then fixed my the Higgs ...


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The 4 generators of $SU(5)$ are not all "equivalent". In general, the generators of the group/algebra satisfy a defining equation of the form $$[T^i,T^j]=f^{ijk} T^k$$ so depending on the structure constants $f^{ijk} $,it is possible for example that $$[T^1,T^2]\neq[T^2,T^3]\quad\text{etc,}$$ so it is important which generators are broken. In terms of ...


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The mechanism for "giving mass" to elementary bosons and fermions is different. With bosons, it is related to the gauge symmetry ($SU(3)_c \times SU(2)_L \times U(1)_Y$) which is partially broken (and become $SU(3)_c \times U(1)_{em})$. The unbroken part imposes its associated bosons (gluons and photon) to be massless to respect this symmetry. With ...


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I'm wondering if there is any explanation for why bosons(specifically gauge bosons) can be massless (photon and gluon) but we don't see any fundamental massless fermions. It's because a fermion is a "body", and because "the mass of a body is a measure of its energy content". See Einstein's E=mc² paper. He talks about a body and an electron here. "The ...


2

There is a number of misconceptions in the question. I did not downvote the question, but I will just try to address some of the mistakes. In Quantum Field Theroy (QFT) all fields permeat all space. I am not sure what you mean by "gluon fluctuating field" - there is simply a quantum field for each particle type: not only for gluons, but also for electrons, ...


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Is it possible to decrease the mass of the object? Perhaps surprisingly the answer is yes. All you need to do is it drop it. Then some of the object's mass-energy, which we call potential energy, is converted into kinetic energy, which ends up getting dissipated. You're then left with a mass deficit. The mass of the object is reduced. It is known ...



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