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Wu and Yang (1968) found a static solution to the sourceless SU(2) Yang-Mills equations, (please, see the following two relatively recent articles containing a rather detailed description of the solution: Marinho, Oliveira, Carlson, Frederico and Ngome The solution constitutes of a generalization of the Abelian Dirac monopole. The vector potential is given ...

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The classical reference is Landau & Lifshitz, The Classical Theory of Fields, from the Course of Theoretical Physics. As all Landau & Lifshitz books, masterpieces [in my opinion] full of content but sometimes a little difficult to grasp for beginners.

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No, it doesn't violate the rules of geometry, it violates the rules of Euclidean geometry. Simple conclusion: for an observer fixed to a disk rotating uniformly relative to an inertial frame, the spatial geometry is non-Euclidean; in particular, the ratio of a circle's circumference to its diameter depends both on the circle's diameter and center position. ...

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The classical limit of bosonic quantum mechanical systems with both finite and infinite degrees of freedom is pretty well understood from a mathematical standpoint (with complete rigour, and for quite general quantum states; see the references in the end). With fermions on the other hand, the situation is more involved. The point is essentially that hinted ...

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People spent a lot of time trying to do this kind of thing ca. 1910, i.e., after SR but before quantum mechanics. To make the electrostatic self-energy no greater than the observed mass of the electron, you have to create some kind of model of an electron as an extended object, with a size that's at least on the order of the classical electron radius. You ...

6

What is wrong is the idea that one can actually make the disk rotate; and it will remain perfectly rigid. In reality, what this correct argument shows is that relativity doesn't admit the existence of any perfectly rigid bodies. This is a perfectly basic, settled, and indisputable textbook material that every mature physicist knows. The first sentence of ...

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This question (v1) asks many questions. Let us here make some general remarks, which OP hopefully will find useful. Noether's theorem only needs infinitesimal transformations to work. Hence the important object is not the set $G$ of finite transformations, but rather the set $\mathfrak{g}$ of infinitesimal transformations. In general, the set $\mathfrak{g}... 5 The two quantities don't correspond because they are conserved quantities corresponding to different symmetries. One is a symmetry from shifting your field, the other from shifting space-time itself. Here is what is going on precisely: Let us do a simpler case first: In a particle mechanics system, let's say a free particle with$L = \frac{1}{2}m\dot{x}^2$, ... 5 That formula is for bosons. In the large particle limit the boson quantum field can be approximated by the classical field. Electrons are fermions and by the Pauli exclusion principle you can have only one particle per state. So you cannot go to the large particle limit in the same way as you can for a boson field which can have many particles in each state. ... 4 The Hamiltonian density for any classical field is given by: $$\mathcal{H} = \pi \dot{\phi} - \mathcal{L}$$ where$\pi$is the canonical momentum density: $$\pi(\mathbf{x},t) = \frac{\partial \mathcal{L}}{\partial \dot{\phi}(\mathbf{x},t)}$$ In classical point particle mechanics the Poisson brackets ... 4 I am going to write a short answer, because no one has answered yet. The left hand side should be $$\eta^{\mu \nu} \partial_\mu \phi \partial_\nu \phi.$$ It is important you only sum raised indicies with lowered indices and you never sum two lowered indices together. Summing two lowered indices gives you a result that is not lorentz invariant. The right ... 3 I) Let there be given a local action functional $$\tag{1} S[\phi]~=~\int_V \mathrm{d}^nx ~{\cal L},$$ with the Lagrangian density $$\tag{2} {\cal L}(\phi(x),\partial\phi(x),x).$$ [We leave it to the reader to extend to higher-derivative theories.] II) Assume that a variation of$S$for arbitrary$x$-dependent infinitesimal$\epsilon(x)$takes the ... 3 Different Lagrangian could give the same equation of motion. If you add non-dynamical variables, you could present a Lagrangian without symmetries, while the equation of motion has symmetries. For instance, take the Lagrangian : $$L(x,f) = \frac{f^2}{2}+\dot f ~x$$ Euler-Lagrange equations (applied to$x$and$f$) give : $$\dot f = 0, f = \dot x$$ ... 3 This theorem was corrected in a published paper as asked by Terry Tao (see here and here). Tao conceded that the new version gives a correct theorem and mapping holds in an approximate way (see here) in the limit of a large coupling that is what one needs in the infrared limit. Please, before to give incorrect information, just ask the author. See this ... 3 Oh, I made a mistake: In deriving$\partial_\mu{T^\mu}_\nu=0$, I had assumed$L$doesn't depend on$x$explicitly but solely on$\phi$and$\partial_\mu \phi$, and this is just the condition of translation symmetry! 3 In my opinion it is better to work in an explicit covariant form. In my answer I will use two different definitions, the Greek indexes always run from$0$to$3$and Latin indexes from$1$to$3$and the metric$g_{\mu\nu}$has signature$(-1,1,1,1)$. To translate the expressions to a explicit covariant form we define some timelike vector field$v^\mu$. We ... 3 This is just relativity of simultaneity again. A similar thing happens if you have a bunch of spaceships in a line that fire their thrusters at a fixed time. Different observers will disagree about whether they fired at the same time and will disagree about the spacing. Always in a consistent way. So I'd like to address the concept of geometry by not having ... 3 Well, if you have a term like$\partial_\mu \mathcal{J}^\mu$, the divergence theorem lets you convert it into a surface term upon integrating to find the action, and since variations are assumed to vanish at the boundary, this term goes away. The Euler-Lagrange equations don't change because they come from setting the variation of the action to zero. ... 3 No one talked about representations for the translations because the translations do not change the natural basis for the vectors. When you apply a rotation, you are rotating the natural coordinate system your vectors and tensors are represented in, so fields which are components of those have to change. Translations don't change the natural coordinate basis ... 3 Two real scalar fields$\phi_1$and$\phi_2$satisfying an$SO(2)$symmetry and one complex scalar field$\psi$are equivalent. However, the latter is more convenient because$\psi$and$\psi^\dagger$form the antiparticle pair, while in the real case, you need to change basis from$\phi_1$and$\phi_2$to$\phi_1 \pm i\phi_2$. Once you do this, you just get ... 3 IMHO asking for the intuitive meaning of expressions is in general sterile. The Langrangian$(2.6)$you wrote down is just the most general covariant/renormalisable scalar that can be written down for a scalar field. This is just mathematics, nothing deep behind it. From that Lagrangian we get the Hamiltonian$(2.8)$. Trying to ask for the physical meaning ... 3 OK, perhaps the notation in Ref. 1 is a bit confusing. Let us elaborate on Derrick's No-Go theorem: Derrick's No-Go theorem: For the number of spatial dimensions$D>2$, the only time-independent finite-energy solutions are ground states. In a nutshell, the idea of the proof is to derive a necessary condition by a simple 1-parameter scaling argument.... 2 This condition is due to the fact that for a free massless particle the Pauli-Lubanski vector$W=*(M\wedge P)$must be proportional to the linear momentum (The proportionality factor being the helicity). Thus the condition must be valid to all free massless relativistic field theories. 2 Consider the$4\times 4$matrix$g_{\mu\nu}$with zeroth row$g_{0\nu}$. Now for$i=1,2,3$, add to the$i$'th row the zeroth row times$-g_{i0}/g_{00}\$. This produces the following matrix $$\begin{bmatrix} g_{00} & g_{01} & g_{02}& g_{03} \\ 0 & -\gamma_{11} & -\gamma_{12}& -\gamma_{13} \\ 0 & -\gamma_{21} & -\gamma_{... 2 Well, this might not be exactly what OP is looking for, but the statement in Ref. 1 is in general not correct. That infinitesimal (global) symmetries (of an action) satisfy a Lie algebra does not imply that the corresponding Noether charges must also form a Lie algebra. There could be (classical) anomalies. Example: One example is free Schrödinger theory, ... 2 The classical Lagrangian for the free electron field is,$$ L=\int d^{3}x(i\psi^{\dagger}\frac{\partial \psi}{\partial t}+i\psi^{\dagger}\alpha_{r}\frac{\partial \psi}{\partial x^{r}}-m\psi^{\dagger}\beta \psi) \ . $$The q's are q^{i}(t)\rightarrow q^{(a,x)}\rightarrow \psi^{a}(t,x) and so the velocities are \dot{q}^{i}(t)\rightarrow \frac{\partial \psi^{... 2 In classical field theory, the ground state is also called minimizer (of the energy functional); and just to prove its existence is already a quite difficult task, from a mathematical standpoint (as you can imagine, much more difficult is to write eventually its explicit form). Often you can only have minimizing sequences, i.e. sequences of classical states ... 2 The action of a massless scalar field is given by: \begin{eqnarray} S(\phi)&=&\int{\cal{L}}dt\\ &=&\int d^{4}x \sqrt{-g}\left(g^{\mu\nu}\phi_{,\mu}\phi_{,\nu}\right) \end{eqnarray} Now choosing a tetrad, i.e., a basis of one form at each spacetime point \{e^{a}=e^{a}_{\mu}dx^{\mu}\} we can rewrite the action as: \begin{eqnarray} ... 2 You're in fact wrong in the claim that the equality [J^{\mu\nu}, H] = 0 doesn't hold for all \mu ,\nu. The commutator is exactly zero for \mu ,\nu \in 1,2,3, i.e., for spacelike indices. But it is not zero for the case when \mu = 0, \nu \neq 0, i.e., for boost generator; precisely, for K_{i}\equiv J_{0i} and H\equiv P_{0} we obtain$$ \tag 1 [K_{...

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