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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 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 $L = \frac{1}{2}m\dot{x}^2$, ...

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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 ...

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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 ...

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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 ...

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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 ...

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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!

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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 $$\begin{bmatrix} g_{00} & g_{01} & g_{02}& g_{03} \\ 0 & -\gamma_{11} & -\gamma_{12}& -\gamma_{13} \\ 0 & -\gamma_{21} & ... 2 This condition is due to the fact that for a free massless particle the Pauli-Lubanski vector W=*(M\wedge P) must be proportional to the linear momentum (The proportionality factor being the helicity). Thus the condition must be valid to all free massless relativistic field theories. 2 Different Lagrangian could give the same equation of motion. If you add non-dynamical variables, you could present a Lagrangian without symmetries, while the equation of motion has symmetries. For instance, take the Lagrangian :$$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$$... 2 This question (v1) asks many questions. Let us here make some general remarks, which OP hopefully will find useful. Noether's theorem only needs infinitesimal transformations to work. Hence the important object is not the set G of finite transformations, but rather the set \mathfrak{g} of infinitesimal transformations. In general, the set ... 2 The classical Lagrangian for the free electron field is,$$ 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) \ . $$The q's are q^{i}(t)\rightarrow q^{(a,x)}\rightarrow \psi^{a}(t,x) and so the velocities are \dot{q}^{i}(t)\rightarrow \frac{\partial ... 2 Well, this might not be exactly what OP is looking for, but the statement in Ref. 1 is in general not correct. That infinitesimal (global) symmetries (of an action) satisfy a Lie algebra does not imply that the corresponding Noether charges must also form a Lie algebra. There could be (classical) anomalies. Example: One example is free Schrödinger theory, ... 2 Provided that \mathcal{L} is a Lorentz scalar, the quantity \partial\mathcal{L}/\partial(\partial_{\mu}\phi) has to carry an upper index. Since \mathcal{L} is a function of \phi and \partial_{\mu}\phi, the only object that can give such an index is \partial^{\mu}\phi. Hence \frac{\partial\mathcal{L}}{\partial ... 1 Cool! I'm working on the exact same thing. The way I proved this was since \mathcal{L} and \phi are both Lorentz scalars they must have the same transformation law. Therefore$$\delta \mathcal{L} = - \omega^{\mu}_{\phantom{\mu}\nu}x^{\nu}\partial_{\mu}\mathcal{L}.$$However, note that$$\partial_{\mu} \left( -\omega^{\mu}_{\phantom{\mu}\nu}x^{\nu} ...

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The key insight is that the field configuration that is constant and whose constant value minimizes the potential energy density simultaneously minimizes both the potential energy functional and the kinetic energy functional individually. Let' see how this works: Let $T$ and $V$ be the kinetic and potential functionals whose sum is the Hamiltonian; ...

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For instance, a Lagrangian $L = \partial_i \phi \partial^i \phi + m^2\phi^2$ has the same equation of movement that the Lagrangian $L' = \partial_i \phi \partial^i \phi + m^2(F\phi - \frac{F^2}{2})$. The Euler-Lagrange equation for $L'$ simply give $\Box \phi +m^2F=0$ and $F = \phi$, so we have $\Box \phi +m^2\phi=0$, which are the Euler-Lagrange ...

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The canonical momentum $\pi$ is not the same as certain components of the energy momentum tensor $T$. This can be seen by going over to the Hamiltonian description. gj255's action gives the Hamiltonian, $$H=\frac{1}{2}\int dx ( \rho\phi_{,0}\phi_{,0}-\kappa\phi_{,1}\phi_{,1}) \ .$$ Here the coords of the Hamiltonian formalism are $q^{i}(t)\rightarrow ... 1 This theorem was corrected in a published paper as asked by Terry Tao (see here and here). Tao conceded that the new version gives a correct theorem and mapping holds in an approximate way (see here) in the limit of a large coupling that is what one needs in the infrared limit. Please, before to give incorrect information, just ask the author. See this ... 1 I) Here is at least a partial answer. Assume the following set-up. Let there be given a classical Lagrangian field theory in$d+1$spacetime dimensions, with dynamical field variables$\phi^{\alpha}(x,t)$, and with no explicit time dependence. Action$S[\phi]:=\int \! dt~ L[\phi(t,\cdot)]$. Lagrangian functional$L:=T-V$. Energy functional$E=T+V$. ... 1 It looks like there is a whole wiki page dedicated to this, unless this isn't what you are asking: Higher dimensional gamma matrices. I believe the study of anti-commuting algebras is called Clifford Algebras. Also I think chirality is special to particular dimensions. I don't think you can find, for instance, a$\gamma^5$matrix which anti-commutes ... 1 Exact solutions could not be the right way to understand infrared behavior of Yang-Mills theory. As we know from quantum field theory, we can start with some approximation (weak coupling). With this in mind, it can be proved that the following holds (see http://arxiv.org/abs/0903.2357) for a gauge coupling going formally to infinity$\$ ...

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In my experience (complex, spatially extended systems) we talk about coupling of oscillators. People usually talk about weak coupling because it allows you to treat many oscillators more simply: http://www.scholarpedia.org/article/Phase_model#Weakly_coupled_oscillators Above, a system of coupled two-dimensional oscillators has been transformed into a ...

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