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11

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 ...


5

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 ...


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


4

The Hamiltonian density for any classical field is given by: \begin{equation} \mathcal{H} = \pi \dot{\phi} - \mathcal{L} \end{equation} where $\pi$ is the canonical momentum density: \begin{equation} \pi(\mathbf{x},t) = \frac{\partial \mathcal{L}}{\partial \dot{\phi}(\mathbf{x},t)} \end{equation} In classical point particle mechanics the Poisson brackets ...


4

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 ...


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

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 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

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 ...


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 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

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.


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 $$ ...


1

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 ...


1

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; ...


1

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 ...


1

You cannot prove the existence of a conserved current directly from the equations of motion in general just given the fact that the equations of motion have a continuous symmetry. A counterexample to this is a damped harmonic oscillator, $\ddot{x}+\Gamma \dot{x}+\omega^2 x=0.$ This is invariant under $t\rightarrow t+\delta t$, but doesn't have a conserved ...


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

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 \begin{equation} \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} ...


1

The definition (pay attention to not confuse generic tensor with tensor components) of the antisymmetric gamma tensor is: $$\gamma^{\mu_1 \mu_2 \dots \mu_r}=\gamma^{[\mu_1 \mu_2 \dots \mu_r]}$$ For the highest rank you have $r=D$, so you have to use all the possible indices. For example, in components, you have the identity: $$\gamma^{1 2 3}=\frac{1}{3!} ...


1

Renormalization group is basically a tool which shows us how our theory responds to the scale transformations. Since a classical theory is completely spawned by the action, this action should be invariant under the RG flow. It means that the classical RG flow is completely determined by the scale dimensions of physical quantities which can be derived from ...


1

is my interpretation of the dynamics of the self-force correct and is there a physical or intuitive explanation for this extremely pathological behavior in the presence of a Coulomb potential? Eliezer makes his argument based on the equation with the Lorentz-Abraham-Dirac term. This term was originally (Lorentz) devised as an approximate way to account ...


1

The super-Poisson bracket follows from a super-version of the Dirac-Bergmann or the Faddeev-Jackiw procedure. Diligent care must be taken to achieve consistent sign conventions when dealing with Grassmann-odd variables, see e.g. my Phys.SE answer here. The singular Legendre transformation for fermions is also discussed in my Phys.SE answer here. In OP's ...


1

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 ...



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