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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 ...


6

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.


6

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 ...


6

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. ...


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 ...


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

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

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

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

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

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

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 ...


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

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 ...


2

In my viewpoint, the free complex field theory $$\tag 1 \mathcal{L}=\partial_\mu\phi\partial^\mu\phi*-\frac{1}{2}m^2\phi\phi*$$ is actually equivalent to the free double real field theory $$\tag 2 \mathcal{L}=\partial_\mu\phi_i\partial^\mu\phi_i-\frac{1}{2}m^2\phi_i\phi_i,$$ where $i=1,2$. To see this, simply write $\phi=\phi_1+i\phi_2$ and you can obtain ...


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.


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, ...


1

I think that the concept of a frame of reference of a field is meaningless, since fields permeate all of space. Therefore, it is meaningless to say that a field is 'going somewhere', or 'moving relative to something'. A frame of reference seems to only make sense for an object that is somehow localized (e.g. a particle). Things get slightly interesting when ...


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

I was also involved in this problem for the past few weeks.You can write newtons law of motion for small segment of string and obtain a differential equation.From that equation you can find the normal modes of the string and the general motion of the string is given by a superposition of the normal modes.But i ignored the longitudinal oscillations and ...


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!} ...



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