# Tag Info

An internal symmetry only involves transformations on the fields of a theory, and must act the same independent of the point in spacetime. For example, consider a Lagrangian, $$\mathcal{L} = \partial_\mu \psi^\star \partial^\mu \psi - V(|\psi|^2)$$ for some potential $V$, and complex field $\psi$. The theory has an internal symmetry, namely one which ...
The phrase "the function is spherically simmetrical" means that, if $G$ is an orthogonal transformation (that sends spheres into themselves), then $$f(G\mathbf r, G\mathbf p,t)=f(\mathbf r , \mathbf p, t).$$ If you know $\mathbf r^2$, $\mathbf p^2$, $\mathbf r \cdot \mathbf p$ you can calculate $f$ by taking an orthogonal transformation which maps $\mathbf ... 2 There is at least one philosopher before Plato and he is Anaximander. There are many passages in his works that relate to the concept of symmetry: The basic elements of nature (water, air, fire, earth) which the first Greek philosophers believed that constituted the universe represent in fact the primordial forces of previous thought. Their collision ... 8 The point is that eq. (1.35) should hold off-shell to have a symmetry, while eq. (1.37) may only hold on-shell. [The term on-shell (in this context) means that the Euler-Lagrange equations are satisfied. See also this Phys.SE post.] In other words: On-shell, the action will only change with at most a boundary term for any infinitesimal variation, whether ... 0 I discussed this with a colleague yesterday and I think that I get it. It has to do with the normalisation of the group generators (momentum operators). Within each representation of the translation symmetry we are free to normalise the momentum operator as we like. Then it is natural to choose this normalisation such that all representations have the same ... 1 In this answer we will consider a Lie algebra$L$(rather than a Lie group). Then: If$M$is a manifold, let there be a Lie algebra homomorphism$\rho:L\to \Gamma(TM)$into the Lie algebra of vector fields on$M$. The map$\rho$is called an anchor. If the manifold$(M,\{\cdot,\cdot\}_{PB})$is a Poisson manifold, it is natural to require that the vector ... 4 The basic idea is the following. For the shake of simplicity, I henceforth assume that every function does not depend explicitly on time (with a little effort, everything could be generalized dealing with a suitable fiber bundle over the axis of time whose fibers are spaces of phases at time$t$). On a symplectic 2n dimensional manifold (a space of ... 0 Great question! One way to look at what is going on is to use the Hamiltonian version of Noether's theorem. The Noether procedure generates a conserved charge$Q$associated with the symmetry with parameter$\theta$. It turns out that$Q$is the generator of that symmetry, in the sense that for some function$A$of phase space variables ... 1 Your question seems to contain two parts. First, you're asking how to set up the equations of motion for this coupled system. Second, you are asking how to use symmetry considerations to find the normal modes and frequencies. Let's first answer the bit about symmetry first Normal modes - symmetry Your observation about the reflection symmetry is spot on. ... 1 Consider an element$g$of the symmetry group. Say$g$is represented by a unitary operator on the Hilbertspace $$T_g = \exp(tX)$$ with generator$X$and some parameter$t$. It acts on an operator$\phi(y)$by conjugation $$(g\cdot\phi)(y) = T_g^{-1}\phi(y) T_g = e^{-tX}\phi(y) e^{tX} = \big[ 1 + t[X,\cdot]+\mathcal{O}(t^2)\big]\phi(y)$$ On the other ... 3 "Derivation" of Baryon Number Conservation - Consider the QCD Lagrangian (density) $$\mathcal{L} = \bar{\psi}(i\gamma^\mu D_\mu - m)\psi - \frac{1}{4}G^a_{\mu\nu}G_a^{\mu\nu}$$ where the symbols have their usual meaning. This is invariant under$U(1)$, which is nothing but a multiplication of$\psi$by a global phase factor$e^{i \theta}\$. This is ...