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D. Hilbert derived the (same as Einstein's) equations of general relativity by demanding the invariance (form of symmetry) of the Einstein-Hilbert action under general differentiable coordinate transformations, i.e diffeomorphisms So this is the symmetry associated with General Relativity, also refered to as general covariance. UPDATE: Note that all ...

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After reading the section put forth by @gcsantucci I think I kind of understand what was happening, but I'm eager to hear feedback on this. If you break $SU(2)$ using a doublet you indeed break all the generators and get massive gauge bosons. What was confusing is that a linear combination of generators leaves the vacuum invariant (namely, $\sigma _1 ... 1 I guess instead of me trying to explain, it is way better if you read section "Non Abelian Examples" of Peskin and Schroeder, in chapter 20. It explains exactly what you are asking for. It's actually the predecessor of the Standard model. Georgi and Glashow proposed this model before the SM, because they didn't know about the Z boson. So it is exactly what ... 3 I) First recall the fact that$SL(2,\mathbb{R})\times SL(2,\mathbb{R})$is (the double cover of) the identity component$SO^{+}(2,2;\mathbb{R})$of the split orthogonal group$O(2,2;\mathbb{R})$. This follows partly because: There is a bijective isometry from the split real space$(\mathbb{R}^{2,2},||\cdot||^2)$to the space of$2\times2 $real ... 1 $$[P,H]f(x)=(PH-Hp)f(x)$$ But $$H=P^2/2m+E(x)$$ $$=PE(x)-Hf(x)$$ $$=E(-x)-E(-x)$$ $$=0$$ The parity operator therefore commutes with Hamiltonian. 4 First, let's go back to flat space. The mentioned SU(2) algebra is just a part of the 10-dimensional Poincare algebra - the algebra of isometries of Minkowski space-time. The generators of isometries are called Killing vector fields, you can easily show that these (they come with the Lie bracket, as usual) obey the Poincare algebra. Because isometries (by ... 0 Your action is (manifestly) Lorentz-invariant. It means that there are 6 associated Noether's currents: one for each direction of the Lorentz group - three for boosts and three (which you described) for rotations. The physical interpretation of these currents is the flow of components of angular momentum of the field through space-time. Take the associated ... 1 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 ... 2 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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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