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Noether's theorem states that if a system has a continuous symmetry, there is a quantity related to this symmetry, called the Noether charge, which is conserved. It does not state anything on the fact that adding a constant term to a measurable quantity may or may not change the physical description of the system. Only some physical quantities in fact are ...


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I do not agree with the answer given by @ACuriousMind. @Scardenalli has asked for a compact Ricci-flat Riemannian manifold $M$ having as isometry group $U(1)\times SU(2)\times SU(3)$. This does not imply that $M$ must be a symmetric Riemannian manifold. However, the answer to @Scardenalli's question is still no, and it follows from a classical result in ...


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Classical Lagrangian field theory deals with fields $\phi: M \to N$, where $M$ is spacetime and $N$ is the target-space of the fields. We shall for convenience call $M$ and $N$ the horizontal and the vertical space, respectively. OP is in this terminology essentially asking Q: What is the meaning of horizontal transformations? A: It is a (horizontal) flow ...


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The appropriate definition of symmetry uses infinitesimal quantities, not just small quantities. Thus, in terms of your question, the Lagrangian is symmetric if $dL/d\epsilon=0$ at $\epsilon=0$. In terms of your example (rotation of a 2D harmonic oscillator), we have $$ L \to (1+\epsilon^2) L = L + \mathcal{O}(\epsilon^2) $$ Thus to first order in ...


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It sounds like you are interested in symplectic reduction procedures. On of these methods is that of Routh's procedure to eliminate cyclic variables using a Legendre transform to a reduced-variable Hamiltonian called a Routhian. Forming a variational approach may be difficult for some reduction procedures, however we can view conserved quantities as ...


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While neither the Lagrangian $\mathcal{L}$ nor the action $S$ are invariant under boosts of the form $$\dot{q}(t) \to \dot{q}(t) + v, \quad v \in \mathbb{R},$$ the Euler-Lagrange equations are. The dynamics of the systems are unchanged for any transformation that preserves $\delta S = 0$, i.e. a transformation of the form $$ \mathcal{L}(q, \dot{q}, t) \to ...


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The answer to your question involves the fact that one does not usually know a priori the electric field $\textbf{E}$ (or, for that matter, its direction) of a charge distribution $\rho$. Gauss's law, in integral form, relates the flux of the electric field through some closed surface $S$ to the charge enclosed within the volume bounded by $S$. Precisely, ...


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As your teacher says, it holds for every surface, but a look at the law itself, should clear out why some form of symmetrie is desirable: $$ \iint_S \vec{E} .\mathrm{d}\vec{A}=\iint_S E . \cos\theta . \mathrm{d}A = \frac{Q}{\epsilon_0} $$ Here, $E$ and $\theta$ are position-dependent, so to calculate the integral, you need to take care of a position ...


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As an example, let us suppose that you want to use the Gauss law to evaluate the electric field generated by a body charged in an uniform way. The gauss law tell you that the flux over an arbitrary closed surface around your body is proportional to the total charge: $$\int_{\partial V} \vec{E}\cdot d\vec{S}=\frac{Q}{\epsilon_0} $$ but this is an ...


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I think a classical example is electrical conductivity and resistivity (see Wikipedia), or any physical quantity which is described in the anisotropic case by a tensor (see also elasticity tensor as suggested in the comments by Jon Custer). Consider the Ohm's law in the anisotropic case $$ J_{i}=\sigma_{ij} E_j, \qquad E_i=\rho_{ij} J_j $$ The conductivity ...


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May I ask what text you are reading? My understanding of the stress energy tensor is as follows. The Noether condition is written as,\begin{equation} \partial _\mu \bigg[\frac{\partial \mathcal L}{\partial (\partial _\mu \phi )}\delta \phi +\mathcal L \delta x^\mu\bigg]=0 \end{equation} In the discrete case we can imagine separate infinitesimal time and ...


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Here we shall only discuss general relativistic diffeomorphism-invariant matter theories in a curved spacetime in the classical limit $\hbar\to 0$ for simplicity. In particular, we will not discuss the SEM pseudotensor for the gravitational field, but only the stress-energy-momentum (SEM) tensor for matter (m) fields $\Phi^A$. We emphasize that our ...


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For an antilinear operator, as the antiunitaries and the complex conjugation, the definition of adjoint is changed: $$\langle U^{a*}\psi,\phi\rangle=\overline{\langle \psi,U\phi\rangle}$$ where $a*$ stands for anti-adjoint. It is therefore easy to see that the anti-adjoint of $K$ is $K$ itself (and in general the anti-adjoint of an anti-unitary is ...


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When you integrate the Lagrangian density over a certain region $\Omega$, this is in principle allowed to change and this gives you a "boundary" term in the variation. This is well discussed in, e.g., the book of Goldstein (3rd edition), where the correct proof of the Noether theorem is given.


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The 4 generators of $SU(5)$ are not all "equivalent". In general, the generators of the group/algebra satisfy a defining equation of the form $$[T^i,T^j]=f^{ijk} T^k$$ so depending on the structure constants $f^{ijk} $,it is possible for example that $$[T^1,T^2]\neq[T^2,T^3]\quad\text{etc,}$$ so it is important which generators are broken. In terms of ...



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