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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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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 $$\mathcal{L}(q, \dot{q}, t) \to ... 1 It sounds like you are interested in symplectic reduction procedures. On of these methods is that of Routh's procedure to eliminate cyclic variables using a Legendre transform to a reduced-variable Hamiltonian called a Routhian. Forming a variational approach may be difficult for some reduction procedures, however we can view conserved quantities as ... 0 Yes, it is, but it is not easy to know the invariant quantities beforehand, so the usual way to use it is to write down the Lagrangian in terms of any (non-necessarily invariant) quantities, and then use the principle of least action to find the invariant quantities, Nother currents, etc. 2 Classical Lagrangian field theory deals with fields \phi: M \to N, where M is spacetime and N is the target-space of the fields. We shall for convenience call M and N the horizontal and the vertical space, respectively. OP is in this terminology essentially asking Q: What is the meaning of horizontal transformations? A: It is a (horizontal) flow ... 1 The 4 generators of SU(5) are not all "equivalent". In general, the generators of the group/algebra satisfy a defining equation of the form$$[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,}$$so it is important which generators are broken. In terms of ... 1 When you integrate the Lagrangian density over a certain region \Omega, this is in principle allowed to change and this gives you a "boundary" term in the variation. This is well discussed in, e.g., the book of Goldstein (3rd edition), where the correct proof of the Noether theorem is given. 1 For an antilinear operator, as the antiunitaries and the complex conjugation, the definition of adjoint is changed:$$\langle U^{a*}\psi,\phi\rangle=\overline{\langle \psi,U\phi\rangle}$$where a* stands for anti-adjoint. It is therefore easy to see that the anti-adjoint of K is K itself (and in general the anti-adjoint of an anti-unitary is ... 1 Here we shall only discuss general relativistic diffeomorphism-invariant matter theories in a curved spacetime in the classical limit \hbar\to 0 for simplicity. In particular, we will not discuss the SEM pseudotensor for the gravitational field, but only the stress-energy-momentum (SEM) tensor for matter (m) fields \Phi^A. We emphasize that our ... 1 May I ask what text you are reading? My understanding of the stress energy tensor is as follows. The Noether condition is written as,$$\partial _\mu \bigg[\frac{\partial \mathcal L}{\partial (\partial _\mu \phi )}\delta \phi +\mathcal L \delta x^\mu\bigg]=0$$ In the discrete case we can imagine separate infinitesimal time and ... 1 I think a classical example is electrical conductivity and resistivity (see Wikipedia), or any physical quantity which is described in the anisotropic case by a tensor (see also elasticity tensor as suggested in the comments by Jon Custer). Consider the Ohm's law in the anisotropic case$$ J_{i}=\sigma_{ij} E_j, \qquad E_i=\rho_{ij} J_j $$The conductivity ... 2 I do not agree with the answer given by @ACuriousMind. @Scardenalli has asked for a compact Ricci-flat Riemannian manifold M having as isometry group U(1)\times SU(2)\times SU(3). This does not imply that M must be a symmetric Riemannian manifold. However, the answer to @Scardenalli's question is still no, and it follows from a classical result in ... 2 Noether's theorem states that if a system has a continuous symmetry, there is a quantity related to this symmetry, called the Noether charge, which is conserved. It does not state anything on the fact that adding a constant term to a measurable quantity may or may not change the physical description of the system. Only some physical quantities in fact are ... 0 LO-TO splitting is caused by the long-ranged nature of the Coulomb interaction (i.e. because the Fourier Transform of the Coulomb interaction,4\pi e^2/q^2, is not well-defined at q=0). Also, it occurs near the Brillouin zone center, but not at the exact Brillouin zone center because of retardation effects (i.e. the finite speed of light). At q=0, the ... 0 Why is it said that Gauss's Law is mainly applicable for symmetric surfaces/bodies? Why not for asymmetric surfaces? Nobody says that. Gauß law holds whenever its hypotheses hold. In particular it has nothing to do with the shape of the bodies at hand, rather it is a mathematical theorem relating the flow of a vector field through a surface to the ... 1 As an example, let us suppose that you want to use the Gauss law to evaluate the electric field generated by a body charged in an uniform way. The gauss law tell you that the flux over an arbitrary closed surface around your body is proportional to the total charge:$$\int_{\partial V} \vec{E}\cdot d\vec{S}=\frac{Q}{\epsilon_0} $$but this is an ... 1 As your teacher says, it holds for every surface, but a look at the law itself, should clear out why some form of symmetrie is desirable:$$ \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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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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You can use diffraction of low energy electrons (to study the crystalline surface) or X-ray crystallography (you can get the surface orientation from that also).

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What would happen to light passing through a narrow space between the event horizons of two equal-mass black holes? Would it deviate or follow a straight path? Like iharob and JohnnyMo1 said, the light goes straight. But something else happens to it. See this screenshot of Irwin Shapiro's seminal paper: See where he said the speed of light depends on ...

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These are two different problems. What Jon was saying is correct. However, it does not explain LO-TO splitting. Like Jon said, because you can tell when you are on a Ga or As atom, the degeneracy of the optical modes are lifted at the Gamma point. This is in regards of 3 different optical modes separating. However, the phenomena Cardona is refers to involves ...

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I'm going to assume you mean that the light travels on the precise center line between the holes, as iharob did. This sort of symmetry question is very common in physics. Here's a similar question in classical electrodynamics. "If I place a positive charge at the center of a perfect equilateral triangle of equal negative charges, will it move?" Let's say it ...

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I think the answer is very simple if you ask another question, you are implying that both black holes generate the same gravity and that the light passes exactly through the middle between them, so the question is If the light deviates, where will it deviate to? Since given the conditions it's not possible to give an answer to this question, it means ...

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