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To identify the Noether 4-current with the electric 4-current, one would in principle have to show that the Noether 4-current indeed appears as the source term in Maxwell's equations. The Maxwell equations with sources (Gauss's + Ampere's laws) are derived by adding the Maxwell Lagrangian $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ to a minimally coupled, gauge-...

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I recently spent a lot of time thinking about this stuff and wrote a little document which I put on my website here (under the title "Visualizing the Inverse Noether Theorem and Symplectic Geometry"). So I will begin by first addressing your specific question of how the symmetries of the Hamiltonian and Lagrangian are connected. However, I also want to ...

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In these types of problems, it is very useful to work on a generic spacetime with metric $g$. For instance, we can promote the usual free-field Lagrangian $$S=\int\mathrm{d}^dx\,\mathcal{L}=\int\mathrm{d}^dx\,\partial_{\mu}\phi\,\partial^{\mu}\phi$$ to one on a general curved spacetime as $$S=\int\mathrm{d}^dx\,\mathcal{L}=\int\mathrm{d}^dx\,\sqrt{g}\,g^{\... 4 Eq. (9.1) defines a finite quasi-symmetry. Note that Noether's theorem only needs an infinitesimal quasi-symmetry of the action to produce a conservation law. Townsend assumes that the two Lagrangians L in eq. (9.1) are the same function (although their arguments are obviously different). This is strictly speaking not necessarily if we are only interested ... 4 Momentum and energy are fundamental because they are conserved quantities. The total momentum and the total energy of a closed system do not change with time. In a world where everything seems to change all all the time, it is nice to find that some things do not. Noether’s Theorem is one way to understand why certain quantities like this are conserved. ... 3 The action shown in the question is a functional of \phi, not of x. A change of the integration variable x is just a relabeling of the index set. It does not transform the dynamic variables \phi at all, so no: a change of variable does not correspond to a conserved quantity. More explicitly, if y(x) is a monotonic smooth function of x, then$$ \...

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I don't have a copy of Srednicki's text on hand, so I'll start by transcribing the question into equations so the OP can check whether or not I'm interpreting the question correctly. The current $j^\mu\propto\overline\psi\gamma^\mu\psi$ satisfies $\partial_\mu j^\mu=0$ whenever $\psi$ satisfies the equation of motion $(i\gamma^\mu D_\mu+m)\psi=0$ with $D_\... 3 In field theory, you often consider "Lagrangian densities" which are to be integrated over space-time instead of just over time. For example, where as in the one dimensional case you would write $$S = \int dt L$$ in field theory you would write $$S = \int d^4 x \mathcal{L}.$$ The equation of motion will be a PDE. Noether's theorem, instead of giving ... 3 Angular momentum and linear momentum are fundamentally different quantities. At a basic level, they are generators/charges of two very different symmetries and are thus, conserved, according to Noether theorem, under two very different symmetries. In particular, angular momenta are the generators of the rotational symmetry, i.e.$SO(d)$in a$d-$dimensional ... 3 Time reversal symmetry is a discrete symmetry, whereas spatial translation is a continuous one. I can do a very big or very small spatial translation, but time reversal isn't like that; it's more like the two states of a light switch. Noether's theorem gets conservation laws from continuous symmetries only. Indeed, time translation is a continuous symmetry ... 3 When we say a symmetry is continuous, it is shorthand for saying that the group of symmetries is continuous. For time translations, the group consists of all translations for any a, where$a$is a continuous parameter. There are an infinite number of these group elements, and there are elements of the group which are infinitely close together. For time ... 3 TL;DR: The conclusions surrounding eq. (2.19) are not true in general. First of all, Ref. 1 makes it clear on top of p. 19 that$T^{\mu\nu}$denotes the metric/Hilbert stress-energy-momentum (SEM) tensor, not some other SEM tensor. Now it is true that if$\epsilon=\epsilon^{\mu}\partial_{\mu}$is a Killing vector field (KVF), then the current (2.19) ... 2 Express your equations of motion as $$\dot q= [H,q]\\\\\dot p=[H,p]$$ Note that, on the mass shell, any function$f(q,p)$obeys$\dot f(q,p)=[H,f]$. Now just hit the commutator$[G, ]$on both sides in the equations of motion. Since$[G, H]=0$then$G$commute with the time derivative, i.e.$\dot G=0$. Using the Jacobi identity on the right hand side ... 2 Magnetic charge is a topological charge that is conserved due to the topology of the manifold of solutions to the field equations --- these are distinct from Noether charges like electric charge that arise from symmetries of the action. From https://arxiv.org/pdf/1411.3099.pdf: "Unlike the above mentioned Noether currents, there exist another class of ... 2 The parameter$s$below eq. (\ref{eq8}) is non-standard. Noether's theorem and its Lagrangian formalism do in general not rely on a metric. Nevertheless, we only need eq. (\ref{eq14}), which is indeed correct. Note that the so-called vertical generator$\zeta_{r}\left( \phi_{r}(x),\partial\phi_{r}(x),x\right)in eq. (\ref{eq17}) depends on the field and ... 2 This is fairly stardard QFT material: How to compute the Noether current for a general group of transformations that may involve both the fields and the coordinates \left\lbrace \begin{aligned} x^\mu\to {x'}^\mu &= x^\mu + \varepsilon\,\xi^\mu(x)\,,\\ \phi^i(x) \to {\phi'}^i(x') &= \phi^i(x) + \varepsilon\,t^i[\phi](x)\,. \end{aligned} \right. ... 2 The solutions to the Euler-Lagrange (EL) equations for OP's non-square root LagrangianL$are affinely parametrized geodesics. Noether's theorem yields that$L$is a constant (of motion) along a geodesic, but it is not necessarily$1$. For more information, see e.g. my Phys.SE answer here. The constant is$1$if one picks the parameter$\lambda$to be ... 2 Sweet question. Please allow mo to say that the global infinitesimal transformations that you write a the beginning are nothing more that the infinitesimal forms of the following global$U(1)$transformation. Now, the Lagrangian that you show is trivially invariant under such global transformations, but it is also invariant under the local ones: $$\psi\... 2 If you replace eqs. (4.1.1) and (4.1.2) by your (A) equations you actually kill the extra parameter introduced \varepsilon, rendering the rest of the chapter pointless. The "generator" term comes from group theory, and in page 64 you find a very brief note: "[...] Because a set of such transformations has an identity element and each transformation has ... 2 This issue has come up many times on this site. It's one of those things where the standard textbook presentation is severely lacking. For example, consider translational symmetry. Sometimes this is written as$$x \to x' = x + a, \quad \phi \to \phi'(x') = \phi(x).$$However, you don't implement this symmetry transformation by substituting every x with an ... 2 FWIW, this procedure is known in the physics literature as the Noether procedure, or as gauging a global symmetry. Typically QFT textbooks don't provide a general proof of the Noether procedure but show it for specific theories only. [In Srednicki's case, the original QED Lagrangian density (58.2) is already gauge invariant, so the Noether procedure is very ... 2 A generating function^1 \epsilon G(q,p,t) [where \epsilon is an infinitesimal parameter] can to first order in \epsilon be identified with a generating function \epsilon G(q,P,t) for a type 2 canonical transformation (CT), P=p+{\cal O}(\epsilon), cf. Ref. 1. On the other hand, a CT takes Hamilton's equations into Kamilton's equations, with$$K(Q,... 2 In a nutshell,$a$is an arbitrary but fixed 1-parameter, while$t$is a running time coordinate. One cannot consistently put a fixed parameter equal to a running coordinate. Phrased differently,$a$is here not allowed to depend on$t$. In particular, the 1-parameter family$(\mathbb{R}\ni t\mapsto t+a\in\mathbb{R})_{a\in\mathbb{R}}$of time translation ... 2 You are right, the most general formalization of energy we have is furnished by Noether's Theorem. As you noted, conservation of energy comes from the time invariance of the laws governing the dynamics of the particular system. If we assert spacial invariance we get conservation of linear momentum and if we assert rotational invariance we obtain conservation ... 2 So the important thing is simply that it take the form of a total derivative, even if that symbolically involves the expressions for$\mathbf q$and its derivatives. So for example it could take the form$\mathbf q \dot{\mathbf q},$and that’s fine because it is symbolically a total derivative. Background: the Lagrangian doesn’t know about paths Backing up ... 2 Position and center-of-mass coordinates are not quantized/discrete in conventional QM (excluding lattice formulations), cf. e.g. this Phys.SE post. In contrast, for the reason why angular momentum is quantized, see e.g. my Phys.SE answer here. 2 This isn't a fact about general relativity, it's a fact about special relativity. Based on hundreds of years worth of experiments, we know that the energy-momentum four-vector is conserved locally. This means that the stress-energy tensor has zero divergence. (The stress-energy tensor isn't conserved. What's conserved is the energy-momentum four-vector.) If ... 2 There are at least 2 issues with OP's discussion (v2): One should properly distinguish between total and explicit spacetime derivatives, cf. e.g. my Phys.SE answer here. In particular, an infinitesimal quasisymmetry of the Lagrangian (density), means by definition that the infinitesimal variation is a total (space)time divergence. Note that not all terms ... 2 The 3D spherically symmetric harmonic oscillator $$H~=~\frac{p_x^2+p_y^2+p_z^2}{2m}+ \alpha (x^2+y^2+z^2) ~=~ H_x + H_y + H_z$$ is a separable, Liouville integrable, and in fact a maximally superintegrable system with additionally integrals of motion$H_x$,$H_y$,$H_z$,$L_x$,$L_y$and$L_z\$, i.e. nothing exotic like the Laplace-Runge-Lenz vector. (A 3D ...

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There are at least 2 lessons to be learned from OP's set-up: Noether's theorem is not necessarily about strict symmetry of the action. It is enough if the action has an (infinitesimal) quasi-symmetry, i.e. symmetry up to boundary terms. There's no free lunch. To prove energy conservation, one must use a non-trivial assumption: In this case, that the ...

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