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At the physics 101 level, you pretty much just have to accept this as an experimental fact. At the upper division or early grad school level, you'll be introduced to Noether's Theorem, and we can talk about the invariance of physical law under displacements in time. Really this just replaces one experimental fact (energy is conserved) with another (the ...


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Here's what I perceive to be a mathematically and logically precise presentation of the theorem, let me know if this helps. Mathematical Preliminaries First let me introduce some precise notation so that we don't encounter any issues with "infinitesimals" etc. Given a field $\phi$, let $\hat\phi(\alpha, x)$ denote a smooth one-parameter family of fields ...


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1) If you want a Noether theorem for information, there is no such thing. Trying to obtain it from a symmetry law, by Noether's theorem can't work, simply because information is not a quantity that can be obtained for instance by the derivative of the Lagrangian with respect to some variable. Information is not scalar, vector, tensor, spinor etc. 2) ...


34

I) For a mathematical precise treatment of an inverse Noether's Theorem, one should consult e.g. Olver's book (Ref. 1, Thm. 5.58), as user orbifold also writes in his answer (v2). Here we would like give a heuristic and less technical discussion, to convey the heart of the matter, and try to avoid the language of jets and prolongations as much as possible. ...


34

We usually call equations like $$\frac{d}{dt} \frac{\partial L}{\partial \dot{q_i}} - \frac{\partial L}{\partial q_i} = 0$$ "equations of motion," because they are equations that tell us how the variables of our system (here $q_i$) evolve in time. Indeed, in general, the solution to $n$ second order differential equations involves $2n$ integration ...


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1) Hamiltonian Problem. The Kepler problem has Hamiltonian $$ H~=~T+V, \qquad T~:=~ \frac{p^2}{2m}, \qquad V~:=~- \frac{k}{q}, \tag{1} $$ where $m$ is the 2-body reduced mass. The Laplace–Runge–Lenz vector is (up to an irrelevant normalization) $$ A^j ~:=~a^j + km\frac{q^j}{q}, \qquad a^j~:=~({\bf L} \times {\bf p})^j~=~{\bf q}\cdot{\bf p}~p^j- p^2~q^j,\...


21

1) Off-shell vs. on-shell action. What may cause some confusion is that Noether's theorem in its original formulation only refers to the off-shell action functional $$\tag{1} I[q;t_i,t_f]~:=~ \int_{t_i}^{t_f}\! {\rm d}t \ L(q(t),\dot{q}(t),t), $$ while Feynman's proof [1]$^1$ mostly is referring to the Dirichlet on-shell action function $$\tag{2} S(q_f,...


20

Indeed, nothing is wrong with Noether theorem here, $J^\mu = F^{\mu \nu} \partial_\nu \Lambda$ is a conserved current for every choice of the smooth scalar function $\Lambda$. It can be proved by direct inspection, since $$\partial_\mu J^\mu = \partial_\mu (F^{\mu \nu} \partial_\nu \Lambda)= (\partial_\mu F^{\mu \nu}) \partial_\nu \Lambda+ F^{\mu \nu} \...


19

Action formulation. It should be stressed that Noether's theorem is a statement about consequences of symmetries of an action functional (as opposed to, e.g., symmetries of equations of motion, or solutions thereof, cf. this Phys.SE post). So to use Noether's theorem, we first of all need an action formulation. How do we get an action for a Hamiltonian ...


17

Assume that the Lagrangian density $$\tag{1} {\cal L} ~=~ {\cal L}(\phi(x), \partial \phi(x), x) $$ does not depend on higher-order derivatives $\partial^2\phi$, $\partial^3\phi$, $\partial^4\phi$, etc. Let $$\tag{2} \pi^{\mu}_{\alpha} ~:=~ \frac{\partial {\cal L}}{ \partial (\partial_{\mu}\phi^{\alpha})} $$ denote the de Donder momenta, and let $$...


17

I) Let there be given a local action functional $$\tag{1} S[\phi]~=~\int_V \mathrm{d}^nx ~{\cal L}, $$ with the Lagrangian density $$\tag{2} {\cal L}(\phi(x),\partial\phi(x),x). $$ [We leave it to the reader to extend to higher-derivative theories. See also e.g. Ref. 1.] II) We want to study an infinitesimal variation$^1$ $$\tag{3} \delta x^{\mu}~=~\...


16

The idea of partitioning energy into different forms like "mechanical energy" or "chemical energy" and such is actually arbitrary. More or less by definition, energy is that which is conserved unter time translations by Noether's theorem. If what you call "mechanical energy" has changed, then there is another term in the Noetherian energy that has changed ...


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The 2n constants in the system of second order differential equations (Lagrange equations) are just the arbitrary initial conditions of generalized coordinates and velocities of the system that determine the time development of the system for a special case. They are not conserved quantities of the system. Conserved quantities follow from the functional form ...


15

If the theory is invariant under translations in space, then linear momentum is conserved by Noether's theorem. If the theory is quantum, conservation holds only on the level of the expectation values (because that's the only meaningful level where you can talk about momentum as a number that's conserved in time), but it still holds. There is no way out. ...


14

CPT seems to imply it. You can reverse the system evolution by applying charge, parity and time conjugation, so the information about the past must be contained in the present state. That implies conservation of information by the evolution. This may not be the answer you wanted, because it does not imply unitarity, but it is the only relationship between ...


14

Explaining Noether's theorem to 8th graders may be hard, as even the simplest special case of the theorem requires some calculus to state. However, if we're talking about why one example of an invariance corresponds to one example of a conservation law, there are some calculus-free arguments you can make. I'll mention one example, energy (whose conservation ...


13

While Kepler second law is simply a statement of the conservation of angular momentum (and as such it holds for all systems described by central forces), the first and the third laws are special and are linked with the unique form of the newtonian potential $-k/r$. In particular, Bertrand theorem assures that only the newtonian potential and the harmonic ...


13

Whether your current $j^\mu$ is conserved off-shell depends on your definition of $j^\mu$. If you define it via the Dirac and other charged fields, it will only be conserved assuming the equations of motion. However, if you define $j^\mu$ via $$ j^\mu = \partial^\nu F_{\mu\nu}, $$ i.e. as a function of the electromagnetic field and its derivatives, then $\...


13

The trick is given in equation 4.4 of the attached article: First couple the theory to gravity, (by introducing a metric tensor in the integration measure and for each index raising) obtaining the action: $S = \int_M d^4x \sqrt{-g} \mathcal{L}$ Then vary the action with respect to the metric tensor: $T_{\alpha\beta} = \frac{1}{\sqrt{-g}} \frac{\delta S}{\...


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This answer will mostly follow this excellent (and quite readable!) paper, pointed out to me by Emilio himself, in the exposition. This is another paper that contains similar considerations. For an extended discussion on this and closely related topics, see this chatroom. There are a number of papers which make all kinds of claims about how one can (attempt ...


13

The answer is yes, the essence of Noether's theorem for linear and angular momentum can be understood without using the Lagrangian (or Hamiltonian) formulation, at least if we're willing to focus on models in which the equations of motion have the form $$ m_n\mathbf{\ddot x}_n = \mathbf{F}_n(\mathbf{x}_1,\mathbf{x}_2,...) \tag{1} $$ where $m_n$ and $\...


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Comments to the question (v1): Last thing first. On-shell means (in this context) that equations of motion (eom) are satisfied. Equations of motion means Euler-Lagrange equations. Off-shell means strictly speaking not on-shell, but in practice it is always used in the sense not necessarily on-shell. [Let us stress that every infinitesimal transformation is ...


11

The guiding principle is: "Anomalous symmetries cannot be gauged". The phenomenon of anomalies is not confined to quantum field theories. Anomalies exist also in classical field theories (I tried to emphasize this point in my answer on this question). (As already mentioned in the question), in the classical level, a symmetry is anomalous when the Lie ...


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Lets consider a local symmetry continuous. Look at an infinitesimal element. By that I mean the following transformation of fields: \begin{equation} \phi_a(x)\to\phi_a(x)+F_\alpha[\phi]g^\alpha(x)+F_\alpha^\mu[\phi]\partial_\mu g^\alpha(x)+\ldots\tag{1} \end{equation} Here $\phi_a$ is the whole set of fields, $F$ are some functionals specific to the symmetry,...


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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 ...


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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 ...


11

Assuming that your primary question is about Noether's theorem when $\mathcal{L} \ne T - V$, we can proceed as follows. Denoting the generalized coordinates by $q_i$, we consider a transformation $q_i \rightarrow q_i + \delta q_i$ which leaves the Lagrangian $\mathcal{L}(q_i, \dot{q}_i)$ invariant: $$\delta \mathcal{L} = 0$$ i.e. $$\sum_i \left( \frac {\...


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There is no general algorithm for doing so, and even figuring out how many conserved quantities a system has can be difficult. A famous example is the Toda lattice. When originally proposed by Toda in 1967, this model was believed to be chaotic. It was in fact proven to be integrable (to have too many conserved quantities to be chaotic) in 1974 by Henon. ...


11

Translational symmetry in the sense of the standard formulation of Noether theorems means that the Lagrangian is invariant under the action of the group of spatial translations. This is not the case in your example because $U$ does not admit such invariance. However there is another, more physical, version of the idea of translational invariance for a ...


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