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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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Warning: this is a long and boring derivation. If you are interested only in the result skip to the very last sentence. Noether's theorem can be formulated in many ways. For the purposes of your question we can comfortably use the special relativistic Lagrangian formulation of a scalar field. So, suppose we are given an action $$S[\phi] = \int {\mathcal L}(\... 25 Put into one sentence, Noether's first Theorem states that a continuous, global, off-shell symmetry of an action S implies a local on-shell conservation law. By the words on-shell and off-shell are meant whether Euler-Lagrange equations of motion are satisfied or not. Now the question asks if continuous can be replace by discrete? It should immediately ... 24 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) ... 23 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 ... 19 Mass is only conserved in the low-energy limit of relativistic systems. In relativistic systems, mass can be converted into energy, and you can have processes like massive electron-positron pairs annhillating to form massless photons. What is conserved (in theories obeying special relativity, at least) is mass energy--this conservation is enforced by the ... 19 The symmetry you are asking about is usually called a scale transformation or dilation and it, along with Poincare transformations and conformal transformations is part of the group of conformal isometries of Minkowski space. In a large class of theories one can construct an "improved" energy-momentum tensor \theta^{\mu \nu} such that the Noether current ... 18 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. ... 17 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,...

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Noether's theorem says that symmetries lead to conservation laws, not the converse. Conservation of mass doesn't follow from any of the obvious symmetries of nonrelativistic motion. Those symmetries are translations in space (leading to conservation of momentum), translations in time (conservation of energy), rotations (conservation of angular momentum), and ...

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

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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} \... 13 Here I would like to mention the notion of quasi-symmetry. In general, if the Lagrangian (resp. Lagrangian density) is only invariant up to a total time derivative (resp. space-time divergence) when performing a certain off-shell^1 variation, one speaks of a quasi-symmetry, see, e.g., J.V. Jose and E.J. Saletan, "Classical Dynamics: A Contemporary Approach"... 13 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 ... 13 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 ... 12 It's intuitively clear that the energy most accurately describes how much the state of the system is changing with time. So if the laws of physics don't depend on time, then the amount how much the state of the system changes with time has to be conserved because it's still changing in the same way. In the same way, and perhaps even more intuitively, if the ... 12 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. ... 11 thank you for the nice question. It directly relates to the topics of conformal field theories. I found a very nice thread in another forum where I guess your question has been answered. Nevertheless, I will try to summarize the main points here and maybe add some points. Symmetries in General Relativity In general relativities, symmetries correspond to an ... 11 In basic Lagrangian mechanics (of the sort that is covered in a sophomore-level classical mechanics class), no it doesn't. The reason is that time plays a special role in the basic Lagrangian theory: it's the only independent parameter, which everything else is expressed as a function of. This is related to the fact that the action is the integral of the ... 11 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}, $$i.e. as a function of the electromagnetic field and its derivatives, then \... 11 The trick is given in equation 4.4 of the attached article: First couple the theory to gravity, (by introducing a metric tensor in the integration measure and for each index raising) obtaining the action: S = \int_M d^4x \sqrt{-g} \mathcal{L} Then vary the action with respect to the metric tensor: T_{\alpha\beta} = \frac{1}{\sqrt{-g}} \frac{\delta S}{\... 11 The guiding principle is: "Anomalous symmetries cannot be gauged". The phenomenon of anomalies is not confined to quantum field theories. Anomalies exist also in classical field theories (I tried to emphasize this point in my answer on this question). (As already mentioned in the question), in the classical level, a symmetry is anomalous when the Lie ... 10 You mentioned crystal symmetries. Crystals have a discrete translation invariance: It is not invariant under an infinitesimal translation, but invariant under translation by a lattice vector. The result of this is conservation of momentum up to a reciprocal lattice vector. There is an additional result: Suppose the Hamiltonian itself is time independent, ... 9 Go out and discover those "other explanations" (and accumulate sufficient supporting evidence, of course) and you can laugh at the dark matter specialists. Until then dark matter is the simplest hypothesis on offer that explains multiple observations in one go (galactic rotation curves, cluster dynamics, super cluster dynamics, the bullet cluster, the ... 9 Assume that the Lagrangian density$$\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})} $$denote the de Donder momenta, and let$$...

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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 theory? Well, let us for ...

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