# Tag Info

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Electric charge conservation is a "discrete" symmetry. Quarks and anti-quarks have discrete fractional electric charges (±1/3, ±2/3) electrons, positrons and protons have integer charges.

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OP wrote (v3): Working out the conserved quantity, we get that the $z$-component of angular momentum $L_z = m x(t)\dot{y}(t) - m y(t)\dot{x}(t)$ is conserved for any path $(x(t),y(t),z(t))$. Well, it should be stressed that the conservation law in Noether's theorem is an so-called on-shell conservation law. It does not hold for any curved (off-shell) ...

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Firstly, you ask isn't there more than one symmetry meaning this question has several answers? Yes! In general, a given theory can have all sorts of symmetries, and each of these symmetries leads to its own conserved quantity via Noether's theorem. As for what's going on with Noether's theorem and applying it in general, I'd like to strongly ...

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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.] II) Assume that a variation of $S$ for arbitrary $x$-dependent infinitesimal $\epsilon(x)$ takes the ...

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Define $F(u):= \int_0^u f(s) ds$, so the equations for the field $u(t,\vec{x})$ can be re-written as $$\frac{\partial^2 u}{\partial t^2}-\Delta_{\vec x} u + \frac{dF}{du}=0\::$$ If defining $${\cal L}:= \frac{1}{2}(-\partial_t u\partial_t u + \nabla u \cdot \nabla u) + F(u)\:.$$ this Lagrangian density leads to your field equations. Moreover, as you can ...

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You can only find the Hamiltonian if you do a so-called 'gauge fixing' procedure, since the Dirac field couples (minimally, but uniquely) to a gauge field. To get the Hamiltonian (density) you need to perform the full Dirac constraint analysis and at the end 'gauge fix'. See the books by Sundermayer or Henneaux+Teitelboim for details regarding the ...

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OP, I am new to stackexchange (but a physics veteran) so I am not yet allowed to comment on the post itself, but you should know that the one you chose as the correct answer is only valid for one dimensional curves, and even there it is valid only for a special definition of symmetry that allows for boundary terms (called "quasi-symmetries," as QMechanic ...

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I recommend you "Symmetries" by Griffiths, or "Symmetry" by Roy McWeeny (it's a Dover book). "Geometry, Topology and Physics" by Nakahara is a good option.

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The technical answer is $No$. Surprisingly I think Wikipedia gives the better definition, though I think both authors are trying to say the same thing. Let the action be defined as $S[\varphi]=\int d^4x\ \mathcal L(\varphi(x),\partial_\mu\varphi(x))$ A differentiable symmetry is a symmetry of the functional that does not change the action ...

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Yes, provided one uses the correct notions of symmetry for the action and the lagrangian. The setup. We assume throughout that the action can be written as the integral of a local Lagrangian. Namely, let $\mathcal C$ be the configuration space of the system, then for any admissible path $q:[t_a, t_b]\to \mathcal C$, there exists a local function $L$ of ...

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First some terminology: In general an infinitesimal transformation of a field theory consists of a so-called horizontal infinitesimal transformation $$\delta x^i ~=~x^{\prime i}- x^i$$ of the base manifold, and a so-called vertical infinitesimal transformation $$\delta_0\phi^{\alpha}(x)~=~\phi^{\prime \alpha}(x)-\phi^{\alpha}(x)$$ of the fields. The ...

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I) Firstly, we mention that Noether's Theorem (in its original form) concerns a symmetry of the action $S$, not necessarily the Lagrangian $L$. The relevant notion for the Lagrangian is quasi-symmetry, cf. this Phys.SE answer. II) Secondly, let us assume that the Lagrangian $L=L(q,\dot{q})$ has no explicit time dependence. We would like to use Noether's ...

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I) First it should be stressed that Noether's theorem is not really about Lie groups but only about Lie algebras, i.e., one just needs $n$ infinitesimal symmetries to deduce $n$ conservation laws. II) Secondly, it is straightforward to check (by recalling the proof of Noether's theorem) that Noether's theorem generalizes to supernumber-valued variables, ...

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Paths AB and CD are exactly the same length, with exactly the same start and end times, which means lengths AC and BD must be zero. Mr Feynman: you're talking baloney, as you would say to a social scientist!

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