Noether's theorem: Difference between revisions

Revision as of 02:03, 22 December 2003

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Before we go on, let's give some examples:

* In [[classical mechanics]], M is the one -dimensional [[manifold]] <math>\mathbb{R}</math>, representing time and the target space is the [[tangent bundle]] of [[space]] of generalized positions.

* In [[Field_theory_(physics)|Field Theory]], M is the [[spacetime]] [[manifold]] and the target space is the set of values the fields can take at any given point. For example, if there are m [[real]]-valued [[scalar]] fields, &phi;₁,...,&phi;_m, then the target [[manifold]] is <math>\mathbb{R}^m</math>. If the field is a [[real]] vector field, then the target [[manifold]] is [[isomorphic]] to <math>\mathbb{R}^n</math>. There's actually a much more elegant way using [[tangent bundle]]s over M, but for the purposes of this proof, we'd just stick to this version.

Now suppose there's a [[functional]]

Now suppose there's a [[functional]], <math>S:\mathcal{C}\rightarrow \mathbb{R}</math>, called the [[Action (physics)|action]]. Note it's a [[mapping]] to <math>\mathbb{R}</math>, not <math>\mathbb{C}</math>. This has got to do with physical reasons and does not really matter for this proof.

:<math>S:\mathcal{C}\rightarrow \mathbb{R}</math>,

Now suppose there's a [[functional]], <math>S:\mathcal{C}\rightarrow \mathbb{R}</math>, called the [[Action (physics)|action]]. (Note that it's atakes [[mapping]]values in to <math>\mathbb{R}</math>, notrather than <math>\mathbb{C}</math>.; Thisthis hasis got to do withfor physical reasons, and does notdoesn't really matter for this proof.)

To get to the usual version of Noether's theorom, we need additional restrictions on the [[Action (physics)|action]]. If <math>\phi\in\mathcal{C}</math>, we assume S(&phi;) is the [[integral]] over M of a function of &phi;, its [[derivative]] and the position called the [[Lagrangian]], <math>\mathcal{L}(\phi,\partial_\mu\phi,x)</math>. In other words,

To get to the usual version of Noether's theoromtheorem, we need additional restrictions on the [[Action (physics)|action]]. If <math>\phi\in\mathcal{C}</math>, weWe assume S(&phi;) is the [[integral]] over M of a function of &phi;, its [[derivative]] and the position called the [[Lagrangian]], <math>\mathcal{L}(\phi,\partial_\mu\phi,x)</math>. In other words,

<math>\forall\phi\in\mathcal{C}\, S[\phi]\equiv\int_M d^nx \mathcal{L}(\phi(x),\partial_\mu\phi(x),x)</math>.

:<math>\mathcal{L}(\phi,\partial_\mu\phi,x)</math>

Given [[boundary]] conditions, which is basically a specification of the value of &phi; at the [[boundary]] of M is [[compact]] or some limit on &phi; as x approaches <math>\infty</math> (this will help in doing [[integration by parts]]), we can denote as N the [[subset]] of <math>\mathcal{C}</math> consisting of functions, &phi; such that all [[functional derivative]]s of S at &phi; are zero and &phi; satisfies the given boundary conditions.

called the [[Lagrangian]], depending on &phi;, its [[derivative]] and the position. In other words, for &phi; in <math>\mathcal{C}</math>

<math>\forall\phi\in\mathcal{C}\, S[\phi]\equiv\int_M d^nx \mathcal{L}(\phi(x),\partial_\mu\phi(x),x).</math>.

GivenSuppose given [[boundary condition]] conditionss, which isare basically a specification of the value of &phi; at the [[boundary]] of M is [[compact]], or some limit on &phi; as x approaches <math>\infty</math>; (this will help in doing [[integration by parts]]),. weWe can denote asby N the [[subset]] of <math>\mathcal{C}</math> consisting of functions, &phi; such that all [[functional derivative]]s of S at &phi; are zero and &phi; satisfies the given boundary conditions.

Now, suppose we have an [[infinitesimal]] [[Transformation (mathematics)|transformation]] on <math>\mathcal{C}</math>, given by a [[functional derivative]], &delta; such that

:<math>\delta\int_N d^nx\mathcal{L}=\int_{\partial N}ds_\mu f^\mu(\phi(x),\partial\phi,\partial\partial\phi,...)</math>

for all [[compact]] submanifolds N. Then, we say &delta; is a generator of a 1-parameter [[symmetry]] [[Lie group]].

Now, for any N, because of the [[Euler-Lagrange]] theorem, we have

:<math>

\delta\int_N d^nx\mathcal{L}=

\int_Nd^nx(\frac{\partial\mathcal{L}}{\partial\phi}-

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Since this is true for any N, we have

:<math>

\partial_\mu(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}-f^\mu)=0

</math>.