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ack! many, many , many errors!!!!!!! and no one noticed??? |
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Noether's theorem is an [[on shell]] theorem.
=== An example ===
OK, that was a general proof. Let's look at a specific case. Let's work with a one dimensional manifold with the topology of <b>R</b> (time) coordinatized by t. Let's assume
<math>S[x]=\int dt \mathcal{L}(x(t),\dot{x}(t))=\int dt \left(\frac{m}{2}g_{ij}\dot{x}^i(t)\dot{x}^j(t)-V(x(t))\right)</math>
(i.e. a Newtonian particle of mass m moving in a curved Riemannian space (but not curved spacetime!) of metric g with a potential of V).
For Q, let's consider the generator of time translations. In other words, <math>Q[x](t)=\dot{x}(t)</math>. (Quantum field) physicists would often put a factor of i on the right hand side, but what the heck. Note that <math>Q[\mathcal{L}]=m g_{ij}\dot{x}^i\ddot{x}^j-\frac{\partial}{\partial x^i}V(x)\dot{x}^i</math>. This has the form of <math>\frac{d}{dt}\left[\frac{m}{2} g_{ij}\dot{x}^i\dot{x}^j-V(x)\right]</math> so we can set <math>f=\frac{m}{2} g_{ij}\dot{x}^i\dot{x}^j-V(x)</math>. Then,
<math>j=(\frac{\partial}{\partial \dot{x}^i}\mathcal{L})Q[x]-f=m g_{ij}\dot{x}^j\dot{x}^i-[\frac{m}{2} g_{ij}\dot{x}^i\dot{x}^j-V(x)]=\frac{m}{2}g_{ij}\dot{x}^i\dot{x}^j+V(x)</math>. You might recognize the right hand side as the energy and Noether's theorem states that <math>\dot{j}=0</math> (i.e. the conservation of energy is a consequence of invariance under time translations).
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