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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).
=== Another example ===
Let's still work with one dimensional time. This time, let
<math>S[\vec{x}]=\int dt \mathcal{L}(\vec{x}(t),\dot{\vec{x}}(t))=\int dt \left (\sum^N_{\alpha=1} \frac{m_\alpha}{2}(\dot{\vec{x}}_\alpha)^2 -\sum_{\alpha<\beta} V_{\alpha\beta}(\vec{x}_\beta-\vec{x}_\alpha)\right )</math>
i.e. N Newtonian particles where the potential only depends pairwise upon the relative displacement.
For <math>\vec{Q}</math>, let's consider the generator of Galilean transformations (i.e. a change in the frame of reference). In other words, <math>Q_i[x^j_\alpha](t)=t \delta^j_i</math>. Note that <math>Q_i[\mathcal{L}]=\sum_\alpha m_\alpha \dot{x}_\alpha^i-\sum_{\alpha<\beta}\partial_i V_{\alpha\beta}(\vec{x}_\beta-\vec{x}_\alpha)(t-t)=\sum_\alpha m_\alpha \dot{x}_\alpha^i</math>
. This has the form of <math>\frac{d}{dt}\sum_\alpha m_\alpha x^i_\alpha</math> so we can set <math>\vec{f}=\sum_\alpha m_\alpha \vec{x}_\alpha</math>. Then,
<math>\vec{j}=\sum_\alpha (\frac{\partial}{\partial \dot{\vec{x}}_\alpha}\mathcal{L})\cdot\vec{Q}[\vec{x}_\alpha]-\vec{f}=\sum_\alpha (m_\alpha \dot{\vec{x}}_\alpha t-m_\alpha \vec{x})=\vec{P}t-M\vec{x}_{CM}</math> where <math>\vec{P}</math> is the total momentum, M is the total mass and <math>\vec{x}_{CM}</math> is the center of mass. Noether's theorem states that <math>\dot{\vec{j}}=0</math> (i.e. <math>\vec{P}=M\dot{\vec{x}}_{CM}</math>).
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