Editing Noether's theorem (section)

=== Example 2: Conservation of center of momentum ===
Still considering 1-dimensional time, let

:<math>\begin{align}
  \mathcal{S}\left[\vec{x}\right]
    & = \int \mathcal{L}\left[\vec{x}(t), \dot{\vec{x}}(t)\right]  dt \\[3pt]
    & = \int \left[\sum^N_{\alpha=1} \frac{m_\alpha}{2}\left(\dot{\vec{x}}_\alpha\right)^2 - \sum_{\alpha<\beta} V_{\alpha\beta}\left(\vec{x}_\beta - \vec{x}_\alpha\right)\right] dt,
\end{align}</math>

for <math>N</math> Newtonian particles where the potential only depends pairwise upon the relative displacement.

For <math>\vec{Q}</math>, consider the generator of Galilean transformations (i.e. a change in the frame of reference). In other words,

:<math>Q_i\left[x^j_\alpha(t)\right] = t \delta^j_i.</math>

And

:<math>\begin{align}
  Q_i[\mathcal{L}]
    & = \sum_\alpha m_\alpha \dot{x}_\alpha^i - \sum_{\alpha<\beta}t \partial_i V_{\alpha\beta}\left(\vec{x}_\beta - \vec{x}_\alpha\right) \\
    & = \sum_\alpha m_\alpha \dot{x}_\alpha^i.
\end{align}</math>

This has the form of <math display="inline">\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>\begin{align}
  \vec{j} & = \sum_\alpha \left(\frac{\partial}{\partial \dot{\vec{x}}_\alpha} \mathcal{L}\right)\cdot\vec{Q}\left[\vec{x}_\alpha\right] - \vec{f} \\[6pt]
          & = \sum_\alpha \left(m_\alpha \dot{\vec{x}}_\alpha t - m_\alpha \vec{x}_\alpha\right) \\[3pt]
          & = \vec{P}t - M\vec{x}_{CM}
\end{align}</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:

:<math>\frac{d\vec{j}}{dt} = 0 \Rightarrow \vec{P} - M \dot{\vec{x}}_{CM} = 0.</math>