Editing Noether's theorem (section)

== Examples ==

=== Example 1: Conservation of energy ===
Looking at the specific case of a Newtonian particle of mass ''m'', coordinate ''x'', moving under the influence of a potential ''V'', coordinatized by time ''t''. The [[action (physics)|action]], ''S'', is:

:<math>\begin{align}
  \mathcal{S}[x] & = \int L\left[x(t),\dot{x}(t)\right] \, dt \\
                 & = \int \left(\frac m 2 \sum_{i=1}^3\dot{x}_i^2 - V(x(t))\right) \, dt.
\end{align}</math>

The first term in the brackets is the [[kinetic energy]] of the particle, while the second is its [[potential energy]]. Consider the generator of [[time translation]]s <math>Q = \frac{d}{dt}</math>. In other words, <math>Q[x(t)] = \dot{x}(t)</math>. The coordinate ''x'' has an explicit dependence on time, whilst ''V'' does not; consequently:

:<math>Q[L] =
  \frac{d}{dt}\left[\frac{m}{2}\sum_i\dot{x}_i^2 - V(x)\right] =
  m \sum_i\dot{x}_i\ddot{x}_i - \sum_i\frac{\partial V(x)}{\partial x_i}\dot{x}_i
</math>

so we can set

:<math>L = \frac{m}{2} \sum_i\dot{x}_i^2 - V(x).</math>

Then,

:<math>\begin{align}
  j & = \sum_{i=1}^3\frac{\partial L}{\partial \dot{x}_i}Q[x_i] - L \\
    & = m \sum_i\dot{x}_i^2 - \left[\frac{m}{2}\sum_i\dot{x}_i^2 - V(x)\right] \\[3pt]
    & = \frac{m}{2}\sum_i\dot{x}_i^2 + V(x).
\end{align}</math>

The right hand side is the energy, and Noether's theorem states that <math>dj/dt = 0</math> (i.e. the principle of conservation of energy is a consequence of invariance under time translations).

More generally, if the Lagrangian does not depend explicitly on time, the quantity

:<math>\sum_{i=1}^3 \frac{\partial L}{\partial \dot{x}_i}\dot{x_i} - L</math>

(called the [[Hamiltonian mechanics|Hamiltonian]]) is conserved.

=== 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>

=== Example 3: Conformal transformation ===

Both examples 1 and 2 are over a 1-dimensional manifold (time). An example involving spacetime is a [[conformal transformation]] of a massless real scalar field with a [[Quartic interaction|quartic potential]] in (3&nbsp;+&nbsp;1)-[[Minkowski spacetime]].

:<math>\begin{align}
  \mathcal{S}[\varphi]
    & = \int \mathcal{L}\left[\varphi (x), \partial_\mu \varphi (x)\right] d^4 x \\[3pt]
    & = \int \left(\frac{1}{2}\partial^\mu \varphi \partial_\mu \varphi - \lambda \varphi^4\right) d^4 x
\end{align}</math>

For ''Q'', consider the generator of a spacetime rescaling. In other words,

:<math>Q[\varphi(x)] = x^\mu\partial_\mu \varphi(x) + \varphi(x). </math>

The second term on the right hand side is due to the "conformal weight" of <math>\varphi</math>. And

:<math>Q[\mathcal{L}] = \partial^\mu\varphi\left(\partial_\mu\varphi + x^\nu\partial_\mu\partial_\nu\varphi + \partial_\mu\varphi\right) - 4\lambda\varphi^3\left(x^\mu\partial_\mu\varphi + \varphi\right).</math>

This has the form of

:<math>\partial_\mu\left[\frac{1}{2}x^\mu\partial^\nu\varphi\partial_\nu\varphi - \lambda x^\mu \varphi^4 \right] = \partial_\mu\left(x^\mu\mathcal{L}\right)</math>

(where we have performed a change of dummy indices) so set

:<math>f^\mu = x^\mu\mathcal{L}.</math>

Then

:<math>\begin{align}
  j^\mu & = \left[\frac{\partial}{\partial(\partial_\mu\varphi)}\mathcal{L}\right]Q[\varphi]-f^\mu \\
        & = \partial^\mu\varphi\left(x^\nu\partial_\nu\varphi + \varphi\right) - x^\mu\left(\frac 1 2 \partial^\nu\varphi\partial_\nu\varphi - \lambda\varphi^4\right).
\end{align}</math>

Noether's theorem states that <math>\partial_\mu j^\mu = 0</math> (as one may explicitly check by substituting the Euler–Lagrange equations into the left hand side).

If one tries to find the [[Ward–Takahashi identity|Ward–Takahashi]] analog of this equation, one runs into a problem because of [[anomaly (physics)|anomalies]].