Nöther's theorem is a fundamental result is physics that relates conserved quantities and symmetries. It is given in one several statements, depending on the field of physics one is working in. Most commonly, it is given in the Lagrangian and Hamiltonian formalisms of mechanics.
The benefit Nöther's theorem is very practical: constants of motion make solving mechanical systems easier, but they are really difficult to find. Symmetries, on the other hand, are usually readily apparent. Thus, the theorem gives us a much easier way to identify constants of motion, which in turn can greatly simplify the differential equations that we must solve to find the motion. At heart, the value of Nöther's theorem is that it makes really hard math a little less hard.
> be a continuous transformation of $\dot{q}$. If, for every choice of $q,\dot{q},\alpha$, we see that $L$ is [[Transformation invariance|invariant under the transformation]]: > $$L(\varphi(q,\alpha),\psi(q,\dot{q},\alpha),t)=L(q,\dot{q},t)> determined from the [[conjugate momenta]] $P_{m}$ is a [[constant of motion]]. > (More correctly, $\varphi$ and $\psi$ are single-parameter families of [[diffeomorphism|diffeomorphisms]].) > [!quote]- Proof > Let $L$ be a Lagrangian that's invariant under the transformations $\varphi$ and $\psi$, so > $$L(\varphi(q,\alpha),\psi(q,\dot{q},\alpha),t)=L(q,\dot{q},t)then the dynamical variable
> Now that we have a constraint, we can develop the derivative > $$\begin{align} > \frac{d}{d\alpha}L(\varphi(q,\alpha),\psi(q,\dot{q},\alpha),t)&=\left.{\sum_{k=1}^{n} \left[ \frac{ \partial L }{ \partial q_{k} }(\varphi,\psi,t)\frac{ \partial \varphi_{k} }{ \partial \alpha } +\frac{ \partial L }{ \partial \dot{q}_{k} } (\varphi,\psi,t)\frac{ \partial \psi_{k} }{ \partial \alpha } \right]}\right|_{\alpha=0} \\ > &=\sum_{k=1}^{n} \left[ \frac{ \partial L }{ \partial q_{k} } (q,\dot{q},t)\frac{ \partial \varphi_{k} }{ \partial \alpha } (q,0)+\frac{ \partial L }{ \partial \dot{q}_{k} } (q,\dot{q},t)\frac{ \partial \psi_{k} }{ \partial \alpha } (q,\dot{q},0) \right] > \end{align}Due to invariance, the derivative over must be zero:
> To further compress the equation, we notice that the derivatives of $L$ collectively make a [[total derivative]] of $L$ in time. The same thing happens on $\varphi$ once we notice that > $$\frac{ \partial \psi_{k} }{ \partial \alpha } (q(t),\dot{q}(t),t)=\sum_{l=1}^{n} \frac{ \partial }{ \partial q_{l} } \left( \frac{ \partial \varphi }{ \partial \alpha } \dot{q}_{l} \right)=\frac{d}{dt} \frac{ \partial \varphi }{ \partial \alpha } (q(t),0)We evaluate the functions of and on solutions of the Lagrange equation. We get
0&=\sum_{k=1}^{n} \left[ \frac{ \partial L }{ \partial q_{k} } (q(t),\dot{q}(t),t)\frac{ \partial \varphi_{k} }{ \partial \alpha } (q(t),0)+\frac{ \partial L }{ \partial \dot{q}_{k} } (q(t),\dot{q}(t),t)\frac{ \partial \psi_{k} }{ \partial \alpha } (q(t),\dot{q}(t),t) \right] \end{align}
> (Remember that this is true specifically for $q(t)$ that solve the system). But $\frac{ \partial L }{ \partial \dot{q}_{k} }=P_{l}$ are just the conjugate momenta. We call the entire term in brackets $\mathcal{P}$ and, since its time derivative is zero, it is a constant of motion. > $$\boxed{\frac{d}{dt} \mathcal{P}=0\quad\Rightarrow \quad \mathcal{P}\text{ is a constant of motion}} > for some infinitesimal $\varepsilon\ll 1$. $\delta H=0$ if the Poisson brackets are zero, which means that $H$ is invariant with respect to the transformation. But if the Poisson brackets with $H$ are zero, then $G$ is a constant of motion. Thus $H$ is invariant if and only if $G$ is a constant of motion. ### Conservations Nöther's theorem is truly general and can be applied to any continuous transformation, so long there's a Lagrangian that's invariant under it. However, some classes of transformations are very special: they are known to conserve very important physical quantities and are therefore of particular interest. The big ones are the following three: - Invariance under spatial shifts is conservation of [[linear momentum]]. - Invariance under [[Rotation|rotations]] is conservation of [[angular momentum]]. - Invariance under temporal shifts is conservation of [[energy]]. The first two are axis-dependent: linear momentum is conserved over the axis of the shift, and angular momentum is conserved over the axis of rotation. Time is one dimensional and only has one axis: if a system is time-invariant, it conserves energy. The critical point here is, I suppose, the reverse. We know that the Universe as a whole abides by conservation of energy: clearly, then, it must be time-invariant. ### Examples > [!example]- Harmonic oscillator > We know the [[harmonic oscillator]] has a constant of motion, so let's test the theorem out on it. The Lagrangian is > $$L(q_{1},q_{2},\dot{q}_{1},\dot{q}_{2})=\frac{m}{2}(\dot{q}_{1}^{2}+\dot{q}_{2}^{2})- \frac{m\omega ^{2}}{2}(q_{1}^{2}+q_{2}^{2})We can now state
> $$\begin{pmatrix} > \dot{q}_{1} \\ > \dot{q}_{2} > \end{pmatrix}\mapsto \psi(q,\dot{q},\alpha)=\begin{pmatrix} > \cos \alpha & -\sin \alpha \\ > \sin \alpha & \cos \alpha > \end{pmatrix}\begin{pmatrix} > \dot{q}_{1} \\ > \dot{q}_{2} > \end{pmatrix}=\begin{pmatrix} > \dot{q}_{1}\cos \alpha-\dot{q}_{2}\sin \alpha \\ > \dot{q}_{1}\sin \alpha+\dot{q}_{2}\cos \alpha > \end{pmatrix}The transformations we choose are the rotations in the plane of motion:
q_{1} \\ q_{2} \end{pmatrix}\mapsto \varphi(q,\alpha)=\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}\begin{pmatrix} q_{1} \\ q_{2} \end{pmatrix}=\begin{pmatrix} q_{1}\cos \alpha-q_{2}\sin \alpha \\ q_{1}\sin \alpha+q_{2}\cos \alpha \end{pmatrix}
We then apply these to the Lagrangian:
L(\varphi(q,\alpha),\psi(q,\dot{q},\alpha))&=\frac{m}{2}[(\dot{q}_{1}\cos \alpha-\dot{q}_{2}\sin \alpha)^{2}+(\dot{q}_{1}\sin \alpha+\dot{q}_{2}\cos \alpha)^{2}] \\
&- \frac{m\omega ^{2}}{2}[(q_{1}\cos \alpha-q_{2}\sin \alpha)^{2}+(q_{1}\sin \alpha+q_{2}\cos \alpha)^{2}] \
> Clearly, then, the Lagrangian is invariant under $(q,\dot{q},t)\mapsto(\varphi,\psi,t)$. This grants us permission to invoke Nöther's theorem: > $$\mathcal{P}=\left.{(-q_{1}\sin \alpha-q_{2}\cos \alpha,q_{1}\cos \alpha-q_{2}\sin \alpha)}\right|_{\alpha=0}\begin{pmatrix} > m \dot{q}_{1} \\ > m \dot{q}_{2} > \end{pmatrix}=m(q_{1}\dot{q}_{2}-q_{2}\dot{q}_{1})&=\frac{m}{2}[\dot{q}{1}^{2}+\dot{q}{2}^{2}]- \frac{m\omega ^{2}}{2}[q_{1}^{2}+q_{2}^{2}] \ &=L(q,\dot{q}) \end{align}
> $$\dot{q}_{1}(t)=A_{1}\omega \cos(\omega t+\phi_{1}),\quad \dot{q}_{2}(t)=A_{2}\omega \cos(\omega t+\phi_{2})This must be a constant of motion. Since the system is easy, we might as well double check that this is actually a constant. To do so, we need to evaluate it over the motion, which in turn requires us to know , , and . These are
> Using the trig identity $\sin \alpha \cos \beta-\cos \alpha \sin \beta=\sin(\alpha-\beta)$ we notice that $\omega t$ cancels out and we're left with > $$\mathcal{P}=m\omega A_{1}A_{2}\sin(\phi_{1}-\phi_{2})Plug these into and we see
But this is all time-independent. Clearly then, it is a constant of motion, which confirms that Nöther's theorem was correct all along.
While it's nice to have an abstract constant, it would be nice if we could understand what we're looking at. To do so, notice that is just the third component of a Vector product: