# Noether’s Theorem

I want to do one more thing in classical mechanics before moving on to classical field theory. It is called Noether’s theorem, and it tells us how to find conserved quantities of our system if we know that a certain group of symmetries acts on the system.

Recall our setup. We have some configuration space ${Q}$. We think of this as the smooth manifold of all possible positions our system can take. A point on ${Q}$ corresponds to some configuration. In classical mechanics we also have a Lagrangian ${L:TQ\rightarrow \mathbb{R}}$. Minimizing the integral of the Lagrangian over all paths in the configuration space tells us (given some initial starting configuration) what path our system will take and hence how it changes over time.

Also, we called ${\Gamma}$ the space of paths on ${Q}$. We now define a (one-parameter) symmetry of ${L}$ to be a smooth map ${F:\mathbb{R}\times \Gamma\rightarrow \Gamma}$, usually denoted using “group action” notation as ${s\cdot q=q_s}$ with some special properties. First, ${q_0=q}$ (i.e. the identity element acts as the identity). Second, there is some ${\ell: TQ\rightarrow \mathbb{R}}$ such that ${\displaystyle \delta L=\frac{d\ell}{dt}}$ (for all paths).

Noether’s theorem tells us that given such a symmetry we will get that the quantity ${p^i\delta q_i-\ell}$ is conserved. Conserved in this case means that given any admissible path ${q}$, the time derivative of the quantity along ${q}$ is ${0}$. Or unravelling what that means, as the system evolves in time, the quantity is constant.

If we get away from the symbols for a little bit, then we’ll find that we probably already would have guessed this intuitively. If the symmetry of our Lagrangian is shifting the time ${(s\cdot q)(t)=q(s+t)}$, then this says that our system has the same physics at all points of time. This occurs in our standard example of ${L=\frac{1}{2}m\dot{q}^2-V}$ on ${\mathbb{R}^n}$. Since ${\ell=L}$, Noether’s theorem tells us that the conserved quantity is ${m\dot{q}^2-(\frac{1}{2}m\dot{q}^2-V)=\frac{1}{2}mv^2+V}$. Thus the potential energy plus the kinetic energy is conserved. This is just the total energy! We find that whenever our Lagrangian is invariant under shifting time, we recover the Law of Conservation of Energy.

Another type of symmetry is to consider our free particle in ${\mathbb{R}^n}$. For any vector ${v\in\mathbb{R}^n}$ we can shift spatially along ${v}$. Thus ${q_s(t)=q(t)+sv}$. Certainly our Lagrangian is invariant under any of these shifts. Our conserved quantity in this case is merely ${p_i\delta q^i=m\dot{q}_iv_i=m\dot{q}\cdot v}$ which is just the momentum in the direction ${v}$. Noether’s theorem tells us that if our Lagrangian doesn’t depend on shifts in the ${v}$-direction, then momentum in that direction is conserved. Moreover, this tells us that our free particle has all momentum conserved…and of course this is true! The equation of motion is just moving in a constant direction at a constant speed.

The same thing is true for our free particle when we consider rotational symmetries. We fix some rotation ${A\in \frak{so}(n)}$, and our action is ${q_s(t)=e^{As}q(t)}$. It shouldn’t be surprising now that we have a feel for Noether’s theorem that this gives us conservation of angular momentum.

The symmetries we have above are known as physical symmetries. One could think of it as moving the frame of reference to a different place and then finding out we get all the same answers. These physical symmetries give non-zero conserved quantities and they don’t introduce ambiguities in the equation of motion given sufficient initial data.

There is another type of symmetry known as a gauge symmetry (we are allowing our action now to be ${G\times \Gamma\rightarrow \Gamma}$ for some Lie group ${G}$). When you work out the conserved quantity you will get ${0}$. This is subtler, because our symmetry shouldn’t be thought of as altering the “physical situation” of the setup, but more that it is a symmetry of the mathematics of the situation. This actually does introduce ambiguities in the path our system will take for the simple reason that given sufficient initial data find some solution path ${q}$, then all paths in the orbit of ${q}$ (i.e. paths of the form ${q_s}$ for some ${s\in G}$) are possible choices for the evolution of the system.

I’m not sure if there is a good example of a gauge symmetry for classical mechanics, but certainly there are for classical field theories which is our next topic. Most people are probably familiar with the fact that the standard model has ${U(3)\times SU(2)\times U(1)}$ gauge symmetry.