# 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.

# Classical (Lagrangian) Mechanics

It turns out that because I work with Calabi-Yau varieties I often encounter various ideas and terms from physics. In particular, quantum field theory is a something that comes up a lot. I took a lot of physics as an undergrad, and I’ve pieced together a tiny bit about what is meant by “quantum field theory.” In order to record this somewhere before I forget it, I’m going to blog some stuff. This should be a very short series, because I don’t want to get hung up on it.

The main point is to try to express the idea of quantum field theory in a way a mathematician would understand it. Before we can do that I need to spend a post on classical mechanics. This post is going to present what is done over the course of a semester long undergrad class, so it will go fast. I’ll give you the take away up front. In a mathematically rigorous way one can prove that the “Lagrangian formalism” we’ll look at soon is exactly equivalent to Newton’s law ${F=ma}$.

Suppose we have some particle in space. If it is moving, that motion has something called kinetic energy. For simplicity, we’ll call this a function of time ${K(t)=\frac{1}{2}mv(t)^2}$. The formula isn’t important here. Usually you also have something called potential energy. For example, a ball is on a table. There is the potential energy of falling to the ground. Technically you can figure out the potential the same way you’d find the potential of any vector field (this took me awhile to connect as an undergrad).

Suppose your particle is moving in ${\mathbb{R}^n}$, then we can describe it by a function ${q: \mathbb{R}\rightarrow\mathbb{R}^n}$. There could be some ambient force (gravity, electromagnetic field, etc doesn’t matter). This is a vector field ${F: \mathbb{R}^n\rightarrow \mathbb{R}^n}$. The potential then is just a function ${V: \mathbb{R}^n\rightarrow \mathbb{R}}$ such that ${\nabla F=V}$. This is what we tell our calculus students, so it shouldn’t be surprising. Of course, we must assume our force is conservative for a potential to exist, so we do that.

Now we define ${L=K(t)-V(q(t))}$. This is called the Lagrangian. We define the action over some path ${q:[t_0, t_1]\rightarrow \mathbb{R}^n}$ to be the integral ${S(q)=\int_{t_0}^{t_1}L(t)dt}$. Now we get to the point. If we let our paths vary, then we get a bunch of real numbers by evaluating ${S(q)}$. From standard calculus we could find the minimum. This is the path of least action, and our particle will follow that path if and only if in our system Newton’s law ${F=ma}$ holds.

We could go off and try to describe physically why one would think of this weird formalism. For example, integrating force over distance is the work needed to move the particle from point ${a}$ to point ${b}$. We would expect that the particle will naturally follow the path that requires the least work. This has roughly the same flavor, but takes into account some extra stuff. Whatever the physical reason, it shouldn’t really matter to us, because it is exactly equivalent to the law we all know ought to be true.

In a classical mechanics class you’d probably take many weeks now just being handed various scenarios where you figure out the Lagrangian, and then given some inital starting point figure out the path by taking the derivative ${\delta S(q)}$ and setting it to ${0}$ and solving. Note: for practical purposes this is a little tricky, because the so-called variation of the action involves differentiating with respect to paths. Since we aren’t computing these, we won’t go through this, but the idea of how to do this is to just parametrize your paths in some nice way (think about a smooth ${1}$-parameter homotopy connecting them for the picture).

Now we must generalize a bit. Suppose we have some physical system (maybe a double pendulum for sufficient complicatedness). There’s more than just one particle, and there are constraints for how things can move in relation to eachother. What we do now is consider the space of all configurations. Think of this as the moduli space of all positions the system could ever take. A point in this space ${Q}$ is one particular configuration. Now a path ${q:[t_0, t_1]\rightarrow Q}$ is just a description of how the system evolves over that time period. This configuration space we assume is a smooth manifold.

This means the velocity, which is the time derivative ${\dot{q}(t)\in T_{q(t)}Q}$ is actually a tangent vector now (it was before, but we just made the canonical identification). Let’s pick a starting point and ending point ${a, b\in Q}$. Then we can formalize what we did last time as follows. Define ${\Gamma=\{q:[t_0, t_1]\rightarrow Q : q(t_0)=a, \ q(t_1)=b\}}$ to be the path space (of smooth paths) with those endpoints. Let ${L: TQ\rightarrow \mathbb{R}}$ be a smooth function called the Lagrangian of the system.

Now the action is ${S:\Gamma\rightarrow \mathbb{R}}$ defined by ${S(q)=\int_{t_0}^{t_1}L(q, \dot{q})dt}$. The path that our system will take in the configuration space will be a minimum of ${S}$. Thus to find it we just solve ${\delta S=0}$.

In order to test whether you follow this, a really quick (if you get it, but painful if you don’t) and wonderful exercise is to figure out the equation of motion of a single free particle in ${\mathbb{R}^3}$. What does this mean? Well, there is no force in the system at all, so the potential is ${0}$, and hence ${L=K(t)=\frac{1}{2}mv(t)\cdot v(t)}$. We already know the answer. No force means no acceleration. Thus from basic calculus the answer is that velocity is a constant ${v_0}$ and the path is ${q(t)=a+v_0t}$ where ${a}$ is the initial starting point. Try to get that using the Lagrangian!