## Noether's Theorem, As I Understand It

A simplified, but not yet simple, statement of Noether’s theorem is that for every global symmetry there is an associated conserved quantity. It is this property that I will aim to demonstrate by the use of the Lagrangian formalism of mechanics.

## Lagrangian Mechanics

Starting with Lagrangian mechanics and the definition of the Lagrangian,

$\mathcal{L} = T - V$where $T$ is the kinetic energy of the system and $V$ is the potential energy of the system. The action of the system is then defined to be

$S = \int_{t_1}^{t_2} \mathcal{L}(x, \dot{x}, t) dt$and the principle of least action then tells us that the equations of motion of a system are determined by minimising the action, by finding a point where $\delta S = 0$,

$\delta S = \int_{t_1}^{t_2} \delta \mathcal{L}(x, \dot{x}, t) dt = 0. \qquad \text{(1)}$Using the chain rule, $\delta \mathcal{L}$ can be represented by

$\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial x} \delta x + \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta \dot{x}.$Note that $\delta x$ and $\delta \dot{x}$ are small changes in *functions* rather than small changes in a variable.
Both are functions of time.
$\delta x$ can be written as

where $x(t)$ and $x'(t)$ are functions that have the *same start and end points* but take a different path, ie $\delta x(t_1) = \delta x (t_2) = 0$.
Similarly $\dot{x}$ can be written as

All of these expressions can be used in equation 1,

$\delta S = \int_{t_1}^{t_2} \left( \frac{\partial \mathcal{L}}{\partial x} \delta x + \frac{\partial \mathcal{L}}{\partial \dot{x}} \frac{d}{dt} \delta x \right) dt = 0.$At this point $\delta x$ cannot be factorised out due to having a time derivative taken of it, so the second term in the integral is reordered using integration by parts,

$\int_{t_1}^{t_2} \frac{\partial \mathcal{L}}{\partial \dot{x}} \frac{d}{dt} \delta x dt = \left[ \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right]_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) \delta x dt.$But recall the definition of $x(t)$ and $x'(t)$, they have the *same start and end points*. The first term therefore is zero.
Substituting this back in gives

where we have factorised out the $\delta x$. Since $\delta x$ can be any arbitrary function, the only way for this to be zero in general is for the section in brackets to be zero

$\frac{\partial \mathcal{L}}{\partial x} - \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) = 0.$This is known as the Euler-Lagrange equation.

## Newton’s Second Law

I will now take a moment to use the Euler-Lagrange equation and obtain the Newtonian equation of motion. Firstly using the definition of the Lagrangian,

$\mathcal{L} = T - V = \frac{1}{2} m \dot{x}^2 - V$using the expression for kinetic energy of a single particle at non relativistic speeds. Using this in the Euler-Lagrange equation gives

$\frac{\partial \left(-V\right)}{\partial x} = \frac{d}{dt}\left(\frac{\partial \left(\frac{1}{2}m \dot{x}^2 \right)}{d \dot{x}}\right).$Resolving the left hand side requires knowing that $\frac{\partial V}{\partial x} = -F$, and the right hand side is following through the differentiation,

$F = m \ddot{x} = ma.$The well known result of Newton’s Second Law.

## Noether’s Theorem

Now on to the main attraction, a demonstration of Noether’s theorem. Using the definition given at the start, “for every global symmetry there is an associated conserved quantity”, we will first look at what a symmetry is. A symmetry is a transformation of the path, $x(t)$, that does not change the action, $\delta S = 0$. It is of the general form

$\delta x = x(x) - x'(x) = g(x, \dot{x}).$Unlike previously, this $\delta x$ does not need to be zero at the start and end points. $\delta \dot{x}$ is the time derivative of $\delta x$,

$\delta \dot{x} = \frac{d}{dt} \delta x.$Firstly we consider the variation of the action of the system to be zero,

$\delta S = \int_{t_1}^{t_2} \delta \mathcal{L} dt = \int_{t_1}^{t_2} \left( \frac{\partial \mathcal{L}}{\partial x} \delta x + \frac{\partial \mathcal{L}}{\partial \dot{x}} \frac{d}{dt} \delta x \right) dt = 0.$Then, like before, we rewrite the second term in the integral using integration by parts,

$\int_{t_1}^{t_2} \frac{\partial \mathcal{L}}{\partial \dot{x}} \frac{d}{dt} \delta x dt = \left[ \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right]_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) \delta x dt.$Which is then substituted back in giving

$\delta S = \int_{t_1}^{t_2} \left( \frac{\partial \mathcal{L}}{\partial x} - \frac{d}{dt} \left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) \right) \delta x dt + \left[ \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right]_{t_1}^{t_2} = 0.$Since the system follows a path described by the Euler-Lagrange equation, the first term must be zero as previously shown,

$\delta S = \left[ \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right]_{t_1}^{t_2} = \left[ \frac{\partial \mathcal{L}}{\partial \dot{x}} g(x, \dot{x}) \right]_{t_1}^{t_2} = 0. \qquad \text{(2)}$We can now define a value $Q$ to be

$Q = \frac{\partial \mathcal{L}}{\partial \dot{x}} g(x, \dot{x}).$Since, by equation 2, $Q(t_1) - Q(t_2) = 0$ with $t_1$ and $t_2$ being arbitrary, $Q$ must be independent of time. Therefore for every symmetry there must be a conserved quantity.