# Noether’s theorem

In connection with classical mechanics and its formulation in terms of a least action principle there exists a remarkable theorem, the so called Noether theorem. It states:

For every continuous symmetry of a physical system, there exists a conserved quantity.

The reverse of this statement is also true as we will see later.

However, let’s introduce some general notions first before we tackle the theorem in full glory. The dynamics of a classical system is captured by the action

$$S[\vec{q}(t)] = \int_{t_1}^{t_2} L(\vec{q}, \dot{\vec{q}}, t) ~ dt$$

which will take on a minimal value for the physically occupied path $\vec{q}(t)$. This can be expressed in terms of the Euler-Lagrange equations

$$\frac{\partial L}{\partial q_i} – \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) = 0$$

There is an easy way to deduce some conserved quantities from this. Suppose that the Lagrange function $L(\vec{q}, \dot{\vec{q}}, t)$ does not depend on some $q_j$. Then certainly we have $\frac{\partial L}{\partial q_j} = 0$ and hence due to the Euler-Lagrange equation

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_j} \right) = 0$$

which tells us that the canonical momentum $p_j := \frac{\partial L}{\partial \dot{q}_j}$ is a conserved quantity. By that we mean, that it stays constant along any physical path. This holds also if we reverse the statements: If $p_j$ is conserved, then $\frac{\partial L}{\partial q} = 0$ and $L$ cannot depend on $q_j$. In all these cases one calls $q_j$ a cyclic coordinate.

What does this mean? What is the physical interpretation? For simplicity, we stick to a single degree of freedom here. Suppose a system is translation invariant in the direction of $q$, that is, the action $S$ is minimal for $q(t)$ and $q(t) + \varepsilon$ for any $\varepsilon \in \mathbb{R}$. So both, $q(t)$ and $\tilde{q}(t) = q(t) + \varepsilon$ are solutions of the Euler-Lagrange equation and will induce the same (minimal) value of the action:

$$S[q(t) + \varepsilon] = \int_{t_1}^{t_2} L(q + \varepsilon, \dot{q}, t) ~ dt = \int_{t_1}^{t_2} L(q, \dot{q}, t) ~ dt = S[q(t)]$$

If we divide by $\varepsilon$ and pull everything on one side we get

$$\int_{t_1}^{t_2} \frac{L(q + \varepsilon, \dot{q}, t) ~ – ~ L(q, \dot{q}, t)}{\varepsilon} ~ dt = 0$$

The integrand is the difference quotient which in the limit of $\varepsilon \rightarrow 0$ becomes the derivative:

$$\int_{t_1}^{t_2} \frac{\partial L}{\partial q} ~ dt = \int_{t_1}^{t_2} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) = \frac{\partial L}{\partial \dot{q}}(t_2) ~ – ~ \frac{\partial L}{\partial \dot{q}}(t_1) = 0$$

Here we have used the Euler-Lagrange equation and integrated by parts. So all in all we have

$$\frac{\partial L}{\partial \dot{q}}(t_2) = \frac{\partial L}{\partial \dot{q}}(t_1)$$

Notice that the times $t_2$ and $t_1$ are completely arbitrary, hence $\frac{\partial L}{\partial \dot{q}}$ is a constant over time. Thus $\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) = 0$ and through the Euler-Lagrange equation we again get $\frac{\partial L}{\partial q} = 0$. Obviously, the reverse is also true: When we have a cyclic coordinate $\frac{\partial L}{\partial q} = 0$, the system is translational invariant, because $L$ does not depend on $q$ and hence $q + \varepsilon$ will be a solution, too.

What we have just shown is the following: The presence of a conserved canonical momentum $\frac{\partial L}{\partial \dot{q}}$ implies the presence of a translational symmetry $q \rightarrow q + \varepsilon$ and vice versa. As we will see shortly, a similar statement can be made for all continuous symmetries. But first, let us introduce some quantities that will make the derivation easier.

Suppose we have a continious transformation $T(\varepsilon)$, that transforms one set of solutions $\vec{q}(t)$ into another set of solutions $\vec{q}'(t’)$:

$$T(\varepsilon): \vec{q}(t) \rightarrow \vec{q}'(t) = \vec{q}(t) + \varepsilon \cdot \vec{\Psi}(t) + \mathcal{O}(\varepsilon^2)$$

Notice that we not only allow for a mere coordinate transformation but also for a transformation of time:

$$T(\varepsilon): t \rightarrow t’ = t + \varepsilon \cdot \varphi(t) + \mathcal{O}(\varepsilon^2)$$

The functions $\vec{\Psi}$ and $\varphi$ are the so called generators of the transformation. In fact, we demand $T(\varepsilon)$ to be differentiable. Such transformations or symmetries have a well studied and thorough mathematical description: Lie groups, named after the famous Norwegian mathematician Sophus Lie. However, we will not go much into the details here. The crucial point is that the essential information is already encoded into the generators (negelecting all higher orders). We will see this largely as a mathematical implication of our manipulations of the comming equations but in fact it has a much more deeper meaning.

Let’s quickly repeat what we have done in the case of a translational symmetry to derive the conserved quantity:

1. Equate $S[\vec{q}(t)] = S[\vec{q}'(t’)]$ because both are (the same) minima of the action.
2. Pull everything on one side, divide by $\varepsilon$ and take the limit $\varepsilon \rightarrow 0$.
3. Integrate by parts.
4. Use the Euler-Lagrange equations for $\vec{q}(t)$.
5. Move terms with $t_2$ onto the left hand side and terms with $t_1$ onto the right hand side of the equation.
6. Identify either side of the equation as a conserved quantity because $t_1$ and $t_2$ are arbitrary.

#### 1. Equate $S[\vec{q}(t)] = S[\vec{q}'(t’)]$.

\begin{align} & \int_{t_1}^{t_2} L(\vec{q}, \dot{\vec{q}}, t) ~ dt = \int_{t’_1}^{t’_2} L \left( \vec{q}’, \frac{d \vec{q}’}{d t’}, t’ \right) ~ dt’ = \int_{t’_1}^{t’_2} L \left( \vec{q}’, \frac{d \vec{q}’}{dt} \frac{dt}{d t’}, t’ \right) ~ dt’ \\ & \\ & = \int_{t_1}^{t_2} L \left( \vec{q} + \varepsilon \cdot \vec{\Psi} + \mathcal{O}(\varepsilon^2), \left( \dot{\vec{q}} + \varepsilon \cdot \dot{\vec{\Psi}} + \mathcal{O}(\varepsilon^2) \right) \cdot \frac{dt}{d t’}, t + \varepsilon \cdot \varphi + \mathcal{O}(\varepsilon^2) \right) \cdot \frac{dt’}{dt} ~ dt \end{align}

#### 2. Pull everything on one side, divide by $\varepsilon$ and take the limit $\varepsilon \rightarrow 0$.

\begin{align} 0 & = \frac{S[\vec{q}'(t’)] – S[\vec{q}(t)]}{\varepsilon} \rightarrow \frac{d}{d \varepsilon} \bigg( S[\vec{q}'(t’)] \bigg)_{\varepsilon=0} \end{align}

Thus we get

\begin{align} 0 = \int_{t_1}^{t_2} \frac{\partial}{\partial \varepsilon} \bigg( L(\dots) \bigg)_{\varepsilon = 0} ~ dt ~ + ~ \int_{t_1}^{t_2} L(\vec{q}, \dot{\vec{q}}, t) \cdot \frac{\partial}{\partial \varepsilon} \bigg( \frac{dt’}{dt} \bigg)_{\varepsilon = 0} ~ dt \end{align}

#### 3. Integrate by parts

The first integral yields

\begin{align} & \int_{t_1}^{t_2} \frac{\partial}{\partial \varepsilon} \bigg( L(\dots) \bigg)_{\varepsilon = 0} ~ dt = \int_{t_1}^{t_2} \sum_{i} \frac{\partial L}{\partial q_i} \Psi_i + \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \dot{\Psi}_i ~ – ~ \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \dot{q}_i \varphi + \frac{\partial L}{\partial t} \varphi ~ dt \\ & \\ & = \left[ \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \Psi_i \right]_{t_1}^{t_2} + \int_{t_1}^{t_2} \sum_{i} \left( \frac{\partial L}{\partial q_i} – \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) \right) \Psi_i ~ – ~ \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \dot{q}_i \varphi + \frac{\partial L}{\partial t} \varphi ~ dt \end{align}

The second integral yields

\begin{align} & \int_{t_1}^{t_2} L(\vec{q}, \dot{\vec{q}}, t) \cdot \frac{\partial}{\partial \varepsilon} \bigg( \frac{dt’}{dt} \bigg)_{\varepsilon = 0} ~ dt = \int_{t_1}^{t_2} L(\vec{q}, \dot{\vec{q}}, t) \cdot \dot{\varphi} ~ dt \\ & \\ & = \bigg[ L \cdot \varphi \bigg]_{t_1}^{t_2} ~ – ~ \int_{t_1}^{t_2} \frac{d L}{dt} \cdot \varphi ~ dt \\ & \\ & = \bigg[ L \cdot \varphi \bigg]_{t_1}^{t_2} ~ – ~ \int_{t_1}^{t_2} \sum_{i} \left( \frac{\partial L}{\partial q_i} \dot{q}_i + \frac{\partial L}{\partial \dot{q}_i} \ddot{q}_i \right) \cdot \varphi + \frac{\partial L}{\partial t} \cdot \varphi ~ dt \\ & \\ & = \left[ \left( L ~ – ~ \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \dot{q}_i \right) \varphi \right]_{t_1}^{t_2} ~ – ~ \int_{t_1}^{t_2} \sum_{i} \left( \frac{\partial L}{\partial q_i} \varphi ~ – ~ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \varphi \right) \right) \cdot \dot{q_i} + \frac{\partial L}{\partial t} \cdot \varphi ~ dt \end{align}

#### 4. Use the Euler-Lagrange equations

The first integral becomes

\begin{align} \left[ \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \Psi_i \right]_{t_1}^{t_2} ~ – ~ \int_{t_1}^{t_2} \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \dot{q}_i \varphi ~ – ~ \frac{\partial L}{\partial t} \varphi ~ dt \end{align}

And the second integral becomes

\begin{align} \left[ \left( L ~ – ~ \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \dot{q}_i \right) \varphi \right]_{t_1}^{t_2} ~ + ~ \int_{t_1}^{t_2} \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \dot{q}_i \dot{\varphi} ~ – ~ \frac{\partial L}{\partial t} \varphi ~ dt \end{align}

#### 5. + 6. Move terms and identify the conserved quanity

By combining both integrals we finally get

\begin{align} 0 = \left[ \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \Psi_i + \left( L ~ – ~ \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \dot{q}_i \right) \varphi \right]_{t_1}^{t_2} \end{align}

Hence, the conserved quantity $\frac{d Q}{dt} = 0$ is

\begin{align} Q = \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \Psi_i + \left( L ~ – ~ \sum_{i} \frac{\partial L}{\partial \dot{q}_i} \dot{q}_i \right) \varphi \end{align}

It is also sometimes called the Noether charge.

This site uses Akismet to reduce spam. Learn how your comment data is processed.