# 14.5: Invariants

- Page ID
- 19030

## 14.5.1: Importance of Invariants

Noether's theorem describes the relationship between continuous symmetries of an equation describing an energy conversion process and invariants of the system. The theorem was originally discovered by Noether around 1918 [165] [166]. The importance of this theorem is described in the introduction to the English translation of the original paper [165]. "The well known theorem of Emmy Noether plays a role of fundamental importance in many branches of theoretical physics. Because it provides a straightforward connection between the conservation laws of a physical theory and the invariances of the variational integral whose Euler-Lagrange equations are the equations of that theory, it may be said that Noether's theorem has placed the Lagrangian formulation in a position of primacy."

## 14.5.2: Noether's Theorem

Consider an energy conversion process with a known Lagrangian that satisfies an Euler-Lagrange equation. Assume that we have identified continuous symmetries described by infinitesimal generators. Noether's theorem says that there is a relationship between these continuous symmetries and conservation laws which say that some quantity is invariant. We would like to find the corresponding conservation laws and invariants. If we can find a quantity \(G\) that satisfies,

\[\frac{dG}{dt} = pr^{(\mathfrak{n})}U\mathcal{L} + \mathcal{L}\frac{d\xi}{dt}, \label{14.5.1}\]

then the quantity

\[\Upsilon=\eta \frac{d \mathcal{L}}{d \dot{y}}+\xi \mathcal{L}-\xi \dot{y} \frac{\partial \mathcal{L}}{\partial \dot{y}}-G \label{14.5.2}\]

is an invariant. For Lagrangians with units of joules, the quantity \(G\) also has units joules. In Equation \ref{14.5.1}, \(pr^{(\mathfrak{n})}U\mathcal{L}\) is the prolongation of the infinitesimal generator acting on the Lagrangian where prolongation was defined in Equation 14.4.3.

## 14.5.3: Derivation of Noether's Theorem

We can derive this form of Noether's theorem, and this derivation closely follows the clear and simplified derivation in reference [192]. This theorem is detailed and derived more rigorously in multiple other references [163, p. 208] [164]. For the purpose of this derivation, assume that we begin with an equation of motion that is at most a second order differential equation. However, the ideas generalize to higher order equations too. Also, assume we know the corresponding Lagrangian of the form \(\mathcal{L} = \mathcal{L}(t, y, \dot{y})\). The general approach is to assume that we can find a value of \(G\) defined by Equation \ref{14.5.1}. We will perform some algebra on Equation \ref{14.5.1} to show that choice of \(G\) necessarily implies that \(\Upsilon\) is invariant.

Use the definition of the prolongation to write Equation \ref{14.5.1} in terms of \(\xi\) and \(\eta\).

\[pr^{(\mathfrak{n})}U = \xi \partial_t + \eta \partial_y + \eta^t \partial_{\dot{y}}\]

For a second order differential equation, no more terms are needed because the Lagrangian depends on, at most, the first derivative \(\dot{y}\). Substitute the prolongation acting on the Lagrangian into Eq. \ref{14.5.1}.

\[\frac{dG}{dt} = [ \xi \partial_t \mathcal{L} + \eta \partial_y \mathcal{L} + \eta^t \partial_{\dot{y}} \mathcal{L}] + \frac{d\xi}{dt} \mathcal{L} \label{14.5.4}\]

Consider the continuous transformation described by

\[t \rightarrow \tilde{t} = (1 + \varepsilon \xi + \varepsilon^2 \ldots) t\]

and

\[y \rightarrow \tilde{y} = (1 + \varepsilon \eta + \varepsilon^2 \ldots) y\]

in the limit \(\varepsilon \rightarrow 0\). The Lagrangian \(\mathcal{L} (t, y, \dot{y}) \) of an energy conversion process represents the difference between two forms of energy. The Lagrangian \(\mathcal{L} (\tilde{t}, \tilde{y}, \dot{\tilde{y}}) \) represents the difference between two forms of energy upon the continuous symmetry transformation described by infinitesimal generator \(U\). Qualitatively, the quantity \(\frac{dG}{dt}\) represents the change in \(\mathcal{L} \frac{d\tilde{t}}{dt}\) with respect to \(\varepsilon\) in this limit [192].

\[\frac{dG}{dt} = \frac{\partial}{\partial \varepsilon} \left[\mathcal{L} (\tilde{t}, \tilde{y}, \dot{\tilde{y}} \frac{d\tilde{t}}{dt}\right]\]

Use Eq. 14.4.4 to substitute for \(\eta^t\) in Eq. \ref{14.5.4}.

\[\eta^{t}=\frac{d}{d t}(\eta-\xi \dot{y})+\xi \ddot{y}\]

\[\eta^{t}=\dot{\eta}-\dot{\xi} \dot{y}-\xi \ddot{y}+\ddot{y}=\dot{\eta}-\dot{y} \dot{\xi}\]

\[\frac{d G}{d t}=\left[\xi \partial_{t} \mathcal{L}+\eta \partial_{y} \mathcal{L}+(\dot{\eta}-\dot{y} \dot{\xi}) \partial_{\dot{y}} \mathcal{L}\right]+\dot{\xi} \mathcal{L}\]

We want to express the right side as the total derivative of some quantity, which we call \(G\). With some algebra, we can write this as a total derivative. We will use the definition of the total derivative.

\[\frac{d \mathcal{L}}{d t}=\partial_{t} \mathcal{L}+\dot{y} \partial_{y} \mathcal{L}+\ddot{y} \partial_{\dot{y}} \mathcal{L}\]

\[\partial_{t} \mathcal{L}=\dot{\mathcal{L}}-\dot{y} \partial_{y} \mathcal{L}-\ddot{y} \partial_{\dot{y}} \mathcal{L}\]

\[\frac{d}{d t}\left(\eta \partial_{\dot{y}} \mathcal{L}\right)=\dot{\eta} \partial_{\dot{y}} \mathcal{L}+\eta \frac{d}{d t}\left(\partial_{\dot{y}} \mathcal{L}\right)\]

\[\dot{\eta} \partial_{\dot{y}} \mathcal{L}=\frac{d}{d t}\left(\eta \partial_{\dot{y}} \mathcal{L}\right)-\eta \frac{d}{d t}\left(\partial_{\dot{y}} \mathcal{L}\right)\]

\[\frac{d}{d t}\left(\xi \dot{y} \partial_{\dot{y}} \mathcal{L}\right)=\dot{\xi} \dot{y} \partial_{\dot{y}} \mathcal{L}+\xi \ddot{y} \partial_{\dot{y}} \mathcal{L}+\xi \dot{y} \frac{d}{d t} \partial_{\dot{y}} \mathcal{L}\]

\[\dot{\xi} \dot{y} \partial_{\dot{y}} \mathcal{L}=\frac{d}{d t}\left(\xi \dot{y} \partial_{\dot{y}} \mathcal{L}\right)-\xi \ddot{y} \partial_{\dot{y}} \mathcal{L}-\xi \dot{y} \frac{d}{d t} \partial_{\dot{y}} \mathcal{L}\]

Use these pieces to replace the terms of \(\frac{dG}{dt}\) in brackets.

\[\frac{d G}{d t}= [\xi \partial_{t} \mathcal{L}] +\eta \partial_{y} \mathcal{L}+ [\dot{\eta} \partial_{\dot{y}} \mathcal{L}] - [\dot{y}\dot{\xi}\partial_{\dot{y}} \mathcal{L}]+ \dot{\xi} \mathcal{L}\]

\[\frac{d G}{d t}= \xi\left[\dot{\mathcal{L}}-\dot{y} \partial_{y} \mathcal{L}-\ddot{y} \partial_{\dot{y}} \mathcal{L}\right]+\eta \partial_{y} \mathcal{L}+\left[\frac{d}{d t}\left(\eta \partial_{\dot{y}} \mathcal{L}\right)-\eta \frac{d}{d t}\left(\partial_{\dot{y}} \mathcal{L}\right)\right] -\left[\dot{y} \dot{\xi} \partial_{\dot{y}} \mathcal{L}\right]+\dot{\xi} \mathcal{L}\]

\[\frac{d G}{d t}= \xi\left[\dot{\mathcal{L}}-\dot{y} \partial_{y} \mathcal{L}-\ddot{y} \partial_{\dot{y}} \mathcal{L}\right]+\eta \partial_{y} \mathcal{L}+\left[\frac{d}{d t}\left(\eta \partial_{\dot{y}} \mathcal{L}\right)-\eta \frac{d}{d t}\left(\partial_{\dot{y}} \mathcal{L}\right)\right] - \left[\frac{d}{dt} (\dot{\xi} \dot{y} \partial_{\dot{y}} \mathcal{L}) - \xi \ddot{y} \partial_{\dot{y}} \mathcal{L} - \xi \dot{y} \frac{d}{dt} \partial_{\dot{y}} \mathcal{L}\right] +\dot{\xi} \mathcal{L}\]

Two terms cancel.

\[\frac{d G}{d t} = \xi\dot{\mathcal{L}} - \xi\dot{y} \partial_{y} \mathcal{L} +\eta \partial_{y} \mathcal{L}+ \frac{d}{d t}\left(\eta \partial_{\dot{y}} \mathcal{L}\right)-\eta \frac{d}{d t}\left(\partial_{\dot{y}} \mathcal{L}\right) - \frac{d}{dt} (\dot{\xi} \dot{y} \partial_{\dot{y}} \mathcal{L}) + \xi \dot{y} \frac{d}{dt} \partial_{\dot{y}} \mathcal{L} +\dot{\xi} \mathcal{L}\]

Regroup terms.

\[\frac{d G}{d t}=\left(\partial_{y} \mathcal{L}-\frac{d}{d t} \partial_{\dot{y}} \mathcal{L}\right)(\eta-\dot{y} \xi) +\left[(\xi \dot{\mathcal{L}}+\mathcal{L} \dot{\xi})+\frac{d}{d t}\left(\eta \partial_{\dot{y}} \mathcal{L}\right)-\frac{d}{d t}\left(\xi \dot{y} \partial_{\dot{y}} \mathcal{L}\right)\right]\]

The first term in parentheses is zero because the Lagrangian \(\mathcal{L}\) satisfies the Euler-Lagrange equation.

\[\frac{dG}{dt} = \frac{d}{dt} (\xi \mathcal{L} + (\eta \partial_{\dot{y}} \mathcal{L}) − (\xi \dot{y} \partial_{\dot{y}} \mathcal{L}))\]

\[\frac{d}{dt}[\xi \mathcal{L} + (\eta \partial_{\dot{y}} \mathcal{L}) − (\xi \dot{y} \partial_{\dot{y}} \mathcal{L}) − G] = 0\]

Therefore, if we can find \(G\), then the quantity in brackets \(\Upsilon\) must be invariant.

\[\Upsilon = \xi \mathcal{L} + (\eta \partial_{\dot{y}} \mathcal{L}) − (\xi \dot{y} \partial_{\dot{y}} \mathcal{L}) − G = \text{invariant}\]

## 14.5.4: Line Equation Invariants Example

Let us apply Noether's theorem to some examples. First, consider the line equation \(\ddot{y} = 0\) which results from application of calculus of variations with Lagrangian

\[\mathcal{L} = \frac{1}{2}\dot{y}^2.\]

A continuous symmetry of this equation is described by the infinitesimal generator \(U = \partial_y\) with \(\xi = 0\) and \(\eta = 1\). The prolongation of the generator acting on the Lagrangian is zero.

\[pr^{(\mathfrak{n})}U\mathcal{L} = \eta^t\dot{y} = \dot{y} \left(\frac{d\eta}{dt}\right) = 0\]

Using Eq. \ref{14.5.1}, we see that \(G = 0\).

\[\frac{dG}{dt} = 0 + \mathcal{L} \cdot 0 = 0\]

Next use Eq. \ref{14.5.2} to find the invariant.

\[\Upsilon = \eta\frac{\partial \mathcal{L}}{\partial \dot{y}} = \dot{y}\]

Qualitatively, \(\dot{y}\) represents the slope of the line, so this invariant tells us that the slope of the solutions to the line equation must be constant.

Another continuous symmetry of this equation is described by the infinitesimal generator \(U = t\partial_y\) with \(\xi = 0\) and \(\eta = t\). We can solve for the prolongation of the generator acting on the Lagrangian.

\[pr^{(\mathfrak{n})}U\mathcal{L} = \eta^t\dot{y} = \dot{y} \left(\frac{d}{dt}(t − 0) + 0 \right) = \dot{y}\nonumber\]

We can find \(G\) using Eq. \ref{14.5.1}, and we can find the invariant using Eq. \ref{14.5.2}.

\[\frac{dG}{dt} = \dot{y} + \frac{1}{2}\dot{y}^2 \cdot 0 = \dot{y}\]

\[G = y\]

\[\Upsilon = y - t\dot{y}\]

Qualitatively, this invariant represents the y-intercept of the line, so this invariant tells us that the y-intercept of the solution to the line equation must be constant.

## 14.5.5: Pendulum Equation Invariants Example

Consider the equation describing a pendulum, studied in Problem 11.8. The energy conversion process is described by the Lagrangian

\[\mathcal{L} = \frac{1}{2}m\dot{y}^2 - mg \cos y \]

which corresponds to the equation of motion

\[\ddot{y} = g \sin y.\]

In these equations \(m\) represents the mass, and \(g\) represents the gravitational constants. Both \(m\) and \(g\) are assumed constant here. This equation of motion has only one continuous symmetry described by the infinitesimal generator (U = \partial_t\) with \(\xi = 1\) and \(\eta = 0\). We can use Noether's theorem to find the corresponding invariant.

Use Eq. \ref{14.5.1} to find \(G\).

\[\frac{dG}{dt} = pr^{(\mathfrak{n})}U\mathcal{L} + \mathcal{L} \frac{d\xi}{dt}\]

\[\frac{dG}{dt} = \eta^tm\dot{y} + \eta mg \sin y + \mathcal{L} \frac{d\xi}{dt}\]

\[\frac{dG}{dt} = \eta^tm\dot{y} = \dot{y}m \left(\frac{d}{dt} (\eta - \xi\dot{y}) + \xi\ddot{y} \right)\]

\[\frac{dG}{dt} = \dot{y}m \left(-\frac{d\xi}{dt}\dot{y} - \xi\ddot{y} + \xi\ddot{y} \right) = 0\]

\[G = 0\]

Use Eq. \ref{14.5.2} to find the invariant.

\[\Upsilon = \eta\dot{y} + \xi \mathcal{L} - \xi m\dot{y}\dot{y} - 0 \]

\[\Upsilon = \frac{1}{2} m\dot{y}^2 - mg \cos y - m\dot{y}^2\]

\[\Upsilon = \frac{-1}{2} m\dot{y}^2 - gm \cos y \]

The quantity \(\Upsilon\) is conserved, and it is the Hamiltonian which represents total energy.

Whenever the Lagrangian does not explicitly depend on \(t\), the system contains the continuous symmetry described by the infinitesimal generator \(U = \partial_t\). This infinitesimal generator has \(\xi = 1\) and \(\eta = 0\). From Eq. \ref{14.5.1}, \(G\) must be zero. From Eq. \ref{14.5.2}, the corresponding invariant has the form

\[\Upsilon = \mathcal{L} - m\dot{y}\dot{y}\]

which has the magnitude of the total energy (assuming t is time). This equation is equal to the Hamiltonian of Eq. 11.3.16. Therefore, if an equation of motion contains the symmetry \(\partial_t\), energy is conserved.