Symmetries and conservation laws

Consider a Lagrangian $L(q^I, \dot{q}^I)$ . A symmetry is a transformation

q^I \to (q^I)'(q^I)

(1)

such that

L(q^I, \dot{q}^I) = L((q^I)', (\dot{q}^I)') + \frac{d\Lambda(q,\dot q)}{dt}

(2)

where $(\dot{q}^I)' = \frac{d}{dt} (q^I)'$ . The last term in (1) is a total derivative; thus (as one can see by studying the resulting action) it does not contribute to the equations of motion.

In other words, the action is invariant up to a boundary term.

Example: cyclic coordinates¶

The simplest example is a Lagrangian which does not even depend on soem subset of the coordinates, only their velocities. Consider $L(q^I, \dot{q}^I)$ to be independent of $q^k$ for some $k$ (but still depend on $\dot{q}^k$ ). Then the transformation $q^k \to (q^k)' = q^k + \alpha$ for $\alpha$ a constant real number. Clearly $\dot{q}^k$ is invariant, so the entire Lagrangian is. We call $q^k$ a cyclic coordinate.

What does this mean for the equations of motion? Well, we know that

{\cal F}_k = \frac{\del L}{\del q^k} = 0

(3)

The Euler-Lagrange equations for $q^k$ become:

\dot{p}^k = \frac{\del L}{\del \dot{q}^k} = 0

(4)

In other words, the generalized momentum $p^k$ is conserved.

Some examples:

Consider $L = \half m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - V(x,y)$ . Then $z$ is a cyclic coordinate, and the momentum $p_z = m \dot{z}$ is conserved, as a result of invariance of the action under translations in teh $z$ direction.
A particle in polar coordinates in a central force, $L = \half m (\dot{r}^2 + r^2 \dot{\phi}^2) - V(r)$ . We studied this last time; the cyclic coordinate is $\phi$ and the conserved conjugate momentum os $p_{\phi} = m r^2 \dot{\phi}$ which is the angular momentum. Thus, invariance under rotations implies the conservation of angular momentum.

These suggest a more general story to which we now turn.

Noether’s theorem¶

Noether’s theorem applies to continuous symmetries, that is, to a continuous family of transformations $(q^I)' = (q^I)'(q^I\ ; \alpha)$ where $\alpha$ is some real parameter and $(q^I)'(q^I\ ; 0) = q^I$ . Noether showed that every such famikly of symmetries implied a conservation law.

We will provide a constuctive proof. Let $(q^I)' = q^I + \delta q^I$ . If this transformation is a symmetry, then

\begin{align} L(q^I, + \delta q^I, \dot{q}^I + \delta \dot{q}^I) & = L(q^I \dot{q}^I) + frac{d\Lambda}{dt} \\ = L(q^I, \dot{q}^I) + \delta q^I \frac{\del L}{\del q^I} + \delta \dot{q}^I \frac{\del L}{\del \dot{q}^I} + {\cal O}(\delta q^2) \end{align}

(5)

Working to first order in $\delta q$ , the fact that this is a symmetry means that

\begin{align} \delta q^I \frac{\del L}{\del q^I} + \delta \dot{q}^I \frac{\del L}{\del \dot{q}^I} & = \frac{d\Lambda}{dt}\\ & = \delta q^I \frac{\del L}{\del q^I} + \frac{d}{dt} \left(\delta q^I \frac{\del L}{\del \dot{q}^I}\right) - \delta q^I \frac{d}{dt} \frac{\del L}{\del \dot{q}^I} \end{align}

(6)

Now the first and last terms on the second line are just $\delta q^I$ times the Euler-Lagrange equations, and so vanish if $q^I$ (t)$ satisfies the classical equations of motion. When it does, we are left with

\frac{d}{dt} \left(\delta q^I \frac{\del L}{\del \dot{q}^I} - \Lambda\right) = \frac{d}{dt} \left(\delta q^I p_I - \Lambda\right) = 0

(7)

Thus we have a conserved charge, sometimes called a *Noether charge,

Q = \delta q^I p_I - \Lambda

(8)

for every infinitesimal symmetry transformation $q \to q + \delta q$ .

Examples.¶

We have given two classic examples in the discussion above, of linear and angular momentum, and I invite the student to show that the charges $Q$ associated to translational and rotational invariance lead to the same conserved quantities.
Similarly, one can work in Cartesian coordinates with the infinitesimal rotation $x \to \cos\epsilon x - \sin \epsilon y$ and $y \to \cos \epsilon y + \sin \epsilon x$ . For infinitesimal $\epsilon$ , $\delta x = - \epsilon y$ and $\delta y = \epsilon x$ . For the Lagrangian $L = \half m (\dot{x}^2 + \dot{y}^2) = V(\sqrt{x^2 + y^2})$ , which is clearly invariant under rotations, the associated Noether charge is

Q = \delta x p_x + \delta y p_y = - y p_x + x p_y = L

(9)

where $L$ is the angular momentum in Cartesioan coordinates. I leave it to the student to show that this is the same as the generalized momentum $p_{\phi}$ dreived in polar coordinates.

Another important symmetry is time translation invariance; the symmetry is a shift $t \to t + \eps$ . Consider a Lagrangian which is explicitly time-independent. Then

\begin{align} L(q(t + \eps), \dot{q}(t + \eps)) & = L(q + \eps \dot{q}, \dot{q} + \eps \ddot{q}) \\ & = L(q,\dot{q}) + \eps \left[\dot{q}^I \frac{\del L}{\del q^I} + \ddot{q}^I \frac{\del L}{\del \dot{q}^I}\right]\\ & = L + \eps \frac{d}{dt} L(q,\dot{q}) \end{align}

(10)

The conserved quantity is thus

Q = \eps (\dot{q}^I p_I - L)

(11)

The quantity $H - \dot{q}^I p_I - L$ is known as the Hamiltonian and the energy can be defined as the value of this Hamiltonian. As an example, for $L = \half m \dot{\vec{x}}^2 - V({\vec x})$ , the Hamiltonian is clearly

\begin{align} H & = \dot{\vec x} \cdot \vec{p} - \half m \dot{\vec{x}}^2 + V({\vec x})\\ & = \frac{1}{2m}{\vec p}^2 + V \end{align}

(12)

where in the final line we used $\vec{p} = m\dot{\vec{x}}$ . In this case $H$ is clearly the kinetic plus potential energy. More generally, we define $H$ as the energy of the system, and so the conservation of energy is a consequence of invariance under time translation.