Symmetries and conservation laws

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

(63)\[q^I \to (q^I)'(q^I)\]

such that

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

where \((\dot{q}^I)' = \frac{d}{dt} (q^I)'\). The last term in (63) 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

(65)\[{\cal F}_k = \frac{\del L}{\del q^k} = 0\]

The Euler-Lagrange equations for \(q^k\) become:

(66)\[\dot{p}^k = \frac{\del L}{\del \dot{q}^k} = 0\]

In other words, the generalized momentum \(p^k\) is conserved.

Some examples:

1. 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.

2. 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

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

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

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

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

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

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

(70)\[Q = \delta q^I p_I - \Lambda\]

for every infinitesimal symmetry transformation \(q \to q + \delta q\).

Examples.

1. 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.

2. 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

(71)\[Q = \delta x p_x + \delta y p_y = - y p_x + x p_y = L\]

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.

3. 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

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

The conserved quantity is thus

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

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

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

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.