Symmetry and conservation law

$\def \R {\mathbb R} \def \X {\mathcal X} \def \L {\mathcal L} \def \A {\mathcal A} \def \TqQ {\mathsf{T}_qQ} \def \H {\mathcal H}$ A beautiful property of Lagrangian dynamics is the equivalence between symmetry and conservation law. For instance, in classical mechanics we have:

• conservation of momentum, which comes from translation symmetry;
• conservation of angular momentum, which comes from rotation symmetry; and
• conservation of energy, which comes from time-translation symmetry.

There are many other examples, and this equivalence has a far-reaching generalization in Noether’s theorem (by the same Emmy Noether for whom Noetherian ring is named).

The general statement of Noether’s theorem involves Lie group and so on, but the main idea is actually simple, as we try to illustrate below. Here we follow Susskind’s Classical Mechanics, Lectures 4 and 5.

Lagrangian dynamics

As in classical mechanics, we denote the configuration space of a system by $Q$. (So in our previous notation, $Q = \X = \R^d$ is space, but in general $Q$ can be an abstract configuration space.)

We assume $Q$ is a smooth $d$-dimensional manifold, which means every point $q \in Q$ has a local coordinate system $q = (q_1, \dots, q_d)$ such that the neighborhood looks like $\R^d$. So henceforth we assume $Q = \R^d$ for simplicity, but the main idea is true in a lot more generality.

Recall, from a Lagrangian $\L \equiv \L(q, \dot q, t)$ we can define dynamics via the Euler-Lagrange equation:

$$\frac{d}{dt} \frac{\partial \L}{\partial \dot q} = \frac{\partial \L}{\partial q}$$

which is the first-order stationary condition for the action:

$$\A(q) = \int_{t_0}^{t_1} \L(q_t, \dot q_t, t) \: dt$$

which is the integral of the Lagrangian along $q = (q_t)$, for fixed endpoints $(q_0, t_0), (q_1, t_1) \in Q \times \R$.

For example, the ideal Lagrangian in physics is the difference between the kinetic and potential energy:

$$\L = \frac{m}{2} \|\dot q\|^2 - V(q)$$

where $m > 0$ is the mass of our (fictitious) particle, and $V \colon Q \to \R$ is the potential function.

Calculating the partial derivatives of $\L$, we find the classical formula for momentum:

$$p = \frac{\partial \L}{\partial \dot q} = m \dot x$$

and we define force as the negative gradient of the potential function:

$$F = -\nabla V(q) = \frac{\partial \L}{\partial q}$$

Then the Euler-Lagrange equation above reads:

$$F = \dot p = m\ddot x$$

so we recover Newton’s law of motion, that force is proportional to acceleration.

Symmetry

Now imagine we have a transformation of space

$$q' = q + \epsilon v(q)$$

generated by a vector field $v \colon Q \to \R^d$ (the image is really the tangent space, $v(q) \in \TqQ \cong \R^d$). Here $\epsilon$ is a small (infinitesimal) parameter, so $q’ = q + \epsilon v(q) \in Q$. Note that here we leave time unchanged, $t’ = t$.

Under the transformation above, the velocity $\dot q$ also transforms to

$$\dot q' = \dot q + \epsilon \frac{d}{dt} v(q) = \dot q + \epsilon \: \frac{\partial v}{\partial q}\: \dot q$$

where $\frac{\partial v}{\partial q}$ is the Jacobian matrix of partial derivatives of $v(q) = (v_1(q), \dots, v_d(q)) \in \R^d$.

Furthermore, the Lagrangian $\L \equiv \L(q, \dot q, t)$ changes to

$$\L' \equiv \L(q', \dot q', t) = \L\left(q + \epsilon v(q), \: \dot q + \epsilon \frac{d}{dt} v(q), \: t \right)$$

which by first-order expansion is equal to

$$\L' = \L + \epsilon \left\langle \frac{\partial \L}{\partial q}, \: v(q) \right \rangle + \epsilon \left \langle \frac{\partial \L}{\partial \dot q}, \: \frac{d}{dt} v(q) \right \rangle$$

Therefore, the variation $\delta \L$ of the Lagrangian $\L$ under the transformation $q \mapsto q’$ above is

$$\delta \L = \frac{\L' - \L}{\epsilon} = \left\langle \frac{\partial \L}{\partial q}, \: v(q) \right \rangle + \left \langle \frac{\partial \L}{\partial \dot q}, \: \frac{d}{dt} v(q) \right \rangle$$

Finally, we say the transformation is a symmetry if the variation of the Lagrangian is equal to zero:

$$\delta \L = 0$$

which means the Lagrangian $\L$ stays constant (up to first-order terms) under the transformation.

Conservation law

Now suppose we have a dynamics (curve) $q = (q_t)$ that evolves following the Euler-Lagrange equation

$$\dot p = \frac{\partial \L}{\partial q}$$

where recall $p$ is the momentum variable conjugate to velocity

$$p = \frac{\partial \L}{\partial \dot q}$$

Then we can show that the variation $\delta \L$ of the Lagrangian under the transformation $q \mapsto q’$ above can be written as a time derivative:

$$\delta \L = \langle \dot p, \: v(q) \rangle + \left \langle p, \: \frac{d}{dt} v(q) \right \rangle = \frac{d}{dt} \langle p, v(q) \rangle$$

This may seem odd, because on the left is the change of the Lagrangian in space, whereas on the right is the change of the quantity $C = \langle p, v(q) \rangle$ in time.

But this turns out to be very useful, because now if we assume that the transformation is a symmetry, $\delta \L = 0$, then we automatically have a conserved quantity:

$$\dot C = \delta \L = 0$$

This means the scalar quantity $C = \langle p, v(q) \rangle$ is conserved (stays constant) along the evolution of the curve $q = q(t)$, which follows the Euler-Lagrangian dynamics from the Lagrangian $\L$, which in turn is preserved under the symmetry transformation $q \mapsto q’$.

Thus, from any symmetry we automatically have a conservation law! This may seem magical, or nonsense, but some examples should make it more clear.

Notice, this only works for Lagrangian dynamics. For example, this wouldn’t be true for gradient flow.

Also note, it is as yet unclear what is the significance of the (conserved) quantity $C = \langle p, v(q) \rangle$, which is the inner product between the momentum $p = \frac{\partial \L}{\partial \dot q}$ and the variation in space $v(q) = \frac{1}{\epsilon}(q’ - q)$.

Conservation of momentum from translation symmetry

Consider the translation transformation

$$q' = q + \epsilon v$$

which is generated by the constant vector field $v(q) = v \in \R^d$. Since $v$ is a constant vector, the velocity remains the same, $\dot q’ = \dot q$.

Suppose we have a Lagrangian that has translation symmetry. For example, consider

$$\L = \frac{m}{2} \|\dot q\|^2 - V(\langle \alpha, q \rangle)$$

where the potential energy only depends on the projection of $q$ along some direction $\alpha \in \R^d$, via a scalar function $V \colon \R \to \R$. (For example, if $\alpha = (1,0,\dots,0)$, then the potential only depends on $q_1$.)

Now let’s do the change of variable $q \mapsto q’ + \epsilon v$. Since $\dot q’ = \dot q$, the kinetic energy remains the same. On the other hand, the potential energy changes to

$$V(\langle \alpha, q' \rangle) = V(\langle \alpha, q \rangle) + \epsilon V'(\langle \alpha, q \rangle) \langle \alpha, v \rangle$$

(recall that $V$ is a scalar function, so $V’(x) = \frac{dV}{dx}$ is its derivative). Now notice that if $\alpha$ is orthogonal to the direction of translation $v$, so $\langle \alpha, v \rangle = 0$, then the potential energy also doesn’t change!

Therefore, if $\langle \alpha, v \rangle = 0$, then the Lagrangian $\L$ above is invariant under the translation transformation. Then by our result above, there is a corresponding conserved quantity, which is just the momentum along direction $v$:

$$C = \langle p, v(q) \rangle = \langle m \dot q, v \rangle$$

Thus, conservation of momentum arises from translation symmetry.

Conservation of angular momentum from rotation symmetry

Now consider a two-dimensional system $Q = \R^2$, so $q = (x,y)$.

Consider a rotation transformation on $Q = \R^2$ by a small angle $\theta$. So each point $q = (x,y)$ is mapped to $q’ = (x’, y’)$ given by:

$$\begin{align*} x' &= x \cos \theta + y \sin \theta \;=\; x + y \theta \\ y' &= -x \sin \theta + y \cos \theta \;=\; y-x\theta \end{align*}$$

where we have used the first-order approximations $\cos \theta = 1$ and $\sin \theta = \theta$. So the rotation transformation above is generated by the vector field

$$v(x,y) = \begin{pmatrix} y \\ -x \end{pmatrix}$$

and notice that now we have dependence between the $x$ and $y$ variables, since $v_x = y$ and $v_y = -x$.

Under this transformation, the velocity $\dot q = (\dot x, \dot y)$ also transforms to

$$\begin{align*} \dot x' &= \dot x + \dot y \theta \\ \dot y' &= \dot y - \dot x \theta \end{align*}$$

which is generated by the time-derivative vector field $\dot v(x,y) = (\dot y, -\dot x)$.

Now suppose we have a Lagrangian that is invariant with respect to rotation. For instance, consider the central force problem where the force (and hence the potential) only depends on the radial distance. Concretely, the Lagrangian is

$$\L = \frac{m}{2} \|\dot x\|^2 + \frac{m}{2} \|\dot y\|^2 - V(x^2 + y^2)$$

for some scalar potential function $V \colon \R \to \R$.

Intuitively, it is clear that the Lagrangian doesn’t change when we rotate the coordinates. Formally, we can also check it by calculating the variation $\delta \L$ as before:

$$\begin{align*} \delta \L &= \left\langle \frac{\partial \L}{\partial q}, v(q) \right \rangle + \left \langle \frac{\partial \L}{\partial \dot q}, \frac{d}{dt} v(q) \right \rangle \\ &= \left\langle \begin{pmatrix} -2x V'(x^2+y^2) \\ -2y V'(x^2+y^2) \end{pmatrix}, \begin{pmatrix} y \\ -x \end{pmatrix} \right\rangle + \left\langle \begin{pmatrix} m\dot x \\ m\dot y \end{pmatrix}, \begin{pmatrix} \dot y \\ -\dot x \end{pmatrix} \right\rangle \;= 0 \end{align*}$$

so indeed the Lagrangian has rotation symmetry.

The corresponding conserved quantity is the angular momentum (along the $z$ direction):

$$C = \langle p, v(q) \rangle = \left\langle \begin{pmatrix} p_x \\ p_y \end{pmatrix}, \begin{pmatrix} y \\ -x \end{pmatrix}\right\rangle = p_x y - p_y x$$

Therefore, conservation of angular momentum arises from rotation symmetry.

Conservation of energy from time-translation symmetry

Our last example is the conservation of energy. It is also the most interesting, because (rather than a space transformation as before) its corresponding symmetry transformation is the translation of time!

This also has a distinct character with the previous examples, because now we need to think about spacetime. (As Susskind said in Lecture 5, we are not thinking about relativity yet, but we still need to think about space and time together.)

Concretely, consider the time translation operator by some infinitesimal displacement $\delta > 0$:

$$t' = t + \delta$$

which means “running the experiment at a slightly later time”. Under this transformation, the time differential stays the same, $dt’ = dt$. Moreover, we also leave space the same, $q’ = q$, as well as the tangent vector, $\dot q’ = \dot q$.

Now if we have a Lagrangian $\L \equiv \L(q, \dot q, t)$, it changes to $\L’ \equiv \L(q’, \dot q’, t’)$, which is equal to

$$\L' = \L(q, \dot q, t+\delta) = \L + \delta \frac{\partial \L}{\partial t}$$

Thus, we see that if the Lagrangian is time-independent, $\frac{\partial \L}{\partial t} = 0$, then the Lagrangian is invariant under the time-translation transformation.

So we say that the Lagrangian has time-translation symmetry if it has no explicit time dependence. What is the corresponding conserved quantity?

The situation is a bit murky because our derivation above assumes the transformation is only on space, not time—which is the opposite of our situation now. (That is, we cannot really use the formula $C = \langle p, v(q) \rangle$ since now the vector field $v$ also has a time component $v_t = 1$, so we should include the momentum $p_t = \frac{\partial \L}{\partial \dot t}$ conjugate to time, but this is weird because $\dot t = \frac{dt}{dt} = 1$, so maybe $p_t = \L$?)

But let’s try to compute something. For example, what is the time derivative of the Lagrangian $\L(q,\dot q)$ along the evolution of its own Euler-Lagrange dynamics $q = (q_t)$? It is:

$$\begin{align*} \frac{d\L}{dt} = \left\langle\frac{\partial \L}{\partial q}, \dot q \right\rangle + \left\langle\frac{\partial \L}{\partial \dot q}, \ddot q \right\rangle \end{align*}$$

Now recall we define momentum $p = \frac{\partial \L}{\partial \dot q}$, so the Euler-Lagrange equation says $\frac{\partial \L}{\partial q} = \dot p$. Then the time derivative above becomes:

$$\frac{d\L}{dt} = \langle \dot p, \dot q \rangle + \langle p, \ddot q \rangle = \frac{d}{dt} \langle p, \dot q \rangle$$

So we see that the Lagrangian is not conserved, but its rate of change is equal to the quantity $\langle p, \dot q\rangle$, which notice is similar to the conserved quantity $C = \langle p, v(q) \rangle$ above with $v(q) = \dot q$.

But now, if we move everything to the right-hand side, we have a conservation law:

$$\frac{d}{dt} \big( \langle p, \dot q \rangle - \L \big) = 0$$

and lo and behold, the Hamiltonian

$$\H = \langle p, \dot q \rangle - \L$$

appears as the conserved quantity, and this is what we usually call the energy. For example, for the ideal Lagrangian $\L = T - V$, the Hamiltonian is the total energy $\H = T + V$.

Therefore, conservation of energy arises from time-translation symmetry, which concretely means the Lagrangian is time-independent.

And as noted above, time translation has a slightly different nature than just space transformation, especially because we are talking about dynamics over time itself. For example, it feels like time should ultimately be the “exponential” of space, but that’s for another day.