Noether's Theorem in a Nutshell

John Baez

March 12, 2002

Noether's theorem is an amazing result which lets physicists get conserved quantities from symmetries of the laws of nature. Time translation symmetry gives conservation of energy; space translation symmetry gives conservation of momentum; rotation symmetry gives conservation of angular momentum, and so on.

This result, proved in 1915 by Emmy Noether shortly after she first arrived in Goettingen, was praised by Einstein as a piece of "penetrating mathematical thinking". It's now a standard workhorse in theoretical physics.

These days, students often first meet this theorem in a course on quantum field theory. That can make it seem more complicated than it really is. It works for classical field theory, not just quantum field theory. And it also works for the classical mechanics of a point particle! The proof looks a lot easier in this context - but it contains all the basic ideas which show up in the more fancy versions.

Of course the proof uses Lagrangians, but a proof can't help using the concepts which the theorem is about. In other words: if someone claims Noether's theorem says "every symmetry gives a conserved quantity", they are telling a half-truth. The theorem only applies to certain classes of theories. In its original version it applies to theories described by a Lagrangian, and the Lagrangian formalism does most of the work in proving the theorem. There is also a version which applies to theories described by a Hamiltonian. Luckily, almost all the theories studied in physics are described by both a Lagrangian and a Hamiltonian. But let's just do the following simple special case:

Suppose we have a particle moving on a line with Lagrangian L(q,q'), where q is its position and q' = dq/dt is its velocity. (I'll always use the symbol ' to stand for time derivatives.)

The momentum of our particle is defined to be

p = dL/dq'
The force on it is defined to be
F = dL/dq
The equations of motion - the so-called Euler-Lagrange equations - say that the rate of change of momentum equals the force:
p' = F
That's how Lagrangians work!

Next, suppose the Lagrangian L has a symmetry, meaning that it doesn't change when you apply some one-parameter family of transformations sending q to some new position q(s). This means that

d/ds L(q(s), q'(s)) = 0.
Then I claim that
C = p dq(s)/ds
is a conserved quantity: that is,
C' = 0.
Proof - Take the time derivative of our supposed conserved quantity using the product rule:
C' = p' dq(s)/ds + p dq'(s)/ds
Next, use the equation of motion of our particle and the definition of momentum to rewrite the p' and p terms in this equation:
C' = dL/dq dq(s)/ds + dL/dq' dq'(s)/ds
Finally, use the chain rule to recognize that the right side of this equation is
d/ds L(q(s), q'(s)) = 0!
QED.

Of course there's nothing special about the particle being on a line. It's easy to generalize to a particle or bunch of particles in higher-dimensional Euclidean space or even a manifold. It's also easy to generalize to field theories. The idea is always the same.

Experts will note that the above simplified version assumes our symmetry is time-independent: that's how I could slip the time derivative past the derivative with respect to s, when going from

(dq(s)/ds)'
to
dq'(s)/ds
Of course, like any good magician, I did this step so quickly that you might not have noticed it! However, it's an important step, and to handle time-dependent symmetries, we need a somewhat fancier version of Noether's theorem. A simple example comes when you consider the conserved quantities associated to Lorentz transformations! But that's another story.

For more on Emmy Noether and her theorem, try this Wikipedia article. By the way, Emmy Noether proved a lot of other cool theorems and introduced a lot of important ideas into mathematics, especially group theory and ring theory.


© 2002 John Baez
baez@math.removethis.ucr.andthis.edu

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