Why is Noether's theorem important?

I am just starting to wrap my head around analytical mechanics, so this question might sound weird or trivial to some of you.

In class I have been introduced to Noether's theorem, which states that if the Lagrangian function is invariant under a continuous group of transformations then it's possible to find a conservation law.

But a Lagrangian system with $n$ degrees of freedom obeys the Euler-Lagrange equations, which are: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{q_i}}-\frac{\partial L}{\partial q_i} = 0$$

for $i = 1, ..., n$.

This equation represents a system of $n$ second order differential equations that already has $2n$ arbitrary constants in its general solution.

This constants are obviously to be preserved in time, so they actually represent conservation laws.

So my question is, what is the utility of a theorem that tells you under what conditions it is possible to find a conservation law, if we already know from the Euler-Lagrange equations that a Lagrangian system has $2n$ conservation laws?

• Of course they are equivalent for the considered case. I'm sorry if it sounded that I meant that you wrote something wrong (or you were mislead by your teacher). What I meant is that the theorem appears to be trivial in the case of Lagrangian mechanics because, as you noted, it is easy to verify from the EL eqs. Indeed the concept of invariant continuous transformations (symmetries in a sense) make the theorem more general. – PML Feb 26 '18 at 0:14
• Btw, I wouldn't worry that the concepts in the wiki article are difficult to understand. If you continue to study physics (and mathematics) you'll understand what is there, crystal clear. – PML Feb 26 '18 at 0:19
• Related: physics.stackexchange.com/q/8626/2451 , physics.stackexchange.com/q/21572/2451 and links therein. – Qmechanic Feb 26 '18 at 7:48
• If you solve Euler-Lagrange equation for a particle under a constant gravitational field (may it be the classical free falling problem), you’ll get for the y axis the following solution: $y= \frac{1}{2}gt^2+c_1t+c_2$ After applying the initial conditions to the problem, you’ll realize that $c_1$ is just the initial velocity $v_0$ and $c_2$ the initial height $h_0$. Therefore, $c_1$ and $c_2$ are velocities and positions. They are not known to be “conserved” quantities. – J. Manuel Feb 26 '18 at 10:22
• This is so obvoius. It's important because if we didn't have it, we'd need to come up with a Noether one. – Magoo Feb 26 '18 at 13:46

We usually call equations like

$$\frac{d}{dt} \frac{\partial L}{\partial \dot{q_i}} - \frac{\partial L}{\partial q_i} = 0$$

"equations of motion," because they are equations that tell us how the variables of our system (here $$q_i$$) evolve in time. Indeed, in general, the solution to $$n$$ second order differential equations involves $$2n$$ integration constants (or initial conditions) in the solution. However, most people would not call these integration constants "conservation laws." In general usage, a "conserved quantity" $$Q$$ is a function of the configuration variables (here $$q_i$$ and $$\dot q_i$$) that does not change in time when the configuration variables evolve according to the equations of motion:

$$\frac{d}{dt} Q(q_i, \dot q_i) = 0.$$

Note that $$Q(q_i, \dot q_i)$$ does not depend on $$t$$ explicitly; it only depends on $$t$$ insofar as $$q_i$$ and $$\dot q_i$$ do. However, an initial condition depends on $$q_i$$, $$\dot q_i$$, and $$t$$. You need to know $$t$$ in order to know "how far to turn back the clock" to find the initial position and velocity.

A slick "proof" of Noether's theorem goes as follows. Say you have some differentiable group of transformations that leave your Lagrangian invariant. Imagine changing a path in configuration space by an infinitesimal group action, using a tiny number $$\varepsilon$$. For example, an infinitesimal translation in the $$x$$-direction in 3D space ($$i = 1, 2, 3$$) would be given by

$$q_1 \to q_1 + \varepsilon$$ $$q_2 \to q_2$$ $$q_3 \to q_3$$ $$\dot q_i \to \dot q_i$$

and an infinitesimal rotation in the $$xy$$-plane would be given by

$$q_1 \to q_1 + \varepsilon q_2$$ $$q_2 \to q_2 - \varepsilon q_1$$ $$\dot q_1 \to \dot q_1 + \varepsilon \dot q_2$$ $$\dot q_2 \to \dot q_2 - \varepsilon \dot q_1$$ $$q_3 \to q_3$$ $$\dot q_3 \to \dot q_3$$

Under these transformations, the Lagrangian $$L(q_i, \dot q_i)$$ will not change its value. In other words, the change in the Lagrangian can be expressed as

$$\delta L(q_i, \dot q_i) = \varepsilon A(q_i, \dot q_i)$$

where $$A = 0$$ if the group action is a symmetry. Here is the slick part: now imagine that the parameter $$\varepsilon$$ is time-dependent, i.e. $$\varepsilon(t)$$. For our above two actions, the transformations would then become

$$q_1 \to q_1 + \varepsilon$$ $$\dot q_1 \to \dot q_i + \dot \varepsilon$$ $$q_{2} \to q_{2}$$ $$q_{3} \to q_{3}$$ $$\dot q_{2} \to \dot q_{2}$$ $$\dot q_{3} \to \dot q_{3}$$

and

$$q_1 \to q_1 + \varepsilon q_2$$ $$q_2 \to q_2 - \varepsilon q_1$$ $$\dot q_1 \to \dot q_1 + \varepsilon \dot q_2 + \dot \varepsilon q_2$$ $$\dot q_2 \to \dot q_2 - \varepsilon \dot q_1 - \dot \varepsilon q_1$$ $$q_3 \to q_3$$ $$\dot q_3 \to \dot q_3$$

(where the extra term above comes from the product rule when differentiating by $$t$$).

Now, $$\varepsilon(t)$$ and $$\dot \varepsilon(t)$$ are both tiny numbers that change paths in configuration space. That means that, just doing a first order Taylor expansion, the change in $$L$$ under these transformations can be expressed as

$$\delta L = \varepsilon A + \dot \varepsilon B$$

where the $$A$$ is the same $$A$$ as before, meaning $$A = 0$$ if the transformation is a symmetry. Now, on actual paths, $$\delta S = 0$$ for any tiny variation we make to our path. (That is just the principle of least action.) That includes our tiny group action variation. Therefore, on actual paths,

$$0 = \delta S = \int \delta L dt = \int \dot \varepsilon B dt = - \int \varepsilon \dot B dt.$$

(In the last step we integrated by parts and imposed boundary conditions $$\varepsilon = 0$$ on the boundary of integration.)

Therefore, if $$\delta S$$ is to be $$0$$ for any $$\varepsilon$$, we must have

$$\dot B = 0$$

so $$B$$ is a conserved quantity. Note that if our transformation wasn't a symmetry, then $$A \neq 0$$ and

$$\dot B = A$$

meaning that $$B$$ would change in time and not be a conserved quantity. This concludes the proof that symmetries give conservation laws, and also instructs you how to find said conserved quantities.

Now this is all nice and interesting. Symmetries imply conservation laws. In a sense, we have understood where "conserved quantities" come from (symmetries). Conserved quantities are very useful in physics because they usually make analyzing the system much easier. For example, even in intro physics, the conservation of momentum and energy are always used to make solving for the motion of a particle much easier. In more complicated examples, like for example a gas of many particles, the evolution of the system is far too complicated to ever hope to describe. However, if you know a few conserved quantities (like energy, for example) you can still get a pretty good idea of how the system behaves.

In quantum field theory, quantum fields are also governed by Lagrangians. However, it is often difficult to figure out exactly what the Lagrangian of quantum fields should be based off of experimental data. Something that is straightforward to ascertain from experimental data, however, are conserved quantities, like charge, lepton number, baryon number, weak hyper change, and many others. Experimentalists can figure out what these conserved quantities are, and then theorists will cook up Lagrangians with symmetries that have the right conserved quantities. This greatly aids theorists in figuring out the fundamental laws of physics. Considerations of symmetries and conserved quantities historically played a large role in piecing together the standard model, and continue to play a crucial role in theorists trying to figure out what lies beyond it.

EDIT: So, to answer your question proper, any system of differential equations will have integration constants (A.K.A. initial conditions). However, from equations of motion derived from a Lagrangian (and all known physical laws can be written with Lagrangians) we have extra symmetries that have important physical meaning. Furthermore, the exact solutions to differential equations are usually impossible to solve for any moderately complex system. Therefore, finding initial conditions is usually a waste of time, while Noether's theorem is easy to use.

• Why does $\varepsilon$ small imply $\dot\varepsilon$ small? Is it an assumption on $\varepsilon$? – anderstood Feb 26 '18 at 22:21
• Well, seeing as you can choose the variations $\varepsilon$, you can indeed impose that $\dot \varepsilon$ is small. Furthermore, another (perhaps slightly more rigorous way) to do calculus of variations is to take the variation in the path to be $\varepsilon \eta(t)$, where $\eta$ is not infinitessimal, and $\varepsilon$ is constant. As long as $\dot \eta$ is bounded above, $\varepsilon \dot \eta$ will shrink as $\varepsilon \to 0$. – user1379857 Feb 26 '18 at 22:29
• For anyone else interested in details, $A$ does not actually need to be $0$ in order for the transformation to be a symmetry, $A$ need only be a total time derivative, say $A = \dot C$. In such a case, $B - C$ would then be the complete conserved quantity. A good example of this is time translations. Under a time translation $q \to q + \varepsilon \dot q$, our "$A$" would be $A = \dot L$, and $B - L$ turns out to be the energy. – user1379857 Feb 26 '18 at 22:37

The 2n constants in the system of second order differential equations (Lagrange equations) are just the arbitrary initial conditions of generalized coordinates and velocities of the system that determine the time development of the system for a special case. They are not conserved quantities of the system. Conserved quantities follow from the functional form of the Lagrange function.

Noether's theorem is very important, as it covers the "big" conservation laws:

Time invariance implies energy conservation.

Translation invariance (homogeneity of space) implies momentum conservation.

Rotational invariance (isotropy of space) implies angular momentum conservation.

Moreover, it depends on what you consider your system: consider a ball in a box in gravity. It is time independent, and energy is conserved, but you need to add the potential energy to the kinetic energy. Momentum is not conserved, and that is because the potential energy is height dependent. Note that space is invariant under horizontal translations, and in fact, the horizontal components of momentum are conserved (well, until it hits a symmetry breaking wall).

Likewise, the angular momentum of Foucault's pendulum is not conserved--there is a preferred direction in space defined by the Earth's axis.

More complicated things like local gauge symmetry lead to EM and QCD, and of course the gauge symmetry breaking mass terms in the Weak interaction was one of the motivations for the Higgs boson.

There are also discrete symmetries that lead to discrete conservations laws--Parity, for example.

protected by Qmechanic♦Feb 26 '18 at 7:25

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