Thursday, July 08, 2010

Symmetries and level-appropriate teaching

This fall I'm going to be teaching honors introductory mechanics to incoming undergraduates - basically the class that would-be physics majors take.  Typically when we first teach students mechanics, we start from the point of view of forces and Newton's laws, which certainly parallels the historical development of the subject and allows students to build some physical intuition.  Then, in a later class, we point out that the force-based approach to deriving the equations of motion is not really the modern way physicists think about things.  In the more advanced course, students are introduced to Lagrangians and Hamiltonians - basically the Action Principle, in which equations of motion are found via the methods of variational calculus.  The Hamiltonian mechanics approach (with action-angle variables) was the path pursued when developing quantum mechanics; and the Lagrangian approach generalizes very elegantly to field theories.  Indeed, one can make the very pretty argument that the Action Principle method does such a good job giving the classical equations of motion because it's what results when you start from the path integral formulation of quantum mechanics and take the classical limit.

A major insight presented in the upper division course is Noether's Theorem.  In a nutshell, the idea is that symmetries of the action (which is a time integral of the Lagrangian) imply conservation laws.  The most famous examples are:  (1) Time-translation invariance (the idea that the laws of physics governing the Lagrangian do not change if we shift all of our time parameters by some amount) implies energy conservation.  (2) Spatial translation invariance (the laws of physics do not change if we shift our apparatus two feet to the left) implies conservation of momentum.  (3) Rotational invariance (the laws of physics are isotropic in direction) implies conservation of angular momentum.  These classical physics results are deep and profound, and they have elegant connections to operators in quantum mechanics.

So, here's a question for you physics education gurus out there.  Does anyone know a way of showing (2) or (3) above from a Newton's law direction, as opposed to Noether's theorem and Lagrangians?  I plan to point out the connection between symmetry and conservation laws in passing regardless, but I was wondering if anyone out there had come up with a clever argument about this.  I could comb back issues of AJP, but asking my readers may be easier.  

17 comments:

Tahir said...

So, what is Newton's second law? It's just F = m*a = m*(dvelocity / dt). And for essentially all cases of interest in classical physics (except perhaps friction, but that's another story, and as far as I know friction isn't even really easy to deal with as regards Lagrangians and Hamiltonians...) we may write the force as the negative gradient of a potential (conservative forces) F = - grad V.

So, what do we mean by spatial-positional independence of the laws of physics? I am under the impression that that implies that the gradient of V would vanish, otherwise theres a spatial dependence in the potential energy and the laws of physics are not invariant under a spatial translation. Thus, grad V = 0, and thus F = - grad V = 0 = d (momentum)/dt so momentum is a constant in time.

In Newtonian terms, torque = d(angular momentum)/dt = r cross product Force = (again, for all cases of interest) - r cross (grad V). I suppose now we figure that if r cross (grad V) is not zero, that would imply that there is a component of grad V that is perpendicular to displacement from the origin. Intuitively it seems to easy to realize (from picture and diagrams) that this would mean that the physics is different at different angles relative to your origin. Thus, just as with the linear momentum - translation symmetry case, in order for the physics to be angle independent, you HAVE to have (grad V) parallel to the displacement relative to your origin, meaing r cross F has to be zero, meaing d(angular momentum)/dt has to be zero.

-Tahir

Tahir said...

In fact, now that I think about it, it is pretty easy to reason r cross (grad V) = 0 if and only if the physics is isotropic. Because when you say r cross (grad V) = 0, it is equivalent to saying that there is no gradient in any direction perpendicular to the outward radial vector, as I have discussed in the previous post. So what does this mean? Any spatial dependence of the potential, and thus the physics, has to be in the outward radial direction -> its isotropic.

-Tahir

Paul said...

Awww, Hannon's not teaching 111? It was that class that made me (partially) switch over from Chemistry to Physics. Is he going to be teaching any other classes this year?

In any case, best of luck with that class. I'm sure you'll do well.

Tahir said...

Sorry to bug you again...but I can simplify everything I talked about with a concise summary.

Linear Momentum:

Assuming translation invariance is equivalent to assuming V is not a function of position.

Thus, Force = d*(momentum)/dt = - gradient V = 0, so linear momentum is conserved.

Angular Momentum:

Assuming rotational isotropic invariance is equivalent to assuming the potential is only a function of radial distance V = V(r) and not angle. By simple calculus, this means grad V has only a radial component parallel to radial displacement r.

Thus, Torque = r x Force = - r x grad(V) = 0 = d(angular momentum)/dt, so angular momentum is clearly conserved.

What I like about this approach is that it connects more to how we approach the problems in Lagrangian and Hamiltonian mechanics. Namely, you look and see if your potential, and thus your Lagrangian (or Hamiltonian) has any positional and/or angular dependence. If it has none, then that's an immediate sign that momentum and/or angular momentum is conserved by Noether's theorem, respectively. You don't even have to do a calculation, just look at the form of the potential and your good to go.

- Tahir

Joe Renes said...

Looking over my old notes from ug days, I see that we did not specifically discuss the connection between translation invariance and conservation of momentum. The argument we used treats all forces as being due to some other object(s), and conservation of momentum follows immediately from the 2nd and 3rd laws and assumes of course that these laws hold regardless of the location of the objects.

Similarly, I think one can say that rotation invariance means that the force on one object due to another must be along the line joining them, which results in conservation of angular momentum since the lever arm and force point in the same direction.

These arguments do seem more casual about the connection between symmetry and conservation than Noether's theorem, but perhaps that's because we're using Newton's laws in the first place. That is, the causality is backwards: momentum conservation from translation symmetry is not something that should come out of Newton's laws, rather it is something that goes in to coming up with them in the first place. Thus, when we use the argument above and just start from Newton's laws, we're showing that this connection was inherent in Newton's laws all along.

Vincent said...

Why does translation invariance imply momentum conservation?

I ask because as I understand it translation invariance is true in general, but momentum conservation is only true in the classical limit of c=inf, and I don't see a very simple way to say that because we can move our experiment arbitrarily and have it act out the same, we must therefore have the momentum conserved.

On a side note, isn't momentum conservation just an axiom of newtons laws, with angular momentum following as consequence?

CarlBrannen said...

Hmmm. I show my algebra based associate degree introductory physics students that conservation of momentum immediately follows from F=ma and "for every action there's an opposite and equal reaction"; one replaces a by change dv/dt, (without calculus). I wonder if that is related.

Doug Natelson said...

Tahir - Thanks for your remarks. I'd already been thinking about conservative force fields (gradients of scalar potentials) as an example. However, that doesn't really get at the deeper issue. In some sense, the deep question is, how does translational invariance imply Newton's 3rd law (which is really critical to momentum conservation for multiparticle systems)? The forces involved in interparticle interactions in general don't have to be described as the gradients of scalar potentials....

Paul, yes, after more than a dozen years, Jim is going to teach 301 instead of 111 in the fall. He'll still do 112 in the spring, though.

Vincent, momentum is always conserved, even in the relativistic (c != infinity) case, as long as momentum is properly defined. Your response makes my point though - the connection between translation invariance and momentum conservation isn't obvious.

Joe, I think your last paragraph is insightful - I need to think more about this.

Tahir said...

Hmm...I guess I see how you'd like something a bit more intuitive and physical. Forgive me, I've been thinking about it from another point of view, and please allow me to present it in a way that I hope better fits what you are looking for. In fact, now that I think about it in that way, what Joe says seems spot on, and I hope the following will convince you of that.

So when we say a systems of particles in mechanics is translationally invariant, we are saying, if we had a big box that hosted the entire system, and we shifted the box and the entire system's coordinates over by some arbitrary position, and then looked at it from this new position, the physics would be the same. Essentially, we could freeze time, put the system in the box, shift it, turn time back on, and we wouldn't notice a thing had changed if we were in the box, per se. For a truly translationally invariant system, this should be the case no matter what the box, as long as it enclosed everything in the system in question.

Now, imagine if the physics were to indeed change were we to move that box and everything in it to another position. That would imply that there is no translational invariance. It would also imply that there has to be some external variable at play that is affecting the box and the system as a whole. Otherwise how could the physics change if we moved our box? There has to be SOMETHING causing the change.

And saying that there has to be something there is equivalent to saying, in Newtonian terms, that there's got to be a NET EXTERNAL FORCE on the box on the system. That net external force is the external variable acting on the box and whole system that is the only way to change physics if you shift the box by some displacement.

So if we are saying that the system's physics is translationally invariant, that's saying there can't be a net external force on the system, so the net force on the system is ENTIRELY DUE TO INTERNAL FORCES BETWEEN THE PARTICLES. Because as is known, the net force on an arbitrary system of particles is the sum of any external forces on the system as a whole, and all internal forces between all particles acting together.

So the above argument shows that translational invariance implies there is no external force and the net force is entirely internal, in a nutshell.

Momentum conservation, by the 2nd law, is equivalent to saying the net force is zero, meaning all internal forces cancel, in other words, for every action there has to be an equal and opposite reaction. Thus, as I see it, Newton's third law and momentum conservation from translational symmetry are essentially two sides of the same coin, so to speak. This coin is a postulate that Newton came up with, and if I understand correctly it is not derivable from Newton's 2nd law itself.

Sorry for my wordiness.

-Tahir

Joe Renes said...

I was led to my final comment by reading a bit about the history of Newton's laws in Julian Barbour's book "The Discovery of Dynamics". Reading a bit more carefully now, it seems that Newton came upon the idea of the third law from studying collisions, and his notion of the third law (which wasn't then yet precisely formulated) is very much bound up with the conservation of momentum.

Here's a quote from Newton's Waste Book, written sometime between 1664 and 1666, regarding two colliding objects p and r (page 508 of Barbour):
"119. If r presse p towards w then p presseth r towards v [w and v are opposite directions of course]. Tis evident without explication.
120. A body must move that way which it is pressed.
121. If 2 bodys p and r meet the one the other, the resistance in both is the same for soe much as p presseth upon r so much r presseth on p. And therefore they must suffer an equall mutation in their motion [momentum]."

Of course, none of that has to do with momentum conservation from translational invariance, but I think Tahir's argument works nicely to show that statement is essentially the same as the third law. If I understood the reasoning correctly, we can get momentum conservation from translational invariance via the 2nd and 3rd laws, or we can get the 3rd law via the 2nd law and the statement that momentum conservation is implied by translation invariance.

There is one possible objection I thought of: why shouldn't we allow the possibility of a constant force acting on all particles? The system is translationally-invariant, inasmuch as the force is the same everywhere, but of course momentum is not conserved. Momentum minus force times time is conserved though.

This situation is a bit subtle in analytic mechanics as well, because now the Lagrangian isn't translationally invariant, even though it "should" be. Actually, it is, once you remember the caveat that the Lagrangian is only defined up to gauge terms which are total time derivatives of a function of position and time. If we use the function Ft, then the symmetry is restored (the gauge term can depend on the translation) and Noether's theorem leads to the correct conservation. Not sure where I'm going with this!

Tahir said...

Hmm...

Joe's counterexample of a constant force is an interesting one. On the surface, it appears that indeed such a system should be translationally invariant.

I am not sure if I am correct with this reasoning, however I wonder if we could argue that in reality, the system is not truly translationally invariant if subject to a constant force.

I guess I am imagining the example of a constant force due to a constant electric field created by an ideal infinite parallel plate capacitor. Then if your system of particles, with charge per se, were sandwiched in the middle of that ideal capacitor, it would indeed be feeling a constant force. However, if you translate the system beyond the edge of the parallel plate capacitor, then the net electric field and force is zero, so in that sense the translational invariance is broken.

I guess (and here, I confess, I am kind of waving my hands and am not sure) you could imagine expanding that parallel plate capacitor's walls out ot plus and minus infinity so that the uniform electric field becomes ever more present "everywhere" -> but there would kind-of always still be a translation beyond an edge that breaks the physical invariance...

-Tahir

Joe Renes said...

@Tahir: I agree! In principle the constant force scenario has the limitations you mention, since ultimately we want to talk about the degrees of freedom which are responsible for the force in the first place. If we assume that we can do this, then we're back to the case of no external force and can proceed as before.

This seemed like an interesting point not because I expected it to wreck the arguments made so far, but because it's a weird case of translational invariance that doesn't quite fit the usual conception, especially in Lagrangian mechanics. Possibly a good homework question, or even more cruelly, a question for an oral exam!

Tahir said...

Yes, indeed Joe, such a thing would indeed be quite fun to orally examine a student on. I can just imagine professors torturing graduate students preparing for their oral qualifying candidacy exam...hehehehe. Ironic that I find such torture amusing since I myself am taking said exam in a few months.

So the picture is really coming together now. In Lagrangian /Hamiltonian mechanics, Noether's theorem leads to the conservation of p - Ft. We can reconcile this with simple Newtonian intuition by, as you said, figuring that when we talk about "translationally invariant system", our system HAS to include the external degrees of freedom responsible for the force. Thus, the total momentum of the system, which is the relevant conserved quantity, is not just the p of the particles, but also the -Ft, which is the additional momentum of the additional force-creating degrees of freedom in the system.

I think we should co-write an article for the American Journal of Physics, Joe. What do you think? ;-)

-Tahir

QuasiNewton said...

It's the generating momentum that takes you from configuration A to B that are related by the corresponding symmetry operations. Without explicit time dependence, the change in energy only depends on the change in generating momentum. If there is no change in energy, there is no change in momentum either. (This is a mere rephrasing the "cyclic coordinates" idea of Lagrangian mechanics, but there might be no need to introduce the Euler-Lagrange equation, if one simply explains that e.g. an angular momentum rotates a system and that all resulting changes in the system can be characterized by the angular momentum).

Doug Natelson said...

Thanks to all of you for a vigorous discussion! I'm traveling right now, or I would post more.

Joe Renes said...

@Tahir -- send me an email and let's discuss! =)

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