Some physics without variational principle ?

1. Jul 6, 2006

lalbatros

I am in search for some part of physics that could not be derived from a variational principle.

For the small part of physics I know (CM, Schrödinger), everything can be derived from a variational principle. I would like to know if this is a deep fingerprint of physics or a general consequence from mathematics (like if all DE would admit a VP).

Michel

2. Jul 6, 2006

Triss

There is a link between ordinary differential equarions, partial differential equations and a variational principle; and it all comes together in Hamilton-Jacobi theory. See for example http://xxx.lanl.gov/abs/quant-ph/0210140

3. Jul 8, 2006

actionintegral

The variational principle is a bit over-rated. It is basically a minimization of an integral, which is a sum. A minimization of a sum is the minimization of the summands, which is 0. Such a summand to be minimized is lagrange's equations, or newton's law.

No deep mystic principles there.

4. Jul 8, 2006

arildno

Navier-Stokes equation, for example.
(It is NOT proven that it cannot be derived from a variational principle, but no such principle has been found).

5. Jul 8, 2006

lalbatros

actionintegral,

Of course, physics doesn't give any place for mystic principles.

But what do you mean by overrated? How would you rate a principle of physics?
You could rate the least action principle as "undergraduate physics", but would it be overrated for this reason?

If all physical laws could be derived from a least action, how would you rate the "least action principle"? And would you be interrested to know why this?

If some physics could be proved incompatible with any LAP, what would be your reaction?

In addition, understanding physics also covers such questions!

I posted a related question:

Michel

Last edited by a moderator: Apr 22, 2017
6. Jul 8, 2006

masudr

Well, a fair bit of graduate physics comes from an LAP. Take general relativity, for example.

7. Jul 10, 2006

actionintegral

Hi lalbatros,

Thank you for your thoughtful reply. By "overrated" I mean that it contains no new information that we didn't have already. When I read that the Least Action Principle is equivalent to f=ma, the least action principle "lost it's luster" for me. Ditto for the Lagrangian formulation for discrete mechanics. Convenient, yes; Powerful, certainly; but no new information.

8. Jul 10, 2006

nrqed

However, notice that it is possible to extend the least action principle to cover quantum mechanics (in which case all paths contribute, not just the one that extremizes the action and the contribution of paths away from the classical one is controlled by Planck's constant) whereas it is not possible to use F=ma as a basis to get to quantum mechanics. So in that sense I would say that the least action principle is more fundamental than F=ma because it can be generalized to include QM.

Just my two cents

9. Jul 10, 2006

actionintegral

Yes, I've heard about this - also that General Relativity is based upon a variational principal. At this time, I don't have enough knowledge to say anything with confidence.

Nevertheless if I learn more about these two applications, and I discover that they are simply re-working a local phenomenon (much like the integral form and the point form of maxwell's equations say the same thing, but in different forms), then I will say that the essence is not with the variational principle per se.

10. Jul 10, 2006

actionintegral

This is a fascinating example. Feynman's example of how the Least Action Principle is really just a recasting of Newtons's laws doesn't seem to cover phenomena that are global in nature. I.e., where the value of the integrand at point a depends on the value of the integrand at another point. I'll have to reread that with arildno's example in mind...

11. Jul 10, 2006

lalbatros

actionintegral,
nrqed,

Have you noticed that the Schrödinger equation can also be derived from a variational principle?

Concerning the Navier-Stokes equations, the situation is indeed more interresting. Either we have here an counter-example to the generality of variational principles. Or we *just* need to find the variational principle involved. And there is a third way: as a macroscopic equation, I could argue that anyway it derives from a microscopic variational principle. However, it may not be a necessity.

Still we cannot say anything about the possiblity that all differential equation might derive from a VP.

I agree with you that if this game only amounts to rederiving known differential equations from a VP, we could not learn a lot. If VPs really apply to all differential equations, this would probably be the case: why my question. In the opposite situation it might be more instructive and useful to investigate.

Michel

Last edited: Jul 10, 2006
12. Jul 10, 2006

actionintegral

Is that the feynman sum over paths thing?

13. Jul 10, 2006

nrqed

Yes, of course. I did not mean to imply that Schroedinger's equation cannot be derived from a variational principle.

My point was that the transition from classical to quantum physics is much more transparent and physically clear in the path integral approach, which is based on the variational principle.

The "diff eq" approach is to postulate that position and momenta are promoted to operators wich act on this totally new quantity, the wavefunction. And *then* one can derive this from a lagrangian involving that weird wavefunction.

On the other hand, the variational principle says that instead of having only the extremum path, all paths with contributions controlled by h bar, and a rule for the coherent combination of different paths to give an amplitude, etc etc.

My point was that the transition classical mechanics <-> QM is more transparent (imho) in the variational principle/path/integral/sum over paths approach.

Of course one can obtain Schroedinger's equation using the "canonical" approach and *then* show how it can be reived from a lagrangian. But this does not show very well the correspondence QM <-> CM , as opposed to the "variational approach".

In my humble opinion.

I agree that this is quite interesting. I did not know about that.

Regards

Patrick

14. Jul 10, 2006

masudr

I feel the integral over paths is the "proper" quantum principle, as opposed to the arbitrary Lagrangian that gets you the Schrodinger equation, by demanding an extremum.

15. Jul 10, 2006

Gokul43201

Staff Emeritus
Indeed it is! And I just realized this now.

Working backworks from the SE (or its complex conjugate) you arrive at the all too intuitive (well, only in retrospect, for me) action integral,

$$E = \int d^3x \left( \frac{\hbar^2}{2m} \nabla \psi ^* \nabla \psi + \psi ^* V \psi \right)$$

The SE and its complex conjugate are simply the equations of the variational problem to extremize the above integral subject to the constraint $\int d^3x \psi ^* \psi = 1$...or so it appears (aka unless I screwed up the calculation).

Nice!

16. Jul 10, 2006

nrqed

EDIT: I see that you gave the energy (I thought you were providing the action). How did you get this expression? Did you get the action first?

Hey Gokul!

You mean the action S, right? I think you are missing $\psi^* \partial_t \psi$.

What a coincidence.
I have posted something exactly on this yesterday on the quantum forum:

I'd like to know how you proceeded to finding your lagrangian.
It is the usual expression given in textbooks but I came up with a different expression which, as far as I can tell, yield the Schroedinger equation as well. However my lagrangian does not have the invariance under a phase shift and it can't be right because the conjugate momemtum to psi is psi itself! So I know there is something wrong with it but if I just apply the Euler-Lagrange eqs, it *does* yield Schroedinger's equation!
So there is something I am obviously missing about how to get a lagrangian starting from the eoms.

How did you get yours? because the eom is first order in time, this seems tricky to me. I would really like to see how you tackled this problem!

Regards

Patrick

Last edited: Jul 10, 2006
17. Jul 10, 2006

maverick280857

Okay me still learning QM but looking at your post, I thought you would be interested in seeing this:

http://www.ece.rutgers.edu/~maparke...hapts/Ch05-Dynamics/Ch05Sec05SchrodEqLagr.pdf

Warning: could be trivial for you

Last edited by a moderator: Apr 22, 2017
18. Jul 10, 2006

Gokul43201

Staff Emeritus
Yes, nrqed, that is the energy in that expression - my error there. All I did was (attempting to construct a variational problem out of the SE) start with the TISE, multiply by $\psi ^*$ and integrate over all space. You have a surface term that vanishes in this limit, and you are left with the integral for energy that I wrote above.

Now it looks to me like the SE (and its conjugate) would become the Euler equations of a variational problem - one where you must extremize the above energy integral subject to the normalization constraint.

PS : These are the first bits of thought that I've put into this problem and need to spend more time thinking about it to make sure I understand what I'm doing and so I can then move on to more stuff. I have not seen the standard textbook expression for the Lagrangian before now - I imagine it's the one in maverick's link.

19. Jul 11, 2006

nrqed

I see. So you said that the integral over all space of $\psi^* \partial_t \psi$ is zero?
I am a bit confused. The euler-Lagrange eqs are applied to the lagrangian (density), right? Not to the energy. But I would like to see how you would derive the equation from the minimization of the energy!

And I am a bit perplexed by using the normalization condition as a constraint. I have never seen this done in a variational principle to obtain Schrodinger's equation! I am not saying it's wrong but it would be a completely different derivation than everything I have seen before.
Schrodinger's equation itself does not presuppose any normalization condition on the wavefunction, it's something that must be added by hand afterward. SO I am a bit confused at using this as a constraint. If it is a constraint than why is the wavefunction unconstrained in Schroedinger's equation itself?

Yes,and I give the expression in my post on the quantum thread.