Significance of the Lagrangian

[alexepascual]

Robphy:
I see your point about the non-uniqueness of the Lagrangian.
I think a good example would be an arbitrary change in the potential function. The zero of the potential function itself is arbitrary to start with. So we should be able to make a change in it without this reflecting any changes in physical reality (and we are). What about changing to a frame of reference where the origin of the space coordinates is different or is moving with respect to the first frame? That would also change the value of the Lagrangian.
I'll keep in mind your suggestion about Arnold or Abraham-Marsden. Right now I have Lanczos and Goldstein and I'll try to get the most out of them for the next few days. When I go to the library to renew Lanczos I'll see if I can get the books you suggest. (I don't know about Goldstein, but I think I can get some more out of Lanczos) If you have not seen that book I recommend it.
Name of the book: "The variational principles of mechanics"
On a different note: Something that I have been thinking about but I don't have any confirmation yet is that in a typical motion of a particle under a potential, you'll have some exchange between the potential and kinetic energies. I was thinking that by minimizing the time integral of the Lagrangian, you might be indirectly minimizing this exchange. (I may be wrong about this). But I think it is more probable that minimizing action may be equivalent to choosing the path with less constraint.

We may have to address the assumption that L is a function of q, q-dot, and possibly t, but not q-doubledot and higher derivatives.

I think we can safely assume this and not worry for the moment about that possibility.

 

[marlon]

OK, maybe I was not being fair with my last statement as the cup is really in a local minumum of it's potential energy. But I think my observation about dissipative/non-dissipative processes still holds.

quite right, alexepascual, thanks for the example and instructing me on my misconception

i stand corrected

regards
marlon

 

[Coelum]

V.I. Arnold on Lagrangian and Hamiltonian

Alexepascual,
let me summarise what I've read in Arnold's "Mathematical Methods of Classical Mechanics" (2nd ed. by Springer).
1- If you want a formulation of the motion as a stationary point of a functional, then you should compare Newton's equations of dynamics with the Euler-Lagrange equations: it's not difficult to see that the function to "stationarise" must be in the form L=T-U, since T is function of r' only and U is function of r only (page 59);
2- Now, if you want a set of first-order ODEs instead of 2nd-order ODEs, you have to apply a Legendre transformation, and get H=pq'-L (page 65) to replace the role of L; the case H=T+U is not always true - although very important (page 66).
Hope it helps,

Francesco

 

[alexepascual]

Francesco,
While point 2 of your post I guess is a very important topic, at the moment I am trying to get a deeper understanding of the Lagrangian before exploring the Legendre transformation.
With respect to point one:

it's not difficult to see that the function to "stationarise" must be in the form L=T-U, since T is function of r' only and U is function of r only (page 59);

I have seen statements like that "It's not difficult to see..", "obviously..", etc. which make me wonder what the author was thinking when making these remarks. What is obvious is that it was easy for him to see, but that doesn't mean that it'll be easy for the reader to see. He could have taken a little time to explain, just in case the argument is not so obvious to the reader.
It is not obvious to me why the function to "stationarize" must be in the form L= T-U. I don't have Arnold, but probably he doesn't explain that point any further than what you wrote in your post.
If you understand what his logic is, I would appreciate your clarifying it for me.
Thanks Francesco,
-Alex-

 

[Coelum]

I have seen statements like that "It's not difficult to see..", "obviously..", etc. which make me wonder what the author was thinking when making these remarks. What is obvious is that it was easy for him to see, but that doesn't mean that it'll be easy for the reader to see. He could have taken a little time to explain, just in case the argument is not so obvious to the reader.
It is not obvious to me why the function to "stationarize" must be in the form L= T-U. I don't have Arnold, but probably he doesn't explain that point any further than what you wrote in your post.
If you understand what his logic is, I would appreciate your clarifying it for me.
Thanks Francesco,
-Alex-

Alex,
let me cite Arnold and add a few comments. I'll do my best to render the mathematical notation.

We compare Newton's equations of dynamics

d/dt(mx')+dU/dx=0

with the Euler-Lagrange equation

d/dt(dL/dx')-dL/dx=0

where L is a funtion of both x and x'. Now, the simpler approach is to define L as the sum of two functions, one depending on x' only and the other depending on x only:

L(x,x')=L1(x)+L2(x').

Then the Euler-Lagrange equation becames

d/dt(dL2/dx')-dL1/dx=0

which we can compare to Newton's equation above. The comparison suggest that

dL2/dx'=mx'

and

dL1/dx=-dU/dx.

The integrals are easy (I'll omit the constant of integration that is irrelevant for the variational formulation):

L2=mx'x'/2=T

and

L1=-U,

so

L=L1+L2=T-U.

Hope it helps.

Francesco,
While point 2 of your post I guess is a very important topic, at the moment I am trying to get a deeper understanding of the Lagrangian before exploring the Legendre transformation.
-Alex-

I don't know of an easier way to move from the lagrangian representation to the hamiltonian. It's not difficult. It's explained very well by Arnold (pages 61-65). Unfortunately, Goldstein seems not to cover it. I guess you cand find it in many standard textbooks. However, if you have many doubts like this (like I had and still have) I warmly recommend you to get your copy of Arnold's masterpiece - provided your mathematical background is up to it or you (like me) are willing to upgrade it were required. Be warned: it's not easy but worth every minute you'll spend on it.



Francesco

 

[alexepascual]

Francesco,
Thanks a lot for your explanation. I'll have to go over it. I started school last monday and I'll be also teching two labs. So I'll be kind of busy. I am taking mechanics (we'll use Marion & Thornton and Goldstein) and Matehmatical methods (Arfken and also Boas).
I found in Goldstein (second edition) a short explanation on the Legendre transformation (pg.339). Lanczos talks a lot about it. With respect to my comment, I didn't mean that I was looking for an alternative explanation to the Legendre transformation. I just meant that I want to go one step at a time, and I am lookin for more insights into the Lagrangian before I explore the transition to Hamiltonian mechanics.
I guess your post may be giving me that insight I was looking for, but I'll have to go over it several times and think about it.
My mechanics professor told us yesterday that we are going to be solving a lot of problems, that that's why we are physicists. I dissagree with that and I am a little dissapointed. I am into physics because I want to gain a better understanding of how the universe works and not just to solve problems.
Well, that was really off the subject of this thread but I was feeling a little frustrated and felt like telling someone about it.
I am affraid that now that I'll be studying mechanics, I'll be so busy working out problems that i won't have enough time to explore and gain a deep understanding of the fundamental concepts.
I hope once I get organized with my lab teaching I'll have more time to do some study on my own.
Thanks again Francesco,
-Alex-

 

[pervect]

Turin:
Your observation is very interesting. I didn't think about inflection points. Woundn't this support my point that Hamilton's principle does not represent an attempt by Nature to obtain an economy in a certain quantity?.

Since any function which yields the proper equations of motion can be and is called a Lagrangian, one has a lot of choice when writing down a Lagrangian for a specific system.

I have a suspicion that when actually writing the Lagrangian in the form L = T-V, the principle of least action actually is a minimization principle. But I don't have a proof of this, and my intuition might well be wrong. I would be interested in seeing a counterexample, if anyone has one to offer, though.

Using Newtonian mechanics, for instance, with T=.5 m v^2, I think it's fairly obvious (?) that the intergal of L*dt will be increased by making the velocity arbitrarily high.

 

[alexepascual]

Pervect,
It's been a while since your post. I have been really busy with school (including classical mechanics). It looks like since I started school I stopped learning, as we just do a lot of algebra an there are no discussions on what it all means.
With respect to your idea about T-V being minimized, I think I have read some place that possible paths may be obtained by maximizing the lagrangian, but that that would be kind of an "unstable" path.
I visualize a ball running downhill on a ridge of a mountain. Most of the time, the ball will fall over the side of the ridge. But build the mountain (Disneyland type) and design the ridge with a very particular shape, it might be possible for the ball to stay on the ridge. (in order for this to work the ridge woud habe to be rounded). I realize we have a probelm with constraints here, but I think it roughly illustrates a case. I have not worked the problem out to see if the integral of the Lagrangian really is maximum for the path that goes along the ridge though. If it isn't then my example won't be valid.
I was just trying to think about it and I see that something that complicates the picture is that In Hamilton's principle we are going from one point to another fixed point. In this case, when the ball falls on the side of the ridge, it ends up somewhere else. But I guess you could have gutters on both sides of the ridge running prallel to the ridge and having these gutters merge with the top of the ridge at the end of the path. If the ball fell on one of the gutters from the start, it is clear that that path on the bottom of the gutter would have the lowest hamiltonian action. So it is possible that if the ball follows the ridge it will have maximum action.

 

[Antonio Lao]

I have the gut feeling that there is more to it than meets the eye.

the significance of the Lagrangian is its inherent natural form for a local gauge invariance. The Hamiltonian is not locally gauge invariance leading to problems with infinities which require renormalizations.

 

[alexepascual]

Thanks Antonio,
I'll keep that in mind. I know that gauge invariance is a very important topic, but my exposure to it in undergraduate studies has been very brief. I think I only saw it in E&M and I don't remember much about it except that the vector potential had something to do with it.
I am looking forward to studying gauge invariance in connection with classical mechanics.
Thanks again for your input.
-Alex-

 

[spacetime]

To me it appears to be quite similar to the law of conservation of energy.

You take the differences of T and V all over the path and then sum all the values up. For the actual path, they have to be zero since energy didn't come from anywhere and didn't go anywhere.

spacetime
www.geocities.com/physics_all/index.html

 

[alexepascual]

Spacetime,
Your explanation sounded appealing because in the Lagrangian approach, conservation of energy is actually assumed. But it fails when you realize that the integral of L=T-V does not necessarily vanish but assumes a stationary value with respect to nearby paths. Let's say we have a particle going from point A to point B in straight line and the potential is zero. Then the Lagrangian would contain only the kinetic energy and the integral would clearly not be zero.
What do you think?

 

[spacetime]

Yeah alexepascual

you are right. I am trying to figure it out. But I think it has something to do with energy conservation. Let me see if I can find an answer.

spacetime
www.geocities.com/physics_all/index.html

 

[alexepascual]

While you are at it, let me give you some additional information.
I have seen two derivations of Lagrange equations of motion. One starts from Hamilton's principle, which already uses the concept of the Lagrangian function, and the starts from D'Alembert's principle and in that case the Lagrangian comes out of the equations after some manipulation. The first procedure, (from Hamilton's principle) does not add any intuitive understanding as it presuposes something which we would like to understand better (The Lagrangian function and Hamilton's principle). So this approach in a way says: If we accept Hamilton's principle as true, then all these other things (such as the Lagrange equations of motion and Newton's laws) can be derived.
The other approach (from D'Alemebert's principle) uses the idea of forces of inertia which is intuitive enough. The problem I found is that the derivation gets to a point where it uses an integration by parts. At that point the argument becomes hidden in the algebra (or calculus) and fails to be easily followed on an intuitive basis. That is, you can follow on paper and verify that all the algebraic manipulations are correct, but I at least loose track of what each thing means in physical terms so that in the end I do get the Lagrangian function and Hamilton's principle, but can't explain intuitively how I got there.