# I don't understand something about the Lagrangian / action?

1. Oct 29, 2011

### jeebs

We can think of a particle having kinetic and potential energy, T and V.
The Hamiltonian is the sum of these, H = T + V. This seems like a sensible enough quantity to think about.

However, we can also define the Lagrangian as being the difference between these two quantities, L = T-V. However, I do not quite understand why this is an interesting/relevant/useful quantity, I don't understand what Lagrange's motivation was to start from this point.

We can then think about another quantity called the action, defined as
S = ∫L.dt.
The action supposedly takes different values for the different trajectories that a moving particle could follow. Now apparently, classical mechanics postulates that the actual path the particle takes is the one for which the action is minimized.
Why is this the case?

Looking at that integral, that says to me that classically, L should be as small as possible. Since L = T-V, then that is equivalent to saying that in a classical situation, the particle should have its potential and kinetic energies being as similar as possible. Where does this come from? I don't get that at all, why should one expect that? What is the logic behind imposing this condition?

This has been nagging me for a while because I'm blindly following Lagrangian mechanics to solve problems, but I don't understand its foundations. I know that it is a useful method, eg. you can easily derive Newton's laws starting from the Lagrangian, but I need to know what was going through Lagrange's head when he first dreamt it up.

Last edited: Oct 29, 2011
2. Oct 29, 2011

### arildno

From what I know, that answer has not been found in terms of deeper principles.
It has been proven that minimizing the action yields the proper equations of motion, but what the action "in itself" truly is, is another matter.

Having just a glancing familiarity with Lagrangian mechanics from years back, I might well be wrong on that account, but I can't give you that answer.

3. Oct 29, 2011

### jeebs

hah, oh well, I can only hope you're mistaken.

4. Oct 30, 2011

### homeomorphic

Well, first of all, I believe it was Euler who had the idea, rather than Lagrange. One thing you learn right away about the history of math and physics is that many or most of the names attached to things are wrong, or at least misleading.

I am not sure about the history, but I have my own thoughts on the meaning of the Lagrangian.

First of all, the Lagrangian approach is one based more on the concept of energy, rather than force. You can already see in introductory physics how the concept of energy is useful computationally and conceptually. So, Lagrangian mechanics builds on that.

Secondly, to my mind, in order for the whole idea not to be painfully ugly and non-intuitive (which, it could have been when Lagrange came up with it, but I don't know what he was thinking), you ought to have some idea of the geometric meaning of a configuration space. One nice example is the configuration space of a double-pendulum or robot arm. That's a torus, since there are two angles needed to specify the position of the system. That's one of the few that is small enough to be visualized directly. Another nice example is the configuration space of n point particles. That is just R^(3n) because each particle can move in 3 different directions. So, the configuration space is some kind of possibly higher dimensional space such that each point of the space is associated with some physical configuration of the system. If your system was a radio dial, maybe the configuration space would be a semi-circle or something like that. And each point of the semi-circle corresponds to some position on the dial. And you can try to play the same game with a more complicated system. Maybe it's a vehicle or some other mechanical device. Think of all the ways it can move. For each degree of freedom of the system, you have some coordinate that pins it down, so that if I give you a point in the configuration space, there will be some corresponding configuration of the system. Maybe, there's one coordinate for the each wheel, one for the steering wheel, a few for the suspension, one for the break pedal, etc. Part of the point of Lagrangian mechanics is that it's telling you to think of the system as a whole, rather than each piece separately. All the different components are combined into one big thing.

Now, one thing you can say right away is the configuration space isn't telling you the whole story because it only records positions. To fully specify the state of the system, you also need the velocities. This leads to what we mathematicians (and mathematically sophisticated physicists) would call the "tangent bundle" of the configuration space. In the case of a point particle, this is pretty straight-forward to describe. The configuration space here is just R^3, plus you need to specify a velocity vector, which sits in another R^3. So, altogether you get R^6. A six dimensional space. So the tangent bundle of the configuration space is a space for which each point corresponds to a state of the system. If I give you a point in the tangent bundle, you know exactly what the system is doing. Each position coordinate and the derivative of each position coordinate.

This is a really imaginative concept. Wiggling around in the tangent bundle corresponds to the the system wiggling around. So, if we want our little vehicle to do a little dance for us, we can just point our pointer at the tangent bundle and make it wiggle around. Of course, not just any path through the tangent bundle is going to be obeying Newton's laws.

Now, the amazing thing is that these paths can be determined by a single function on the tangent bundle. That's quite something. But how could we possibly dream up such a diabolical idea? What would lead us to expect that such a thing would be possible?

Well, I've blabbered on for long enough, now, so maybe I can continue this discussion later.

5. Oct 30, 2011

### Matterwave

Lagrangian mechanics works almost by tautology. The "correct" Lagrangian, is the Lagrangian which gives you the correct equations of motion. For classical mechanics, this happens to be T-V, but this is not true always (in fact, for many applications, these energies are not well defined).

The Lagrangian for a free relativistic particle, for example, is -mc^2*sqrt(1-v^2/c^2). This is obviously NOT the kinetic energy of the free particle (which would be mc^2/sqrt(1-v^2/c^2)).

You can see, then, really you "pick" the Lagrangian which gives you the correct equations of motion. It just so happens that for classical mechanics, the Lagrangian is always T-V. I have not seen a very good "derivation" of this fact, unless you consider its reduction to Newtonian Mechanics a "derivation".

In the end, which Lagrangian you pick must be dictated by experiment.

But then again, even Newton's laws suffer from this. They are simply laws dictated by experiment. They are not derived. (In fact, Newton's first 2 laws are arguably just definitions for what an inertial reference frame is and what a force is respectively).

You can pick whatever satisfies you.

6. Oct 30, 2011

### bp_psy

The Lagrangian is a function L(q,q',t). The integral does not say anything about a the value of L should be at any point (q,q',t).It says that a physically realizable path of a system from a state A to a state B will minimize that integral.
The Lagrangian can take any value from T to -U along that path. For example a free particle moving in a straight line will have L=(1/2)mq'^2 at all points on the physical path. The harmonic oscillator will have L take values in between (1/2)mq'^2 and (1/2)kq^2.So making any claim about about what T-U will be from the form of the action is not correct.

7. Oct 31, 2011

### homeomorphic

Picking up where I left off...

Why do we expect that the paths in the tangent bundle that follow Newton's laws (or just some laws) should be determined by a function on the tangent bundle?

Well, for the sake of brevity, let me not dwell on this point too much, since it's not the question being asked. Suffice it to say that the calculus of variations tells us that we can integrate the function over paths to get extremal paths.

So, how do we know which Lagrangian to pick?

I don't think you can make a case for T-U experimentally. I mean, you can, but how the hell do you get T-U in the first place? You can't test it if you haven't come up with it yet.

So, how could we argue for it theoretically?

One way is to use the principle of virtual work. This idea is more natural in statics than in dynamics. If you want to use energy methods in statics, rather than forces, you can consider the work that the forces would do and require that the contributions to the total work have to cancel out.

But, as d'Alambert realized, dynamics isn't all that different from statistics, at least if you only have one point particle or consider just the center of mass. Say you have one particle moving along, not necessarily in a straight line. Maybe it's moving in a potential. You can just have your reference frame move along with it and then you are back to statics. However, it's not an inertial reference frame, so you will get some fictitious forces acting. The fictitious force will just be m a, mass times acceleration.

So you have

$-\frac{\partial V}{\partial q} - m \ddot{q}$

as the total force.

Oops.

I accidentally submitted instead of previewing...

8. Oct 31, 2011

### homeomorphic

So, anyway.

In statics, you just have one point at which you can do this imaginary work, but in dynamics, you have a whole path, along which you can do it. So, you vary the path and see how much virtual work is done at each point.

It's beginning to look a lot like Christmas--I mean, calculus of variations.

So, if we vary the path, q(t), the principle of virtual work tells us

$-\frac{\partial U}{\partial q }-m\ddot{q} \delta q = 0$

So if we integrate that over the correct path, it should still be zero.

$\int dt ( -\frac{\partial U}{\partial q }\delta q -m\ddot{q} \delta q ) = 0$

See if you can figure out the rest. I'm pooping out here because I don't like typing this Latex stuff. You just integrate by parts and you get that -U + T is extremal.

That's the "derivation" of L = T- U.

Last edited: Oct 31, 2011