# Principle of Least Action & Euler-Lagrange Equations

1. Jan 30, 2007

### ObsessiveMathsFreak

I'll just throw down some definitions and then ask my question on this one.

In a conservative system, the Lagrangian, in generalised coordinates, is defined as the kinetic energy minus the potential energy.

$$L=L(q_i,\dot{q}_i,t) = K(q_i,\dot{q}_i,t) - P(q_i,t).$$
All $$q_i$$ here being functions of t.

It satisfies the Euler-Lagrange equations in all its generalised coordinates.

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right) - \frac{\partial L}{\partial q_i} = 0 \ \ \ forall i=1,2,\ldots,n$$

The "Action" of a path betwwen two points is the integral of the Lagrangian along that path.
$$A = \int_{t_1}^{t_2} L(q_i,\dot{q}_i,t) dt$$

The Principle of least action states that the path actually taken is the path with with least Action.(The path that minimises the integral). For the path to have least action, the Euler-Lagrange equations are a neccessary condition .

Now here is my question. Are the Euler-Lagrange equations a sufficient condition for the path to have least action? It seems so to me, but can anyone confirm this?

2. Jan 30, 2007

### Roman

hi,

I would say yes, because:
the Euler-Lagrange equations can be derived from the "action" integral.
The solution of Lagrange equations minimises the integral

3. Jan 30, 2007

### cesiumfrog

I think the principle actually says that the physical system will take a path of stationary action (ie. it could be a maximum, or just a local minimum, etc), and that this is equivalent to being a solution of the Euler-Lagrange equations. Hence, no, I don't think the equations are sufficient if you want the least action (especially not the global minimum).

4. Jan 30, 2007

### Crosson

Expanding on what cesiumfrog said, a solution to the Euler-lagrange equations has maximal action, minimal action, or something in between corresponding to an "inflection point".

We can use the Euler-Lagrange equations to find the shortest path between two points A and B on the surface of a sphere. This path is the segment of the "great circle" that passes through A and B. This segment has minimal distance, but the complement of this segment on the great circle has maximal distance, but it also satisfies the euler-lagrange equations.

5. Jan 31, 2007

### ObsessiveMathsFreak

OK, but if we ammend to the Euler-Lagrange equations the condition that is I believe $$\frac{\partial^2 L}{\partial \dot{q}_i^2}>0$$ as a neccessary condition for a path of least action.

Is this condition, along with Euler-Lagrange, sufficient for the path to be of least action?

6. Feb 3, 2007

### coalquay404

The principle of least action is actually a misnomer, at least in so far as it is related to physics. What you're actually looking for are those paths in the configuration space for which the value of the action is stationary, i.e., those paths along which the action takes either its minimum, maximum, or shoulder-point value.

The Euler-Lagrange equations are certainly a necessary condition for the action to have a least value, but they are not a sufficient condition. For the action to have a minimum value along a path, one needs the path to satisfy not only the Euler-Lagrange equations, but also the second-variational inequality.