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## Main Question or Discussion Point

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.

[tex]L=L(q_i,\dot{q}_i,t) = K(q_i,\dot{q}_i,t) - P(q_i,t).[/tex]

All [tex]q_i[/tex] here being functions of t.

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

[tex]\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[/tex]

The "Action" of a path betwwen two points is the integral of the Lagrangian along that path.

[tex]A = \int_{t_1}^{t_2} L(q_i,\dot{q}_i,t) dt[/tex]

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

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

[tex]L=L(q_i,\dot{q}_i,t) = K(q_i,\dot{q}_i,t) - P(q_i,t).[/tex]

All [tex]q_i[/tex] here being functions of t.

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

[tex]\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[/tex]

The "Action" of a path betwwen two points is the integral of the Lagrangian along that path.

[tex]A = \int_{t_1}^{t_2} L(q_i,\dot{q}_i,t) dt[/tex]

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?