# Lagrange versus Euler-Lagrange equation for inverted pendulum

On the wikipedia page of inverted pendulum, in the section "inverted pendulum on a cart" (https://en.wikipedia.org/wiki/Inverted_pendulum#Inverted_pendulum_on_a_cart), the equations of motion will be derived using three different methods. Two of them are from lagrange's equations, that use only kinetic energy and generalized forces, and from Euler-Lagrange equations, that use the lagrangian. In the past I try to found the equations of motion of various dynamic systems, subjected to external forces, and every time I found difficult to include these forces into the system. I found that lagrange equations are simpler to use compared to Euler-Lagrange, and the result is the same. So I would like to ask, why compute the lagrangian? Where is the advantage compared to the use of kinetic energy and generalized forces only? What I'm missing?

## 1 Answer

As you point out: computing the Lagrangian isn't necessary.

The powerful stuff is the following:

• representing the physics taking place in terms of energy
• Using generalized coordinates (when applicable)

You are already doing that. In that sense you are set.

As we know, in physics over time a range of representations has been developed. Transformation from one form to another is a mathematical operation. There is the pair Lagrangian mechanics and Hamiltonian mechanics. The transformation between those is Legendre transformation. Legendre transformation is it's own inverse; applying Legendre transformation twice recovers the original function. That is, the transformation between Lagrangian and Hamiltonian mechanics is bi-directional.

The transformation from force-acceleration representation to energy representation is integration with respect to position.

As we know: when you know the force (entered as generalized force, if applicable), then the expression for the potential energy is obtained by evaluating the integral of the force with respect to position coordinate.

Conversely, to go from energy representation to force-acceleration you take the derivative with respect to position.

So we have that the transformation between force-acceleration representation and energy representation is bi-directional too.

In classical mechanics we can think of the Euler-Lagrange equation as an operator that takes the derivative of the energy with respect to position.

For discussion of why the Euler-Lagrange equation takes the derivative of the energy with respect to position, see my october 2021 discussion of Hamilton's stationary action