# How are the Euler-Lagrange Equations any more coordinate-invariant than Newton's?

In my experience it is often said that the Lagrangian formulation of mechanics can be much much more convenient because the form of the (E-L) equations remains the same whatever coordinates we choose, allowing us to pick as convenience dictates. Often this is exemplified by an example featuring some system constrained to a circle or a sphere, or something.

It is true that the Second Law $$\mathbf{F} = m\ddot{\mathbf{x}}$$ requires us to express the system in special "inertial" coordinates, but this is fixed easily by introducing a connection and writing $$\mathbf{F} = \nabla_{\dot{\mathbf{x}}}\dot{\mathbf{x}}$$. Of course, introducing a connection is conceptually identical with declaring certain curves to be geodesics which is conceptually identical with the original speak about "inertial frames." However, it is coordinate-invariant, and as far as I'm concerned coordinate-invariant is coordinate-invariant is coordinate-invariant.

• The second law doesn't require you to write the equations in intertial coordinate frames. It is a covariant tensor expression and as such is equally valid in any coordinate frame. What Galilean invariance does in Newton's equations is limit which vectors can be forces, much like it limits which Lagrangians are physical and which are unphysical. They are not more coordinate-invariant, they are equivalent. – S V Nov 15 at 20:03
• I am not sure how Galilean Invariance constrains the possible vector fields $\mathbf{F}$? Not to go too high-tech for what may be a simple point, but $\mathbf{F}$'s being a particular section of the Tangent Bundle is a statement made without coordinates. – DPatt Nov 15 at 20:05
• That is explained in Arnold's Mathematical Methods of Classical Mechanics (I think it was chapter 1). – S V Nov 15 at 20:07

Well, both work in any coordinate system, but the form of Newton's law can and does change, while the form of the Euler-Lagrange equations do not change. For a very basic example, consider Newton's second law vs the Euler-Lagrange equations in polar coordinates. Newton's second law changes significantly, and becomes \begin{align} F_r&=m\left(\ddot{r}-r\dot{\phi}\right)\\ F_\phi &= m\left(r\ddot{\phi}+2\dot{r}\dot{\phi}\right) \end{align} This is far different then the simple Cartesian form: \begin{align} F_x &= m\ddot{x}\\ F_y &= m\ddot{y} \end{align} Meanwhile, the Euler-Lagrange equations stay precisely the same in form: \begin{align} \frac{\partial L}{\partial r}&=\frac{d}{dt}\frac{\partial L}{\partial \dot{r}}\\ \frac{\partial L}{\partial \phi}&=\frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}} \end{align} compared to \begin{align} \frac{\partial L}{\partial x}&=\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}\\ \frac{\partial L}{\partial y}&=\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} \end{align} So, you'll get the same results no matter what (both are equivalent), but the Euler-Lagrange equations are a whole lot easier to use then Newton's law in certain coordinate systems.
• Hmmm ... the thing is that in the variational forumation you've hidden any completity (or simplicity) assocaited with your choice of generalized coordinates in the symbol $L$ but you exhibit it explicitly in the Newtonina formulation. – dmckee Nov 15 at 23:57