Why Euler-Lagrange equation does not depend on the second derivative of the function?

Question

Why Euler-Lagrange equation does not depend on the second derivative of the function? I.e. why it's $L[q, \frac{dq}{dt}]$ but not $L[q, \frac{dq}{dt}, \frac{d^2q}{dt^2}]$, neither not $L[q, \frac{dq}{dt}, \frac{d^2q}{dt^2}, \frac{d^3q}{dt^3}]$?

Answer 1

The reason that the Euler-Lagrange equation is usually cited for Lagrangians depending only on the first order derivatives is due to convenience in the original setting - in classical mechanics, the state of a system depends only on the current configuration and the instantaneous rate of change - this is enough to predict its future motion. Thus it's convenient to formulate the mathematics in this restricted setting. As described in the wikipedia page that Scott linked in the comments, there's no mathematical difficulty in generalising to functionals involving higher-order derivatives. You just end up with a more complicated Euler-Lagrange equation: a generic $n$th order Lagrangian gives you a $2n$th order Euler-Langrange equation.

One example of a higher-order Lagrangian in physics is the Einstein-Hilbert action for general relativity, which is defined in terms of the second derivatives of the metric - however in this very special case the third and fourth order terms in the Euler-Langrange equation cancel out, leaving you with the second-order Einstein equation.