Deriving the Euler-Lagrange Equation From the Principle of Least Action
By Jon Willcutt
Classical Mechanics assumes that nature is lazy- it doesn’t like to convert energy from potential to kinetic or vice versa. Lagrange quantified that with the Principle of Least Action.
Suppose that at a certain time, we know where a particle is. We know the potential well that in which it sits, but we get distracted. After a specific time later, we find the particle at another location. What path could it have possibly taken when we were distracted?
This is where the Principle of Least Action comes in- we assume it took the one that minimizes the amount of energy transferred.
In classical mechanics, the Lagrangian L is given as the difference between the kinetic energy T and the potential energy U, or that L = T - U. If we refer to the position as q and thus the velocity as q’, then clearly the Lagrangian is just a function of q, q’, and t. Saying that we are minimizing the Action S over the given time interval is equivalent to minimizing the following integral:
Minimizing this integral seems impossible at first- what does it even mean? For most Lagrangians, no matter how optimal of path we take, even the most optimal path, will have a lower-bound value that we cannot improve upon. Perhaps it is more fruitful to consider how much the action changes if we deviate off of the optimal path. We don’t yet know what this path is, but we will refer to it as Q. So we can say that any path q the particle could have taken is the optimal path plus a little nudge off of it, or:
We can substitute this into the equation for the Action and expand the Lagrangian via its Taylor series as follows:
Although the first term is clearly just the minimal action, we don’t know many things about the nudge off of the optimal path. Luckily, we do know exactly where we stated and ended, meaning the nudge is zero at the end points. With this in mind, we can simplify the action integral by integrating out little nudge by the following integrating by-parts and substitution:
By the fundamental theorem of calculus, we can say that the last term is just the integrand evaluated at the end points, but we know this is zero since our nudge is zero at both points:
We can also assume that we’re close to the best path and thus our nudge is “small”. This means that we are saying that the integral only matters to first order. So for the difference between our action and the optimal action to be zero, this clearly means that the integral must evaluate to zero. But we’re assuming that the nudge off the path still might not be zero since it’s arbitrary. Therefore, the integrand must be zero for all t. This final insight gives us the Euler-Lagrange equation in its most common form:
