## Minimizing Quantity Part 5: Principle of Stationary Action

Last post, I hinted that finding the path that minimizes quantities is very fundamental to physics. You can formulate an alternative to Newton’s laws using it. With Newton’s second law, you use F=ma in order to build a differential equation that models an object’s motion. In something called Lagrangian mechanics, you do it by using the Euler Lagrange equation on a functional called the Lagrangian. The Lagrangian is the kinetic energy minus the potential energy. In a Lagrangian, kinetic energy is T and the potential is V, so L=T-V. By using the E-L equation on the Lagrangian, you find the path in which a quantity S called action is stationary. It looks like this:

1) $S=\int^{t2}_{t1} L(t, q, \dot{q}) dt$

2) $0= \frac{\partial L}{\partial q} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}})$

This is called the principle of stationary action, and it is another way to formulate the laws of mechanics. And the above statement is equivalent to Newton’s second law. First of all, kinetic energy is:

3) $T=\frac{1}{2} m \dot{q}^2$

Since kinetic energy is represented by the velocity, the partial derivative of it over position is zero, so:

4) $\frac{\partial L}{\partial q}=\frac{dV}{dq}$

Since the potential of the force is defined as the integral of the force over position, and energy is conserved, the negative of its reverse is true:

5) $-\frac{dV}{dq}=F$

As for the right component of the right and side of equation 2, since the potential does not depend on the velocity, that is zero. So:

6) $\frac{d}{dt} (\frac{\partial}{\partial \dot{q}} \frac{1}{2}m \dot{q}^2) = \frac{d}{dt} m \dot{q}$

7) $\frac{d}{dt} \dot{q} = m \ddot{q}$

Inserting 7 and 5 into 2 gets us Newton’s second law as expected:

8) $F=m \ddot{q}$

What an amazing result! This is telling us that every object moves in such a way that it minimizes the quantity called action. At the same time, it does seems suspicious, doesn’t it? What is this Lagrangian and action and how did anyone think of doing things this way? And we have Newton’s second law, why would we ever want to use this Lagrangian mechanics? It is, after all, going to get us the same results.

Well, all of these scientific principles involving variational calculus were done by people like Maupertuis, D’Alember, Lagrange, Euler, Fermat, Hamilton, etc who believed that nature did things in ways that was the most efficient. The fact that Fermat’s principle worked to explain the path of light stoked their ambition. The Lagrangian itself comes out of something called D’Alembert principle. D’Alembert principle basically puts all forces in a standing still reference frame. Objectss that are being accelerated are to be made seem standing still by adding a pretend force that opposes such acceleration, creating a non inertial frame of reference. Here is a material I found which explains more about the Lagrangian and how it comes out of D’Alembert.

As for why we would ever use Lagrangian mechanics, well, it allows us to use other kinds of coordinate systems instead of just the cartesian one. Some problems are easier in alternate coordinate systems. You also don’t have to deal with a bunch of equations built from the various forces. As an example, I will show you how we can do the pendulum problem using this method, and we shall do it in polar coordinates, which consists of angle and radius from the center of the coordinate. I will label the radius l and the angle theta. First of all, here is the setup, with height being zero at the pendulum’s lowest point:

Now, instead of drawing the forces, we have to determine the kinetic and potential energy:

1) $T=\frac{1}{2}mv^2$

In order to transform it for polar coordinates, you have to remember that in a rotating motion:

2) $v=\dot{\theta} l$

Plugging equation 2 back in the first one:

3) $T=\frac{1}{2}ml^2 \dot{\theta}^2$

Potential energy is much simpler:

4) $V=mgh=mgl(1-cos(\theta))$

So the Lagrangian is:

5) $L=\frac{1}{2} ml^2 \dot{\theta}^2 - mgl(1-cos(\theta))$

Use the following Euler Lagrange equation:

6) $\frac{\partial L}{\partial \theta} - \frac{d}{dt}( \frac{\partial L}{\partial \dot{\theta}})=0$

Doing the partial derivatives gets us:

7) $-mgl sin(\theta) - ml^2 \ddot{\theta} = 0$

Dividing both sides by -ml^2 gets us:

8) $\ddot{\theta} + \frac{g}{l} sin(\theta) = 0$

That is indeed the equation of motion for the pendulum. Further simplification is possible by having the angle close to zero. After all, I don’t feel like dealing with the elliptical integral of  the first kind, which I don’t even know how to use. The linear expansion of sin(x) is x, so:

9)  $\ddot{\theta} + \frac{g}{l} \theta = 0$

See? It gets us the same results, but instead by using concepts of energy and variational calculus.

The five parts posts on minimizing, or as you have eventually learned, finding the stationary points, is a taste of what variational calculus can bring into physics. As you can tell, a lot of problems involving extremizing behavior that are hard to grasp becomes possible with these new set of tools, and there is the added bonus of having another framework of analyzing classical mechanics problem via the Lagrangian. Now, I know that for many of you, these series of posts are pretty useless for your personal life. For most of you it is probably true, and these posts were mostly done in order to learn about variational calculus physics myself, and crystallize what I have learned. Despite all that, hopefully you got something out of this, whether your interest in physics has been piked, or your mind got blown away by the principle of stationary action (it did with mine!) or you learned new ways to do certain physics problems.