# Harmonic Oscillator #2 - Lagrangian Formulation

A few days ago we looked at what a Lagrangian actually is. The short of it is that it's the kinetic energy minus the potential energy of a given mass*. More importantly, if you construct the classical action by integrating the Lagrangian over the time (see the previous link for a more full explanation) you'll find that the actual trajectory is the one that minimizes the action.

It turns out that the way to find the path that minimizes the action is pretty easy. In honor of two of its pioneers, we call it the Euler-Lagrange equation. If you can find a solution that satisfies the Euler-Lagrange equation for your problem, you have determined the trajectory of your system. The equation is this: Here x is position and x-dot is velocity. We're working with a 1-d harmonic oscillator, so that one coordinate is enough to describe the system. In more dimensions you just write this equation down more times, with (say) y and z replacing x. The lagrangian for our system is: Do we all know enough calculus to plug that into the Euler-Lagrange equation? If you're not familiar with partial derivatives (the script d), that means to differentiate with respect to that variable, treating all the other variables as constant. For instance, the partial derivative of the Lagrangian with respect to t would be 0 because t doesn't explicitly appear. The d/dt derivative (which is what's in the E-L equation) is not zero, because x and x-dot are functions of t. Anyway, plugging in, we get: Which is exactly the equation we got when we did the harmonic oscillator using force-based methods rather than this potential energy based method.

So why all that extra trouble? In this case, just to see how things work. But in other cases, this is our only option. There are plenty of times (for instance, a roller coaster on tracks) where the forces involved (such as the tracks on the car) are so blisteringly complicated as to be practically impossible to solve. However, it's easy to write the relation between the track height and the potential energy, and the Lagrangian formulation can automatically give us a much simpler differential equation to solve. We can do this because the Euler-Lagrange equations don't even care if we use x, y, and z as our coordinates. We can use radial coordinates, spherical coordinates, hyperbolic coordinates, or any ridiculous purpose-built coordinates for our problem. We don't even have to modify the form of the equation. And that makes our lives ridiculously simpler. Which I'm all for.

*This is true in classical mechanics. It's not true in relativistic mechanics even if you use relativistic kinetic energy. That's more complicated and we'll worry about it later.

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Great post. But there's a small typo on the third equation. It should be mx(double dot) ...

This might be confusing to someone who is learning about Lagrangians from this post.

By Andy Kuziemko (not verified) on 02 Dec 2009 #permalink

The Lagrangian description has another very important advantage: it translates well to a quantum-mechanical setting. In fact, the principle of least action appears quite beautifully as a classical limit of Feynman path integrals.

Lagrangian or Hamiltonian descriptions also often make plain conserved quantities, which can be a big advantage in both theoretical and numerical analyses. On the other hand, when constraints can't be described in terms of conserved quantities (e.g. a sphere rolling on a plane), they kind of fall down.

Is there a good explanation for why the "principle of least action" so often actually comes out as the *least* action, rather than just any old stationary point, or the most action? (I suppose one also has to make the constraint that you can't get your extremum by running up against some sort of boundary.)

It's not true in relativistic mechanics even if you use relativistic kinetic energy. Oh boy! Every failure of Newtonian physics is a gem of excitement. All the fun is in the footnotes.

Are you familiar with the common undergraduate "skier on a snowball" mechanics problem? If not, it's described here: http://www.physicsforums.com/archive/index.php/t-80176.html

I'm not much of a physicist, so I don't have a lot of experience in mechanics past the intro undergrad level, but this problem has always been intriguing to me. Specifically, I understand how to solve this problem, but what if you were to change the shape of the ground to a parabola, or some other shape? In general, how can you determine when a moving rigid body loses contact with the curved ground it's traveling on?

I've never read anything about the Lagrangian and the action before, but there something about this formulation that makes me wonder if it simplifies solving these snowball-type problems. Is that true at all?

By Alan B (not verified) on 03 Dec 2009 #permalink

@Alan B: In undergraduate physics classes you first learn to work out things in terms of forces, then you learn to use conservation laws (energy, in particular) to solve problems where the forces are hopelessly messy. Lagrangian methods generally take this another step further by letting you find other conserved quantities and make use of them. But for the skier on a bumpy track problem, conservation of energy is enough: first assume the skier stays on the track at all times. Then conservation of energy gets you her speed at any position, and the track itself gets you the direction. In principle, you then extract the vertical acceleration. If this vertical acceleration is upward, or downward by an amount less than g, the skier stays on the track; if it's larger than g the skier would be off the track. So she comes off at the first place the vertical acceleration needed to follow the track reaches g downward.

The actual details are a bit messy, since conservation of energy gets you the velocity as a function of position, not of time, and to get the acceleration you need to take the time derivative of the velocity. But some clever calculus can save you, since dv/dt = dv/dx dx/dt and dx/dt is just the horizontal component of the velocity. With sufficient effort I'm pretty sure the answer could be phrased in terms of the height and curvature of the track.

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