# What is an effective potential in classical mechanics?

What is an effective potential in classical mechanics? I have read the wikipedia article and David Tong's lectures notes, but I didn't understand how an effective potential simplifies a situation or calculation, and why the ordinary potential won't suffice.

• This is very context dependent as there are a lot of times when an effective potential is useful. Usually one introduces an effective potential when a physical effect whose origin is not a potential can be described adequately as if it simply gave rise to a potential (at least in some variables in some regime of parameter space). This description is usually simpler because potentials are easy to work with. But it would really help to have more context. Commented Jul 8, 2015 at 15:24
• @Andrew :I am talking about effective potential when describing orbits in classical mechanics.
– Paul
Commented Jul 8, 2015 at 15:34

## 1 Answer

It isn't necessary to introduce the effective potential in orbital mechanics but it is really useful.

Let's say we have a particle moving in a central gravitational potential. Newton's laws give you a vector equation of motion $$m \ddot{\vec{x}} = - \nabla U$$ where $U = - G M m /r$. In a general coordinate system this is a complicated set of three coupled differential equations.

We want to simplify and decouple these equations as much as possible. So we work in spherical coordinates. I'll leave the derivation of the angular parts of the equation of motion to the textbook, let's focus on the radial part. The left hand side of Newton's law becomes

$$m \ddot{\vec{x}} \cdot \hat{e}_r = m \frac{d^2}{dt^2}\left(r \hat{e}_r\right)\cdot\hat{e}_r = m\ddot{r} - m \dot{\theta}^2 r = m \ddot{r} - \frac{L^2}{m r^3}$$ In the last line, I've used the fact that we know that the angular momentum $L = m \dot{\theta} r^2$ is conserved. Similarly, $$-\nabla U \cdot \hat{e}_r =- \frac{G m M}{r^2}$$ So, putting it together, $$m \ddot{r} - \frac{L^2}{m r^3} = -\frac{G m M}{r^2}$$

So now it's a matter of interpretation--we can now think of $r$ as the coordinate of a particle living in one dimension. We have effectively add a term to newton's law for $r$ that has no derivatives on it. Why not call that a potential? In other words, why not rearrange the above equation so that it looks more like a simple 1D mechanics problem $$m \ddot{r} = - \frac{G m M}{r^2}+ \frac{L^2}{m r^3} = - \frac{d}{dr}\left( - \frac{GmM}{r} + \frac{L^2}{2 m r^2} \right)$$ This is an extremely useful picture, because we all know how a particle moves in a potential! Given an angular momentum, you can plot the potential and immediately see where the stable circular orbits are (the minima of the potential). You can also qualitatively see how there is a barrier to approaching the object too closely (which makes sense--if you have angular momentum you wouldn't expect a head on collision).

This is pretty non-trivial: you have computed a two dimensional problem (finding a circular orbit, or even oscillations around that orbit). In other words, the problem was much simpler than it originally appeared (you didn't have to solve three arbitrary coupled differential equations, just one simple one with a potential), and we take advantage of this by using an effective potential. This kind of trick shows up all over the place in physics.

• Can you give me some more examples besides celestial mechanics where the effective potential concept is useful? I have just learned this technique myself and used it to study oscillations of a planet in a stable orbit, and I found your answer to be very helpful. Commented Mar 7, 2017 at 16:17