# Equations of motion from Lagrangian and metric

1. Mar 29, 2012

### Ai52487963

Disregard. I done figured it out.

1. The problem statement, all variables and given/known data
Find equations of motion for the metric:

$$ds^2 = dr^2 + r^2 d\phi^2$$

2. Relevant equations

$$L = g_{ab} \dot{x}^a \dot{x}^b$$

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}^a} \right) = \frac{\partial L}{\partial x^a}$$

3. The attempt at a solution

So I know to just cycle the free index in the lagrangian from r to phi to get the corresponding equations of motion, but my phi solution isn't coming out right. For r, I get

$$L = \dot{r}^2 + r^2 \dot{\phi}^2$$

which gives

$$\ddot{r} - r\dot{\phi}^2 = 0.$$

For phi, I get:

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\phi}} \right) = \frac{\partial L}{\partial \phi}$$

which yields:

$$\frac{d}{dt}\left( 2r^2 \dot{\phi} \right) = 0$$

The solution lists the other equation of motion as:

$$\ddot{\phi} + 2 \frac{\dot{r}}{r}\dot{\phi} = 0$$

I'm having trouble bridging the last bit. If I apply the time differential operator on the dotted phi, I can see where the double dotted phi comes from, but does that mean I have to do some kind of wacky product rule business on the r as well?

edit: got it, nevermind. For those who google this years from now and hate how people always say they got it without showing how they got it, here's the step I was too tired to see:

$$\frac{d}{dt}\left( 2r^2 \dot{\phi} \right) = 0$$

apply product rule

$$2r^2 \ddot{\phi} + 4 r \dot{r} \dot{\phi} = 0$$

divide by $$2r^2$$

gives

$$\ddot{\phi} + 2 \frac{\dot{r}}{r} \dot{\phi} = 0$$

Just goes to show if you stare at it long enough it can come to you. Plus making lengthy tex-based posts on PP can help flesh out whatever misunderstandings you may have!

Last edited: Mar 29, 2012