# Euler-Lagrange eqns confusion

1. May 23, 2010

### wglmb

I have a Lagrangian $$L = \frac{R^2}{z^2} ( -\dot{t}^2 +\dot{x}^2 +\dot{y}^2 +\dot{z}^2)$$ and I want to find the Euler-Lagrange equations $$\frac{\partial L}{\partial q} = \frac{d}{ds} \frac{\partial L}{\partial \dot{q}}$$
I'm fine with the LHS and the partial differentiation on the RHS, but when it comes to the $$\frac{d}{ds}$$ I'm not sure which coordinates I'm supposed to consider as a function of s.

Is it all of them (ie t, x, y, z, and their derivatives)
Or is it only the one I'm doing the equation for (so for the z-equation that's just z and its derivative)?

2. May 23, 2010

### George Jones

Staff Emeritus
Consider the RHS of the z equation in two steps. First step: What is

$$\frac{\partial L}{\partial \dot{z}}?$$

3. May 23, 2010

### wglmb

$$\frac{2R^2}{z^2}\dot{z}$$

4. May 23, 2010

### George Jones

Staff Emeritus
So

$$\frac{d}{ds} \frac{\partial L}{\partial \dot{z}} = \frac{d}{ds} \left( \frac{2R^2}{z^2}\dot{z} \right) = ?$$

5. May 23, 2010

### wglmb

Well this is it - I don't know what should be considered a function of s.

If it's just z-dot then $$\frac{2R^2}{z^2}\ddot{z}$$

If it's z-dot & z then $$\frac{2R^2}{z^2}\ddot{z} - \frac{4R^2}{z^3}\dot{z}^2$$

6. May 23, 2010

### George Jones

Staff Emeritus
If $z$ weren't a function of $s$, then $\dot{z}$ would always be zero.

7. May 23, 2010

### wglmb

oops, haha good point. Thanks.