Confused about Euler-Lagrange Equations and Partial Differentiation

[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)?

 

[George Jones]

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

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

 

[wglmb]

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

 

[George Jones]

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

So

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

 

[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

 

[George Jones]

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

 

[wglmb]

oops, haha good point. Thanks.