# Euler Lagrange of this

1. Dec 14, 2014

### PhyAmateur

If metric is $$ds^2 = -f(x)dt^2 + g(x)dx^2 + 2l(x)dxdt$$
Then we have this Lagrangian:

$$L= \frac{1}{2}(-f(x)\dot{t}^2 + g(x)\dot{x}^2 + 2l(x)\dot{x}\dot{t}).$$

The Euler-Lagrange equation for $$t$$ is:

since $$t$$ is not there in the Lagrangian then $$\partial L/ \partial t=0$$
This implies that $$\frac{d}{d\tau}\frac{\partial L }{\partial \dot{t}}= 0$$

so $$\frac{\partial L }{\partial \dot{t}}$$ is a conserved quantity we call energy and I got it equal to $$-f\dot{t} + l\dot{x}$$ where my professor only got it $$E= -f\dot{t}$$

Am I mistaken somewhere?

2. Dec 14, 2014

### Matterwave

Looking at this another way, we know that since the metric does not depend on $t$ that $\partial_t$ is a Killing field, meaning $E=g_{ab}u^a(\partial_t)^a=-fu^t+lu^x=-f\dot{t}+l\dot{x}$ so it looks like you are correct and your professor is missing the second factor...

3. Dec 14, 2014

Thank you!