Euler Lagrange of this

  • Thread starter PhyAmateur
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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
Matterwave
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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
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Thank you!
 

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