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Euler Lagrange of this

  1. Dec 14, 2014 #1
    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. jcsd
  3. Dec 14, 2014 #2


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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...
  4. Dec 14, 2014 #3
    Thank you!
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