Does a covariant version of Euler-Lagrange exist?

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  • Thread starter JuanC97
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  • #1
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Hello everyone.

I've seen the usual Euler-Lagrange equation for lagrangians that depend on a vector field and its first derivatives. In curved space the equation looks the same, you just replace the lagrangian density for {-g}½ times the lagrangian density. I noticed that you can replace partial derivatives for covariant ones and easily arrive to the same result (just take into account the formula for contraction of indices in the christoffel symbols and you get this result in no more than 2 or 3 lines).

The last comment ensures there's a covariant version of Euler-Lagrange for lagrangians with dependences on first order derivatives but I may ask... what happen with dependences on second derivatives in the lagrangian? - I tried to arrive at a similar result (just changing partial derivatives for covariants) but I didn't find it and I need it for my thesis.

Guys, to be concise: Have you seen the covariant version of this equation? How can I find it?
(It is supposed to look like this https://pasteboard.co/H8AyvWS.png)
 
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Answers and Replies

  • #2
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Is it possible to express the last term (partial derivative of the lagrangian wrt a double covariant derivative) as:
(partial derivative of the lagrangian wrt a double partial derivative of the field) + (other terms - maybe involving Christoffel symbols)?

Some thing like chain rule or similar?
 
  • #3
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I've arrived to this conclusion:
https://pasteboard.co/HaR8xFf.png

but I was expecting just the underlined terms to be equal to zero, not with the Christoffel term.
I don't see how would the last term be identically zero (if possible) but what do you think guys?
 

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