1

When obtaining the Euler Lagrange equations for a scalar field with higher order derivatives in curved space is it the same to use $$ -\partial_\nu\partial_\mu\frac{\partial \sqrt{-g} \mathcal{L}}{\partial\partial_\nu\partial_\mu\phi}+\partial_\mu\frac{\partial \sqrt{-g} \mathcal{L}}{\partial\partial_\mu\phi}-\frac{\partial \sqrt{-g} \mathcal{L}}{\partial\phi}=0 $$ or $$ -\nabla\nu\nabla\mu\frac{\partial \mathcal{L}}{\partial\nabla\nu\nabla\mu\phi}+\nabla\mu\frac{\partial \mathcal{L}}{\partial\nabla\mu\phi}-\frac{\partial \mathcal{L}}{\partial\phi}=0~~? $$

For a canonical kinetic term it's easy to see that it's true.

Qmechanic
  • 201,751
marRrR
  • 143
  • No, that is clear. Here there are terms like ∇ν∇μϕ which are clearly different than ∂ν∂μϕ. My question is if both ways of obtaining the equations of motion are equivalent. If we don't have higher order derivatives is trivial. I'm wondering about the case with higher order derivatives. – marRrR Apr 13 '16 at 20:03
  • Possible duplicates: http://physics.stackexchange.com/q/239974/2451 and links therein. – Qmechanic Apr 13 '16 at 23:19
  • Since I don't have enough points I can't comment in the answer in the other question but I wonder when is it possible to replace the total space-time derivatives with covariant ones? Also, using your equation doesn't give the klein-gordon equation, there's an extra term that appears from the third term on your total spacetime derivative – marRrR Apr 13 '16 at 23:50

0 Answers0