In order to derive the equation of motion for a general action of the form

$$ A = \int L \ dv$$

where the lagrangian function $L = L ( \ T^{a ... b}_{\quad c...d} \ , \nabla_{\nu} T^{a ... b}_{\quad c...d})$ depends only upon a tensor field and its covariant derivative.

I tried to derive the Euler Lagrange equations of motion but I'm stuck with the variation of the covariant derivative.

In S.Hawking-Ellis The Large Scale Structure of Spacetime, pag. 65, it is stated that $\delta \ \nabla \ T = \nabla \ \delta \ T $ for a given variation of the tensor fields $T \longrightarrow T + \delta T$

Now for simplicity consider the case of a single vector fields. First of all how do I define $\delta T$? And then how to prove that the variation commutes with the covariant differentiation?


It appears to be the case that you are considering the motion of fields in a given curved spacetime without considering the backreaction on the spacetime itself, i.e., you are varying the matter fields over the background of a fixed metric. Since the covariant derivative depends on the metric and not on the matter fields, the variation of the matter field simply doesn't change the covariant derivative.

If you were to indeed change the metric, you would need to consider the variation of the covariant derivative. To see this directly, notice that the covariant derivative of the metric must always vanish (for covariant derivatives defined using a metric compatible connection which is what we use in GR). Thus, if the metric changes then the covariant derivative must also change to make sure that the new metric is still covariantly constant. Another way to look at it is to notice that the covariant derivative in GR depends on the Christoffel symbols and the Christoffel symbols would be affected by the variation of the metric. For more explicit discussion on the variation of the covariant derivative when the metric is varied, see this answer.

  • $\begingroup$ You may want to edit that link $\endgroup$ – Nihar Karve Nov 26 '20 at 3:27
  • $\begingroup$ @NiharKarve Indeed, thanks! :) $\endgroup$ – Dvij D.C. Nov 26 '20 at 3:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.