Timeline for Metric variation of Laplace-Beltrami
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 27, 2021 at 14:23 | comment | added | GenericPhysicsStudent | I see, so what I am gathering from your response is that: a) I should always be using $\square=g^{ab} \nabla_a \nabla_b$ and b) should be aware of non-trivial contributions from Christoffels despite considering a scalar field. | |
| Sep 27, 2021 at 14:21 | comment | added | GenericPhysicsStudent | I am calculating the metric variation of the laplace-beltrami operator to get the energy-momentum tensor in flat space, i.e. I vary w.r.t some general metric and evaluate it on flat Minkowski space. Since I am considering the EM tensor of a scalar field, I always assumed to simply put $\square\phi = g^{ab} \nabla_a \nabla_b= g^{ab} \partial_a \partial_b \phi$. Since I have to first take the metric variation, you are probably right that I should also consider the contributions from the covariant derivative (i.e. christoffels). | |
| Sep 27, 2021 at 14:14 | comment | added | Michael Seifert | @GenericPhysicsStudent: Are you using $\partial$ for the covariant derivative rather than the more standard $\nabla$? The covariant derivative of the (inverse) metric vanishes by definition, but it's definitely not the case that the coordinate derivatives of the (inverse) metric vanish. That doesn't even happen in flat space (e.g., the metric for Euclidean space in polar coordinates depends on $r$.) | |
| Sep 27, 2021 at 14:12 | comment | added | GenericPhysicsStudent | Also, $\partial_b g^{bd}$ vanishes in metric theories of gravity if I am not mistaken. So, where does the minus sign then come in ? | |
| Sep 27, 2021 at 13:58 | comment | added | GenericPhysicsStudent | That makes sense to me, but why is $\square = g^{ab} \partial_a \partial_b$ the correct definition ? | |
| Sep 27, 2021 at 13:48 | history | answered | Javier | CC BY-SA 4.0 |