# Non-commutativity of the d'alambert operator acting on the covariant derivative of a scalar field in general relativity

+ 5 like - 0 dislike
720 views

Recently, I saw the following formula for the non-commutativity of the d'Alembert operator $\Box$ acting on the covariant derivative of a scalar field in general relativity, $\Box (\nabla_{\mu}\phi)-\nabla_{\mu}\Box\phi=R_{\mu\nu}\nabla^{\nu}\phi$. How exactly it is derived, considering the metric compatibility and that $\phi$ is a scalar function depending on time?

This post imported from StackExchange MathOverflow at 2017-07-11 20:37 (UTC), posted by SE-user Mikey Mike

recategorized Jul 11, 2017

This looks as if it would be coming from acting with the Hodge Laplacian on $\nabla_{\mu}\phi$ ...

+ 5 like - 0 dislike

Use $$[\nabla_\mu,\nabla_\nu]V^\rho=R_{\mu\nu}{}^{\rho\sigma} V_\sigma$$ to conclude that \begin{aligned} {}[\nabla^\nu\nabla_\nu,\nabla_\mu]\phi&\overset{ \mathrm A}=\nabla^\nu[\nabla_\nu,\nabla_\mu]\phi+[\nabla^\nu,\nabla_\mu]\nabla_\nu\phi\\ &\overset{ \mathrm B}=0+R_{\mu\nu}{}^{\nu\sigma} \nabla_\sigma\phi\\ &\overset{ \mathrm C}=R_{\mu\nu}\nabla^\nu\phi \end{aligned}

Note that in $\mathrm A$ we have used the fact that covariant derivatives commute with contractions, in $\mathrm B$ we have used $[\nabla_\nu,\nabla_\mu]\phi=[\partial_\nu,\partial_\mu]\phi=0$ (assuming a torsion-free connection), and in $\mathrm C$ we have used the definition of $R_{\mu\nu}$ as the contraction of the Riemann tensor.

This post imported from StackExchange MathOverflow at 2017-07-11 20:37 (UTC), posted by SE-user AccidentalFourierTransform
answered Mar 14, 2017 by (480 points)
Thanks, interesting approach from quantum mechanics as operators. I have seen that it works also if we take into account that $\nabla^{\mu}\nabla_{\nu}\phi=\nabla_{\nu}\nabla^{\mu}\phi$ for a scalar field in the first term, the term $\nabla_{\mu}\nabla_{\nu}\nabla^{\mu}\phi$ becomes $\nabla_{\mu}\nabla^{\mu}\nabla_{\nu}\phi$.

This post imported from StackExchange MathOverflow at 2017-07-11 20:37 (UTC), posted by SE-user Mikey Mike
Hi @MikeyMike, Im not sure what you mean by quantum mechanics. All I used is standard differential geometry, with no reference to QM. The use of commutators, $[\nabla_\mu,\nabla_\nu]=\nabla_\mu\nabla_\nu-\nabla_\nu,\nabla_\mu$ is very common when discussing the Riemann tensor. It is ordinary, old-school, differential geometry, not quantum mechanics.

This post imported from StackExchange MathOverflow at 2017-07-11 20:37 (UTC), posted by SE-user AccidentalFourierTransform

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.