Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

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

+ 5 like - 0 dislike
1453 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

asked Mar 14, 2017 in Mathematics by Mikey Mike (25 points) [ revision history ]
recategorized Jul 11, 2017 by Dilaton

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

1 Answer

+ 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 AccidentalFourierTransform (480 points) [ no revision ]
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

Your answer

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):
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$\hbar$ysics$\varnothing$verflow
Then 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).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...