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 ...

  Are covariant derivatives of Killing vector fields symmetric?

+ 2 like - 0 dislike
668 views

I'm reading the Lecture Notes on General Relativity by Matthias Blau, and in section 9.1 (point 1) he writes:

Let $K^\mu$ be a Killing vector field, and ${x^\mu(\tau)}$ be a geodesic. Then the quantity $$Q_K=K_\mu\dot{x}^\mu$$ is constant along the geodesic.

Now, in equation (9.2) he writes $$\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu=\frac{1}{2}(\nabla_\nu{K}_\mu+\nabla_\mu{K}_\nu)\dot{x}^\mu\dot{x}^\nu$$ which vanishes due to Killing's equation.

The thing I don't really understand is why ${\nabla_\nu{K}_\mu-\nabla_\mu{K}_\nu=0}$. I can see that because ${\mu,\nu}$ are dummy indices, then \begin{align}\nabla_\nu{K}_\mu\dot{x}^\mu\dot{x}^\nu&=\nabla_\mu{K}_\nu\dot{x}^\nu\dot{x}^\mu\\&\Longrightarrow\,\nabla_\nu{K}_\mu-\nabla_\mu{K}_\nu=0 \tag{1}\end{align} Is this argument valid? If so, does $(1)$ hold in general for Killing vectors? As I'm almost sure it doesn't, despite the title of my question (all Killing vector fields would be gradient fields), does it happens only along the geodesic? If so, what interpretation would this condition have?

This post imported from StackExchange Physics at 2014-05-04 11:40 (UCT), posted by SE-user Pedro Figueroa
asked Apr 26, 2014 in Theoretical Physics by Pedro Figueroa (85 points) [ no revision ]
He does not say that antisymmetric part must vanish, in fact it does not in general. What he says is that only symmetric part contributes to the expression. Indeed, if you have a vector, $V^\nu$, and a tensor $T_{\mu\nu}$, then only symmetric part of $T$ contributes to $T_{\mu\nu} V^\mu V^\nu$. In his case, $T=\nabla K$ and $V^\nu=\dot{x}^\nu$

This post imported from StackExchange Physics at 2014-05-04 11:40 (UCT), posted by SE-user John

1 Answer

+ 2 like - 0 dislike

A Killing vector $K^\mu$ is defined as a vector Lie derivative of metric along which vanishes. \begin{equation} \mathcal{L}_K g_{\mu\nu}=0, \quad \Longrightarrow \nabla_\mu K_\nu+\nabla_\nu K_\mu=0. \end{equation} I guess there is no need to write derivation of this equation explicitly as you can find it everywhere.

Concerning you question about the antisymmetrisation. Lets start with the expression $$ \begin{align}\nabla_\nu{K}_\mu\dot{x}^\mu\dot{x}^\nu&=\nabla_\mu{K}_\nu\dot{x}^\nu\dot{x}^\mu\\&\Longrightarrow\,(\nabla_\nu{K}_\mu-\nabla_\mu{K}_\nu)\dot{x}^\mu\dot{x}^\nu=0 \tag{1}\end{align} $$ In this form the last equation is trivial as we contract an antisymmetric tensor with a symmetric one $\dot{x}^\mu\dot{x}^\nu$. However, you can not derive from here the equation $$ \nabla_\mu K_\nu-\nabla_\nu K_\mu=0 $$ since this is true only upon contraction with a symmetric tensor.

Hence, the Killing equation $\nabla_\mu K_\nu+\nabla_\nu K_\mu=0$ that was used by Blau actually comes from the definition at the beginning of this post. The symmetrisation in the expression $\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu$ as was already mentioned by John comes from contracting with the symmetric tensor $\dot{x}^\nu\dot{x}^\mu$. In details: \begin{equation} \begin{aligned} \nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu&=\frac12(\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu+\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu)\\ &=\frac12(\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu+\nabla_\alpha{K}_\beta\dot{x}^\alpha\dot{x}^\beta)\\ &=\frac12(\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu+\nabla_\alpha{K}_\beta\dot{x}^\beta\dot{x}^\alpha)\\ &=\frac12(\nabla_\nu{K}_\mu\dot{x}^\nu\dot{x}^\mu+\nabla_\mu{K}_\nu\dot{x}^\nu\dot{x}^\mu)\\ &=\frac12(\nabla_\nu{K}_\mu+\nabla_\mu{K}_\nu)\dot{x}^\nu\dot{x}^\mu). \end{aligned} \end{equation} Here in the second line I just renamed the indices, in the third the $\dot{x}^\alpha$ and $\dot{x}^\beta$ were permuted and then I renamed the indices again.

This post imported from StackExchange Physics at 2014-05-04 11:40 (UCT), posted by SE-user Edvard
answered Apr 26, 2014 by Edvard (20 points) [ no revision ]

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$\hbar$y$\varnothing$icsOverflow
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
...