• 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.


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


(propose a free ad)

Site Statistics

204 submissions , 162 unreviewed
5,024 questions , 2,178 unanswered
5,345 answers , 22,686 comments
1,470 users with positive rep
815 active unimported users
More ...

  Gauge freedom in the tetrad

+ 3 like - 0 dislike

I'm reading the following paper about Petrov type D space times called "Type D vacuum metrics":


by Kinnersley. I have a question about his choice of gauge.

In particular, starting with the tetrad $\lbrace l^{\mu},n^{\nu},m^{\mu},\overline{m}^{\mu}\rbrace$ where $l^{\mu},n^{\mu}$ are null, he then makes an argument that we can pick a scaling $A$ such that $l^{\mu}\rightarrow Al^{\mu}$ and $n^{\mu}\rightarrow (1/A)n^{\mu}$, and make $\nabla_{l}l=0$ (i.e. $l^{\mu}$ a geodesic vector field). He then picks a coordinate system $(x^{1},r,x^{2},x^{3})$ such that $l^{\mu}=(0,1,0,0)$. That's what he explains at the beginning of his section 2, and it's clear to me.

What confuses me is what he writes at the beginning of section 3C: he says that there is still freedom left if we choose the scaling, call it $A^{0}$, to be independent of $r$. Now, I understand that such choice will not violate the condition $\nabla_{l}l=0$. However, my question is: why doesn't it violate the already chosen $l^{\mu}=(0,1,0,0)$?

What am I missing? Why is one allowed to scale once more?

Thanks for any help.

This post imported from StackExchange MathOverflow at 2015-07-05 20:45 (UTC), posted by SE-user GregVoit
asked Jun 26, 2015 in Theoretical Physics by GregVoit (115 points) [ no revision ]
retagged Jul 5, 2015
Crossposted to physics.stackexchange.com/q/191983/2451

This post imported from StackExchange MathOverflow at 2015-07-05 20:45 (UTC), posted by SE-user Qmechanic
maybe @WillieWong can help?

This post imported from StackExchange MathOverflow at 2015-07-05 20:45 (UTC), posted by SE-user GregVoit

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:
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).
To avoid this verification in future, please log in or register.

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

Your rights