• 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

205 submissions , 163 unreviewed
5,037 questions , 2,191 unanswered
5,345 answers , 22,706 comments
1,470 users with positive rep
816 active unimported users
More ...

  Ernst potential from Kaluza-Klein reduction of axisymmetric space-time

+ 2 like - 0 dislike

Following appendix A of "Ergoregions in Magnetised Black Hole Spacetimes" by G. W. Gibbons, A. H. Mujtaba and C. N. Pope, starting from the Lagrangian

$$\mathcal{L} = \hat{R} - \hat{F}_{\mu\nu}\hat{F}^{\mu\nu},$$ metric and field

$$\mathrm{d}\hat{s}^2 = e^{2\phi} \mathrm{d}s^2 + e^{-2\phi}(\mathrm{d}z + 2\mathcal{A})^2,\\ \hat{A} = A + \chi(\mathrm{d}z +2 \mathcal{A}),$$

after Kaluza-Klein reduction, with Killing vector $K\equiv\partial_z$ corresponding to a spatial dimension, we obtain the reduced lagrangian. After using Lagrange multipliers and dualizing the fields, we obtain $$\hat{F} = -e^{2\phi} \star \mathrm{d}\psi + d \chi \wedge (\mathrm{d}z + 2\mathcal{A}),\quad e^{-2\phi} \star F = \mathrm{d} \psi, \quad F \equiv \mathrm{d} A + 2\chi \mathrm{d}\mathcal{A},$$

where the hatted quantities are 4-dimensional, none of the fields depends on $z$ and $\star$ is the Hodge dual with respect to $\mathrm{d}s^2$. At the end, they define the complex Ernst potential by $d \Phi = i_K(\hat{\star}\hat{F} + \mathrm{i} \hat{F})$, $\Phi = \psi + \mathrm{i}\chi$, where $i_K$ is the interior product by $K$. $i_K \hat{F} = -\mathrm{d} \chi$ is easy to derive, but $\mathrm{d}\psi = i_K \hat{\star} \hat{F}$ has proved to be not so easy.

My question is about this last equation. Considering that $\det \hat{g}_{MN} = e^{4\phi} \det g_{\mu\nu}$, I obtain $$i_K \hat{\star} \hat{F} = - e^{2\phi} \star(F + 2 \mathrm{d} \chi \wedge \mathcal{A}),$$ which is obviously wrong. Could somebody provide some hint about this last equation? It must be easy to derive, but I cannot see the way at the moment.

This post imported from StackExchange Physics at 2015-03-31 11:34 (UTC), posted by SE-user auxsvr

asked Mar 30, 2015 in Theoretical Physics by auxsvr (30 points) [ revision history ]
edited Mar 31, 2015 by Dilaton

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).
Please complete the anti-spam verification

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

Your rights