• 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

198 submissions , 156 unreviewed
4,896 questions , 2,074 unanswered
5,309 answers , 22,537 comments
1,470 users with positive rep
803 active unimported users
More ...

  Persuing geometrical meaning of Wald (7.1.20) to (7.2.21)

+ 3 like - 0 dislike

Suppose the metric $g$ with respect to coordinate $X^{a}=(x^{i},y^{m})$ has the form $g=h_{ij}(y)dx^{i}dx^{j}+k_{mn}(y)dy^{m}dy^{n}$, and $R_{x^{i}}^{x^{i}}=R_{x^{1}}^{x^{1}}+....+R_{x^{n}}^{x^{n}}=0$, then the square root of det of $h_{ij}$ is harmonic in the metric $k_{mn}$, ie,


where $D$ is the connection of $k_{mn}$.


in the page 166 Wald says: the equation (7.1.20) yields (7.1.21),

here the two equations are,




where the metric ansatz is,


it can be checked by some direct calculation and the two are equal to $\frac{\sqrt{\Lambda}_{,z}}{\rho\Omega^{2}\sqrt{\Lambda}}$, up to a factor, and certainly it holds not accidentally, which motivites the foregoing results.


Can it be derived more geometrically using something like foliation?

And what is the intuitive picture?

Here is a proof I got. It's direct, but with less intuition.


There are n Killing vector fields $(A_{i})^{a}$ wrt $x^{i}$ for $g$ is independent of $x^{i}$, which leads to, (notice: no summation here, and $x^{i}$ is a scalar function)


for a antisymmetric (2,0) tensor, the covariant derivative is,

\[\nabla_{c}A^{ca}=\frac{1}{\sqrt{|g|}}\frac{\partial(A^{ca}\sqrt{|g|})}{\partial X^{c}}\]

applying to (\ref{a}),

\[\nabla_{c}\nabla^{c}(A_{i})^{a}=\frac{1}{\sqrt{|g|}}\frac{\partial(\nabla^{m}(A_{i})^{a}\sqrt{|g|})}{\partial y^{m}}\]

we need to calculate $\nabla^{m}(A_{i})^{a}$,

\[\nabla^{m}(A_{i})^{a}=g^{mb}\Gamma_{\ bd}^{a}(A_{i})^{d}=k^{mn}\Gamma_{\ nx^{i}}^{a}\]


\[\Gamma_{\ nx^{i}}^{a}=g^{ac}(g_{cn,x^{i}}+g_{cx^{i},n}-g_{nx^{i},c})=g^{ac}g_{cx^{i},n}=h^{ai}h_{ix^{i},n}\]

(notice: the non-vanishing components of the connection are $\Gamma_{\ xy}^{x},\ \Gamma_{\ xx}^{y},\ \Gamma_{\ yy}^{y}$.)

substitute it into (\ref{eq:b}),

\[\nabla_{c}\nabla^{c}(A_{i})^{a}=\frac{1}{\sqrt{|g|}}\frac{\partial(k^{mn}h^{ai}h_{ix^{i},n}\sqrt{|g|})}{\partial y^{m}}\]

$\nabla_{a}x^{i}\cdot$ is to pick out the $x^{i}$ term of the index $a$, thus summing up (\ref{eq:c}) over $x^{i}-$s, the vanishment of partial trace of Ricci tensor is equivalent to,

\[\frac{1}{\sqrt{|g|}}\frac{\partial(k^{mn}h^{ji}h_{ij,n}\sqrt{|g|})}{\partial y^{m}}=0\]

while for $D^{m}D_{m}\sqrt{|h|}=\frac{1}{\sqrt{|k|}}\frac{\partial}{\partial y^{m}}(k^{mn}\sqrt{|k|}\frac{\partial}{\partial y^{n}}(\sqrt{|h|}))$, the last ingrediant left is,

\[\frac{\partial}{\partial y^{m}}h=hh^{ij}\frac{\partial}{\partial y^{m}}h_{ij}\]

asked Jan 25, 2016 in Theoretical Physics by Reiko Liu (15 points) [ revision history ]
edited Jan 25, 2016 by Reiko Liu

It is the first time I ask question and I'm not sure it's appropriate to put it here, so feel free to give suggestions or vote.

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