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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Spherically Symmetric Metric in Nordstrom Gravity

+ 3 like - 0 dislike
180 views

Nordstrom's theory of gravity postulates that the metric is of the form 

$g_{\mu \nu} = \phi ^2 (x) \eta_{\mu \nu}.$

The field equations, in vacuum are of the form $R=0$, where $R$ is the Ricci scalar.

In the wikipedia article on <a href="https://en.wikipedia.org/wiki/Nordstr%C3%B6m%27s_theory_of_gravitation">Nordstrom gravity</a>, it says that the Ricci scalar of the metric above is given by 

$R = -6 \frac{ \Box \phi}{\phi^3},$

so the field equations should be $\Box \phi=0$.

However, later on in the article, they set $g_{\mu \nu} = \exp (2\psi) \eta_{\mu \nu},$ so that $\phi = \exp(\psi)$, and say that $\psi$ satisfies $\Box \psi=0$. However, this is not compatible with the result above, since $\Box \phi = \Box \exp(2\psi)=0$ does not imply that $\Box \psi =0$. 

Later on, the article says that, for a static, spherically symmetric solution, we have $\nabla^2 \psi=0$, with $\psi$ as above.  Using the usual expression for laplacian in spherical coordinates, we get $\nabla^2 \psi = \frac{d}{dr}(\frac{1}{r^2} \frac{d \psi}{dr})=0$

They then say that the metric is $g_{\mu nu} = (1-m/r) \eta_{\mu nu}$, with spherical coordinates. However, the $\psi$ needed to get this metric is not a solution of laplace's equation. And $\Box \left ( \sqrt{1-m/r} \right )$ does not equal zero. So it satisfies neither of the two field equations (which, as far as I can tell are different). What's going on here?

Anyone have a reference that discusses the spherically symmetric solution to Nordstrom gravity? The usual ones (Misner/Thorne/Wheeler etc) don't discuss the spherical vacuum solution to Norstrom gravity. 

Any help is much appreciated! 

asked Dec 9, 2015 in Theoretical Physics by BobG [ no revision ]
recategorized Dec 9, 2015 by dimension10

If $0=\Box \varphi = \Box \exp(2\psi) = 2\Box \psi \exp(2\psi)$, now since the exponent is never zero, you can deduce that $\Box \psi = 0$.

1 Answer

+ 2 like - 0 dislike

The wikipedia article is simply wrong at a number of points. Instead of correcting all the points in the wiki article, I will just give a brief and correct overview. As good scientific practice, I suggest to verify any claim independent of wikipedia, in this case perhaps in some of the science history articles by J. D. Norton.

Nordström theory is indeed fully equivalent to metric gravity with $$R=24 \pi T\,,\;\; C_{\mu \nu \kappa \lambda}=0$$ Which means that we can characterize it by a conformally flat metric

$$\mathrm{d}s^2 = \Phi^2 (-\mathrm{d}t^2 + \mathrm{d}x^2 + \mathrm{d}y^2+\mathrm{d}z^2)$$

Note that the Newtonian (flat) limit is in fact $\Phi=1$. The Ricci scalar then reads (as I have verified by direct calculation): $$R=-\frac{6}{\Phi^3}(-\Phi_{,tt}+ \Phi_{,xx} + \Phi_{,yy}+ \Phi_{,zz})$$

The vacuum equations can then be written as a simple wave equation in the $t,x,y,z$ coordinates. The $\Phi\equiv \exp(\psi)$ ansatz is interesting because then the Ricci scalar is (also verified by direct calculation)
$$R=-6 \Box \psi$$

Where the $\Box \psi$ is the fully covariant D'Alembertian $(\partial_\mu(\sqrt{-g} \partial^\mu \psi))/\sqrt{-g}$! The $\psi$ ansatz may yield a geometrically elegant form of the equations, but this in fact corresponds to a non-linear problem even in vacuum.

To solve the full vacuum equations, it is better to use the $\Phi$ form of the metric, for which the spherically symmetric, static and asymptotically flat vacuum solution is $\Phi=1-M/r\; (r=\sqrt{x^2+y^2+z^2})$. (It would be interesting to investigate the nature of the singularities at $r=0$ and $r=M$...)

Schwarzschild-like coordinates can be obtained by transforming the radial coordinate as $R=r-M$ to obtain
$$\mathrm{d}s^2=(1+\frac{M}{R})^{-2}(-\mathrm{d}t^2 + \mathrm{d}R^2) + R^2 \mathrm{d}\Omega^2$$

answered Dec 9, 2015 by Void (1,505 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$\hbar$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).
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
...