Spherically Symmetric Metric in Nordstrom Gravity

Question

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! 

Comments

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 $\Box \psi = 0$.

Answer 1

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

$$\mathrm{d}s^2=(1+\frac{M}{R})^{-2}(-\mathrm{d}t^2 + \mathrm{d}R^2) + R^2 \mathrm{d}\Omega^2$$

Comments

@Void How is it sufficient here to fix the scalar curvature and the Weyl tensor but not the Ricci curvature tensor? According to the Ricci decomposition don't we need to fix all three of these in order to fix the metric?