Talk:Nordström's theory of gravitation

Article uses Planck units
Unfortunately, this article uses Planck units which assumes that the gravitational constant G = 1 and the speed of light c = 1. This makes most of the equations in this article useless for practical computations because it is not clear where G and c should appear in the equations. This entire article needs to be re-written using SI units so that it is more comprehensible to non-physicists.

Some expository defects in 14 April 2006 version
Due to my own laziness, in the 14 April 2006 version, the historical exposition could trip up newcomers: If possible I plan to fix these expository defects at some point. ---CH 17:56, 14 April 2006 (UTC)
 * 1) it is largely left to reader to distinguish between wave equation, raising/lowering/trace, wrt $$\eta_{ab}, \; g_{ab}$$
 * 2) it is largely left to the reader to figure out $$\phi = \exp(\psi) \approx 1 + \psi$$, so that small $$\psi$$ corresponds to weak fields, but due to arbitrary additive constants in gravitational potential, in weak fields either of these can be identified with Newtonian potential ($$\psi$$ more natural for comparison with Einstein's later weak-field metric for gtr).

Students beware
I created the original version of this article and had been monitoring it for bad edits, but I am leaving the WP and am now abandoning this article to its fate.

Just wanted to provide notice that I am only responsible (in part) for the last version I edited; see User:Hillman/Archive. I emphatically do not vouch for anything you might see in more recent versions, but alternative gravitation theories attract many cranks, so be careful. Be wary also of external links to outside websites, which may well be cranky, and may present pseudoscience or fringe science as mainstream, which would of course be very misleading.

Good luck in your search for information, regardless!---CH 02:33, 1 July 2006 (UTC)

Error?
The metric


 * $$ ds^2 = (1-m/\rho) \, \left( -dt^2 + d\rho^2 + \rho^2 \, ( d\theta^2 + \sin(\theta)^2 \, d\phi^2 ) \right) $$

under $$r = \rho \, (1 - m/\rho)$$ and $$dr = d\rho$$, $$r+m = \rho$$

does not give
 * $$ ds^2 = (1+m/r)^{-2} \, (-dt^2 + dr^2) + r^2 \, (d\theta^2 + \sin(\theta)^2 \, d\phi^2 )$$

but
 * $$ ds^2 = (1+m/r)^{-1} \, (-dt^2 + dr^2) + r^2 (1+m/r) \, (d\theta^2 + \sin(\theta)^2 \, d\phi^2 )$$ — Preceding unsigned comment added by 195.113.87.138 (talk) 06:12, 21 June 2012 (UTC)


 * I don't think that is correct; I just checked and the form of the metric given in the main text seems to be correct. vttoth (talk) 13:34, 22 January 2013 (UTC)

Filled with factual inaccuracies
See http://www.physicsoverflow.org/34568/spherically-symmetric-metric-in-nordstrom-gravity?show=34572#a34572 203.188.228.228 (talk) 16:23, 10 December 2015 (UTC)

Gravity waves and gravity affecting bodies through empty spacetime
Gravity can propagate in the theory even if it were true that $$ S_{\mu\nu}=0.$$. R itself obeys a wave equation, and one of its solutions is $$ 1-M/r$$ or even $$ e^{i \omega (t-r)}/r$$. Note that the singularity at r=0 is, just like in Newton's theory, a replacement for a matter source. If the source had a distribution, the field would be regular everywhere, and gravity could act across empty space. Of course the Bianci identity would make sure that $$ S_{\mu\nu}$$ is not identically zero, but that would not change the solution. Note that in this theory, gravity waves are monopole wave-- they expand and contract bodies they pass by, not shear them as GR gravity waves do. — Preceding unsigned comment added by 142.103.234.23 (talk) 01:20, 11 February 2023 (UTC)

Really special relativity?
Lately, I have been thinking that general relativity, as usually understood, is an abandonment of special relativity. If we want to preserve special relativity, we should make an assumption such as this &mdash; there is at least one Cartesian coordinate system within which the metric tensor is the product of the Minkowski metric with the exponential of twice the Newtonian gravitational potential (a scalar field).
 * $$ g_{\alpha \beta} = \eta_{\alpha \beta} \exp (2 \psi) $$

in such a coordinate system, we should have
 * $$ g_{\alpha \beta, \rho} = \eta_{\alpha \beta, \rho} \exp (2 \psi) + \eta_{\alpha \beta} \exp (2 \psi) 2 \psi_{, \rho} = 0 + \eta_{\alpha \beta} \exp (2 \psi) 2 \psi_{, \rho} = 2 g_{\alpha \beta} \psi_{, \rho} .$$

From
 * $$ g_{\alpha \beta, \rho} = 2 g_{\alpha \beta} \psi_{, \rho} $$

and the definition of the Christoffel symbol
 * $$ \Gamma^\alpha_{\beta \gamma} = \tfrac12 g^{\alpha \epsilon} \left( g_{\epsilon \beta, \gamma} + g_{\gamma \epsilon , \beta} - g_{\beta \gamma , \epsilon} \right)$$,

we get
 * $$ \Gamma^\alpha_{\beta \gamma} = \tfrac12 g^{\alpha \epsilon} \left( 2 g_{\epsilon \beta} \psi_{, \gamma} + 2 g_{\gamma \epsilon} \psi_{, \beta} - 2 g_{\beta \gamma} \psi_{, \epsilon} \right)$$.

Cancelling a factor of two, and making use of the Kronecker delta, this simplifies to
 * $$ \Gamma^\alpha_{\beta \gamma} = \psi_{, \rho} ( \delta^{\alpha}_{\beta} \delta^{\rho}_{\gamma} + \delta^{\alpha}_{\gamma} \delta^{\rho}_{\beta} - g^{\alpha \rho} g_{\beta \gamma} ) \,.$$

Since both sides have the same transformation law, this also holds in other coordinate systems.

A free-falling test particle would satisfy
 * $$ \frac{d^2 x^{\alpha}}{(d \tau)^2} + \Gamma^\alpha_{\beta \gamma} \frac{d x^{\beta}}{d \tau} \frac{d x^{\gamma}}{d \tau} = 0 \,$$


 * $$ \frac{d^2 x^{\alpha}}{(d \tau)^2} = - \psi_{, \rho} ( \delta^{\alpha}_{\beta} \delta^{\rho}_{\gamma} + \delta^{\alpha}_{\gamma} \delta^{\rho}_{\beta} - g^{\alpha \rho} g_{\beta \gamma} ) \frac{d x^{\beta}}{d \tau} \frac{d x^{\gamma}}{d \tau} = - 2 \frac{d \psi}{d \tau} \frac{d x^{\alpha}}{d \tau} + g^{\alpha \rho} \psi_{, \rho} \,$$

Compare this with the equation in the article. JRSpriggs (talk) 01:22, 27 June 2023 (UTC)