# A Competitor for General Relativity?

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GR stands alone in its ability to pass both weak and strong field tests of gravity fields.

From 1905 to 1915, there was renewed interest in a somehow modified scalar field theory. Here is the action of a scalar gravity model, with $z^{\mu}$ a worldline parameterized by $\lambda$: \begin{align*}I_{sg} =& - m\int d \lambda \sqrt{-\left(e^\Phi \frac{d z_0}{d \lambda}, e^\Phi \frac{d z_1}{d \lambda}, e^\Phi \frac{d z_2}{d \lambda}, e^\Phi \frac{d z_3}{d \lambda}\right)\left(e^\Phi \frac{d z_0}{d \lambda}, e^\Phi \frac{d z_1}{d \lambda}, e^\Phi \frac{d z_2}{d \lambda}, e^\Phi \frac{d z_3}{d \lambda}\right)\eta} \\ =& - m\int d \lambda \sqrt{e^{2\Phi} \left( \frac{d z_0}{d \lambda}\right )^2 - e^{2\Phi} \left( \frac{d z_1}{d \lambda}\right )^2 - e^{2\Phi} \left(\frac{d z_2}{d \lambda}\right )^2 - e^{2\Phi} \left(\frac{d z_3}{d \lambda}\right )^2} \end{align*}

The terms under the square root look just like a metric (completely flat in this case). Physicists knew the $g_{00}$ term from Newton's law. Physicists completely guessed at the $g_{uu}$ terms [see footnote], and that guess was wrong. When a path is varied in the presence of a gravity field, the changes in $g_{uu}$ should be approximately inverse of $g_{00}$.

I am exploring an action (scalar gravity coupling) that at the very least is consistent with experimental tests unknown to physicists prior to 1919: \begin{align*}I_{sgc} =& - m\int d \lambda \sqrt{-\left(\frac{1}{e^\Phi} \frac{d z_0}{d \lambda}, e^\Phi \frac{d z_1}{d \lambda}, e^\Phi \frac{d z_2}{d \lambda}, e^\Phi \frac{d z_3}{d \lambda}\right)\left(\frac{1}{e^\Phi} \frac{d z_0}{d \lambda}, e^\Phi \frac{d z_1}{d \lambda}, e^\Phi \frac{d z_2}{d \lambda}, e^\Phi \frac{d z_3}{d \lambda}\right)\eta} \\ =& - m\int d \lambda \sqrt{e^{-2\Phi} \left( \frac{d z_0}{d \lambda}\right )^2 - e^{2\Phi} \left( \frac{d z_1}{d \lambda}\right )^2 - e^{2\Phi} \left(\frac{d z_2}{d \lambda}\right )^2 - e^{2\Phi} \left(\frac{d z_3}{d \lambda}\right )^2} \end{align*} The terms under the square root look like the Rosen metric which will pass all weak field tests (take the velocity equal to square root of $g_{00}$ to $g_{uu}$ to calculate light diffraction, or get the equations of motion, and use the constants of motion to get the same value). Rosen's bi-metric tensor theory fails strong tests of GR, specifically it allows for dipole modes of gravity wave emission which are not consistent with observations of energy loss in binary pulsars.

In the scalar gravity coupling model, the $g_{00}$ must be exactly the inverse of $g_{uu}$. It predicts slightly more bending that GR to second order PPN accuracy (11.7 versus 11.0 $\mu$arcsecond).

New proposals for gravity are usually pretty easy to shoot down. Do you see a flaw in this one? It looks like the Rosen metric and thus curves spacetime for weak fields (good), and is simpler than a bi-metric theory for strong field tests (good). It sure is simpler than GR. Does GR finally have a real competitor?

[footnote]: Metrics can be written in an unending variety of coordinates. I have written this action presuming a point source and Cartesian coordinates. Had I chosen spherical coordinates, then there would be no gravity term in front of $z_2$ or $z_3$. Note added in response to an issue raised by @Trimok.

This question has been put on hold.

For the record, I certainly do believe I "have specific questions evaluating new theories in the context of established science are usually allowed". There is both an action, and a metric associated with the scalar gravity coupling model. That is the canonical method for doing gravity research. The outcome is specific down to 0.7 $\mu$arcseconds (details below), something established physics like work with strings never achieves. It is a rare, but should be a valuable bird of a model that could respect the strong equivalence principle, meaning that gravity is only about making the interval dynamic.

Given a metric for any model, the post post Newtonian deflection can be calculated using a formula from Epstein and Shapiro (Phys. Rev. D, 22:12, p. 2947, 1980): \begin{align*} d\tau^2 =& \left(1 - 2 \gamma \frac{G M}{c^2 R} + 2 \beta \left(\frac{G M}{c^2 R}\right)^2 \right) dt^2 - \left( 1 + 2 \gamma \frac{G M}{c^2 R} + \frac{3}{2} \epsilon \left(\frac{G M}{c^2 R}\right)^2 \right) dR^2/c^2 \\ \theta_{_{ppN}} =& ~\pi (2 + 2 \gamma - \beta + \frac{3}{4} \epsilon) \left(\frac{G M}{c^2 R} \right)^2 \\ \theta_{_{GR}} =& ~\frac{15}{16} \pi \left(\frac{G M}{c^2 R} \right)^2 \\ \theta_{_{sgc}} =& ~4 \pi \left(\frac{G M}{c^2 R} \right)^2 \end{align*} The difference between the two will be about 6%.

A longer blog on this subject is online. Feel free to contact me if you any technical issues with the proposal, large or small.

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user sweetser
asked Nov 25, 2013
Gee, those matrices look too big to fit inside those square roots. How 'bout some simpler notation? (On the other hand, it looks like you've already done quite a bit of work on this. Is there e.g. an arXiv preprint you can point to to avoid duplication of effort?)

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user Emilio Pisanty
The flaw I see, is that you suppose $z_1=r$, and so, there is certainly not a $e^{2\phi}$ term in front of $z_2$ and $z_3$....

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user Trimok
@Emilio-Pisanty I am not allowed on arXiv :-) I wanted to not use the indexes so one could see exactly what my point is.

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user sweetser
@Trimok, I meant coordinates like t, x, y, z, not spherical coordinates. I am not sure how to say that "smartly".

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user sweetser
@Kyle-Kanos, the proposal does not involve the Ricci scalar, so it looks different than that class of work.

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user sweetser
@Emilio-Pisanty Any model that matches all known observations for weak and strong field tests of gravity would be a candidate to replace GR. That would be the implication of such an accomplishment. I have provided the action. You should be able to spot the Rosen metric in what is written. In the literature the Rosen metric satisfies the weak field tests (easy enough to confirm by yourself by taking the Taylor series and comparing that to the Schwarzschild solution). Since I did not provide the reference nor the formulas for the numbers claimed, I modified the note.

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user sweetser
@Emilio-Pisanty I looked at the meta thread, and am guessing there is a "general theorem that disqualifies your" model. In MTW, they say any proposal for gravity must be a metric theory. A metric tensor requires 10 equations to be solved (less if symmetric). The sgc model has one function and different coupling of that function to time versus space. It is not a metric theory. My reply would be that the scalar gravity coupling model's metric representation is identical to the Schwarzschild metric to PPN accuracy so gets the right about of bending, no calculation needed beyond a Taylor series.

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user sweetser

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How does this theory gel with experimental verifications of GR, e.g., the slow-down of binary systems due to gravitational radiation, as well as gravitational lensing?

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user lionelbrits
answered Nov 26, 2013 by (110 points)
The scalar gravity coupling model looks like it should obey the strong equivalence principle. There is but one field, and the only thing that one field does is make space-time have a non-zero curvature. To quote Will: "This suggests the general conjecture that a theory of gravity predicts no dipole gravitational radiation if and only if it satisfies SEP [Strong Equivalence Principle] to the appropriate order of approximation." The strong equivalence principle eliminates most proposals because people want to "add more stuff". No dipole radiation is what the slow-down data shows.

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user sweetser
I am not familiar with how gravitational lensing per se could be used to test different gravitational theories. Once space-time is curved, there is lensing. This model curves space-time. I don't know the literature on the subject.

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user sweetser
"Once space-time is curved, there is lensing." Well... if I recall correctly most scalar models of gravity don't actually allow light to be deflected. This is probably the biggest reason why they are no longer used.

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user lionelbrits
You do recall correctly: most scalar gravity theories are completely flat. Gravity only rescales flat spacetime, so no lensing. The "twist" done in this model between how the scalar field couples to time versus space means that space-time is curved. The Rosen metric is not flat.

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user sweetser
If it's coupling to time differently than to space, then it isn't a scalar model of gravity. "Scalar" means it transforms as a scalar under coordinate changes. How does your "scalar" transform under say, Lorentz boosts?

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user lionelbrits
Scalar can be used 2 ways. First, scalar means that it depends on 1 and only 1 value. Second, scalar transforms under a Lorentz boost like a number, so it is an invariant. Scalar gravity as done in the literature is scalar in both senses of the word. This scalar gravity coupling model is a scalar in the sense of the first definition, not the second. One cannot think of the exp(Phi) separately from the tensor it is coupled to, in this case the 4-vector z'. Together they remain a 4-vector whose invariant square depends on where they are in a gravitational field.

This post imported from StackExchange Physics at 2014-03-05 14:56 (UCT), posted by SE-user sweetser

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