Can a scalar field model gravity? How accurate would be the results? Are there any difficulties with such a model?

Newtonian gravity can be described by the equation: $$\nabla^2 \phi = 4 \pi \rho G$$ where $\rho$ is the mass density, $\phi$ is the gravitational potential, and G is the universal gravitational constant. Of course, one of the shortcomings is that it is not consistent with special relativity. General relativity handles this shortfall but is a tensor theory. However,the above equation can be modified as follows: $$(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{{\partial t}^2}) \phi = 4 \pi \rho G$$ This would be consistent with special relativity. Would this model give results close to experiment? Would it give correct results in some observed situations where Newtonian gravity fails? Why or why not?

• Oct 15 '13 at 19:09

There is indeed a scalar field model of gravity, in fact Einstein originally tried that before settling on a spin 2 description. Scalar gravity is called Einstein-Nordstrom gravity, here is a link to wikipedia: http://en.wikipedia.org/wiki/Nordstr%C3%B6m%27s_theory_of_gravitation. At the nonlinear level it amounts to using $R$ in Einsteins equations instead of $G_{\mu\nu}$.

What you wrote down was indeed guessed. The problem is that $\rho$ actually isn't relativistically invariant--energy and momentum get mixed under boosts--so you really need to use $T$, the trace of the stress energy tensor. You also need to have the gravitaional sector to have nonlinear interactions, because gravity carries energy and so it couples to itself. So you can generalize what you wrote, that is the Einstein-Nordstrom theory.

While scalar gravity does reproduce the Newtonian limit, the Newtonian limit is easy to get. The problems all amount to the fact that the graviton is a spin 2 particle, not a spin 0 particle.

For example, scalar gravity cannot couple to light. This is because a scalar (spin-0) can only couple to the trace of the stress energy tensor $T$, but Maxwell's equations are famously conformally invariant at the classical level and so $T=0$. This violates Einstein's equivalence principle (which one reason why Einstein wouldn't have liked it). It also is empirically ruled out (which is a great reason for us to rule it out, though Einstein didn't have those experiments when he was developing GR).

Scalar gravity also has completely different properties for gravitational waves: it has one helicity 0 polarization instead of 2 helicity 2 ones. This would change the output of radiation from a binary pulsar system, for example.

Another consequence is that Birkhoff's theorem is no longer true. A scalar mode can be sensitive to overall changes of scale in an object--a spherically symmetric object with time varying radius $R(t)$ will radiate in scalar gravity, but will definitely not radiate in GR.

• Thank you for the informative post. In a scalar field theory of gravity is an interpretation of gravity as the curvature of spacetime necessary? Also, experimentally, what are the failures of a scalar theory of gravity? Also, how about a vector theory of gravity,similar to Maxwell's equations?
– guru
Oct 2 '13 at 16:10
• All of those things I mentioned are experimental failures: We know light couples to gravity from many sources, including cosmology and gravitational lensing. The GR formula for graviational radiation from binary pulsars has been tested very well on e.g. the Hulse Taylor pulsar, the scalar gravity theory would give a different answer. Maybe we haven't directly tested Birkhoff's theorem, I have to think about it, but I'd be surprised if you could get away with violating it by a lot. Vector gravity doesn't work because forces coming from vector fields are not universally attractive (contd) Oct 2 '13 at 16:13
• ... which is a non-trivial result of QFT. The curvature interpretation comes out of trying to write down a consistent non-linear theory. This is the field theory way of looking at GR (developed by people like Weinberg, Deser, Feynman; Feynman explains it well in his 'lectures on gravitation'): the only consistent way to have the graviational field couple to its own stress energy tensor leads you to writing down curvature invariants. You have a little more freedom for scalar gravity then for tensor gravity, but you still are ultimately lead to curvature by the symmetries of the problem. Oct 2 '13 at 16:16
• Ah, also it's important to point out that scalar, vector, and tensor refer to the graviational POTENTIAL. So Maxwell's theory is a vector theory because you have a vector potential, not because the electric and magnetic fields are vectors (and actually from a relativistic perspective the electric and magnetic fields are properly considered as components of a tensor). The force on a particle is always a vector, regardless of whether the underlying potential is a scalar, vector, or tensor. Oct 2 '13 at 16:24
• The Nordstrom theory of gravitation fulfills the conditions of the equivalence principles (the weak, the Einstein and the strong ones). All objects, massless, massive or self-gravitating's, fall with the same acceleration. The absence of light bending does not imply violation of the equivalence principle. In the Nordstrom theory, free-fall term (g_00) and space curvature terms (g_ii) cancel out in the computation of the light bend amplitude. Aug 22 '14 at 22:02

There are several ways that scalar gravity fails, but the most dramatic one is that a scalar theory of gravity does not predict that light will bend in a gravitational field.