In Brans-Dicke gravity, we have the following gravitational action:

$S_{BD} = \frac{1}{16\pi}\int d^4 x \sqrt{-g} [\phi R - \frac{\omega}{\phi}(\nabla \phi)^2] + S_m[g_{\mu \nu}; \Psi_n]$,

with $S_m$ the matter action for fields $\Psi_n$, and $\phi$ the long range scalar field.

In Misner, Thorne and Wheeler's "Gravitation" (section 39.1), they observe that "the scalar field does not exert any direct influence on matter; its only role is that of participant in the field equations that determine the geometry of spacetime". This is clear.

However, if one undergoes the conformal transformation

$g_{\mu \nu} = e^{2\alpha \phi}\tilde{g}_{\mu \nu}, \quad \phi = e^{-2 \alpha \psi}, \quad \alpha = \sqrt{\frac{4\pi}{2\omega + 3}}$,

the action then takes the form

$S_{BD} = \frac{1}{16\pi}\int d^4 x \sqrt{-\tilde{g}} [\tilde{R} - \frac{1}{2}(\nabla \psi)^2] + S_m[e^{2\alpha \psi}\tilde{g}_{\mu \nu}; \Psi_n]$.

In this Einstein frame then we explicitly see that matter now couples to both the scalar field as well as the new metric. Hence we have some form of fifth force being mediated by $\phi$, resulting in a violation of the strong equivalence principle. However, the comments from Gravitation seem to contradict this. But since classically all physical observables are independent of the choice of frame, then this fifth force should still emerge regardless. How then do we see this effect in the Jordan (initial) frame of reference, where the coupling of $\phi$ to matter is not explicit?


1 Answer 1


Well, I am not working on modified gravity theories. But this question seems to be understandable. In the Einstein frame, as an observable, you see "Gravity from GR and the fifth force'' and in the Jordan frame you see "Gravity from Brans-Dick theory''. The fifth force is just combined with Einstein gravity in the second formulation. For local observers, there is no difference in the experimental results. They are just two interpretations. Gravity can be decomposed in different ways into geometry and force.


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