The field equations of the Brans-Dicke gravity are

$$\Box\phi = \frac{8\pi}{3+2\omega}T$$ $$G_{ab} = \frac{8\pi}{\phi}T_{ab}+\frac{\omega}{\phi^2} (\partial_a\phi\partial_b\phi-\frac{1}{2}g_{ab}\partial_c\phi\partial^c\phi) +\frac{1}{\phi}(\nabla_a\nabla_b\phi-g_{ab}\Box\phi)$$

In general relativity, the singularity in the Schwarzchild black hole seems to be considered either a 'delta function' like mass source term, or the "edge"/boundary condition of a manifold. Either way, the equations can't tell us how the "infinite curvature"/singularity evolves and therefore we are forced in some sense to "impose it" as a boundary condition either way.

However, in Brans-Dicke gravity the effective gravitational coupling constant appears to be sourced by $T$. So it appears in a true vacuum $T \rightarrow 0$, $\phi \rightarrow 0$, and therefore no Schwarzchild solution is possible regardless of the value of the parameter $\omega$.

But I've heard that Brans-Dicke is equivalent to General Relativity at least in some limit, so what is going on here? Brans-Dicke seems to really care precisely how we deal with singularity, while General Relativity doesn't care what is behind the event horizon.

1) Can Brans-Dicke gravity support black hole vacuum + curvature singularity solutions?
2) Does it force us to interpret the singularity a certain way, and if so does the Brans-Dicke $\longrightarrow$ General Relativity limit tell us the singularity should be interpretted as non-vacuum to make sense?

  • $\begingroup$ I don't agree that the "true" Brans-Dicke vacuum requires $\phi\rightarrow 0$. In the Brans-Dicke theory, the field $\phi$ is a part of the gravitational field as much as $g_{ab}$ is. Also, the limit at which it approaches GR is $\omega \rightarrow \infty$, so be careful about saying what is possible. $\endgroup$ Commented May 18, 2011 at 11:47
  • $\begingroup$ I'm probably oversimplifying, but if it was partial derivatives instead of covariant derivatives then it seems obvious to me that $\phi$ is only specified up to an additive constant. So the equations you give don't fully specify the results, so something is missing. Can someone respond with an answer at least giving what all the Brans-Dicke equations are? $\endgroup$
    – Ginsberg
    Commented May 29, 2011 at 23:05

2 Answers 2


The answer to your first question is yes. If you have only trace-free matter (e.g. only electromagnetic fields or plain vanilla vacuum with no cosmological constant) then the first of your field equations is solved by constant $\phi$, which then plays the role of Newton's constant, as evident from the second field equation, which is then nothing but GR coupled to some traceless energy momentum tensor (or vacuum GR). Therefore, any solution to the vacuum Einstein equations is also a solution to the Brans-Dicke equations (but not vice versa, of course). This contains in particular black hole and curvature singularity solutions.

The answer to your second question depends on which vacuum you choose. If you are in a constant dilaton vacuum (as in my answer to question 1) and have only trace-free matter then you recover GR and its singularities, with no essential change of interpretation. Otherwise you can view the dilaton just as yet-another matter source as far as singularities are concerned. There may be some qualitative changes as compared to GR if you choose generalized Brans-Dicke models with potentials where the dilaton does not obey the energy conditions that you need in the proofs of singularity theorems. But otherwise there is again no essential change in the story of singularities. Another complication arises if you define singularities not through unboundedness of some curvature invariants, but through completeness properties of geodesics (as is done commonly). Then you have to specify if your geodesics couple to the metric $g$ appearing in the Brans-Dicke action, or to some conformally related metric $\Omega(\phi)g$, with a $\phi$-dependent conformal factor. If $\Omega$ has some zeros or singularities this can change the singularity structure, even if the solution looks formally the same as in GR. You can then convert loci that are singular in GR into loci that are regular in Brans-Dicke theory and vice versa.

  • $\begingroup$ I agree with Ginsberg, an equation seems to be missing. The equations don't allow one to solve for the constant $\phi$ solution, and so as written those equations are not predictive since there is no unique solution. Do people commonly add in $\phi=c^4/G$ as a boundary condition at infinity? How is this handled? $\endgroup$
    – John
    Commented Jun 2, 2011 at 21:16
  • 2
    $\begingroup$ Brans Dicke theory has two fields, the metric and the scalar field. Therefore, varying this action with respect to these two fields gives two sets of equations, the (generalized) Einstein equations and the scalar field equation. Both are displayed in the question, so no field equation is missing. Note that you never get unique solutions from bulk field equations. You always have to provide boundary and/or initial conditions. This is what fixes the constant mode of the scalar field in the present example. $\endgroup$ Commented Jun 29, 2011 at 14:19
  • $\begingroup$ @John: The inverse proportionality of $\phi$ to $G$ is really just an interpretation of the form of the field equations; note that there is no actual Newtonian coupling constant $G$ anywhere in the equations. $\endgroup$
    – user4552
    Commented Jul 29, 2011 at 14:56

Daniel Grumiller's answer is great. I will just take this from a slightly different angle in an attempt to shed more light.

In GR, it makes sense to imagine that something local, like a black hole, can be separated from what's going on far away. As boundary conditions, we just say that spacetime should be asymptotically flat, and away we go. This is essentially because of Newton's shell theorem for a $1/r^2$ field. If we think the distant universe is spherically symmetric, then we don't expect it to have any effect on us locally other than imposing things like cosmological curvature and expansion of space, which don't have any significant effect on bound systems.

But in B-D gravity, the field $\phi$ falls off like $1/r$, and therefore the contribution to $\phi$ from the distant universe is not expected to cancel, or even to be small. In fact, if you try to make a cosmological model in B-D gravity with a nonvanishing density of massive particles, $\phi$ will generically diverge. Basically the only way to keep it from diverging is to have a big bang cosmology.

So although Daniel Grumiller is absolutely right that the Schwarzschild vacuum is a perfectly legitimate B-D spacetime, it is in some sense not what we generically expect a B-D spacetime to be like. Generically we expect it to be a big bang cosmology with some value of $\phi$ set by the cosmological conditions.


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