The action for the Brans-Dicke-Jordan theory of gravity is $$ \\S =\int d^4x\sqrt{-g} \; \left(\frac{\phi R - \omega\frac{\partial_a\phi\partial^a\phi}{\phi}}{16\pi} + \mathcal{L}_\mathrm{M}\right). $$

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

where $\\G_{ab}$ is the standard Einstein tensor and $\mathcal{L}_\mathrm{M}$ is the matter Lagrangian density. It's supposed to be obvious that this reduces to general relativity in the limit of infinitely large $\omega$. How so? You may assume that the trace $\\T$ of the stress energy tensor $\\T_{ab}$ is not zero.


1 Answer 1


It's obvious because the scalar kinetic terms acquire an infinite action for any value of the derivative, forcing the derivative to be zero and the scalar to be constant. This is just a general limiting property of a scalar field.

If you consider the action

$$ \int A |\nabla\phi|^2 + J(x)\phi(x)d^3x $$

Where J is a (classical) source, the response to the source is through the propagator:

$$ {J(k)J(k')\over A k^2}$$

You can either absorb the constant A into the field, which makes the coupling to the source vanish as $A\rightarrow \infty$, or else keep the coupling constant, in which case the field response vanishes, as reflected in the vanishing propagator.

The end result is that a large A clamps the field to be constant, the field doesn't respond to sources, and you find the Brans-Dicke theory has a constant $\phi$, in which limit the action and equations of motion become the same as GR.

You might be concerned about the $\phi$ in the denominator, but this is entirely due to the unfortunate conventions Brans and Dicke chose. $\phi$ should be thought as making small fluctuations around a constant value. It is best to call $\phi$ by the name $e^{\phi}$, so that the singularity at zero field is pushed away to $-\infty$. This is the dilaton convention in string theory.

  • $\begingroup$ Ok, Ron, so the trick is in realizing that large $\omega$ merely drives $\phi$ to a constant value, not to zero. [Only the piece of $\phi$ sourced by $\\T$ goes to zero.] And because the general solution to $\Box\phi=0$ is a wave with a null wave vector $\\k_a$, inserting this into the Lagrangian causes $\partial_a\phi\partial^a\phi$ to be multiplied by $\\k_ak^a$ = 0, making this term zero just as for constant $\phi$. Is that about right? $\endgroup$
    – Belizean
    Commented May 20, 2012 at 2:30
  • $\begingroup$ Only problem I see is that you'd still have nonconstant solutions to $\Box\phi = 0$ in the $\phi R$ term. How do you argue that these solutions should be ignored? The same reason we ignore gravitational waves in working out the metric around a star? $\endgroup$
    – Belizean
    Commented May 20, 2012 at 2:47
  • $\begingroup$ @Belizean: These solutions are travelling waves, and they can't be sourced at infinite omega--- nothing would emit them. If you put them in by hand, they still have an effect. $\endgroup$
    – Ron Maimon
    Commented May 20, 2012 at 20:23
  • $\begingroup$ That's reasonable. The solutions to $\Box\phi = 0$ are mathematically permitted but physically ignored, because there would be nothing to cause them. $\endgroup$
    – Belizean
    Commented May 24, 2012 at 19:12

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