Scalar equation of Brans-Dicke theory ($(2\omega +3)\square \Phi =8\pi GT$)

Assume four spacetime dimensions, $\Phi \ne 0$, constant $\omega$, no scalar potential and matter independent of $\Phi$. In the stated action convention, scalar variation gives $\Phi R+\omega {\Phi }^{-1}(\nabla \Phi {)}^2-2\omega \square \Phi =0$. The trace of the metric tensor equation gives $\Phi R+\omega {\Phi }^{-1}(\nabla \Phi {)}^2+3\square \Phi =8\pi GT$. Their subtraction proves the displayed wave equation with $T=g^{ab}{T}_{ab}$. The degenerate coupling of Brans-Dicke theory needs separate treatment before dividing by its coefficient.

At $\omega =-3/2$, the combined scalar equation of Brans-Dicke theory becomes the matter consistency condition $T=0$. It does not determine $\square \Phi$ by division, since the wave coefficient has vanished. The original metric tensor and scalar Euler-Lagrange equations must be retained. Introducing scalar-dependent matter or a scalar potential changes the consistency condition.