The Lagrangian in general relativity is written in the following form: $$ \begin {aligned} \mathcal {L} & = \frac {1} {2} g ^ {\mu \nu} \nabla \mu \phi \nabla \nu \phi-V (\phi) \\ & = R + \frac {16 \pi G} {c ^ {4}} \mathcal {L} _ {\mathcal {M}} \end {aligned} $$$$ \begin {aligned} \mathcal {L} & = \frac {1} {2} g ^ {\mu \nu} \nabla_{\mu} \phi \nabla_{\nu} \phi-V (\phi) \\ & = R + \frac {16 \pi G} {c ^ {4}} \mathcal {L} _ {\mathcal {M}} \end {aligned} $$
with $g^{\mu\nu}: $ the metric
$ \phi: $ non-gravitational scalar field
$ R $: Ricci scalar
$ \mathcal {L} _ {\mathcal {M}}: $ Lagrangian of the density of matter
By replacing the gravitational constant $ G $ by its new definition, $\dfrac {1} {\varphi (t)}, $ How can I prove that the Lagrangian within the framework of Brans-Dicke becomes :
$$ \mathcal {L} = \varphi R + \frac {16 \pi} {c ^ {4}} \mathcal {L} _ {\mathcal {M}} - \omega_ {BD} \left (\frac {\varphi_ {, i} \varphi ^ {, i}} {\varphi} \right) $$$$ \mathcal {L} = \varphi R + \frac {16 \pi} {c ^ {4}} \mathcal {L} _ {\mathcal {M}} - \omega_ {BD} \left (\frac {\varphi_{, i} \varphi ^ {, i}} {\varphi} \right) $$
?
Any track/clue/help is welcome
