The Lagrangian in general relativity is written in the following form: $$ \begin {aligned} \mathcal {L} & = \frac {1} {2} g ^ {\mu \nu} \nabla_{\mu} \phi \nabla_{\nu} \phi-V (\phi) \\ & = R + \frac {16 \pi G} {c ^ {4}} \mathcal {L} _ {\mathcal {M}} \end {aligned} $$

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 :

$$ \mathcal {L} = \varphi R + \frac {16 \pi} {c ^ {4}} \mathcal {L} _ {\mathcal {M}} - \omega_ {BD} \left (\frac {\varphi_{, i} \varphi ^ {, i}} {\varphi} \right) $$


Any track/clue/help is welcome

  • $\begingroup$ Are you asking how the $\frac {\varphi_ {, i} \varphi ^ {, i}} {\varphi}$ term arises? $\endgroup$ Commented Oct 30, 2020 at 3:22
  • $\begingroup$ @NiharKarve . Yes, I talk about this term which appears magically. $\endgroup$
    – user87745
    Commented Oct 30, 2020 at 9:04
  • $\begingroup$ Considering we promote $1/G$ to a scalar field, it is natural to provide it with a kinetic term $\endgroup$ Commented Oct 30, 2020 at 12:09

1 Answer 1


Some preliminary consideration: In a uniform expanding universe the gravitational potential fulfills in atomic units the equation

$$\frac{GM}{R} \propto 1$$

with $c=1$ and $R$ is the reciprocal Hubble parameter and $M$ the total mass of the visible universe. Now the assumption of Brans was that not only the geometry defines gravitation, but also the mass. So the gravitational constant should be depending on the mass. When you write this equation like

$$ \phi= \frac{1}{G} \propto \frac{M}{R} \propto \sum_i \frac{m_i}{r_i}$$

you can interpret this as $G$ depending on the contribution of all matter to the inertial reaction. Here $G$ is assumed to be a function of $\phi$ so that the variation of both quantities is assumed to be equal. Then you get a roughly equal equation to that one above when assuming a inhomogeneous wave equation

$$ \square \phi \propto \rho$$

with a scalar mass density $\rho$. The concept of this kind of wave equation is that the quantity on the right hand side describes a so called source function. These source functions describes the effect of the sources, here the mass, on the medium carrying the waves. This approach arises from the implementation of Machs principle which leads to the assumption that not only the geometry of the Rieman manifold, but also the mass have an effect on the gravitation.

Outgoing from the Einstein Lagrangian

$$\mathcal{L} = R + \frac{16\pi G}{c^4} \mathcal{L}_{\mathcal{M}},$$

the approach is to replace the gravitational constant by the scalar field $\phi$. For that the Einstein Lagrangian is multiplied by $G^{-1}=\phi$:

\begin{align} \mathcal{L}_{\text{JBD}} :&= \frac{\mathcal{L}}{G} = \frac{R}{G} + \frac{16\pi}{c^4} \mathcal{L}_{\mathcal{M}} \\ &= R \phi + \frac{16\pi}{c^4} \mathcal{L}_{\mathcal{M}}. \end{align}

Because of the previous considerations we want a wave equation for $\phi$. So we add a term $\mathcal{L}_\phi \propto \phi_\mu \phi^\mu$:

$$\mathcal{L}_{\text{JBD}}=R \phi + \frac{16\pi}{c^4} \mathcal{L}_{\mathcal{M}} + \mathcal{L}_{\phi}(\phi, \phi_\mu).$$

Because of unit consistency you can show that $\mathcal{L}_\phi$ has to be second order in space-time and first order in $\phi$. The $\phi$ in the denominator of $\mathcal{L}_\phi$ arises, because in physics the coupling constants describing the strength of fundamental forces are chosen dimensionless. Because $\omega$ describes a coupling between the mass and the scalar field this quantity should also be dimensionless. With these assumptions you should be able to verify the desired equation. Keep in mind that $\omega$ is only a dimensionless constant without further meaning in this derivation.

  • $\begingroup$ Thanks for your quick answer. Unfortunately, I don't understand very well how you add and under which conditions the term $\mathcal{L}_{\phi}(\phi, \phi_\mu)$. Moreover, in a first time, you talk about the Lagrangian, and after, you talk about the Lagrangian density. So for all of this, could you detail a little more your reasoning ?, I make confusions for the moment. Best regards $\endgroup$
    – user87745
    Commented Oct 29, 2020 at 15:03
  • $\begingroup$ I've reworked my answer. I hope it is more understandable now. $\endgroup$
    – enco909
    Commented Oct 30, 2020 at 1:04
  • $\begingroup$ Thanks for your reworking. 1) Why do we write $\square \phi \propto \rho$ ? Indeed, a wave equation should be written as : $\square \phi=0$. 2) Moreover, I don't understand why you write $\mathcal{L}_\phi \propto \phi_\mu \phi^\mu$ whereas thre is, in my initial formula, the term : $- \omega_ {BD} \left (\frac {\varphi_ {, i} \varphi ^ {, i}} {\varphi} \right)$, could you detail please ? $\endgroup$
    – user87745
    Commented Oct 31, 2020 at 10:13
  • $\begingroup$ In short a wave equation that not equals zero is called an inhomogenous wave equation. To your second question the reasoning for this choice of the Lagrangian is that you want $\omega$ to be a dimensionless constant. I will discuss this further in a rework of the answer later this day. $\endgroup$
    – enco909
    Commented Oct 31, 2020 at 12:00
  • $\begingroup$ Thanks a lot. I didn't know the concept of inhomogenous wave equation and its characteristics. It's fine from your part to detail latter your answer. $\endgroup$
    – user87745
    Commented Oct 31, 2020 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.