# Scalar field in a curved space-time

We know that action of a scalar field in a curved space-time in Jordan frame is given by:

$$\int \left( \frac{1}{2} M^2 R -\frac{1}{2} \partial_\mu \phi \partial^\mu \phi –V(\phi) -\frac{1}{2} \xi \phi^2 R \right) \sqrt{-g}\ d^4x$$

1. Why does this action break the equivalence principle?

2. It is said in the literature that when coupling to gravity is minimal $$\xi = 0$$, then M is the Planck scale.

Firstly Planck mass is given by $m_p = \sqrt{\frac{\hbar c}{G}}$. I cannot see any $\hbar$ in this equation, so where is the Planck scale coming from?

Secondly why does M correspond to Planck mass only when $\xi = 0$ ?

Any ideas are appreciated.

• Which literature? – Qmechanic Apr 20 '17 at 13:40
• @Qmechanic for instance please look at this one link when they discuss about equation (1). – Ramtin Apr 21 '17 at 13:05
• So, in natural units $\hbar=c=1$, the reduced Planck mass is defined by $m_p=(8\pi G)^{-1/2}$. Since in the standard Einstein-Hilbert action, $R$ should be multiplied by $1/16\pi G=m_p^2/2$, it is clear why $M$ should be identified with $m_p$ in the limit $\xi\to 0$. There is no need for $\hbar$. I hope that clarified it! – Bob Knighton Apr 22 '17 at 12:19

If you have a matter field $\psi$ which is coupled minimally to the curved background, and a field $\phi$ which is coupled non-minimally, objects made of $\psi$ will move on different free-fall trajectories than $\phi$. This is a direct violation of the weak equivalence principle.
But the fact that the field $\phi$ is coupled non-minimally also means that you can observe its behaviour and essentially measure the local value of $R$. This is a violation of the Einstein equivalence principle even if you do not have a reference field $\psi$.
Now for the question of $M$: The reason why $M$ is the Planck mass is directly dependent on the context you are considering! The paper you link talks about $\phi$ being the Higgs field undergoing electroweak symmetry breaking which drives the inflation. Thus, $\phi$ reaches an essentially constant value $\phi_0$ in the post-inflationary era.
That is, the gravitational part of your action reduces effectively to $$S_\mathrm{grav} = \int \sqrt{-g} \frac{1}{2}(M^2 - \xi \phi_0^2)R \mathrm{d}^4 x$$ in the post-inflationary era.
However, we measure experimentally right here and right now in the post-inflationary era that the gravitational term in the action is, at least phenomenologically, $R/(4\pi G_\mathrm{N})$ where $G_\mathrm{N}$ is Newton's gravitational constant. In Planck units this term is written as $M^2_\mathrm{p} R/2$. This means that if we want to fit the respective term in the action you give in the postinflationary era, we must fulfill $$M^2_\mathrm{p} = M^2 - \xi \phi_0^2$$ This means that if we impose the phenomenological constrain, $M$ will never be equal to $M_\mathrm{p}$ for a non-minimally coupled field with symmetry breaking. The magnitude of $\phi_0$ depends on the details of the symmetry breaking and then imposes the range of $\xi$ which do not take $M$ too far from $M_\mathrm{p}$. However, if $\xi=0$, the phenomenological constraint gives immediately $M=M_\mathrm{p}$.