# General Relativity Lagrangian

The Lagrangian density for the $h=h^{00}$ term of the Einstein gravity tensor can be simplified to: $$L=-\frac{1}{2}h\Box h + (M_p)^ah^2\Box h - (M_p)^b h T$$ The equations of motion following from this Lagrangian looks roughly like (I didn't calculate this, they are given in the problem): $$\Box h = (M_p)^{a}\Box(h^2)-(M_p)^bT$$ For a point source $T=m\delta^3(x)$, solve the equation for h to first order in the source T, with $M_p=\frac{1}{\sqrt{G_N}}$. This result should reproduce the Newtonian potential.

Attempted Solution:

So to first order, we can drop the $h^2$ term and we are left with $$\Box h = -(M_p)^bT$$ $$h = \frac{1}{\Box} (-(M_p)^bT)=-(M_p)^b\frac{1}{\Box} (T)$$ where $\frac{1}{\Box}$ is the propagator (Green function) associated with the field. Based on some calculations and properties of the Green function I got $$h=-\frac{M_p^b m }{4 \pi r}$$ I am pretty confident of my calculations so far. Now, to actually get Newton potential I need $M_p^b=4\pi G$. It is not mentioned, but I assume $G_N=4\pi G$ so the only thing I have to show is that $b=-2$ to reproduce the classical result. I just don't get that value... I tried to do a dimensional analysis of the Lagrangian, and I have $[L]=4$, $[\Box] = 2$ so $[h]=1$. As T is the stress energy tensor $[T]=4$ and $[M_p]=1$ so we are left with $b=-1$ I just don't know where I am missing a factor of 2. Also, assuming my calculations above were wrong, the whole time $(M_p)^b$ was just a constant so that should be the same, regardless of the rest of the solution. So I guess I am doing something wrong with the dimensional analysis. Can someone help me please? Thank you!

With the given normalization, the definition of $h$ is
$$g_{\mu\nu} = \eta_{\mu\nu} + \frac{h_{\mu\nu}}{M_P},$$
so that the Newtonian potential is $\phi = g_{00} = h_{00}/M_P$, $b=-1$, and all is well.
• @Alex It just comes from the Einstein-Hilbert Lagrangian. You can see from dimensional analysis that to get a canonically normalized kinetic term for $h$ (i.e. the $1/2 h \square h$ term) you need to put a mass in the definition. – Javier Sep 15 '18 at 20:18