0
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you so much! Now it works out right. But where is this coming from? $\endgroup$ – Alex Marshall Sep 15 '18 at 20:08
  • $\begingroup$ @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. $\endgroup$ – Javier Sep 15 '18 at 20:18
  • $\begingroup$ Thank you so much! I posted a part 2 of this, if you could help me with that, too: physics.stackexchange.com/questions/428991/… $\endgroup$ – Alex Marshall Sep 16 '18 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.