# Newtonian limit of Schwarzschild metric

• I
If I am asked to show that the tt-component of the Einstein equation for the static metric
##ds^2 = (1-2\phi(r)) dt^2 - (1+2\phi(r)) dr^2 - r^2(d\theta^2 + sin^2(\theta) d\phi^2)##, where ##|\phi(r)| \ll1## reduces to the Newton's equation, what exactly am I supposed to prove?

## Answers and Replies

PeterDonis
Mentor
2019 Award
Compute the Einstein tensor for this metric and plug it into the Einstein Field Equation; take its 0-0 component; and show that the resulting equation reduces to Newton's equation.

dextercioby
Homework Helper
I think it leads to Poisson's equation for the gravitostatic potential. ## \nabla^2 \phi (\mathbb{r}) = -\rho ##

Compute the Einstein tensor for this metric and plug it into the Einstein Field Equation; take its 0-0 component; and show that the resulting equation reduces to Newton's equation.
I have got ##G_{tt} = - \frac{2(-1+2\phi)(\phi + 2 \phi^2 +r \phi')}{(r+2r\phi)^2}##
How do I proceed from here? I am getting a first derivative of ##\phi## instead of second derivative.

PeterDonis
Mentor
2019 Award
I have got ##G_{tt} = - \frac{2(-1+2\phi)(\phi + 2 \phi^2 +r \phi')}{(r+2r\phi)^2}##

Yes, this looks ok.

I am getting a first derivative of ##\phi## instead of second derivative.

Yes, but remember that the Einstein tensor has two pieces: ##G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R##, where ##R_{\mu \nu}## is the Ricci tensor, and ##R## is the Ricci scalar. So the EFE is really ##R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = 8 \pi T_{\mu \nu}##. You might try calculating the two pieces separately to see if there are second derivatives there.

Also, you might take a look at Carroll's online lecture notes on GR, chapter 4, which has a discussion of this calculation.

Yes, but remember that the Einstein tensor has two pieces: ##G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R##, where ##R_{\mu \nu}## is the Ricci tensor, and ##R## is the Ricci scalar. So the EFE is really ##R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = 8 \pi T_{\mu \nu}##. You might try calculating the two pieces separately to see if there are second derivatives there.

I don't understand how this would affect anything. Even if Ricci tensor and Ricci Scalar have second derivatives, what ultimately matters is this particular sum.

PeterDonis
Mentor
2019 Award
Even if Ricci tensor and Ricci Scalar have second derivatives, what ultimately matters is this particular sum.

Not necessarily. Take a look at Carroll's notes. The short version: there is an alternate way of writing the EFE, which moves the trace term from the LHS to the RHS:

$$R_{\mu \nu} = 8 \pi \left( T_{\mu \nu} - \frac{1}{2} g_{\mu \nu} T \right)$$

For the case under discussion only the 0-0 component of this equation is significant, and only ##T_{00}## is significant in the trace ##T##.