# Newtonian limit of Schwarzschild metric

• I
• dwellexity
So the EFE can still be written in this form, and the only difference is that the trace term is now on the RHS. 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:$$R_{\mu \nu} = 8 \pi \left( T_{\mu \nu} - \frac{1}{2} g_{\mu \nu} T \right)$$

#### dwellexity

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?

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

PeterDonis said:
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.

dwellexity said:
I have got ##G_{tt} = - \frac{2(-1+2\phi)(\phi + 2 \phi^2 +r \phi')}{(r+2r\phi)^2}##

Yes, this looks ok.

dwellexity said:
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.

PeterDonis said:
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.

dwellexity said:
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##.