Newtonian limit of Schwarzschild metric

[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?

 

[PeterDonis]

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]

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

 

[dwellexity]

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]

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.

 

[dwellexity]

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]

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 $T$.