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The metric for the weak field approximation to gravity is given by $ds^2 = -(1-\Phi(r))dt^2 + (1+\Psi(r))\left(dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right)$

When $\Phi(r)=\Psi(r)$, e.g. when we take a spherically symmetric source and solve the equations of motion, then the Ricci scalar is

$R=\frac{1}{2 r (\Phi (r)-1)^2 (\Phi (r)+1)^3}\Bigg[2 r \left(-3 \Phi (r)^3+\Phi (r)^2+3 \Phi (r)-1\right) \Phi ''(r)\\ +r \left(3 \Phi (r)^2-4 \Phi (r)+5\right) \Phi '(r)^2- 4 \left(3 \Phi (r)^3-\Phi (r)^2-3 \Phi (r)+1\right) \Phi '(r)\Bigg]$

and the Kretschmann curvature scalar $K=R^{\mu\nu\rho\sigma} R_{\mu\nu\rho\sigma}$ is

$K= \frac{\Phi '(r)^2 \left(r \Phi '(r)+2 \Phi (r)+2\right)^2}{2 r^2 (\Phi (r)-1)^2 (\Phi (r)+1)^4} +\frac{\Phi '(r)^2 \left(r \Phi '(r)+4 \Phi (r)+4\right)^2}{4 r^2 (\Phi (r)+1)^6}\\ +\frac{2 \left(r (\Phi (r)+1) \Phi ''(r)+\Phi '(r) \left(-r \Phi '(r)+\Phi (r)+1\right)\right)^2}{r^2 (\Phi (r)+1)^6} +\frac{\left(\Phi (r) \Phi '(r)^2-\left(\Phi (r)^2-1\right) \Phi ''(r)\right)^2}{\left(\Phi (r)^2-1\right)^4}$

Now both of these are divergent where $\Phi(r)=1$ (as long as $\Phi'(r)\neq 0$). $\Phi(r)=1$ corresponds to the event horizon if we take this to be a black hole metric. This is perhaps to be expected, as the weak field approximation is only valid for $\Phi\ll 1$. However, I was wondering what the physical reason is for this metric is to have a divergence in general at the event horizon.

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  • $\begingroup$ My answer here might be relevant to this. $\endgroup$
    – John Rennie
    Commented Apr 5, 2017 at 15:20
  • $\begingroup$ Thanks, @John Rennie you say there that you find a coordinate singularity at $\phi=1$ but I find that this is actually a curvature singularity. Do you get the same result? $\endgroup$
    – supercoolphysicist
    Commented Apr 6, 2017 at 12:23
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    $\begingroup$ Ah, OK, sorry I skimmed your question in a hurry and misunderstood it. No, I don't find a curvature singularity at the horizon because there isn't one there. I would have to dig out my copy of Mathematica and calculate the Kretschmann scalar to see what is going on, but the obvious suggestion is that there's an error in your calculation somewhere. If you put in $\Phi=-GM/r$ do you get the Schwarzschild Kretschmann scalar $48M^2/r^6$? $\endgroup$
    – John Rennie
    Commented Apr 6, 2017 at 12:58
  • $\begingroup$ Yes, when I put in $\Phi=\frac{GM}{r}$ into the above metric, then I find $K=(4 (GM)^2 (240 (GM)^6 + 32 (GM)^5 r - 104 (GM)^4 r^2 - 72 (GM)^3 r^3 + 43 (GM)^2 r^4 - 16 M r^5 + 12 r^6))/(r^2 (r-2 (GM) )^4 (2 (GM) + r)^6)$ which becomes $K=48GM/r^6$ at large distances, i.e. where the terms of $r$ with the largest power dominate. Can I ask why you think that this metric would not have a curvature singularity at the event horizon, as it is not the same as Schwarzschild unless $r\gg 1$. $\endgroup$
    – supercoolphysicist
    Commented Apr 6, 2017 at 16:32
  • $\begingroup$ Intuitively, if one of the Riemann tensors with a t index is non-zero, then when we find the curvature tensor we will multiply by $g^{tt} =1/(1-Phi)$ which is divergent at $\Phi=1$. $\endgroup$
    – supercoolphysicist
    Commented Apr 6, 2017 at 22:26

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