# Radial reparametrization ansatz in Schwarzschild metric derivation

The standard derivation of Schwarzschild solution (and Birkhoff's theorem) seem to begin with the most general spherically symmetric static metric

$$ds^2 = -U(\rho) dt^2 + V(\rho) d\rho^2 + W(\rho) \rho^2 ( d\phi^2 + \sin^2(\phi) d\theta^2) .$$

Then they proceed to postulate that a radial reparametrization $$r^2 = \rho W(\rho)$$ is always possible, and dispose of the $$W(\rho)$$ factor:

$$ds^2 = -A(r) dt^2 + B(r) dr^2 + r^2 ( d\phi^2 + \sin^2( \phi ) d\theta^2).$$

This seems like a very strong imposition on the solution, as it seems to impose that the $$W(\rho)$$ has to be monotonic (no shrink or expansion of the throat allowed). In fact the Morris-Thorne wormhole solution seems to never want to impose such restriction on the metric, otherwise there would be no throat.

In fact, I could find more palatable an argument that there can always be a radial reparametrization for which $$V(\rho)$$ (not $$W(\rho)$$) can be set to one. That seems a more reasonable simplification.

Is there a proper derivation of the Schwarzschild metric somewhere without this seemingly arbitrary ansatz? I couldn't find any

• well, what is r anyaways? a bunch of rulers? Commented Jun 11 at 3:14
• I think the best way to see it is by defining the radius in a geometric way: because of the spherical symmetry, you can define r in terms of the area of the 2-sphere. So, r would be the radius at which $A = 4 \pi r^2$ takes a particular value. This would correspond to your second metric. Note that the area is a coordinate/diffeomorphism-invariant concept. Commented Jun 11 at 5:46
• @Guliano that would be baking the conclusion that area grows monotonically with some parameter into your assumption. What if area could've taken a given value at two or more values of r? those solutions are ruled out by the assumption Commented Jun 11 at 6:46
• Einstein's equations are local. We don't care whether $W(\rho)$ is a monotonic function everywhere. As long as there is some local region where $W(\rho)$ is monotonic, we can define the coordinate $r$ in that local region and solve Einstein's equations there (which then gives the Schwarzschild metric). Commented Jun 11 at 7:38
• @UnkemptPanda that would mean the area of the 2-sphere is growing in one area and shrinking in another one. I think this would require some nonzero stress-energy tensor, like e.g. a nonzero cosmological constant in a certain patch. This would break the assumption that we are looking at a vacuum solution. Commented Jun 11 at 8:43

If we start from the Ansatz

$$ds^2 = -e^{\alpha(\rho)} dt^2 + e^{\beta(\rho)} d\rho^2 + e^{\gamma(\rho)} ( d\phi^2 + \sin^2(\phi) d\theta^2) ,$$ (This is equivalent with the OP's original Ansatz as long as $$U$$, $$V$$, and $$W$$ are nonzero. If they were the metric would be degenerate. So this is a safe assumption.)

then the vacuum Einstein equations $$R_{tt}=0$$ and $$R_{\rho\rho}=0$$ yield

$$\alpha''(\rho) = -1/2 \alpha'(\rho) \left[\alpha'(\rho)-\beta'(\rho)+2\gamma'(\rho)\right]$$ $$\gamma''(\rho) = -1/2 \gamma'(\rho) \left[\alpha'(\rho)+\beta'(\rho)-\gamma'(\rho)\right]$$

The second equation implies that if $$\gamma'(\rho)=0$$ for any value of $$\rho$$ than $$\gamma$$ is constant.

If we set $$\gamma(\rho) = \log(r_0^2)$$, then $$R_{\phi\phi}=0$$ implies

$$e^{\beta(\rho)} = 0,$$

i.e. the metric is degenerate, which is not allowed. Consequently, we conclude that $$\gamma'(\rho)\neq 0$$ and that $$\gamma$$ is a monotonic function of $$\rho$$. We can now safely make the coordinate transformation $$r^2 = e^{\gamma(\rho)}$$, and return to the usual derivation of the Schwarzschild metric.

• Thanks for the answer! Where are the $R_{tt} = 0, R_{\rho \rho}=0$ coming from? my guess is that you are using staticitity to obtain them, but make it more explicit. Commented Jun 11 at 13:45
• They are the components of the Ricci tensor. The vacuum Einstein Equation is simply the Ricci = 0 Commented Jun 11 at 14:15
• Also, note that use of the imperative in your comment is rather rude. Commented Jun 11 at 14:15
• for $R_{\rho \rho} = 0$ I am getting instead $\gamma''(\rho) = \frac{\beta'(\rho)}{4}( 2 \gamma'(\rho) + \alpha'(\rho) ) -\frac{1}{4} \alpha'(\rho)^2 -\frac{1}{2} \alpha''(\rho) - \frac{1}{2} \gamma'(\rho)^2$. It is weird that I am getting both $\alpha$ and $\gamma$ 2nd derivatives on $R_{\rho \rho}$ Commented Jun 11 at 14:44
• You can eliminate the $\alpha''$ using $R_{tt}=0$ Commented Jun 11 at 14:50