# Schwarzschild solution derivation and determinant of the Schwarzschild metric tensor

At some point we are looking for the solution of the Einstein field equation for the Schwarzschild gravitation from

$$ds^2 = A(r)dt^2 +B(r)dr^2+r^2d\theta+r^2\sin^2(\theta)d\phi^2\tag{1}$$

I am reading Wikipedia and the obtained solution is

$$A(r)B(r)=K \quad \mathrm{with} \quad A(r)=(1-\frac{1}{Sr})\tag{2}$$

Then requiring $$r \to \infty$$ to approach Minkowski one gets $$K=-1$$. Therefore, as $$A(r)B(r)=-1$$ then the determinant of the Schwarzschild metric is the same as the determinant of the Minkovski metric

$$det(g)=det(\eta)=-1 \tag{3}$$

My question is: why we can't look for the solution from this $$ds^2 = A(r)dt^2 +A^{-1}(r)dr^2+r^2d\theta+r^2\sin^2(\theta)d\phi^2\tag{4}$$

instead of (1)? I guess the answer is: because we do not know "a-priory" that $$det(g)=-1$$.
Is so, then could we theoretically get $$det(g)=-4$$?

Or re-phrasing: do we know a-priori that $$det(g)=-1$$ ? If so then why we can't start looking for the solution in the form of (4)?

• This is derived in my Phys.SE answer here. Commented Aug 20, 2020 at 17:38
• For the $-1$ check this: inspirehep.net/literature/756421 Commented Aug 20, 2020 at 17:55
• @Qmechanic - I do not see any connection to the Birkhoff's theorem. I am talking about the classical derivation of the Schwarzschild static metric. I am aware of a few different derivations (not only Wikipedia's one that I cited). Commented Aug 20, 2020 at 18:03
• @ApolloRa thanks. Looks similar but is off-topic. Commented Aug 20, 2020 at 18:07
• I think my question is very related to physics.stackexchange.com/q/363602 Commented Aug 20, 2020 at 20:04

You could, of course, look for a solution of the form (4). In this case you can know you will find one since we know that Schwarzschild is a solution of that form.

When you try to solve a PDE via ansatz (i.e. guessing), you can try whatever you want. Not every possible ansatz will lead to a solution though. There's an art to choosing one that's general enough to (a) admit at least one solution and (b) capture an interesting range of possible solutions while (c) keeping the ansatz specific enough that you get a useful result to you in context of the question you're answering.

In contrast to both (1) and (4) as labeled in your question and illustrating my point (c), you can always guess that there's some function $$g_{ab}(x^c)$$ that satisfies the physical equations, but that's a useless guess from a practical point of view because you've remained completely general.

If you're following a derivation given by someone who already knows the answer, they are typically either following a form that they think had pedagogical value, reflects a historical perspective, or fits some set of conditions given by the context in which they are solving the problem. Possibly a combination of these factors.

In this case you've got here, if you choose the form (4), you can get back to Schwarzschild, but you won't know if $$B(r)=A^{-1}(r)$$ was necessary to find a solution because you built that constraint into your ansatz. By starting with $$B(r)$$ general, you still get back to Schwarzschild and you derived the additional fact about the determinant along the way.

• Thanks! It answers - at least partially. I thought that we have some math constraints on $det(g)$ prior deriving the metrics so we may start with (4). Commented Aug 20, 2020 at 19:45
• The determinant is a function of space and time. So you have the condition that you used at $r \rightarrow \infty$, but I'm not sure what would have let you know that it had be the same in the strong-field regions, at least not for sure. Of course you could have guessed that it would be or set out to find any solution that happened to have that property. @Eddward Commented Aug 20, 2020 at 19:48
• Well, "determinant of the metric isn't a property of space, it's a property of the coordinates you've chosen. " at physics.stackexchange.com/questions/363602/… Commented Aug 20, 2020 at 20:11
• My point was that it doesn't have to be a constant and the condition that you used only gives the value at infinity. @Eddward Commented Aug 20, 2020 at 20:31
• Most differential equations require the "intelligent guess" method! Even plain old integration does not have a complete analysis I think, so there's some trial and error there too. Commented Aug 20, 2020 at 22:56