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In this paper a proof of Birkhoffs theorem is provided. However, there is one step that I am quite suspicious of, so I would like to know what you think about it. The relevant part of the paper for this question starts at Theorem 1 (page 3) and ends at equation (11) (page 4).

Within the paper, they consider the metric (equation (5) therein)

$$ds^2 = F(u,v) \, du^2 + 2X(u,v) \, du dv + Y^2(u,v) \, d\Omega^2 \tag{5}$$

with yet unspecified coordinates $u,v$. By Einsteins vacuum field equations they obtain the additional equation $X = \xi(u) \, \partial_v Y$ (see the two sentences starting above equation (11) therein). With this equation and the fact that $dY = (\partial_u Y \, du + \partial_v Y \, dv)$ they get the metric

$$ds^2 = \Big( F - 2 \, \partial_u Y \Big) \, du^2 + 2 \xi \, du dY + Y^2 \, d\Omega^2 \tag{$\ast$}$$

So far so good. Now within equation (11) they claim that by simple coordinate transformation, one obtains a metric of the form

$$ds^2 = F \, du^2 \pm 2 \, du dv + v^2 \, d\Omega^2 \tag{11}$$

However, this is what surprises me. Comparing equation ($\ast$) and equation (11), it seems obvious, that $Y = \pm v$ must hold, otherwise the coefficient for the round metric on the unit 2-sphere would differ. But then $dY = \pm dv$ and the metric ($\ast$) becomes

$$ds^2 = \Big( F - 2 \xi \Big) du^2 + 2\xi \, du dv + v^2 \, d\Omega^2$$

So it does not coincide with what is claimed here. Furthermore the "transformation" $Y = v$ seems a bit odd to me as $Y$ is claimed to be a function of $u$ and $v$ and this "transformation" would just neglect the dependency on the coordinate $u$.

So what is meant here and what did I misunderstand. I do not believe that there is a mistake in the paper as Birkhoff's theorem is deduced correctly, so I assume I must be mistaken somewhere.

Update 1: I corrected the mistake pointed out by "Andrew" in the comments.

Octavius
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1 Answers1

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As already pointed out by the user @Pipe in a comment, you have a typo in your equation labelled ($\ast$). In the paper you reference it reads,

$ds^2 = \Big( F - 2 \partial_u Y \Big) \, du^2 + 2 \xi \, du dY + Y^2 d\Omega^2$,

without the $\xi$ in the $du^2$ term.

Having said that, I think the presentation in the paper could be a bit clearer. Equation ($5$) is presented in a footnote as following from a specific coordinate transformation, which includes taking $v=R$ (the radial coordinate). If you then take $Y(u, v)$ as the identity function on $v$, i.e. $Y(u, v)=v$ equation ($11$) follows immediately

If you are worried about the algebra, note that $Y=v$ implies that $\partial_u Y = 0$ and $\partial_v Y = 1$, while the total differential $dY = dv$.

edit

In response to your comment: In general you are free to choose whichever coordinates you like, although some will be more convenient. There is no reason that $Y$ has to depend on $u$ specifically. Perhaps you could explain why you think it should?

Martin C.
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