If one considers the Schwarzschild metric

$$ \text d s^2 = -V(r)\text d t^2 + \frac{1}{V(r)}\text d r^2 + r^2 \text d \Omega^2\;,\qquad V(r) = 1-\frac{2m}{r}\;, $$

and introduces the Eddington-Finkelstein coordinates

$$ v = t+f(r)\;,\qquad u = t-f(r)\;, \qquad f'(r) = \frac{1}{V(r)} \;. $$

the coordinate singularity at $r=2m$ vanishes and the metric reads:

$$ \text d s^2 = -V(r)\text d v\text d u + r^2\text d \Omega^2\,. $$

However, by integrating $f'$ to get $f$ I would calculate $$ f(r) = \int \frac{1}{V(r)}\text d r = \int \left( 1+\frac{2m}{r-2m} \right) d r = r + 2m\ln|r-2m| \;. $$

But in books and lecture notes I always read $\ln(r-2m)$ instead of $\ln|r-2m|$. Isn't that wrong? In my opinion this makes a difference.

For example when I introduce the Kruskal-Szekeres coordinates $$ \tan V = \text e ^{\alpha v}\;,\qquad \tan U = -\text e ^{-\alpha u}\;,\qquad \alpha = \frac{1}{4m}\, $$ I get $$ \tan V \tan U = -\text e ^{\frac{r}{2m}} |r-2m| $$ instead of $$ \tan V \tan U = -\text e ^{\frac{r}{2m}} (r-2m) $$ But these two cases lead to different Penrose diagrams.

So why is it wrong to to use the absolute value of $r-2m$ in the logarithm?

PS: When it is about the Reissner-Nordstrom metric all books and notes I find surprisingly use the magnitude in the argument of the logarithm. But why not in the case of Schwarzschild?

  • $\begingroup$ I think the EF coordinates are defined differently depending on if you are inside or outside of the horizon, thus eliminating the need for the absolute value. $\endgroup$ – Ryan Unger Feb 11 '15 at 19:20
  • $\begingroup$ As far as I know this is not the case. Because the expression for the Kruskal-Szekeres coordinates $\tan V \tan U = -\text e ^{\frac{r}{2m}} (r-2m)$ does not cause any problems inside or outside the black hole (so even when you don't use the absolute values...). $\endgroup$ – thyme Feb 12 '15 at 15:53

The issue here is that the Schwarzschild coordinates are divided into two disconnected patches by the coordinate singularity at $r=2m$. There is no physical connection and the two pieces can be viewed as separate solutions.

The Eddington-Finkelstein coordinates take the outer solution and extend it beyond the horizon, but inside the horizon it is different from the Schwarzschild solution beyond the coordinate sigularity.

A simple mathematical reason for the absence of the absolute value would be that the coordinate change would not be a bijection including the absolute value as the function $v(r,t)=t+r+2m \log|r-2m|$ is not injective on $r\in[0,\infty)$. As stated above, the change of variables is only supposed to map the outer part of the spacetime $r\in(2m,\infty)$ into the Eddington-Finkelstein coordinates.

The confusing part is that using $v(r,t)= t+r+2m \log|r-2m|$ can indeed also be used to map the inner part of the Schwarzschild coordinates into the Eddington-Finkelstein coordinates, but because we have to require the change of coordinates to be one-to-one we can then only define it for $r\in(0,2m)$.

So the change of coordinates $v(r,t)= t+r+2m \log|r-2m|$ can be used separately for the inner and outer sections of the Schwarzschild coordinates to map them into Eddington-Finkelstein coordinates, but not both sections simultaneously, i.e. the Eddington-Finkelstein coordinates can be seen an extension of both the inner and the outer Schwarzschild solution but they cannot incorporate both at the same time.

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  • $\begingroup$ Great answer...i was about to write the same......upvoted! $\endgroup$ – GRrocks Mar 26 '15 at 3:14

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