Coordinate singularity in Schwarzschild solution

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SUMMARY

The discussion centers on the coordinate singularity in the Schwarzschild solution of a black hole, specifically addressing the line element given by ds²=(1-2m/r)dτ²+(1-2m/r)⁻¹dr²+r²(dθ²+sin²(θ)dφ²). Participants clarify that the apparent singularity at r=0 is likely a misstatement, and the correct focus should be on r=2m. By treating τ as an angular coordinate, which ranges from 0 to 2π, the singularity can be shown to be a coordinate singularity rather than a physical one.

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Confused Physicist
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Hi! I have the following problem I don't really know how to approach. Could someone give me a hand?

The line element of a black hole is given by: ds^2=\Bigg(1-\frac{2m}{r}\Bigg)d\tau ^2+\Bigg(1-\frac{2m}{r}\Bigg)^{-1} dr^2+r^2\Big(d\theta ^2+\sin^2(\theta)d\phi ^2\Big)

It has an apparent singularity at ##r=0##. By making ##\tau## an angular coordinate, show that this singularity is a coordinate singularity (not physical) and find the period of ##\tau## that makes it possible. (consider expanding the metric functions about ##r=2m##).

Thanks for the help!
 
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Confused Physicist said:
The line element of a black hole

Note that this is not the usual Schwarzschild metric; there is no minus sign in front of the first term on the RHS. (If there were, the question would not make sense because ##\tau## would not be able to be treatd as an angular coordinate.)
 
@Confused Physicist I have moved this thread to the homework forum. You will need to at least show some attempt at a solution. You could start by taking the hint in the parenthetical in the OP.
 
Confused Physicist said:
It has an apparent singularity at ##r=0##. By making ##\tau## an angular coordinate, show that this singularity is a coordinate singularity

Are you sure the problem statement says ##r = 0## and not ##r = 2m##?
 
PeterDonis said:
Are you sure the problem statement says ##r = 0## and not ##r = 2m##?

Thanks PeterDonis, I will post my future questions in the homework forum. I've been trying to squeeze my head around it, but I haven't posted my attempt because I literally don't have a decent one.

Yes, the problem says ##r=0##, but you're right. I believe it's a mistake and it should say ##r=2m##.
 
Last edited:
PeterDonis said:
Note that this is not the usual Schwarzschild metric; there is no minus sign in front of the first term on the RHS. (If there were, the question would not make sense because ##\tau## would not be able to be treatd as an angular coordinate.)

What does it mean to treat ##\tau## as an angular coordinate? Is it a specific change of variable?
 
Confused Physicist said:
What does it mean to treat ##\tau## as an angular coordinate?

It means it only covers the range ##0## to ##2 \pi## instead of ##- \infty## to ##+ \infty##.
 

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