Coordinate singularities and coordinate transformations

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I have a metric of the form [tex] ds^2 = (1-r^2)dt^2 -\frac{1}{1-r^2}dr^2-r^2 d\theta^2 - r^2 sin^2\theta d\phi^2 [/tex]

A singularity exists at [tex] r=\pm 1 [/tex]. By calculating [tex] R^{abcd}R_{abcd} [/tex] i found out that this singularity is a coordinate singularity.

I found the geodesic equations for radial photons and performed the eddington-finkelstein coordinate transformation.

my metric turns out to be:

[tex] ds^2 = (1-r^2)dudv-r^2 d\theta^2 - r^2 sin^2\theta d\phi^2 [/tex]

where u and v are the constants of integration for the outgoing and incoming radial photons.

I also took a further coordinate transformation where T=(u+v)/2 and X = (u-v)/2

and the metric takes the form:

[tex] ds^2 = (1-r^2)dT^2-(1-r^2)dX^2-r^2 d\theta^2 - r^2 sin^2\theta d\phi^2 [/tex]

However for [tex] r=\pm 1 [/tex] the metric still does not behave properly.

Do you suggest any other coordinate transformation? Thank you.
 

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  • #2
Dale
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In what way does it not behave properly? There does not appear to be a singularity. (I didn't check your math)
 
  • #3
bcrowell
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A singularity exists at [tex] r=\pm 1 [/tex]. By calculating [tex] R^{abcd}R_{abcd} [/tex] i found out that this singularity is a coordinate singularity.

First I wanted to point out a couple of things about this part of your post.

One is that just because a particular curvature scalar is not singular, that doesn't mean that there's no singularity there. Only the converse is true: if some curvature scalar *does* blow up, then the singularity *isn't* just a coordinate singularity. Also, note that the Kretchmann invariant isn't the only curvature scalar that can be constructed from the Riemann tensor.

The other is that I doubt that it makes sense to refer to r=-1. I'm pretty sure your solution can't be continued to negative r.

Anyway, getting back to your question, one thing I would suggest is that before you start doing coordinate transformations that are analogous to the ones that we know are helpful in the case of the Schwarzschild metric, I would suggest seeing if you can transform to coordinates such that at large r, the metric is flat. When you calculated the Kretchmann invariant, how did you find that it depended on r and t? This would help in figuring out if there is an asymptotically flat part of your spacetime somewhere. If you can find a relation between r and t such that the Kretchmann invariant vanishes, then presumably that is where asymptotically flat infinity is hiding. Note that r is timelike for large r.

Is this a vacuum solution with zero cosmological constant? If so, then I think Birkhoff's theorem says it has to be the Schwarzschild solution written in unusual coordinates, and you can probably locate the Schwarzschild singularity by looking for a relation between r and t such that the Kretchmann invariant blows up.
 
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  • #4
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1. Actually, yes you're right i can't take r = -1.
2. I do have a cosmological constant + a gauss-bonnet term.
3. To detect the singularities, i look at my metric and check at what values of r it might blow up. Then, i look at the kretchmann invariant to check whether the singularities are real or coordinate. IS there any other way to find the singularities of my metric?
 
  • #5
Ich
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Your metric is de Sitter. There are horizons, but no singularities.
 
  • #6
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Your metric is de Sitter. There are horizons, but no singularities.

Can you explain more? :)
 
  • #7
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so what i have is only points at which the r coordinate becomes timelike and the t coordinate becomes spacelike. This is what happens at r=1. But, i do not have real singularities similar to r=0 in schwarzschild. right?

if i want to think of some test particle that exists in my spacetime, it would not have any "odd" behavior?
 
  • #8
Ich
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