Talk:Kruskal-Szekeres coordinates
From Wikipedia, the free encyclopedia
 |
This article is within the scope of WikiProject Physics, which collaborates on articles related to physics. |
| Stub | This article has been rated as Stub-Class on the assessment scale. |
| ??? | This article has not yet received an importance rating within physics. |
|
Help with this template This article has been rated but has no comments. If appropriate, please review the article and leave comments here to identify the strengths and weaknesses of the article and what work it will need. |
|
 |
This article has been automatically assessed as Stub-Class by WikiProject Physics because it uses a stub template.
|
Contents |
[edit] Expert needed?
I see a note that this page needs the attention of an expert. I suppose that it lacks the format commonly found in Wikipedia pages which begins with a high level introduction, followed by substantive content.
I am not an expert, but a borderline specialist. I find the page useful and I am trying to verify some of the technical detail.
It is a bit confusing that there are two variables 'r'—one from classic Schwarzschild metric, and one implicitly defined for the Kruskal-Szekeres line element. Perhaps they are the same.
--NormHardy 20:12, 13 August 2006 (UTC)
The two r are the same (unless I'm misunderstanding your question) — the Schwarzschild coordinate. JanBielawski 21:39, 28 September 2006 (UTC)
[edit] Question, maybe a suggestion
How can it be seen that the Kruskal-Szekeres metric has an horizon? It could be useful to indicate that. Similarly, in the article on Schwarzschild metric, the horizon seems obvious because of the apparent singularity. But what is the real way to look for an horizon in a metric?
(Partial) answer. It depends what kind of horizon you are looking for. Event horizons are often tricky to find generally. In stationary spacetimes they coincide with Killing horizons, which are easy to find if you know what the Killing vectors are since the norm of a vector is an invariant. In general spacetimes, one (slow) way to find the event horizon numerically is just to propagate lots of null shells through the metric and see where they end up. Apparent horizons are also reasonably easy to locate. Pick a spacelike hypersurface, choose a compact two-surface within the hypersurface and see if the null normals have zero expansion. In Schwarzschild (in any coordinates) the event horizon can be found by looking at where the expansion of radially outgoing null geodesics is zero.
[edit] Too much math talk ?
Topics like these make me wonder, to some the math may be clear, but why not a description of what the scientist had in mind perhaps some images??? Words are a language to me, math is only rules but in itself it doesnt form a story to me, while wiki is suposed to explain. —The preceding unsigned comment was added by 82.217.143.153 (talk) 01:46, 14 January 2007 (UTC).
[edit] Dimensionality of the singularity
On 6 May 2008, TimothyRias (talk · contribs) removed the sentence "Note that in these coordinates the curvature singularity is in fact represented by a curved line, that is, it is one dimensional.". He justified this by saying that "dimensionality of the singularity is pretty much undefined. If anything it is either 0 or 3 dimensional". The curvature is finite except at the singularity which occurs at r=0. As one approaches r=0, the contributions of variations in the co-latitude and longitude to the line element become zero because of the factor of r2 in r2dΩ2. As the article correctly says the "curvature singularity is given by the equation UV = 1". Thus one of U or V may be varied freely (with the other determined thereby) while remaining at the singularity. Thus three dimensions have either been constrained or rendered null while one dimension remains. In other words, the singularity is one dimensional. JRSpriggs (talk) 17:21, 7 May 2008 (UTC)
