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Both positive and negative curvature?by anorlunda
Tags: curvature 
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#1
Jan1313, 06:45 AM

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What happens to the Reimann tensor at the event horizon of a black hole? Do some of the 24 components become zero or infinite?
What happens to parallel transport of a vector on the surface of an event horizon that is different than on a surface outside the event horizon? I'm newly educated on general relativity and hopeful that these new tools may enable me to see things differently. 


#2
Jan1313, 07:14 AM

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#3
Jan1313, 07:17 AM

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In standard Schwarzschild coordinates there are many zero components throughout the spacetime and there are some infinite components in the limit as you approach the horizon. In Kruskal Szekeres coordinates there are also many zero components throughout the spacetime but there are no infinite components in the limit as you approach the horizon. All curvature invariants are finite at the horizon. Furthermore, the larger the black hole the smaller the curvature invariants at the horizon. For a supermassive black hole there could be less tidal gravity than at the surface of the earth. 


#4
Jan1313, 07:24 AM

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Both positive and negative curvature?
[tex]{R^{\theta}}_{r \theta r}={R^{\phi}}_{r \phi r}=\frac{M}{(2Mr)r^2}[/tex] goes to ∞ as r goes to 2 M. 


#5
Jan1413, 10:19 AM

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#6
Jan1413, 10:36 AM

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#8
Jan1413, 02:13 PM

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My bad  I shold have calculated them rather than going frommemory.



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