Schwarzchild metric spherically symmetric space or s-t?

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This is probably a stupid question, but, is the Schwarzchild metric spherically symmetric just with respect to space or space-time?

Looking at the derivation, my thoughts are that it is just wrt space because the derivation is use of 3 space-like Killing vectors , these describe 2-spheres, and ##S^{2}## spheres foliate ##ℝ^{3}##...

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Timelike and spacelike separation are always different, and lightlike motion is different from a "diagonal line" in space. I don't see how anything could have a spherical symmetry in spacetime. The Schwarzschild metric is just symmetric in space.
 
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bcrowell
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Yes, it's just spatial. There is no useful notion of spherical symmetry in the sense of 3-spheres, because there is no useful notion of four-dimensional rotation. The closest thing we have to a rotation mixing space and time is actually a boost, and boosts are not the same as rotations.
 
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Thanks. And why the 3 spatial dimensions, why not 2 spatial and 1 time?
 
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bcrowell
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Thanks. And why the 3 spatial dimensions, why not 2 spatial and 1 time?
Could you clarify what the question is?
 
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Could you clarify what the question is?
Sorry, so we have a notion of 2-spheres that foliate 3-d space, but not 3-spheres that foliate 4-d space; why is the spherical symmetry of the Schwarzschild metric spatial , not, say 2 space dimentions and 1 time dimension?
 
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The time dimension is different from the spatial dimensions.
Did you ever see anything symmetric in time?
 
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The time dimension is different from the spatial dimensions.
Did you ever see anything symmetric in time?
Sorry I don't understand the question, Aren't flat space-time and de-sitter space-time maximally symmetric?
 
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PeterDonis
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we have a notion of 2-spheres that foliate 3-d space, but not 3-spheres that foliate 4-d space
Sure we do; there are 4-d spacetimes (such as closed FRW spacetime) that are foliated by spacelike 3-spheres.
why is the spherical symmetry of the Schwarzschild metric spatial , not, say 2 space dimentions and 1 time dimension?
Because "spherical symmetry" means "there is a set of Killing vector fields closed under commutation and with closed integral curves". A spacetime that was "spherically symmetric in time" would have to have a set of such Killing vector fields with one of them being timelike. I'm not aware of any such spacetime, but even if there is one, it certainly is not Schwarzschild spacetime; Schwarzschild spacetime does have a timelike Killing vector field (at least, it does outside the horizon), but that Killing vector field does not have closed integral curves and its commutator with the spherical symmetry Killing vector fields (which are spacelike) is zero.

Aren't flat space-time and de-sitter space-time maximally symmetric?
"Maximally symmetric" does not mean "all the dimensions are the same". It means that those spacetimes have the maximum possible number of Killing vector fields (in 4-d spacetime, that number is 10). It does not mean that all of those Killing vector fields are the same as the ones that define spherical symmetry.

If you'll notice, I've thrown a bunch of jargon at you; that was deliberate. If you don't understand the terms I used above, I strongly suggest looking them up and taking the time to understand them. They are critical to a proper understanding of what "symmetry" in general and "spherical symmetry" in particular mean, and I think that understanding will help to answer your questions.
 
  • #10
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Sorry I don't understand the question, Aren't flat space-time and de-sitter space-time maximally symmetric?
Okay, to be more precise: Did you ever see anything symmetric, but not constant in time?
Or even worse, something that is symmetric in a time/space plane. How would such a concept even look like? Motion through time is always different from motion through space, so you always have a way to break the symmetry.
 

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