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Are Finkelstein/Kruskal interior black hole solution compatible with Einstein's GR? 
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#73
Dec412, 09:55 PM

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PF Gold
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The definition of manifold has no requirement of geodesic completeness. A 2sphere minus a pair of antipodal points is a riemanninan manifold, for example. There is a requirement that it be a collection of open sets, but there is no completeness requirement. As I said, if you don't impose some other defining condition for the manifold, you see the SC metric corresponds to two patches, and if you want the most complete solution you have to investigate further. However, there is nothing wrong (mathematically  physically is another matter) with declaring r> Rs as a boundary condition. (Physically, I stand by my argument that the boundary condition r> Rs leads to a violation of the equivalence principle for classical GR; but for mathematical GR, it's not a problem). 


#74
Dec412, 10:04 PM

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PF Gold
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#75
Dec412, 10:10 PM

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I'm with PAllen on this one. Since the exterior Schwarzschild coordinates only include the portion of the manifold outside the horizon the horizon is a boundary and the Riemann curvature at the boundary is a boundary condition. This boundary condition can be characterized by a single parameter, usually denoted M.



#76
Dec412, 10:14 PM

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PF Gold
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#77
Dec412, 10:18 PM

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#78
Dec412, 10:31 PM

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I would call it a boundary condition because you need to know the value on that edge in order to solve the differential equation. That is what a boundary condition is. But, you need to know more than just that there is an edge, you need to know the value of your function at the edge.
Sorry, I am rambling, I think I will go sleep. 


#79
Dec412, 10:45 PM

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If the actual, physical object is a disk with no hole, but I restrict attention to an annular region, then no, that "boundary condition" does not affect the solution; I have to solve the equations for a disk with no hole, and then restrict attention to the portion of the solution that I'm interested in. If I solve the equations with the "boundary conditions" included, I will get the wrong solution, the one for a disk with a hole, not the one for a disk with no hole where I'm not looking at a region in the center. Or let's consider this case: I'm modeling a capacitor. Consider two different possibilities: (1) The actual, physical capacitor has finitesized plates, and I'm interested in their entire area. My solution will then have edge effects because of the finite plate size; in other words, the boundary condition affects the solution. (2) The actual, physical capacitor has extremely large plates, but I'm only interested in a small area near the center. I impose a "boundary condition" of a small plate area; but if I then solve the equation using that condition, I will get the wrong answer: my solution will have edge effects, and there aren't any in the small area I'm consideringthe edge effects are way out in a different place that I'm not modeling. To get the right answer, I have to ignore the "boundary condition" and instead solve the equations with a much larger "real" plate size, and then restrict my solution to the small area I'm interested in. (What I'll really do, of course, is solve the equations assuming infinitely large plates, but that's just a shortcut to make the math easier since I know it will give the same answer as the more complicated procedure that's strictly correct.) The "boundary condition" here does *not* affect the solution, which is why I would prefer *not* to call it a "boundary condition". The parallel with the case of Schwarzschild spacetime should be clear. If I impose a "boundary condition" r > Rs, that doesn't change any of the actual geometric invariants at Rs or as r > Rs. Those invariants are all given by the solution with no boundary condition at the horizon, the one I derive purely from the EFE with the assumptions of spherical symmetry and vacuum plus the coordinate conditions on the line element. The boundary condition adds nothing to the solution; it's just a way of restricting attention to a portion of the complete manifold. Contrast this with, for example, a solution for a static, spherically symmetric star. Here there is a boundary condition that does affect the solution: there will be some radial coordinate r at which the spacetime is no longer vacuum. That does affect the geometric invariants. 


#80
Dec412, 11:01 PM

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PF Gold
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I'm looking at it this way:
1) If I want SC coordinates as one or two patches of a more complete manifold, there are only boundary conditions at infinity; further, I need to do further analysis to find the complete manifold consistent with no boundary conditions except at infinity. 2) If I want SC exterior coordinates to represent a complete GR solution (not a patch of a more complete solution), then I need additional boundary conditions. I must posit that r > Rs, and that the metric remains vacuum and spherically symmetric at this boundary. (How do I know the value of Rs beforehand? It doesn't matter. I can try (1), and let this suggest (2).) For mathematical GR, (1) leads to the unique complete Kruskal manifold; (2) leads to one exterior region considered as a solution (manifold) unto itself. In either case, any of dozens of popular coordinates can be used (along with infinite others). Physically, I can argue that (2) is absurd even on local grounds, and (1) is also absurd for different (global) reasons. To get an idealized, physically plausible model of something, the simplest is OS collapse, incorporating part of one interior and one exterior sheet of the complete Kruskal manifold in its 'late' stage. 


#81
Dec412, 11:08 PM

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#82
Dec512, 06:00 AM

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#83
Dec512, 06:13 AM

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Obviously, if you use the "nonconductive" boundary condition in a situation where the boundary is "conductive" then you will get a wrong answer, but that is because you used the wrong boundary condition, not because what you used isn't a boundary condition nor because you needed to solve a bigger problem. 


#84
Dec512, 06:48 AM

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#85
Dec512, 09:09 AM

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PF Gold
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[Edit: being more specific, suppose I don't know the form of the solution, and don't know how much information needs to be specified on the boundary to get a unique solution. So, since I am assuming vaccuum, I know the Ricci scalar is identically zero. I make a guess  let's set an arbitrary constant value of the Kretschmann scalar for the inner boundary, and further require staticity. Then, I find a unique solution (as long as I don't pick K too large; if I do, I can't meet staticity). Or, don't impose staticity. Then, you find for given boundary r, if you pick K above a critical value, you have a horizon in your solution at some greater r than your boundary. Without the extra boundary condition, you find a one parameter family of solutions, of more general 'shape'  the Kruskal geometries. You can then use Newtonian behavior at weak field as a condition to define the free parameter physically.] 


#86
Dec512, 09:27 AM

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PF Gold
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But again, this is really a matter of terminology, now with the term "solution"; I am using it to mean the Schwarzschild line element, without specifying M; you (and PAllen; I'll respond to his post separately) are using it to mean the Schwarzschild line element, *plus* the specification of M. With that usage of the term, I agree with what you're saying; the line element itself only gives a "partial solution". 


#87
Dec512, 09:30 AM

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PF Gold
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The only possible issue here is that, as I said in my response to DaleSpam, until you've got the partial solution (the line element, which gives you the curvature components explicitly in terms of M), you don't know how to specify the boundary condition that picks out one particular geometry from the infinite family that the line element describes. But that seems like a minor issue and I don't have a handy term for it anyway, so I'll stop here. 


#88
Dec512, 09:44 AM

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#89
Dec512, 10:00 AM

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PF Gold
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If the discussion is about what Einstein said, then yes, asking for Coke is perfectly reasonable. But if the discussion is about what's currently used in GR, then a client who keeps asking for Coke even after everybody has pointed out repeatedly that the discussion is really about Pepsi would seem a little weird. 


#90
Dec512, 10:35 AM

PF Gold
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Anyhow, I'll stay out of this now. 


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