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Does anyone know of an example, preferably a simple one, that can be used to demonstrate that we can have a curvature singularity without a singularity in the Kretschmann scalar?
bcrowell said:Does anyone know of an example, preferably a simple one, that can be used to demonstrate that we can have a curvature singularity without a singularity in the Kretschmann scalar?
load(ctensor);
ct_coords:[t,x,y,z];
u:bessel_j(0,exp(t-z))^2;
lg:matrix([1,0,0,0],
[0,-u*exp(-2*exp(t-z)),0,0],
[0,0,-u*exp(2*exp(t-z)),0],
[0,0,0,-1]);
cmetric();
ricci(true);
lriemann(false);
uriemann(false);
exptdispflag:false;
scurvature();/* scalar curvature */
rinvariant(); /* Kretchmann */
load(ctensor);
dim:2;
ct_coords:[t,x];
u:1/(1+exp(t));
lg:matrix([u,0],
[0,-u]);
cmetric();
ricci(true);
lriemann(true);
uriemann(true);
exptdispflag:false;
scurvature();/* scalar curvature */
rinvariant(); /* Kretchmann */
PAllen said:Is this really a curvature singularity or just a case of coordinates describing a manifold that can be readily treated as a submanifold of one that has no geodesic inompleteness?
PAllen said:Well, for starters, while the upper index curvature tensor blows up as t→∞, I'm thinking (in my head, no full calculation), that the lower index does not blow up at t approaching infinite. This suggests t infinite is perfectly regular and only geodesically incomplete by chopping.
I'm guessing Schwarzschild.bcrowell said:What does SC stand for?
SC stands for Schwarzschild as PWiz guessed.bcrowell said:@PAllen: What does SC stand for?
BTW, google books won't let me access the link you posted in #2.
What is a strong curvature singularity? Does it mean the same thing as a curvature singularity? Is it defined in the link from #2?
This would indicate that the tidal forces experienced by the object approach infinity as the object reaches the end of its geodesic (i.e., as it approaches the point at which the object's worldline will be maximally extended). I can't see how the Riemann tensor/contraction of the Riemann tensor doesn't blow up here.PAllen said:is forced by compression/tension to have zero volume on on approach to the singularity.
I don't get this. A singularity isn't a point on the manifold - rather, a manifold is singular if it has one or more geodesics of finite affine length. You can't "deal away" with a singularity if a manifold is maximally extended.loislane said:manifolds in GR are defined with any singularity cut out(otherwise we wouldn't even have a smooth manifold in which doing calculus)
Also, wouldn't it be a good idea to define singularities only for maximally extended spacetimes? This way, you avoid cases like the Schwarzschild one for the EH, and you don't have to concern yourself with continuation issues.PAllen said:At the moment, I am suspicious that the at least the weak singularities mentioned in this thread are removable by continuous extension.
Determining whether a manifold is maximally extended wrestles with the same question as the incompleteness definition of a singularity. Things become much harder if one does not demand global analyticity. For example, there are many non-analytic, smooth extensions of the exterior SC manifold that are topologically distinct from the Kruskal manifold (which is only unique as a globally analytic maximal extension). Some years ago, I posted papers (whose main subject was what is the correct formal statement of Birkhoff's theorem) describing 'odd' smooth manifolds formed by gluing patches of Kruskal together in unusual ways to end up with everywhere spherically symmetric manifolds that are topologically distinct from Kruskal.PWiz said:Also, wouldn't it be a good idea to define singularities only for maximally extended spacetimes? This way, you avoid cases like the Schwarzschild one for the EH, and you don't have to concern yourself with continuation issues.
t t - 1 2 t 2 t t - 1
R= %e (%e + 1) R1= %e (16 %e + 32 %e + 16) R2= 0 R3=
4 t 4 t 3 t 2 t t - 1
%e (1024 %e + 4096 %e + 6144 %e + 4096 %e + 1024) M3=
4 t 4 t 3 t 2 t t - 1
%e (576 %e + 2304 %e + 3456 %e + 2304 %e + 576) M4= 0
2 t 2 t t - 1
W1= %e (24 %e + 48 %e + 24) W2=
3 t 3 t 2 t t - 1
%e (288 %e + 864 %e + 864 %e + 288) M1=
3 t 3 t 2 t t - 1
%e (96 %e + 288 %e + 288 %e + 96) M2=
4 t 4 t 3 t 2 t t - 1
%e (576 %e + 2304 %e + 3456 %e + 2304 %e + 576) M5=
5 t
(%i + 1) %e
5 t 4 t 3 t 2 t t - 1
(3456 %e + 17280 %e + 34560 %e + 34560 %e + 17280 %e + 3456)
test_end.mac
A curvature singularity is a point or region in spacetime where the curvature becomes infinite. This means that the laws of physics as we know them break down, making it impossible to predict the behavior of matter and energy at this point.
The Kretschmann scalar is a mathematical quantity used in general relativity to measure the curvature of spacetime. It is calculated by combining the components of the Riemann curvature tensor, which describes the local curvature of spacetime.
A well-behaved Kretschmann scalar is one that remains finite at a curvature singularity. In contrast, a regular Kretschmann scalar would become infinite at a curvature singularity, indicating a breakdown of the laws of physics. A well-behaved Kretschmann scalar is desirable as it allows for a more consistent and predictable description of spacetime.
Yes, it is possible for a curvature singularity to have a well-behaved Kretschmann scalar. In fact, some theoretical models, such as the Kerr black hole, predict the existence of such singularities. However, the existence of these singularities is still a topic of debate and further research is needed to fully understand them.
If a well-behaved Kretschmann scalar is confirmed to exist, it would have significant implications on our current understanding of space and time. It would suggest that the laws of physics are consistent and predictable even at the extreme conditions of a curvature singularity. This could lead to new insights and advancements in our understanding of the universe and its fundamental laws.