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A black hole itself (the singularity) is 1 dimensional - a single infinitesimal point. The event horizon around it is roughly a sphere. The diagrams that you ar

A black hole itself (the singularity) is 1 dimensional - a single infinitesimal point. The event horizon around it is roughly a sphere. The diagrams that you are talking about are a visual metaphor that represent 3D space as a 2D plane and the 3rd dimension standing in for the influence of gravity. IRL, spacetime has 3 spatial dimensions, not 2, and gravity is not a dimension but a force. It's hard to visually represent gravitational distortion of 3 dimensional space without a 4th spatial dimension to do it with, so textbook diagrams use a 2D plane.

pflats 7y ago
I think you've got a typo here: points should be 0-dimensional.
- curlypaul924 7y ago
  Are you sure it's a typo? https://physics.stackexchange.com/a/194947 https://physics.stackexchange.com/a/194947 (I'm not a physicist.)
  - checkyoursudo 7y ago
    I believe the previous poster is referring to a typo following from this: >A black hole itself (the singularity) is 1 dimensional - a single infinitesimal point. In Euclidean geometry, A cube is 3 dimensions. A plane is 2 dimensions. A line is 1 dimension. A point is ...
  - codethief 7y ago
    Not sure why you're getting downvoted, this is a legitimate question. However, in this particular case, I think the final sentence of the accepted answer on Physics.SE, namely that > In this diagram the singularity is a line in spacetime i.e. a one dimensional object in spacetime. is wrong or at least very misleading – the answer does (correctly) say that asking for the "dimensionality of a singularity […] is a meaningless question because the spacetime geometry is undefined at a singularity".
- cuspycode 7y ago
  A point is 0-dimensional, but in a space-time diagram the time dimension is added, which makes it 1-dimensional. Penrose diagrams for black holes assume spherical symmetry, so all of space is represented by a single radial coordinate, which makes it possible to display such diagrams in 1+1=2 dimensions.
  - codethief 7y ago
    A singularity doesn't have a dimension. It is a portion of spacetime that is missing, not a point or set of points. We can't define its dimensionality, either.(×) What we can say is that the singularity in Schwarzschild black holes is spacelike. ×) Counterexample: Consider the manifold M := R³\B, where B is the closed unit ball, equipped with the standard Euclidean metric. This manifold is certainly not Cauchy-complete and we can reach the singularity at r=1 in finite time. Now, if we had to define the dimension of the singularity, what dimension n should it have? n=2 (a sphere)? Maybe. At least we could extend M by the unit sphere to make it complete. But could the singularity also be a point (i.e. n=1)? Yes, certainly. By diffeomorphism invariance, we could simply find new coordinates and map R³\B to R³\{0}, so the singularity would suddenly become a point. So, as you can see, interpreting the singularity as a point or set of points that have a topological dimension doesn't work.
antidesitter 7y ago
> A black hole itself (the singularity) is 1 dimensional - a single infinitesimal point. A point is 0-dimensional. A line is 1-dimensional. The singularity is 1-dimensional (more precisely, a ring) if the black hole is rotating. [1] [1] https://en.wikipedia.org/wiki/Ring_singularity https://en.wikipedia.org/wiki/Ring_singularity
pavelrub 7y ago
The singularity is likely a mathematical artifact of the fact that GR is insufficient to describe black holes. In reality (quantum gravity) they probably do not exist.
- lostmsu 7y ago
  I would not be so sure. Its presence might simply indicate, that black hole's inner volume is infinite.
  - codethief 7y ago
    Could you elaborate on your definition of volume here (are you talking about spatial volume or spacetime volume?) and how the curvature going to infinity at the singularity should imply its infiniteness? My thought process here is the following: The inside of an (eternal) black hole carries four (Schwarzschild) coordinates t, r, theta, phi – r now being timelike and confined to the interval (0, 2M) and t now being spacelike and being any real number. That is, depending on when (at what time t) you cross the event horizon, you end up at a different point in space. The singularity at r=0 is then a point in your future which, like your own death, you cannot actually see but which you will nevertheless hit in finite proper time. So in this sense I'd say the volume is very finite (if we disregard the (trivially unbounded) spacelike coordinate t which, as mentioned before, simply corresponds to the time of entering the BH).
    - codethief 7y ago
      I should add: 1. In the interior of the BH, the determinant of the metric is bounded since the Schwarzschild factors in the metric cancel out. So the volume measure doesn't do anything crazy as one gets closer to the singularity and boundedness of coordinates implies boundedness of the volume. Again, I'm disregarding the spacelike t coordinate because to me the relevant fact is that all matter reaches the singularity in finite proper time, so while we could theoretically stack lots of (actually, an infinite amount of) (massless) cubes inside a black hole, they would soon all get crushed. 2. Of course the situation is slightly different if we're talking about a growing black hole whose mass (and, therefore, radius) increases as we throw matter into it.

pflats 7y ago

I think you've got a typo here: points should be 0-dimensional.

curlypaul924 7y ago
Are you sure it's a typo? https://physics.stackexchange.com/a/194947 https://physics.stackexchange.com/a/194947 (I'm not a physicist.)
- checkyoursudo 7y ago
  I believe the previous poster is referring to a typo following from this: >A black hole itself (the singularity) is 1 dimensional - a single infinitesimal point. In Euclidean geometry, A cube is 3 dimensions. A plane is 2 dimensions. A line is 1 dimension. A point is ...
- codethief 7y ago
  Not sure why you're getting downvoted, this is a legitimate question. However, in this particular case, I think the final sentence of the accepted answer on Physics.SE, namely that > In this diagram the singularity is a line in spacetime i.e. a one dimensional object in spacetime. is wrong or at least very misleading – the answer does (correctly) say that asking for the "dimensionality of a singularity […] is a meaningless question because the spacetime geometry is undefined at a singularity".
cuspycode 7y ago
A point is 0-dimensional, but in a space-time diagram the time dimension is added, which makes it 1-dimensional. Penrose diagrams for black holes assume spherical symmetry, so all of space is represented by a single radial coordinate, which makes it possible to display such diagrams in 1+1=2 dimensions.
- codethief 7y ago
  A singularity doesn't have a dimension. It is a portion of spacetime that is missing, not a point or set of points. We can't define its dimensionality, either.(×) What we can say is that the singularity in Schwarzschild black holes is spacelike. ×) Counterexample: Consider the manifold M := R³\B, where B is the closed unit ball, equipped with the standard Euclidean metric. This manifold is certainly not Cauchy-complete and we can reach the singularity at r=1 in finite time. Now, if we had to define the dimension of the singularity, what dimension n should it have? n=2 (a sphere)? Maybe. At least we could extend M by the unit sphere to make it complete. But could the singularity also be a point (i.e. n=1)? Yes, certainly. By diffeomorphism invariance, we could simply find new coordinates and map R³\B to R³\{0}, so the singularity would suddenly become a point. So, as you can see, interpreting the singularity as a point or set of points that have a topological dimension doesn't work.

antidesitter 7y ago

> A black hole itself (the singularity) is 1 dimensional - a single infinitesimal point. A point is 0-dimensional. A line is 1-dimensional. The singularity is 1-dimensional (more precisely, a ring) if the black hole is rotating. [1] [1] https://en.wikipedia.org/wiki/Ring_singularity https://en.wikipedia.org/wiki/Ring_singularity

pavelrub 7y ago

The singularity is likely a mathematical artifact of the fact that GR is insufficient to describe black holes. In reality (quantum gravity) they probably do not exist.

lostmsu 7y ago
I would not be so sure. Its presence might simply indicate, that black hole's inner volume is infinite.
- codethief 7y ago
  Could you elaborate on your definition of volume here (are you talking about spatial volume or spacetime volume?) and how the curvature going to infinity at the singularity should imply its infiniteness? My thought process here is the following: The inside of an (eternal) black hole carries four (Schwarzschild) coordinates t, r, theta, phi – r now being timelike and confined to the interval (0, 2M) and t now being spacelike and being any real number. That is, depending on when (at what time t) you cross the event horizon, you end up at a different point in space. The singularity at r=0 is then a point in your future which, like your own death, you cannot actually see but which you will nevertheless hit in finite proper time. So in this sense I'd say the volume is very finite (if we disregard the (trivially unbounded) spacelike coordinate t which, as mentioned before, simply corresponds to the time of entering the BH).
  - codethief 7y ago
    I should add: 1. In the interior of the BH, the determinant of the metric is bounded since the Schwarzschild factors in the metric cancel out. So the volume measure doesn't do anything crazy as one gets closer to the singularity and boundedness of coordinates implies boundedness of the volume. Again, I'm disregarding the spacelike t coordinate because to me the relevant fact is that all matter reaches the singularity in finite proper time, so while we could theoretically stack lots of (actually, an infinite amount of) (massless) cubes inside a black hole, they would soon all get crushed. 2. Of course the situation is slightly different if we're talking about a growing black hole whose mass (and, therefore, radius) increases as we throw matter into it.