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by bawana
1754 days ago
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Lots of discussion here about higher dimensional matrices. The argument is that the dot product is an operation on two arguments, therefore any multidimensional matrix multiplication can be broken down into multiple two dimensional operations. The image offerred is that two vectors define a plane so any two dimensional operation is valid. But how about curved space? Suppose I have two vectorsin curved space-they look like two droopy arrows. How does one calculate this dot product? I guess the the local curvature has to be taken into account. In spherically shaped space, this is a constant, but what about irregularly bent space-like that near an amorphous blob of dark matter? Suppose you are trying to calculate electrostatic potentials or do anything with Maxwell's equations? Currently, we define the curvature of space with gravity or the deflection of light past massive objects. Is there a way to measure the curvature of space locally? Can the curvature of space alter the piezoelectric potential generated by a crystal for example and allow its measurement- I am thinking that a miniscule deformation on a large macroscopic object is multiplied many fold on a property that is distributed over the atomic level and happening in parallel on all the atoms of a crystal. |
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No, they don't. In math, curved spaces are modeled through so-called manifolds which locally look like ℝ^n. In particular, at any point p of the manifold there's a tangent space, i.e. a linear space (higher-dimensional plane) tangent to the manifold at p. Vectors at p are just vectors in that linear space. So they are "straight" not "droopy".
On the tangent space of each point p you can now define an inner product g(p). The resulting family of inner products g is called a (Riemannian) metric on the manifold[0] and describes how lengths (of vectors) and angles (between vectors) can be measured at each point.
> Currently, we define the curvature of space with gravity or the deflection of light past massive objects. Is there a way to measure the curvature of space locally?
Yes, there is. In fact, the curvature[1] of a (Riemannian) manifold is a purely local quantity – it's basically the second derivative of the metric g, so it describes how the notion of length changes (more precisely: how the change in length changes) as you go from a point p to neighboring points.
There are other ways to express what curvature is, e.g. by locally parallelly transporting[2] a vector along a closed curve (and making that curve smaller and smaller) which basically measures how the notion of straight lines changes locally. (Though, since a line being "straight" means "locally length minimizing" this brings us back to the notion of length and, thus, the metric.)
Alternatively, if the manifold has dimension 2, there's a particularly simple way of looking at and interpreting curvature, see [3].
In any case, curvature being used to model gravity is entirely separate from that idea.
[0]: Provided this inner product "varies smoothly" as you move from p to a neighboring point q.
[1]: https://en.wikipedia.org/wiki/Riemann_curvature_tensor
[2]: https://en.wikipedia.org/wiki/Parallel_transport
[3]: https://en.wikipedia.org/wiki/Gaussian_curvature