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  What is the benefit of using simplicial manifolds in the regularization of gravity?

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In the Causal Dynamical Triangulation (CDT) approach to quantum gravity, space-time is regularized by a triangulated (simplicial) manifold. All configurations of the theory at a given scale are given by simplicial manifolds of the same topology with a fixed edge length. The curvature on each of these configurations is not a field on this lattice but is instead intrinsic to the geometry of the manifold and is located along the lower dimensional simplices. I was just wondering what makes regularization of gravity this way more ideal than say a square lattice with the curvature a field on the lattice?  I'm guessing that there's some sort of universality and that perhaps it could be equivalently be modeled on a standard lattice. 

asked Jun 15, 2015 in Theoretical Physics by coarsegrained (60 points) [ no revision ]

1 Answer

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The reason for using a dynamical grid rather than a square grid is the same as why people working in fluid dynamics use adaptive grids rather than square ones - they adapt much better to the solution. Even if one starts out with a regular grid, different regions very soon need different resolution, which can be captured numerically (without unduly large work) only with an adaptive grid.

To get the same numerical accuracy with a square grid one would need an much finer (indeed typically extremely fine) lattice spacing.

A second reason is that in general relativity one wants to consider topologies different from the topology of Minkowski space - and possible even changes in topology. This cannot be done by a square lattice. On the other hand, a triangulation of a manifold completely describes its topology, hence is the appropriate tool for the discretization of general manifolds. A manifold in itself has not yet metric propoerties such as curvature, but these are given by fields defined on lower-dimensional subsimplices. This generalizes lattice gauge theory where the gauge field is given on the edges of the lattice, rather than on its vertices.

answered Jun 16, 2015 by Arnold Neumaier (15,787 points) [ revision history ]
edited Jun 17, 2015 by Arnold Neumaier

I did not get it - is it simply about numerical calulations of a given theory or it is another analytical formulation of the theory (another theory)?

@VladimirKalitvianski: Regularization always means approximation: Each way of regularization defines a different approximate theory for the same putative true theory. A good regularization is one where the accuracy suffers little while the complexity of the calculations is reduced to a tractable amount.

So it is about another theory, as I suspected.

@VladimirKalitvianski: Not really. It is about approximately the same theory. Just as when we do calculations with real numbers on a computer, we accept that it computes only approximations to the exact results. I

n physics this is always done since very few theories are exactly solvable. Any perturbative scheme for solving an equation defines strictly speaking an approximate theory only, but it is approximately the same theory.

Arnold, I understand everything you wrote. The question is not here. First, it looks like a new (discrete) theory with a specific discretization. I guess triangulation allows to satisfy the necessary "boundary conditions" for each cell, which cannot be done for a "square" cells because of, for example, lacking an extra information needed to put in "extra" points (a "square" cell can also be adaptive). As well, no curvature exists on each cell; instead the "internal" geometry is "plane", which distinguishes this physical model from others. Such a modeling can be useful for numerical purposes too, but I think the simplified physics (plane geometry) makes it possible to easier analyze properties of solutions of such a theory.

A triangulation specifies the topology, which in general relativity is not necessarily trivial. See my updated answer.

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