Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

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

+ 2 like - 0 dislike
1395 views

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

+ 2 like - 0 dislike

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.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverflo$\varnothing$
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...