It is often claimed in quantum gravity circles (I'm thinking of LQG in particular) that it is "not possible" to discretize spacetime. The issue, as I understand it, is that should one attempt to assign each point on a spatial lattice a value of e.g. the spatial metric $\gamma_{ab}$, the lattice spacings (both across time and space) will not be coordinate-invariants. This is pretty obviously true.

I work in numerical relativity, however, and we do precisely this all the time, maintaining parameters during the simulation such that the lattice spacings are indeed functions of spacetime. Should we seek to express quantities in a new "frame" we just transform those functions.

There appears to be no particular difficulty with this approach. One might, I suppose, run into trouble if a particular transformation were sought that brought previously irrelevant ultraviolet modes into the simulation domain, but this can be dealt with in a problem-dependent way by just choosing a high enough resolution at the outset. I am presumably, therefore, misunderstanding something about the difficulty the LQG people are getting at. What is it?

  • $\begingroup$ I discretized space, and it works out to have non-zero intrinsic curvature because the tessellation simplices (e.g. the tetrahedra) cannot all be regular. github.com/sjhalayka/4d_universe $\endgroup$ Dec 23, 2021 at 19:13

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I've never heard of such claim, especially taken into account that LQG is based on a particular discretization of space on graphs (the spin network basis), while the evolution of the spin network state under the Hamiltonian constraint operator can be modeled by a discretization of spacetime on 2-complexes (spinfoams). LQG theorists often say that in quantum gravity, spacetime is fundamentally discrete, and this is not a feature of a particular approach, but rather of a whole range of background-independent nonperturbative models for quantum gravity.

In numerical relativity, however, you deal with classical theory, and thus a question may arise of how to preserve Lorentz invariance on the lattice, which is impossible in general. But do not be confused: in a quantum-mechanical model with discrete basis of the space of quantum states there is room for continuous, operator-valued transformations. Discreteness of space and existence of minimal possible length does not go against Lorentz invariance in the realm of the quantum theory of spacetime.

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    $\begingroup$ I remember that someone from the LQG circle said that the discrete spectrum of area/volume operators does not mean the spacetime is discrete. For me the 'miminal' spatial unit reminds me of the isoholonomy problem so that the Planck length is just the minimal path length to achieve a given holonomy ($\pi$) which is used to bring us from one state to another orthogonal state. From S. Lloyd's computational universe, I do not see why the 'computation' of qubits to build the spacetime must be discrete. $\endgroup$
    – XXDD
    Oct 22, 2016 at 15:33
  • $\begingroup$ @X.Dong I don't understand your first statement. Could you provide a reference? $\endgroup$ Oct 23, 2016 at 1:22

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