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Taking a grid of 'evenly-spaced' space-time points. e.g. at integer values of (x,y,z,t). Now do a Lorentz boost on this grid. We end up with a grid of points which are much closer together.

It is bothering me that the grid of points only looks evenly distributed in certain frames (related by spacial rotations) but looks different under Lorentz boosts. The points that looked like neighbours no longer look like neighbours.

When working out results in lattice QCD, does this mean that the results you get don't exhibit Lorentz symmetry. And how can we be sure in the limit the Lorentz symmetry is conserved?

Edit: I just came across this animation from John Baez on time crystals. enter image description here

Which seems to suggest that for certain lattices like a triangular lattice, the Lorentz transformation can preserve the Lattice form. Although looking at the middle point, it's neighbours are continually changing. And Lattice QCD is based on calculations of nearest neighbours (discrete derivative). I guess one could reformulate the calculations to all neighbours a unit Minkowski-distance away. But this would increase the number of neighbours a lot for large volumes of space-time. (and does this obey the concept of `locality'??)

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  • $\begingroup$ If your points are in fact evenly spaced points in "space-time" then the interval between them WILL in fact be invariant under a boost. That's just special relativity! Though in general that's not how such lattices are defined. $\endgroup$ – R. Rankin Mar 28 '19 at 22:28
  • $\begingroup$ Well exactly. They are evenly space if we think of it as Euclidean 4D space instead of Minkowski 3+1D space-time. Interestingly, taking a set of random points and then Lorenz boosting them, the density of the boosted points is the same. So I think this is only a problem with regular lattices. As far as I know there's no Minkowski equivalent of a regular lattice. $\endgroup$ – zooby Mar 28 '19 at 22:39
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Lattice theories are not invariant under Lorentz transformations. They are usually defined in Euclidean signature (imaginary time) so the symmetry woud be ${\rm SO}(N)$ rather than ${\rm SO}(N-1,1)$ --- but apart from this the lattice breaks the rotational symmetry down to a discrete subgroup of 90 degree rotations. It's only the in the continuum limit (i.e close to a critical point) that full rotational invariance is restored.

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  • $\begingroup$ Yes, that's true. Although one might say that under a small rotation, it still looks like a lattice. And it doesn't alter which neighbouring points appear nearer (in the Euclidean sense). $\endgroup$ – zooby Mar 28 '19 at 22:41
  • $\begingroup$ A belated note that the restoration of continuous SO(d) rotational symmetry in the continuum limit is guaranteed if the lattice discretization is sufficiently symmetric (e.g., hypercubic). arXiv:0804.1145 provides an example a less-symmetric discretization for which (in essence) the speed of light needs to be fined-tuned in order to restore SO(d) in the continuum limit. $\endgroup$ – David Schaich May 18 '19 at 14:31
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Yes, a Lorentz boost breaks the symmetry of the lattice. And it creates a different lattice. We in fact use this! On the lattice your momenta can only be multiple of $2\pi/L$ where $L$ is the spatial extent. When you boost, $L$ changes and so do your momenta. Looking at an interacting system from two pions you will see that the moving system has a different total energy (of course). But when you boost that back such that you are in the center of mass frame (at the cost of a distorted lattice) you will find that the energy is different from a two pion state where the total momentum was zero! This way you can create different interacting energies even though the individual momenta are quantized by the enclosing box.

The group theory will get pretty messy though, and you need to be careful with the reduced symmetries.

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