Timeline for Why are infinite order Lagrangians called 'non-local'?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2019 at 23:10 | comment | added | MannyC | @VladimirKalitvianski In the lattice we consider local all operators that contain spins at finite distance from each other. That's because in the continuum limit (in the RG sense), they all come to a point. Non-local operators on the lattice are those that extend over an infinite region, roughly speaking. | |
| Aug 23, 2011 at 2:12 | comment | added | Revo | I also didn't get the reason why or where "non-local" terminology came from. Regarding the number of points needed to define derivatives, if the points are closer and closer to each other as we go to higher derivatives, they will be "local", don't they? | |
| Aug 21, 2011 at 5:47 | vote | accept | WIMP | ||
| Aug 21, 2011 at 5:47 | comment | added | WIMP | Yes, okay, thanks. I still find it confusing to call that 'non-local' but if that's what they mean, then, well, that's what they mean. (I believe what you say is true only for a finite number of derivatives, but then that wasn't my question.) | |
| Aug 16, 2011 at 17:25 | comment | added | Daniel Grumiller | No, the first derivative is not non-local, it is just "less local" than the zeroth derivative. As I said, I would no call any higher derivative theory "non-local", only those where you need infinitely many derivatives - and thus infinitely many lattice points. | |
| Aug 16, 2011 at 16:01 | comment | added | Vladimir Kalitvianski | I do not like a discretization ("lattice") explanation. According to it, the first derivative is non local. | |
| Aug 16, 2011 at 14:01 | history | answered | Daniel Grumiller | CC BY-SA 3.0 |