# Will an eventual discreteness of spacetime have consequences in the light of Noether's theorem?

No doubt Noether's theorem holds for the symmetry of translations in space and time. But what if we zoom in on very small lengths and times, and spacetime maybe becomes discrete? Will this have consequences?

• For a discrete version of Noether's theorem, see physics.stackexchange.com/q/8518/2451 . For discrete spacetime, see physics.stackexchange.com/q/9720/2451 , physics.stackexchange.com/q/33273/2451 and links therein. – Qmechanic May 16 '17 at 21:37
• Except for lattice models "very small length and times" appear in quantum gravity proposals in the form of minimal expectation values for some "spacetime" observables, not in the literal form. So "zooming in" will not make much sense, at small values they can no longer be interpreted as length and time, see Baker. The observables themselves are not discrete, and there will probably be some noncommutative version of conservation laws for them related to symmetries. We can't say more until we know how exactly spacetime emerges. – Conifold May 16 '17 at 22:30

First note that a momentum $p$ is associated with a de Broglie wavelength $\lambda = h/p$. If there is a lattice spacing $a$ then only wavelengths longer than $a$ are meaningful, and so there is a maximum possible momentum $h/a$.
The consequence of the discrete translation symmetry is that momentum is still conserved, but only up to multiples of $h/a$. (We sometimes refer to it as quasimomentum.) A scattering event where momentum is conserved only modulo $h/a$ is called umklapp. This is in fact a very similar situation to that of phonons in a solid or the tight-binding model for electrons in a crystal lattice.
Does this have any consequences? Assuming the lattice spacing $a$ is of order the Planck length, then the maximum momentum $h/a$ is far too large to constrain any current experiments. And the energy required for an umklapp process to occur would also be of order the Planck scale.