I'm aware of Noether's theorem and symmetries leading to conserved quantities. Excuse me if this is a straightforward result in the underlying math but as a newb to the math, I want to ask, are infinitesimals absolutely necessary for translational invariance and a corresponding conserved quantity such as momentum? Why can't a discrete or grid space-time exhibit conservation laws?

  • 2
    $\begingroup$ Note that the anisotropy is probably the biggest issue. We observe angular momentum to be conserved, so how do you make a discrete grid isotropic? $\endgroup$ Sep 26, 2016 at 1:21
  • $\begingroup$ That helped. Can we also show that energy too, is not conserved? $\endgroup$ Sep 26, 2016 at 4:03
  • $\begingroup$ From what i understand, a non infinitesimal amount of motion cannot be curved, so any body moving around in circular/elliptical paths won't be stable and would either fall off its orbit or fall towards the center. Is that it? $\endgroup$ Sep 26, 2016 at 4:42
  • $\begingroup$ Shouldn't it also follow that in a non continuous space-time, no matter how small the distance between two points is, we should, over time, observe bodies falling out of orbits, so theoretically an electron would fall out of its orbit without any external forces, given enough time? $\endgroup$ Sep 26, 2016 at 4:49

1 Answer 1


I am a newbie to the math as well, but let's take one of the conservation laws.

If a physical system behaves the same regardless of how it is oriented in space, its Lagrangian is rotationally symmetric: from this symmetry, Noether's theorem dictates that the angular momentum of the system be conserved, as a consequence of its laws of motion. The physical system itself need not be symmetric; a jagged asteroid tumbling in space conserves angular momentum despite its asymmetry. It is the laws of its motion that are symmetric.

I honestly don't know if this is a valid argument against your suggestion, because I have not got the math background, but the very first line of Wikipedia's article on Noether's work says:

Noether's (first) theorem states that every differentiable symmetry of the action of a physical system has a corresponding conservation law. 

That being the case, allied with the derivative with respect to time implied in the angular momentum example above then this, to my (very naive) understanding of the continuity of a function, would suggest a problem with modelling spacetime as a set of discrete points, rather than a differentiable continuum.

  • $\begingroup$ I think I get the point now. While it's impossible to model discrete points in space preserving angular momentum, we can make them sufficiently small and close together, so that deviations in angular momentum observed at each point are too little for measurement? $\endgroup$ Sep 26, 2016 at 2:11
  • $\begingroup$ Well I know that there is evidence for no discreteness at really small scales, from space probes, there are questions on this site regarding that. What you are describing sounds like a Wilson lattice idea from particle physics. And also a bit like loop quantum gravity. Personally I don't have a problem with discrete points, there is just no evidence for them......good question though . regards $\endgroup$
    – user108787
    Sep 26, 2016 at 2:39
  • $\begingroup$ Aye, I've read answers to similar questions. Ultimately they all boil down to the fact that it's continuous "as fast as we can tell ", if there is discreteness it has to be below the Planck scale, as a 2009 Fermi experiment falsified predictions of any discrete model larger than a scale that is well smaller than the Planck scale. However we could keep making claims that the effects only show up at even smaller scales than what we probed, and so on. Thus making these models unfalsifiable, in a sense. $\endgroup$ Sep 26, 2016 at 3:29
  • $\begingroup$ But I intuitively feel like there must be some experiment that would show deviations from continuous models regardless of scale, if space-time were indeed, discrete $\endgroup$ Sep 26, 2016 at 3:31
  • $\begingroup$ my intuition was right ;-) physics.stackexchange.com/a/282877/128779 $\endgroup$ Sep 28, 2016 at 8:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.