arildno said:
It isn't just that simple. is it vanesch?
After all, there are numerous examples where the (naively) associated difference equation shows quite a different behaviour than the differential equation.
Well, that would only indicate that the "continuous" model is a bad model, but it wouldn't make the taking of the derivative within the continuous model problematic.
BTW, if we could NEVER approach any discrete system more or less reliably with a continuous model, then most fluid dynamics, and most continuum mechanics, most elasticity calculations would be wrong etc... because we all know that matter is not a continuum, and consists of atoms on this level.
Of course, you are right that *sometimes* there's a problem in the continuum model that arises from the lower-lying discreteness. But there's no *mathematical* error in the continuum model because it doesn't describe correctly, in all circumstances, a discrete nature. Within a continuum model, the stress tensor is defined mathematically on scales far below the atomic dimensions for instance. That doesn't make physical sense because we know that the model isn't applicable there, but it is mathematically perfectly defined.
Furthermore, might it be that the divergent integrals appearing within quantum mechanics are symptoms that using continuous models for essentially discrete (and finite) phenomena generates a number of "problems" precisely because the modelling of the world as a continuum (or inter-related continua) is actually physically dead wrong?
You might ALSO have a mathematically problematic "continuum" model of course. But the question of the OP was, as I understood it: "how can we take a time derivative (in, say, Newtonian mechanics), if we somehow "know" (we don't, actually, but suppose we did) that time is discrete ? "
My answer is simply: WITHIN the model of Newtonian mechanics, dynamical variables are real functions of t in R, and you can take the derivative in that model. Whether or not this makes physical sense has nothing to do with the question of whether the mathematical operation is well-defined or not.
And in as much as the model is "applicable", the derivative is also "applicable" (up to the accuracy of the model in this case, and with the caveat you pointed out, that *sometimes* a microscopic discrete behavior is different from the smoothed-out continuum model that one uses for it).