As far as I have understood, the case is that there is nothing that argues that time or space is continuous, but at the same time we must assume this in order to be able to calculate derivatives or integrals with respect to these, how can we justify this?
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22$\begingroup$ Because it works? $\endgroup$– Jon CusterCommented Oct 19, 2020 at 23:57
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2$\begingroup$ Some questions on whether spacetime is discrete or continuous: physics.stackexchange.com/q/9720/123208 & physics.stackexchange.com/q/33273/123208 & links therein. $\endgroup$– PM 2RingCommented Oct 20, 2020 at 1:10
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2$\begingroup$ @Labbsserts Whether a model "works" or not is decided only by how well it agrees with or predicts experimental outcomes. I don't know of any experiments that suggest a discrete model of spacetime "works better" than a continuous one. $\endgroup$– d_bCommented Oct 20, 2020 at 1:17
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3$\begingroup$ While AFAIK there's no actual evidence that space & time are not continuous, you might think about why can we use continuous functions to model the behavior of entities that indisputably ARE discontinuous, e.g. particles in fluid dynamics. Or FTM the fact that calculus is based on the limits of discrete intervals :-) $\endgroup$– jamesqfCommented Oct 20, 2020 at 16:41
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2$\begingroup$ Related: physics.stackexchange.com/q/579311/195139, physics.stackexchange.com/q/568833/195139 $\endgroup$– SandejoCommented Oct 20, 2020 at 18:02
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2 Answers
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Let's say space is really a lattice with spacing $\Delta x$. It turns out that this idea has more trouble with experiment than you might think, but we can plow ahead for the purposes of this question.
You might propose replacing integrals in physics with discrete sums over individual lattice points, to take a concrete example let's think about the work needed to move a particle from point $A$ to point $B$ \begin{equation} W = \int_A^B \vec{F} \cdot {\rm d} \vec{x} \rightarrow \sum_{i=1}^N \vec{F}(\vec{x}_i) \cdot \hat{e}_{i,i+1} \Delta x \end{equation} where $i=1,2,...,N$ labels the lattice points that the particle follows in going from $A$ to $B$ and $\hat{e}_{i,i+1}$ is a vector pointing from the lattice space at $i$ to the lattice point at $i+1$.
If $\Delta x$ is small enough so $N$ is large enough, these two quantities will be quite close (since in the limit of infinite $N$ the two quantities are actually exactly the same). To see a difference (if there is one) we need to probe distances of the same order or smaller than $\Delta x$, or else have a large precision to tell the difference between these two expressions.
Here's the point. No one has ever found any disagreement between experiment and theory that can be attributable to the failure of the continuum limit. If there is such a $\Delta x$, it must be so small that it is a very good approximation to use integrals instead of sums over lattice in all experiments done to date. You can think of the LHC as probing energy scales of order 1-10 TeV, which amounts to $10^{-18}-10^{-19}$ meters -- so $\Delta x$, if it is nonzero, must be smaller than this.
There are other problems with having a lattice, but this is already a powerful argument that the world is at least effectively continuous at the scales we can probe.
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6$\begingroup$ most excellent analysis. clear and concise. $\endgroup$ Commented Oct 20, 2020 at 3:21
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$\begingroup$ @Andrew, do you mean by subtleties things like the breaking of continuum symmetries, and fermion doubling problem? $\endgroup$ Commented Oct 21, 2020 at 6:55
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$\begingroup$ @spiridon_the_sun_rotator Yes, exactly! $\endgroup$– AndrewCommented Oct 21, 2020 at 8:03
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$\begingroup$ But when you go to model things on a computer, you are back to using lattices (usually). $\endgroup$ Commented Oct 22, 2020 at 6:19
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$\begingroup$ @KevinKostlan Of course, but an important check on your numerical results is that the answers are sufficiently converged that you are confident that the answers do not depend on the discretization (to some precision). $\endgroup$– AndrewCommented Oct 22, 2020 at 11:31
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This is a comment, as Andrew's answer is adequate for the problem.
I want to point out , which is not clear in your question, the difference between mathematical modeling and the object modeled.
When modeling an object mathematically one can use continuous variables by the function of mathematics. If the object modeled has discontinuities, the mathematics will model it with continuous variables. A crystal lattice for example can be modeled in continuous (x,y,z) variables with discontinuities in the mathematical model where the atoms are.
In the case of space-time there are really two concepts overlapping: the mathematical variables of space and time to be used in modeling it, and the object modeled, i.e. space and time. The mathematical variables to model a latticed space-time can be continuous, and give the functional form of the latticed physical space-time. The $x$ in Andrews answer is still a continuous variable used to model a latticed space.
