# Do we know, or at least have a strong argument for the fact that for a given time interval, we can always find a smaller time interval? [duplicate]

## Motivation:

In Biology, when, for example, biologists try to model the population dynamics of a population, they say:

Let $$N: \mathbb{R}^{nn} \to \mathbb{R}^{nn}$$ be a function that represents the number of individuals in a given population.

And with that they construct some differential equations, which models the interactions of the population with the environment, other populations, etc..However, for a given time $$t$$, if $$N(t)$$ is not an integer, say 1.5, it does not mean anything physically - we cannot have 1.5 member in a population. For that reason, even though this model have lots of successes, it is clear that we can make improvements. In fact, it has been already done. They define $$N: \mathbb{N} \to \mathbb{N}$$, and they use difference equations for describing the changes in the number members in population. Turns out, in this way, we can explain real phenomenas much better (citation needed, but a quick google research can reveal this).

Similarly, in Newton's time, they thought they could move as fast as they can, but it turned out, we cannot have a velocity bigger than $$c$$, which changed everything.

## Explanation:

We always model time with $$\mathbb{R}$$, which is a dense set; meaning given any two different real numbers $$a$$ and $$b$$, we can always find a third real number $$c$$ s.t $$a

Now, as far as our measurements imply, we can always measure and look at smaller time intervals - or at least this is what I know. However, as the history of physics have shown, this is not the case with distance, when we look smaller distances - or objects with smaller sizes - the physics get weird, and I'm not sure that quantum physics suggest that we cannot see smaller sizes after at some point.

## The actual question:

So, my question is that, is there any argument why modelling time as a real number in any physical theory is sensible?

Note the relation of this question to the motivation given in the beginning .

Edit:

I'm directly going to quote what I have written in the comments:

I do not ask whether time is something or else, but rather, is there any motivation why we use a dense field to model it, and is there any counter experimental observations, or any other models who use discreet time intervals in Physics.

Moreover, I'm talking about Physics in here, not about Biology.

• Well, aside from the fact that physics works quite well assuming time is a real number... – Jon Custer Oct 9 '18 at 12:38
• Related: physics.stackexchange.com/q/35674/2451 , physics.stackexchange.com/q/64101/2451 , and links therein. – Qmechanic Oct 9 '18 at 12:38
• Possible duplicate of Is time continuous or discrete? – knzhou Oct 9 '18 at 12:51
• @JonCuster Read my question again. Do you think the actual question is related to Biology ? – Our Oct 9 '18 at 12:54
• Any physical theory which is built on differential equations needs at least a notion of differentiability which in turn requires a notion of continuity. I'm unclear if you could build such a thing on top of a set which is only countable (say the rationals): perhaps you could. (I suspect you can't build one on top of a set which is not dense in the reals.) If you can't then you either need to use something built on the reals or you're wandering off into weird maths (ie assuming the continuum hypothesis is false). – tfb Oct 9 '18 at 12:57

## 2 Answers

We use continuous time when it's convenient, and discrete time when it's convenient.

Continuous time is better for analytic calculations because you can use calculus, along with everything in math that grew out of it (differential equations, linear algebra, Lie theory, differential geometry, topology, etc.). If you give up continuous time, then all you can do is stuff like addition and subtraction; the discrete analogous of calculus (finite differences) is much messier. However, digital computers can fundamentally only do stuff like addition and subtraction, so it's more convenient to use discrete time for them.

Of course nobody knows what time "really" is. For all we know, it could be that we're living in a simulation, run by hand by a guy in an infinite desert moving around rocks. However, it is true that for every experimentally confirmed model of physics we have today, the observations are completely consistent with continuous time, and they can be computed with arbitrary accuracy by a computer using discrete time. So I can't give you a philosophical argument for why continuous time is "inherently" better, but really, no physicist actually cares.

Conversely, situations where time actually is discrete, such as time series data in biology, may sometimes be modeled to good accuracy using continuous time equations. This generally works as long as the sampling frequency is high compared to the frequency of what you're trying to model.

This answer seeks to explain my comment to the question that 'analysis fails'.

$$\newcommand{\R}{\mathbb{R}}$$​If we model time as some set $$S \subset \R$$ which is not dense in $$\R$$, then there are at least two times in $$S$$, $$t_1$$ and $$t_2$$ with $$t_1 < t_2$$ such that $$\forall t \in (t_1, t_2), t \notin S$$. What this means is that there must be at least one finitely-sized gap: if there isn't then it's obvious that $$S$$ is dense in $$\R$$.

What this in turn means is that the usual definition of continuity in $$\R$$ does not carry across to $$S$$. It's possible to put a topology on $$S$$ based on the metric on $$\R$$, with essentially the same definition of its base: $$\forall t \in S, B(t,r) = \left\{t' \in S, \left|t'-t\right| \lt r\right\}$$, but the definition of continuity you get from this is different than the definition of continuity you get in $$\R$$.

In particular there will be functions which are continuous on $$S$$ but discontinuous on $$\R$$: as an example consider $$f:\R\to\R$$ and its restriction $$f\vert_S$$ defined such that $$f(t_1) = f(t_2)$$ but $$f$$ is discontinuous everywhere in $$(t_1, t_2)$$. This is continuous on $$S$$ but not on $$\R$$.

What this really means is that because $$S$$ has gaps in it, arbitrarily bad things can sneak into those gaps which show no trace in $$S$$ but do in $$\R$$.

As an example, if $$S = \left\{n\alpha, \alpha \in \R, n \in \mathbb{Z}\right\}$$, then the topology you get on $$S$$ is the discrete topology, and all functions in this topology are continuous!

So continuity in $$S$$ and continuity in $$\R$$ are different, and therefore everything built on continuity is different. I'm not sure you can build analysis in $$S$$, but if you could it would be different to analysis in $$\R$$ because it would rely on different definitions of continuity. And analysis underlies differential equations and differential equations underly physics.

[Below here is opinion.]

So if you try and work in $$S$$ then physics as we understand it falls completely. You would need to build some completely new theory.

Well, you could do what we actually do when running simulations on computers: use some discrete notion of time and space with a process which is set up so that it approximates the actual differential equations we want to model. That's perfectly reasonable for a computer model, but as a proposal for a theory of physics I think it's very questionable: why is the theory approximating some other theory which uses differential equations if it's meant to be a basic theory of nature? Why not just use that other, 'deeper' theory built on $$\R$$ in the first place? And, because of the gaps, I think it's always possible to construct situations in the deeper theory which the discrete theory will miss / disagree with by sneaking behaviour into the gaps (I don't have anything like a proof of this, but it seems right to me).

And that's a real problem I think: we do have theories built on differential equations which work extremely well experimentally: if things were really discrete, why would they model the differential equationy theories so well?

An interesting (to me, anyway) question is what happens with a set (the rationals, say) which is dense in $$\R$$ but is still countable. None of the above argument works, since there are no gaps, but something smells bad about such a set.
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