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Consider the following well-known function: $$ \operatorname{sinc}(x) = \begin{cases} \sin(x)/x & \text{for } x \ne 0 \\ 1 & \text{for } x =0 \end{cases} $$ In physics, the sinc function has applications with for example spectrography. Mathematically speaking, there would be no objection against an alternative like this: $$ \operatorname{suck}(x) = \begin{cases} \sin(x)/x & \text{for } x \ne 0 \\ 0 & \text{for } x =0 \end{cases} $$ But in physics such an alternative proposal would be void of applications. It is silently assumed that $\operatorname{sinc}(x)$ is continuous at $x=0$. Physicists do not even think about a discontinuous alternative.

The sinc function is only an example of a far more general claim, uttered by one of my heroes, the great mathematician L.E.J. Brouwer. It is (not very well) known as Brouwer's Continuity Theorem, grossly stating that every real-valued function is continuous. More precisely, as quoted from Strong Counterexamples: "In intuitionistic mathematics, the Brouwer Continuity Theorem states that all total real functions are (uniformly) continuous on the unit interval".

Real valued, physical quantities have uncertainties. That is one of the fundamental properties of physics. And it isn't just due to quantum considerations. Take an average metal bar. It has no exact length. There are for example temperature fluctuations (atoms in motion) which will cause the bar's length to fluctuate. This effectively means that any real number in physics is accompanied with an uncertainty, an error, often denoted as $\delta$ or $\varepsilon$.

Consider the classical mathematical definition of continuity of a function. All numbers are assumed to be real-valued. A function $f(x)$ is said to be continuous at $x=a$ if and only if for all $\varepsilon > 0$ there exists a $\delta > 0$ such that if $|x-a| < \delta$ then $|f(x)-f(a)| < \varepsilon$, where it may be that $x \ne a$.

A physical interpretation of this might be formulated as follows: an error in a continuous function can be made as small as desired by adapting the error in the function's argument accordingly. Due to the errors, $|x-a|<\delta$ is physically the same as $x\approx a$ ($x$ equals $a$ approximately) and this can be said for $f(x)$ and $f(a)$ as well. So we can even write $\; x\approx a \,\Longrightarrow\, f(x)\approx f(a)\;$ , as a (sloppy) definition of continuity. The latter formulation is even closer to Brouwer's Continuity Theorem, if we replace the $\,\approx\,$ by a common equal sign: $\; x=a \,\Longrightarrow\, f(x)=f(a)\;$ , expressing the idea that a function is continuous where it really is .. a function!

Now consider again the above suck function. Whatever small it might be, inevitably there is an error in the argument, meaning that $x=0$ should actually be replaced by an interval $|x-0| < \delta$. There are values $x\ne 0$ in that interval, though, and $\,\lim_{x\to 0} \operatorname{suck}(x) = 1$. Hence, physically speaking, $\operatorname{suck}(0) = 1\,$ and $\operatorname{suck}(0) = 0\,$ must be true at the same time. Which is impossible. IMHO this is the reason why $\operatorname{suck}(0) = 1\,$ is involved automatically in physics, resulting inevitably in our old friend the sinc function and nothing else.

I'm well aware of the fact that this way of physical reasoning does not involve all sorts of continuity that mathematicians might think of. So the question is what sorts of continuity are sensitive to the automatism that is present in the sinc function and what sorts of continuity are distinct from this. It's a somewhat vague question, but I am a humble physicist by education and I do not know of any better way to formulate it.

EDIT. A far more simple example of a function with the same sort of "automatism" as with the $\operatorname{sinc}$ function is given by: $$ f(x) = \begin{cases} (x^2-1)/(x-1) & \text{for } x \ne 1 \\ 2 & \text{for } x=1 \end{cases} $$ Which is physically the same as $\,f(x) = x+1$ . A counter example is the function $\,g(x) = 1/x$ , much like the one given by snulty . So it seems that some singularities are "essential" (physically speaking) while others are not. Can someone be more specific? Because I find it a can of worms, as is exemplified by related Q&A in MSE and elsewhere:

Han de Bruijn
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    Any bad behavior that can be removed essentially "is" removed by physical considerations. For example, sinc, from a physical point of view, is not the function you wrote. It is (under some normalization assumptions) the Fourier transform of the indicator function of an interval symmetric about zero. This Fourier transform is defined by an integral, which in some sense automatically removes any "removable" bad properties. – Ian Jun 17 '16 at 21:10
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    (Cont.) To put it another way, the "physical version" of a function $f : [a,b] \to \mathbb{R}$ is $\frac{d}{dx} \int_a^x f(y) dy$. That is, it is the value that you would asymptotically observe with arbitrarily localized (but not infinitely localized) measurement. It could happen that this limit actually fails to exist; for instance that would happen if the underlying $f$ had a non-removable discontinuity. Then that is a "real" singularity which must somehow be explained. – Ian Jun 17 '16 at 21:10
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    (Cont.) This might be a mathematical model of a process which is not quite discontinuous but still has quantitatively bad behavior, like a very large Lipschitz constant or something to this effect. For example, a shock in the solution of a hyperbolic equation perhaps doesn't really exist (maybe there is some small dissipation or something that regularizes it away), but in the real world there may be almost a jump, meaning that there is some scale separation in the real world. Modeling this scale separation by a jump is useful even if it is not real. – Ian Jun 17 '16 at 21:12
  • What leaps to my mind is that, due to error as you describe it, for physical applications, one ought to consider two functions equivalent if they are equal almost everywhere. So, $\operatorname{suck}(x)$ isn't too remarkable, being equal almost everywhere to a continuous function $\operatorname{sinc}(x)$. (This is also what happens in the setting of Fourier analysis, where one doesn't work on functions, but rather almost-everywhere equivalence classes of functions. Of course, it is often convenient to pretend this is not the case) – Milo Brandt Jun 17 '16 at 21:18
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    I think the short answer is that most of physics (and engineering) lives in $L^2[\mathbb{R}]$, where almost everywhere equivalence is king. In the case of single slit diffraction, you're getting back a wave function which is a $\Psi=\mbox{sinc}$, but the part you actually care about is how $\Psi$ projects onto other wave functions, or what $\int_a^b |\Psi|^2dx$ is. – Alex R. Jun 17 '16 at 21:19
  • @Dr.MV: These can be removed with a change of gauge, right? So they are not properties of spacetime but choice of coordinates (see my answer) – Alex R. Jun 17 '16 at 21:28
  • @AlexR. I had not intended to post that comment as it was incomplete - but my "not-so-smart-phone" added it upon my shutting it down. – Mark Viola Jun 17 '16 at 22:01
  • @Ian I had not intended to post that comment as it was incomplete - but my "not-so-smart-phone" added it upon my shutting it down. – Mark Viola Jun 17 '16 at 22:01
  • @AlexR. How is the discontinuity of the tangential electrostatic field across a double dipole layer removed by a change of gauge? – Mark Viola Jun 17 '16 at 22:14
  • @Dr.MV: I might have read your previous comment to quickly. In this case it's an actual singularity, which i suppose could be somewhat resolved by assuming the field breaks down or behaves differently as you get closer to the atomic layer – Alex R. Jun 17 '16 at 22:31
  • @AlexR. Physical reasoning aside, this mathematical model, which is a macroscopic field model, does embed this singularity. Similarly, the surface charge density and associated potential and electric field at the edge of a perfect (surface) conductor are singular. I suppose the question is really about "reality" and models of "reality. "While models might have singularities, are those due to only imperfections in the model?" – Mark Viola Jun 17 '16 at 22:40
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    If you write \mbox{sinc} instead of \operatorname{sinc} then you don't automatically get proper spacing in things like $a\operatorname{sinc}b$ and $a\operatorname{sinc}(b)$, coded as a\operatorname{sinc}b and a\operatorname{sinc}(b). Notice the space to the right of $\operatorname{sinc}$ is different in those two expressions. Also, I edited for proper use of "cases". $\qquad$ – Michael Hardy Jun 18 '16 at 01:18
  • @MichaelHardy: Thanks for the editing. It's worthwhile to have some better tools in my MathJax bag now and I'm going to use them :-) – Han de Bruijn Jun 18 '16 at 09:57
  • @Ian: If you would be so kind to copy and paste your (Cont.) comments into a single, ever lasting answer, I would really appreciate the effort. – Han de Bruijn Jun 18 '16 at 10:28
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    "a humble physicist" who spent years trolling the sci.math newsgroup. – Gerry Myerson Jun 18 '16 at 11:31
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    @GerryMyerson: Apparently, this user prefers to keep an air of mystery about them. – Han de Bruijn Jun 18 '16 at 15:01

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The canonical example of this is the apparent singularity that arises in spherical coordinates when you pass around the earth only to find that your longitude has gone from $0$ to $180$. Or for example the singularity that arises in the Laplacian with spherical coordinates. These are all non-physical and are a consequence of choosing a coordinate system.

A prime example of this kind of thing comes up in general relativity where you'll see singularities in your metric. In many situations these singularities are actually artifacts of the coordinate system chosen, see here for example: https://physics.stackexchange.com/questions/223549/coordinate-singularity-in-metric

here's a wonderful read on this subject: "What is a Singularity - Geroch" and the takeaway quote:

The presence or absence of a coordinate singularity is not a property of the spacetime itself, but rather of the physicist who has chosen the coordinates by which the spacetime is described.

However sometimes these singularities point at failings of a given theory, like ultraviolet catastrophes for example. See here. In particular if a singularity exists in all coordinate systems (i.e is diffeomorphism invariant), only then can we conclude that perhaps this is a failing of our current theory. The point is that nature should not care about our choice of coordinate system.

Alex R.
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This is a cleaned up version of some comments of mine on the original question.

Some bad behavior is removable, some is not. The behavior that is removable in some sense "already is", from the physicists' perspective. For example, sinc can be thought of as (a multiple of) the Fourier transform of the indicator function of some interval symmetric about zero. This is really only uniquely determined up to a.e. equivalence, so we may freely choose our "favorite" representative to be "the function".

To put it another way, we can identify the "physical version" of a function $f$ as $\frac{d}{dx} \int_a^x f(y) dy$. This is what you would get by averaging $f$ over smaller and smaller intervals containing $x$. It can happen that this limit doesn't exist. In this case $f$ "really does" have a singularity, and that must be addressed somehow for $f$ to have any physical meaning.

Because singularities almost never really exist in nature, usually the answer is that there is some gap between the model and the reality. For example, it could be that the "physical" equation has a "regularizing" term with a small coefficient that you are neglecting. In this case or similar cases, the "real" function might not have any actual discontinuity but there could be a scale separation at the position that your equation predicts a discontinuity. Understanding this scale separation, even if only through a somewhat unrealistic model, is useful.

Ian
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  • If we have a measuring device with rectangular aperture $\sigma$, then I would rather think that the "physical version" of a function $f(x)$ is $\int_{x-\sigma/2}^{x+\sigma/2} f(y),dy / \sigma$, in concordance with the reference list in my EDITed question. How does that compare with your point of view? – Han de Bruijn Jun 19 '16 at 15:09
  • @HandeBruijn Except for endpoints, my definition is the result of taking $\sigma \to 0$ in your definition. – Ian Jun 19 '16 at 15:12
  • Your answer encompasses the most relevant issues, I think, but I still do not understand everything. (1) does the presence of your integral mean that measure theory comes in to play and that effectively we are talking about measurable functions? And (2) what exactly is the meaning of scale separation? Do you have a good reference for it? I've added another writeup to my list in the question: Critical Mass Flow. And I'm wondering if that is relevant w.r.t. scale separation. – Han de Bruijn Jun 20 '16 at 10:21
  • @HandeBruijn 1. To make sense of this sort of thing you do need to assume measurability. But assuming ZFC is consistent, nonmeasurable functions "practically" don't exist, in the sense that none can be explicitly constructed. So in some sense there is no loss here. – Ian Jun 20 '16 at 14:23
  • @HandeBruijn 2. You can look up any number of references on "multiscale analysis". What I am imagining in this particular case is some profile. For ease of visualization, let's say it is a cylindrically symmetric density of a red fluid in a cylinder. Suppose that for the most part varies in a macroscopic way on macroscopic length scales (the typical case). So maybe the average is 1 g/mL and it varies by 0.1 g/mL over a scale of a meter or so, over most of its domain. – Ian Jun 20 '16 at 14:25
  • @HandeBruijn (Cont.) But then there is some interface where it changes by 0.1 g/mL over a scale of a micron. This interface is not infinitely thin; indeed the thinnest it could possibly be is one atomic radius, and in most situations it is thicker than that. Yet it is convenient to model it as infinitely thin. – Ian Jun 20 '16 at 14:27
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Discontinuous functions are fairly common.

What's the magnitude of the force between two point charges, or particles which can be considered point charges,

$$F=\frac{kq_1 q_2}{r^2}$$

Where $q$'s are the charges, $k$ is constant, and $r$ is the distance between them.

This is quite clearly discontinuous when the distance is is zero, and diverges as the particles get arbitrary close to each other. Same thing happens for Newtonian gravity.

In fact this causes a problem when taking a Fourier transform of the potential, which is usually 'fixed' by either physically giving the photon a mass and letting the mass go to zero afterward, or just admitting we can't know with arbitrary precision the exact form of the potential, our experiments aren't good enough, so they add an extra $e^{-ar}$ into the potential and let $a\to 0$ after the calculation.

The other thing that happens is when you want to thing about point sources, so physicists introduce the Dirac delta 'function', which I believe is more properly described as a distribution, can be obtained as the discontinuous limit of continuous functions.

snulty
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  • Than you think? Actually, I don't think so and I'm aware of the point you are making: see the latest EDIT of the question. (+1) anyway .. – Han de Bruijn Jun 19 '16 at 14:02
  • @HandeBruijn The title of the question was "Is any real-valued function in physics somehow continuous", and I just gave an example of where there was no obvious fix to a discontinuity, unlike the examples you give where you take a continuous function and make it discontinuous. I'll modify my answer to remove the "than you think". – snulty Jun 19 '16 at 19:41
  • That is an artifact of the same question Alex R said. This is the Green's function for the Laplacian in those chosen coordinates. The use of that potential is always as a distribution which gets smoothed out by your actual charge distribution, which out of laziness sometimes you make strongly peaked at the origin. – AHusain Jun 19 '16 at 20:14
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Being continuous or not is often a matter of referential: origin, orientation of axes and units, or scale.

When discontinuity appears, the source is often in the simplification of the model, think about reducing a moving object to its center of mass.

But would physics exists without discontinuity? If an equation is continuous, and differentiable, it seems to me that derivative discontinuities often arise with a sufficient order. Is that because the time could be discrete?

  • Doesn't "discrete" or "continuous" depend upon the way we are "looking" at things? In kinetic gas theory, a gas consists of discrete molecules. In fluid dynamics, it is a continuous medium. Both points of view are quite useful, dependent upon the (technical) application at hand. I have no idea how this example compares with (space-)time, though. – Han de Bruijn Jun 19 '16 at 15:18
  • @Han de Bruijn This is what I hide under the notion of scale. I have a background in image processing with multiscale techniques, and the definition or perception of an object depends in the scale of observation and sampling. From afar, one sees objects. Closer, one sees pixels. Due to our sensors and uncertainty, I suspect we never can be sure things are continuous. Should I expand in one direction? – Laurent Duval Jun 19 '16 at 15:27