Almost every physical equation I can think of (even though I don't actually feel comfortable beyond the scope of classical mechanics and macroscopic thermodynamics, as that's enough for dealing with everyday's engineering problems) is expressed assuming continuous domains at least for one variable to range over; that is, the real and complex number sets are ubiquitously used to model physical parameters of almost any conceivable system.

Nevertheless, even if from this point of view the continuum seems to be a core, essential part of physical theories, it's a well-known property that almost all of its members (i.e. except for a set of zero Lebesgue measure) are uncomputable. That means that every set of real numbers that can be coded to compute with -as a description of a set of boundary conditions, for example-, is nothing but a zero-measure element of the continuum.

It seems to me that allowing that multitude of uncomputable points, which cannot even be refered to or specified in any meaningful way, makes up an uncomfortable intellectual situation.

I wonder if this continuum-dependent approach to physics can be replaced by the strict use of completely countable formalisms, in a language which assumes and talks of no more than discrete structures. What I'm asking is if the fact that we can only deal with discrete quantities, may be embedded in physical theories themselves from their conception and nothing more being allowed to sneak in -explicitly excluding uncomputable stuff; or if, on the other hand, there are some fundamental reasons to keep holding on to continuous structures in physics.

  • $\begingroup$ Related: physics.stackexchange.com/q/9720/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Jul 27, 2012 at 20:36
  • $\begingroup$ You might become interested in the frontier studies of twistors , summarized at motls.blogspot.com/2012/07/… with links to the talk of Arkani-Hamed . Counting over complex quantities. $\endgroup$
    – anna v
    Commented Jul 30, 2012 at 6:43
  • $\begingroup$ Space and time are modelled as continuous and it works - the infinity of points is infinity of possible locations, not all of which get realized. Those that are realized can still be uncountable. Is there something uncomfortable about that? $\endgroup$ Commented May 4, 2019 at 19:28

4 Answers 4


I tend to think there is no way or even point in getting rid of the continuum. Even if an unbelievable breakthrough in physics happened to fulfill that desire, I bet old continuum theories would be more preferable "for dealing with everyday's engineering problems".

However, that's the point of my answer (which is more like an extended comment), continuum entities exhibit or can be modeled via more discrete structures. The trick is to employ more algebra.

For example when you (or your CAS) are calculating derivatives symbolically you are just applying some simple algebraic rules. You aren't calculating limits, messing with infinitesimals, instead you are finding the algebraic derivative.

Moreover, finding antiderivatives in this setting also has nothing to do with calculus. Mathematica or any other CAS uses sophisticated algorithm originated in algebra coupled with lots of heuristics to find indefinite integrals.

Or for example when multiplying or adding polynomials you don't treat them as functions $\mathbb R \to \mathbb R$.

$$\pi x^2 + 1$$

is not a very continuous object — it's just a polynomial in $\mathbb Z[x, \pi]$, it can be encoded as $[((0,0),1),((2,1),1)]$.

Differential geometry (the part calculus on manifolds) is just like calculus, but it is often presented using lots of algebra.

Of course, underlying physical interpretation is continuous, but treating "continuum" objects doesn't always require dealing with their continuity. Ultimately, you wrote the equations without having to count $\mathbb R$.

UPD As for the computability, discrete doesn't mean computable. See for example Hilbert's tenth problem — there is no algorithm to find integer roots of a polynomial with integer coefficients. Of course, if you consider such problem as discrete enough. So computability problem may arise almost everywhere and its relation to physics is indeed very interesting.

  • $\begingroup$ This is a nice answer, but it addresses it from the point of view of current practice when dealing with continuous structures. This may point that my concern isn't shared at all by the physics community, or maybe because I didn't express it in a clear way. Whatever the case, of course we can treat polynomials as finite dimensional vector fields and use linear algebra to compute with them, without ever noticing the continuum over which its variables range over. I suppose that's a pragmatic point of view, which may judge pointless that what I'm really refering to. $\endgroup$
    – Mono
    Commented Jul 25, 2012 at 15:30
  • $\begingroup$ Think of it as a case for Occam's Razor: why would we assert that such an infinite amount of points "are out there", if we can't possibly refer to them? We have no intuition about, and no possible way to reach the full continuum. Even if I'm taking too much a philosophical stance with this, it makes me think of "here be dragons" warnings at the corners of Middle Age's maps. $\endgroup$
    – Mono
    Commented Jul 25, 2012 at 15:30
  • $\begingroup$ "Why would we assert that such an infinite amount of points "are out there"" Frankly, for me it is sufficient they are in my mind, I can deal with them to a certain extent. I enjoy the beauty of my perception of the world, I don't care if it is wrong provided it doesn't contradict with what I see/know. And I don't see/know if continuum exists in Nature or not. $\endgroup$
    – Yrogirg
    Commented Jul 25, 2012 at 16:08
  • $\begingroup$ "And I don't see/know if continuum exists in Nature or not". Not only you don't see it, you can't possibly see it. So we are asserting something which asks for a huge load of confidence, and we should be suspicious, at least without some very compelling argument which explains why our theoretical tools (computable routines) need it to be there. That's the base of scientific skepticism. There's nothing wrong with your answer, it's just you find non-contradiction with experience enough to accept a formalism. It's a pragmatic stance, and I suspect most of the physics community thinks like you. $\endgroup$
    – Mono
    Commented Jul 26, 2012 at 22:18
  • $\begingroup$ No, no, Mono, don't take scientific skepticism too serious, it is far not the main engine and essence of science. And the "pragmatic stance" --- it is just an excuse to tell philosophers so they wouldn't bother somebody with questions. It's mostly my beliefs and general agreement with experience (mostly personal) that drive my acception. $\endgroup$
    – Yrogirg
    Commented Jul 27, 2012 at 4:25

This is an interesting question. I don't know if it's answerable in a very strict sense, and this is also why you might see alot of comments. This is basically another one, which got too long.

Firstly, username Yrogirg makes some good points in that many mathematical theories which are modeled using the set of reals (and I think I even used these terms in the technical sense here) entail many facettes which are really agebraic in nature. You can easily implement derivations by representing the quantities of interest as distinct symbols and translating the associated abstract rules of computation to manipulation of these symbols.

You formulate the physical problem and then its sulution by taking a look at the structure of the theory and you deduce the way to get from the object "boundary/starting conditions" $\mathcal{B}$ to the object "expectation value of observable" $\langle A \rangle$ (and stating the boundary for the system is also really only providing input which has been taken from another observable). Everything in between $\mathcal{B}\rightarrow\langle A \rangle$ is computational in nature in the sense above, so in view of your question one probably only has to bother about the real-ness of the objects on the left and the right. Measuring is fundamentally about comparing two things and as the rationals $q=\frac{n}{m}$ lie dense in the reals and so no human can tell the difference, I see no reason to be worried.

The same point is made if I say that if you define a curve to be perfectly smooth in the mathematical sense and then draw a picture of it by successively approximating it using finite lines. The experiment distinguishablity of the real and the approximated curve, for each level of sophistication, can be overcome by enough time and data storage.

The point you raise regarding the null-measure of the reals is not a problem as, with respect to the observables, I don't see why you'd need to manually integrate over points. Again, the numbers express or store information about a comparison. You might for example compare areas (how often does the first fit into the second?) or shades of gray (what is darker/makes the measuring device react stronger?), and if you "compare points" you really express distances, which means you'll compare lengths. Here you get into the whole unit business.

As a remark, although I think it's not very welcomed on this site, there actually are some papers (in subfields, in which some peolpe with names are involed too) which contain statements like

"What is needed is a formalism that is (i) free of prima facie prejudices about the nature of the values of physical quantities—in particular, there should be no fundamental use of the real or complex numbers..."

To sum it up, I think your first sentence "Almost every physical equation I can think of is expressed assuming continuous domains at least for one variable to range over" is exactly what it is. A method of representation. The method of getting at the results is implied by the mathematical relations associated with the symbols.

I personally don't take anything in physical theory particularly literally, but it's a useful and accessible language in any case, so that doesn't really influence how you do physics.


I am not an expert, so you should take my answer with grain of salt. I think the main thing to realize is that precisely because celestial mechanics/hydrodynamics can be formulated as continuum theories, they are "easy" to discretize and solve numerically. With the advent of computers this has become even more obvious than it has been before. It is in fact so easy to solve Newtons equations numerically that it is done in one of the first of Feynman's lectures for example. Differentiability is a promise that

$$x(t+\delta t) = x(t) + v(t)\delta t$$

will be a good discretization, if you take a small enough $\delta t$. This view of analysis is of course not emphasized until you take lectures on numerical analysis. From this viewpoint the real numbers are just a polite fiction, part of "Cantor's paradise" that mathematicians won't leave as Hilbert said.

To summarize: I think it is misguided to attempt a reformulation of continuum physics, the continuum theories lead to consistent discretized versions, that can be proven to approximate them to arbitrary precision.

  • $\begingroup$ Regarding the last sentence, the statement for there being "any discretized version" included is eighter a tautology (if the word "version" is interpreted to be a priori considered included) or I see no real justification for that claim. On the other hand, I don't know in which sense the rule for breaking up a series (or coming up with any other rules relating differentials) is "included" in the idea of the reals. I never understood how to rigorously formulize things like "$\gg$", do you have an idea? $\endgroup$
    – Nikolaj-K
    Commented Jul 26, 2012 at 21:57
  • $\begingroup$ @NickKidman I modified the last sentence, hopefully that is not bad form. $\endgroup$
    – orbifold
    Commented Jul 26, 2012 at 22:13
  • $\begingroup$ @orbifold: The problem is that your last sentence is sometimes not true. If you have Newtonian points orbiting with Newtonian gravity you can get blowups for a small number of points in finite time, where the particles go faster and faster on more elliptical orbits, until at a finite time they are infinitely fast. It is likely that many differential equation systems lead to hypercomputation. In the real world, adding a little bit of thermal or quantum jitter removes this property of point particle differential equations, as does smoothing out things to continuous fields with smoothing dynamics $\endgroup$
    – Ron Maimon
    Commented Jul 27, 2012 at 5:50
  • $\begingroup$ @RonMaimon You are right, I am not terribly satisfied with my answer myself. The idea I tried to convey was that numerical analysis has been very successful in simulating classical physics, because they deal with differentiable functions, those functions look boring locally and differential equations can therefore be discretized successfully. The "therefore" sweeps the whole field of numerical analysis under the rug and you are right that there are many subtle problems, energy conservation for example. $\endgroup$
    – orbifold
    Commented Jul 27, 2012 at 16:00

As the incredible difficulty to finding solutions to the three dimensional Ising model proves beyond doubt, discrete problems are fundamentally just as hard as continuum problems, in many cases probably even more so. More interestingly, physically relevant solutions of the higher dimensional Ising model correspond to modes of certain continuous equations, and the actual "beef" of the theory is in finding these intricate correspondences between systems that look very different on the surface, and yet, share very fundamental properties.

  • $\begingroup$ The question is not about feasability, it is about the possibility of formulating everything with discrete systems. $\endgroup$
    – ACuriousMind
    Commented Aug 11, 2014 at 18:00
  • $\begingroup$ The answer to that is a non-trivial yes, see a good textbook about discretization and numerical methods for the solution of ordinary and partial differential equations and general operator problems. My follow-up questions to that would, of course, be... "Why?" and "Is this even a physics question?" $\endgroup$
    – CuriousOne
    Commented Aug 11, 2014 at 18:45
  • $\begingroup$ A plain reference to numerical methods is not an actual answer, because in that case the mathematical object modeled by the fundamental equations still has the cardinality of the continuum, and this continuous object is what the theoretician holds in his mind when conceiving nature. The question aims to the plausibility of a discrete model reflected in the physics, not in the resolution methods adopted (which will always be discrete and computable, just because it cannot be otherwise). $\endgroup$
    – Mono
    Commented Aug 12, 2014 at 20:27

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