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## How can simple physical “proofs” of mathematical facts be made rigorous?

Mark Levi's The Mathematical Mechanic is a book of examples of how physical reasoning can be used to solve mathematical problems; another couple of examples is in this blog post at Concrete Nonsense. Many of these proofs rely on conservation laws and other physical facts which, at least as I understand it, can be made rigorous by appealing to Noether's theorem. So these proofs themselves ought to be presentable in a mathematically rigorous way. But how?

There are many examples, so I will focus my question on two, one of which is in Levi's book and one of which I recently encountered on a friend's homework.

• Problem: prove the Pythagorean theorem. Solution: imagine a fish tank in the shape of a triangular prism. The triangle is a right triangle with side lengths $a, b, c$; call the vertices of the two copies of the triangle $A, B, C$ and $A', B', C'$, where the right angle is at $C$ (resp. $C'$). Drive a rod through $BB'$ and fill the tank with water. This system is at equilibrium (by conservation of... angular momentum?), so the net torque through the rod is zero. On the other hand the net torque through the rod is proportional to $c^2 - a^2 - b^2$.

• Problem: on each face of a tetrahedron, not necessarily regular, place a vector pointing normal to and out from the face with magnitude equal to the area of the face. Prove that the sum of these vectors is zero. Solution: fill the tetrahedron with an ideal gas. This system is at equilibrium (by conservation of... energy?), so the net force exerted by the gas on the tetrahedron is zero. On the other hand the net force exerted on each face is proportional to its area and points normal to and out from each face.

As you can see, I'm not even totally clear what conservation law(s) I'm invoking at the crucial step of the physical arguments. I would really appreciate mathematical insight into what's going on here.

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Another interesting physical argument is for the Fermat point minimizing distances to the vertices of a given triangle: tie three weights to strings which are joined at a point; fix each string to pass through small loops at the vertices. Hold the triangle flat in midair. As the weights will assume the lowest potential energy possible, they will minimize the sum distances to the vertices. (Note: there are some serious problems with the rigor of my explanation here, most notably uniqueness of a minimum and some conditions on the angles of the triangle. The general solution is similar though.) – drvitek Sep 16 2010 at 6:10
I think these "proofs" are bad examples for the idea of "physical proofs". You use some complicated ideas to explain some simple things ... I didn't check but I believe you use the Pythagorean theorem to make sense of angular momentum. – Helge Sep 16 2010 at 12:22
@Helge: maybe the first example isn't great. For me the surprise is that, as Igor mentions in his answer, "the real world" is in some sense doing an oracle computation, and I would like to understand how this oracle computation works mathematically independent of whether the proof it provides is a "good proof" or not. – Qiaochu Yuan Sep 16 2010 at 12:52

Both proofs in the question are of the following type. Suppose you want to prove a certain identity among numbers $X$, say $f(X)=0$. If you can find another function $g(X,Y)$ such that $f(X)=g(X,Y)$ for some $Y$. Then, proving that $g(X,Y)=0$ for this choice of $Y$, independent of $X$, automatically proves $f(X)=0$.

In the case of the Pythagorean theorem, $X=(a,b,c)$ are the dimensions of the triangular base of the water tank, $Y$ are all the other details of the physical setup and $g(X,Y)$ is the total torque about $BB'$ exerted on the tank. As long as $Y$ contains the fact that there are no torques acting on the tank other than the different parts of the tank acting on each other, physical law dictates that the total torque must be zero, hence $g(X,Y)=0$. The Kut-the-knot page linked from one of jc's comments shows why $g(X,Y)$ is proportional to $c^2-a^2-b^2$.

The case of the tetrahedron is similar, $X$ are the dimensions of the tetrahedron and $g(X,Y)$ is the total force acting on the gas container. The physical situation $Y$ is set up such that no forces act on the container external to the container itself, hence physical law dictates that the total force must be zero.

Now, what is it that allows us to conclude something about the value of $g(X,Y)$ in each case, given only $Y$ and independent of $X$? One way to think about it is to consider the physical world as a giant computer or oracle, $W$. If you set up a physical situation $Z$ and measure some observable quantity $G$, the world will give you a numerical answer $W(G,Z)$. The goal of physics, of course, is to build mathematical models of $W$, so that we don't have to consult the oracle every time we need an to know the value of $W(G,Z)$. But a mathematical model is not necessary if we are happy with just consulting the oracle when needed. Still, observation of the answers that we get from $W$ allows us to identify certain regularities in its output: these are the laws of physics.

Thus, supposing that we can find $G$ and $Z=(X,Y)$ such that $g(X,Y)=W(G,Z)$, the kind of of proof I outlined in the first paragraph can be carried out, appealing to the laws of physics as properties of the oracle $W$. To convert this kind of proof into a sequence of logical deductions, as a usual proof should be, requires a mathematical model for $W$, at least a restricted one that suffices for the purposes of making deductions about $g(X,Y)$. In both cases the model that does the job is plain old Newtonian mechanics of particles, solid bodies and fluids (if desired, solid bodies and fluids may be seen as a limiting cases of large assemblies of particles). The equality $g(X,Y)=f(X)$ follows from direct application of Newton's second law (an axiom of this mathematical model) and the identity $g(X,Y)=0$, independent of $X$, follows from Newton's third law (another axiom).

Personally, I think the appeal of such "physical" proofs is the fact that they rely only on some properties of $W$, and not on the details of a mathematical model for it. However, the validity of the proof only follows if one can show that a detailed mathematical model exists.

Hope this helps.

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Dear Qiaochu,

This is an interesting question. One way to solve the second question is by applying (some appropriate version of) Stokes's theorem. You are integrating a unit normal over the surface of the tetrahedron; it is a closed surface (or perhaps more relevantly, it is a boundary), so the answer is zero.

This interpretation of the second problem is essentially how one would mathematize the physics of flux, pressure, and so on, and, at least based on my experience in undergraduate physics classes, this is in fact how physicists will think about it. Of course, the converse applies, namely, that when confronted with various surface integrals and so on, physicists have the ability to think of them in physical terms, as computing a flux or the like, and this gives an insight into their meaning.

As for the first problem, I'm a little worried that if one mathematizes it too much, the proof will vaporize, because the mathematics of computing moments of inertia and so on is surely going to involve many applications of Pythagoras's theorem. On the other hand, at least formally, I would guess that one can again set it up as some kind of Stokes's theorem type computation.

I guess the point of view that I am suggesting is that these conservation-law type arguments are mathematized by integral formulas of the Stokes's theorem type.

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Thanks for the answer; Stokes' theorem definitely seems like the "right" way to think about the second question. For the first question my vague conception is that the crux of the problem is its rotational symmetry and there is some unspoken assumption about this symmetry which powers the proof. If someone could point that assumption out for me, that would be enough of a resolution, I think. – Qiaochu Yuan Sep 16 2010 at 2:46
I wasn't able to understand the first solution until I found this more lengthy writeup cut-the-knot.org/pythagoras/… and now I think the principle at work is the same as in the second solution - that is, the pressure of the water in the tank pushes differently on the different faces of the tank - the Pythagorean theorem is just Stokes theorem in this geometry. – jc Sep 16 2010 at 3:02
Dear jc, This is what I had in mind with my oblique comment at the end of my second-to-last paragraph. If you think about how the tank would hang, you can see that the pressure on the faces of length a and b is opposite to that on the face of length c, which is where the signs in the expression $c^2 - a^2 - b^2$ are coming from. – Emerton Sep 16 2010 at 3:18

Though I'm not sure I understand the solutions yet, the assertion "this system is at equilibrium" perhaps should be more properly interpreted as "an equilibrium configuration of the aforementioned system exists" (and depending on how closely you want to model the physics, showing this could be incredibly nontrivial). The rest of the solutions then appear to follow from properties of systems in equilibrium.

As an example of what I mean by nontrivial, a silly question that has always bugged me about ideal gases is that if they are noninteracting, how could they possibly equilibrate? The resolution in my mind to this is that despite talk about ergodicity, statistical mechanics in practice is not about dynamics, but rather about properties of configurational averages. But this is probably too far afield.

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 That's an interesting perspective. I am willing to interpret both solutions as limiting arguments about a large collection of particles exerting small amounts of force through collisions; that seems like a reasonable model of the physics. – Qiaochu Yuan Sep 16 2010 at 2:06 Thanks, great question! The derivation of the force from a gas can be found here en.wikipedia.org/wiki/… . As this derivation requires Newton's 3rd law, the physical symmetry of relevance for the final part of the second solution is conservation of momentum. I still can't say anything about equilibration, but I'll edit if I think of anything. – jc Sep 16 2010 at 2:21