Examples of using physical intuition to solve math problems

Question

For the purposes of this question let a "physical intuition" be an intuition that is derived from your everyday experience of physical reality. Your intuitions about how the spin of a ball affects it's subsequent bounce would be considered physical intuitions.

Using physical intuitions to solve a math problem means that you are able to translate the math problem into a physical situation where you have physical intuitions, and are able to use these intuitions to solve the problem. One possible example of this is using your intuitions about fluid flow to solve problems concerning what happens in certain types of vector fields.

Besides being interesting in its own right, I hope that this list will give people an idea of how and when people can solve math problems in this way.

(In its essence, the question is about leveraging personal experience for solving math problems. Using physical intuitions to solve math problems is a special case.)

---

These two MO questions are relevant. The first is aimed at identifying when using physical intuitions goes wrong, while the second seems to be an epistemological question about how using physical intuition is unsatisfactory.

Answer 1

The first and second laws of thermodynamics allow you to recover the inequality between the arithmetic and the geometric means: Bring together n identical heat reservoirs with heat capacity $C$ and temperatures $T_1,\ldots,T_n$ and allow them to reach a final temperature $T$. The first law of thermodynamics tells you that $T$ is the arithmetic mean of the $T_i$. The second law of thermodynamics demands the non-negativity of the change in entropy, which is

$$ Cn \, log(T/G) $$

where $G$ is the geometric mean. It follows that $T > G$.

I believe this argument was first made by P.T. Landsberg (no relation!).

Answer 2

Here is a proof of Pick's area theorem $\mu(P)=i +{b\over2}-1$ "using physical intuition": Assume that at time 0 a unit of heat is concentrated at each lattice point. This heat will be distributed over the whole plane by heat conduction, and at time $\infty$ it is equally distributed on the plane with density 1. In particular, the amount of heat contained in $P$ will be $\mu(P)$. Where does this amount of heat come from? Consider a segment $e$ between two consecutive boundary lattice points. The midpoint $m$ of $e$ is a symmetry center of the lattice, so at each instant the heat flow is centrally symmetric with respect to $m$. This implies that the total heat flux across $e$ is 0. As a consequence, the final amount of heat within $P$ comes from the $i$ interior lattice points and from the $b$ boundary lattice points. To account for the latter, orient $\partial P$ so that the interior is to the left of $\partial P$. The amount of heat going from a boundary lattice point into the interior of $P$ is a half, minus the turning angle of $\partial P$ at that point, measured in units of $2\pi$. Since the sum of all turning angles for a simple polygon is known to be one full turn, we arrive at the stated formula.

Answer 3

Theorem: Every permutation in $S_n$ is a product of transpositions.

Proof: If I number cups from 1 to $n$ and set them down in a row on the table in a mixed-up order, even a child could put the cups back into their natural order by exchanging cups two at a time using the left hand and right hand. QED

I learned this example from Ryan Kinser.

Answer 4

Polya's Induction and Analogy in Mathematics has a chapter on this, along with some great examples. It's not just physical intuition influencing mathematics; it's more a powerful synergy between physical and mathematical intuition. I'll summarize some of it:

1. Suppose we have two points A and B on the same side of some line L in the plane. What's the shortest path from A to L and then to B? The solution is obvious once we reflect one of the points (and its segment of the path) across L. That solution seems tricky in the abstract, but it's very intuitive if we imagine a reflecting ray of light and think about looking at things in a mirror.

2. Now suppose A and B are on different sides of L, and a particle moves from A to B, and its speed is different on the two sides of L. What's the shortest path (in time)? (This problem is to a refracting ray of light as the previous one is to a reflecting ray.) It turns out this can be solved by reducing it to a physical problem involving a system of weights and pulleys at equilibrium. I won't try to describe it here, but it might be fun to try to reinvent it.

3. Now let's take a serious math problem: what plane path minimizes the time an object takes to move from point A (at rest) to point B, assuming constant gravity? (This is the famous "brachistochrone" problem.) By conservation of energy, the speed of the object at a point on the curve depends only on its height (defined relative to its starting point and with respect to the direction of gravity). Thus, we're led to consider light moving in a very particular heterogeneous refracting medium, where the index of refraction depends in a specific way on the height. To find the path taken by light, we simply apply the law of refraction to this medium to obtain a differential equation for the path, which we can then solve.

The interplay between mathematical and physical intuition is very interesting here. The first problem is mathematical, but in trying to solve it, it's natural to draw an analogy to optics. The second problem is suggested by optics, but we solve it by analogy with mechanics. The third problem is basically mechanical, but we solve it by analogy with optics, and we actually use the solution to the second problem!

Answer 5

For electrical network intuition/applications to random walks see the beautiful little book of Doyle and Snell http://arxiv.org/abs/math/0001057

Answer 6

A paradigmatic example is Riemann's original "proof" of his mapping theorem in complex analysis. He gave an heuristic argument using Dirichlet's principle which was motivated by electrostatics in the plane.

Answer 7

Mark Levi's book The Mathematical Mechanic is full of elementary and beautiful examples of this kind. Some examples are also given in this blog post by Yan Zhang.

A classic example is a "proof" that there exist non-constant meromorphic functions on a compact Riemann surface, which I think is due to Klein: see this MO question.

Answer 8

Archimedes gave exact proofs as well as mechanically motivated explanations for results like the quadrature of the parabola or the volume of spheres.

Answer 9

The October 2015 issue of American Mathematical Monthly has an article by Tadashi Tokieda with the following title and abstract.

A viscosity proof of the Cauchy-Schwarz inequality

Abstract. The Cauchy–Schwarz inequality for positive quadratic forms has many proofs. This note gives a new derivation that looks unusual at first, but is natural in retrospect, interpreting the quadratic form as kinetic energy and the inequality as dissipation in a viscous flow.

Tokieda also has several articles applying physical intuition to mathematical problems, such as:

• Mechanical ideas in geometry, American Mathematical Monthly 105 (1998) 697–703.

• Applying physics to mathematics (in Japanese)

Answer 10

Read the following paper for some striking examples.

MR2587923 Atiyah, Michael; Dijkgraaf, Robbert; Hitchin, Nigel Geometry and physics. Philos. Trans. R. Soc. Lond. Ser. A Math. Phys. Eng. Sci. 368 (2010), no. 1914, 913–926.

Answer 11

I like this one.

Problem: Take an arbitrary tetrahedron. Draw 3 lines through the middles of opposing edges. Draw 4 more lines through vertices and intersections of medians of opposing faces. Prove that all 7 lines intersect in one point.

Solution: Place equal weights in the tetrahedron's vertices. The intersection point of all 7 lines is the centre of mass. The 7 lines correspond to different ways of computing the centre of mass based on the elementary physics observation that if you consider a set of objects as a union of 2 subsets the common set of mass is going to lie on the line connecting the centers of mass of the 2 subsets.

Answer 12

My favorite example: it's not too hard to show---see here for example---that if $U$ is a unitary matrix, then $\left|\operatorname{Per}(U)\right|\le1$, where Per denotes the matrix permanent function,

$\operatorname{Per}(U) = \sum_{\sigma \in S_n} \prod_{i=1}^n u_{i,\sigma(i)}$.

However, one can also give an immediate "physics proof" of the same inequality, as follows. Given any $n\times n$ unitary matrix $U$, one can set up a quantum optics experiment where $n$ identical photons are generated in separate input ports and pass through a network of beamsplitters; then the total amplitude for a single photon to appear in each of $n$ output ports, with no "bunching" of multiple photons in the same port, is equal to $\operatorname{Per(U)}$. (Intuitively, this is because photons are bosons, so you need to sum over all $n!$ possible ways that they could be permuted, with each permutation contributing an amplitude that's a product of the transition amplitudes for the photons considered individually.)

OK, but the probability of a measurement outcome is just the squared absolute value of its amplitude, and probabilities can never exceed $1$. Therefore $\left|\operatorname{Per}(U)\right|\le1$.

Answer 13

Let $X$ be a random variable taking on $n$ distinct values with probabilities $p_1,\dots,p_n$. The entropy of $X$ is defined by $H(X)=\sum p_i \log_2(1/p_i)$. An early theorem is that $H(X) \leq \log_2(n)$, and here's a physical proof. Place a point with mass $p_i$ at $(x_i,y_i)=(1/p_i,\log(1/p_i))$. The center of mass $$(\bar x,\bar y) = \frac{\sum (m_ix_i, m_iy_i)}{\sum m_i} = (n,H(X))$$ of the $n$ points must lie in the convex hull of the points (this is the physical intuition part). But since $y=\log(x)$ is concave, the convex hull is completely below (or on) the curve $y=\log(x)$. That is, $H(X) \leq \log_2(n)$.

Answer 14

I gave an answer based on surface tension (which I did not invent) to the napkin ring problem

Answer 15

Assume you want to prove a parabolic strict maximum principle, that is: Given an initial datum that attains everywhere a value $\ge 0$ but is not identically zero the solution of the corresponding linear heat equation is $>0$ everywhere and for all time $t>0$.

One possibility is the following: One shows that (because the semigroup that yields the solution is analytic) the above property is equivalent to irreducibility of the semigroup.

Now, irreducibility of the semigroup is (at least in my eyes) a very physical property of a diffusion process: It essentially says that you cannot expect particles to consistently hit a certain point and being reflected if that very point is not part of the boundary at all (say no potential-generated barriers).

(Indeed you can formalize this latter reasoning, but in my opinion even the introduction of the notion of irreducibility in the semigroup theory is perfectly justified by physical reasons.)

Answer 16

Let A, B, C be three points in the plane and assume that P minimizes the sum of the distances to A, B, C. One can proof that then the angles APB, BPC, CPA are all 120 deg. Physically we can imagine a table and three strings knoted together at P on the table and going through holes at A B C respectively. At each string a (unit) weight is attached. At equilibrium the potential energy and therfore the sum of the distances of the knot to A, B, C is minimal. On the other hand at the equilibrium the forces on the knot sum up to zero which is why the angles must be 120 degree.