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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.

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You have two links to only one question. –  J. M. Nov 22 '10 at 0:45
    
J.M.: Thanks. Fixed the link. –  Jack Lemon Nov 22 '10 at 1:12
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Another relevant question is about translating a physical argument into a mathematical one. –  Ramsay Nov 22 '10 at 2:17

13 Answers 13

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

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

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

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

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Warmest congratulations for this cool example. –  Georges Elencwajg Nov 22 '10 at 13:10
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Does this argument basically say conversely that the first law of thermodynamics conversely boils down to the same kind of convexity as in the arithmetic-geometric mean? –  Phil Isett Sep 8 '11 at 2:38

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.

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Beautiful! How does one make this rigorous? Gauss-Bonnet? –  Qiaochu Yuan Nov 22 '10 at 14:11

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

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Nice example. More generally, the whole interplay between algebraic graph theory and electric networks. Say, you are given a tree-shaped electric network and the value of the current on each wire, and want to reconstruct the potential in each node. Of course, you can do so only up to a reference (ground) potential: This is equivalent to the fact that the range of the oriented incidence matrix of the graph has codimension $|V|-1$. –  Delio Mugnolo Nov 15 '13 at 9:32

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!

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I assign question one as a calculus problem with coordinates given for the two points. Of course, everyone tries to solve this using calculus, but the algebra of finding the zeros of the derivative of the length function is a little tricky. Then when someone asks me to solve the problem, I show them the conceptual solution. –  Steven Gubkin Nov 22 '10 at 12:59
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It's nearly trivial if you find the minimum of the square of the length rather than the length (they are related by a monotonic function, hence they have the same minima). –  Chris Taylor Apr 1 '11 at 11:19

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.

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+1: You beat me to it -- I heard Mark Levi give a number of talks along those lines while I was at Penn State, and they always ranked among the more interesting talks I've attended. –  Vaughn Climenhaga Nov 22 '10 at 1:33
    
This is officially now one of my all time favorite books and not only am I inspired to write a similar and more advanced text someday,but to use this book and others of its ilk in my teaching! –  Andrew L Nov 22 '10 at 6:28
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+1: You beat me too! (yet why giving +1 to whom beats you?) ;-) –  Pietro Majer Nov 22 '10 at 13:23

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.

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

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Franz: Do you know any sources that detail this? –  Jack Lemon Nov 22 '10 at 20:25
    
I seem to remember that this is described in "Archimedes. What did he do besides cry eureka?" by Sh. Stein. –  Franz Lemmermeyer Nov 23 '10 at 6:53

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.

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If we affix long strings between the cups and the table, we can get a similar statement about braids. –  S. Carnahan Sep 14 at 0:15

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.

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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, $H(X) \leq \log_2(n)$.

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

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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.

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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.)

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protected by Scott Morrison Nov 15 '13 at 7:42

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