What is digital system and analogue system?

What are some problems/proofs in mathematics that can and have been solved by finding an analogue in a physical system?

• For example, what 3-Dimensional structure has the smallest surface area for a given volume? A sphere. An analogue found in the physical world is a water drop, which will typically assume a symmetric round shape. The amount of water in a drop is fixed, https://en.wikipedia.org/wiki/Surface_tension forces the drop into a sphere which minimizes the surface area of the drop (and hence the energy). This question is a specific case to

• Answer:

  There are several instances of varying levels of sophistication where physical insight has led to a better understanding of how a mathematical result might be proved. I'll categorise some of the better known examples into 'elementary' (meaning it can be fully understood with only knowledge of high school physics), 'intermediate' (requires some familiarity with undergraduate physics for full appreciation) and 'advanced' (requires too much of heavy-duty machinery to be properly elucidated upon in this answer). Elementary The Fermat point of a triangle is a point in the interior such that the sum of the distances to the vertices is the least possible. It turns out that the angle subtended by the sides of the triangle at the Fermat point is 2Ï€/32\pi/3. Leibniz came up with a lovely demonstration of this that involves straightforward physical reasoning. Place the triangle on a table-top, drill a hole at each vertex, thread a string through each of the holes and tie the three ends on the table-top onto a knot. Now procure three identically weighted bobs, tie them onto the the ends of the strings below the table and let everything snap into place. In equilibrium, the net force on the knot must be zero, and so the angles between the strings which carry the same tension must be 2Ï€/32\pi/3. At the same, we know that at (stable) equilibrium the potential energy is at a minimum, which means the the sum of the lengths of the portions of the string below the table must be maximised. Or in other words, given that the sum of the total lengths of the strings is fixed, the sum of the lengths of the strings on the the table-top is minimised. Q.E.D.                                                         *** Given an ellipse and any point on it, the normal to the ellipse at the point bisects the angle between the lines joining the point to the foci of ellipse. Since the sum of the distances from any point on the ellipse to the foci is, by definition, constant, Fermat's principle (the path taken by light between two points is such that it requires the least* time) dictates that since the time taken to travel from one focus to the other via a point on the ellipse is constant, light takes all possible paths to do so. In other words, the ellipse reflects light emitted at a focus in any direction onto the other focus. It trivially follows from the laws of reflection at a surface that the normal to the ellipse at the point bisects the angle between the lines joining the point and the foci of ellipse. Q.E.D.                                                         *** The area vectors of the faces of any polyhedron always sum to zero. To see why this must be true, picture a vial of homogeneous isotropic fluid floating about in space. Because it's in free fall, there is no pressure gradient and the pressure throughout fluid is a (nonzero) constant. Now picture a tiny fluid element inside this vial that is shaped like the polyhedron in consideration. The force on a face is minus the pressure times the area vector, so the net force is minus the pressure times the sum of all the area vectors. But the net force is zero since the whole thing is in free fall, and hence, the sum of the area vectors of the faces is zero. Q.E.D.                                                         *** Given a convex polyhedron and a fixed point in its interior, we can always find at least one face such that the orthogonal projection of the point onto the plane of the face lies in the interior of the face. If this wasn't the case, we could build a physical model the polyhedron in a way such that all its mass is virtually concentrated at the point in question. Setting the model on level ground would cause it to keep tipping over till Hell freezes over. Q.E.D. Intermediate Let's have a shot at the Fundamental Theorem of Algebra, which states that any polynomial with complex coefficients has at least one root, whence it follows that a polynomial of degree nn as nn roots, counting multiplicity. At the onset, I must admit I'm cheating a bit here as traditionally one uses FToA (or more powerful results such as Bézout's Theorem in algebraic geometry) to rigorously prove what we are going to take as our starting point. But then, the point of physical 'proofs' isn't to establish something rigorously but rather to base mathematical ideas on our physical intuition. So, beware of a considerable amount of handwaving. Consider a state with total spin JJ. We know from quantum mechanics that since the spin operator ^J2\hat J^2 commutes with the operator for a particular component of spin, say ^Jz\hat J_z, we can have simultaneous eigenstates, so that any JJ-spin state contains 2J+12J+1 'substates' corresponding to a quantum number MM ranging from −J-J to +J+J in steps of 11. In the language of quantum computation, this might be referred to as a qu-2J+12J+1-it. The idea we are going to exploit is that a qu-2J+12J+1-it might be built up from an unordered collection of 2J2J qubits. Classically, this makes sense — since each qubit possesses an up state and a down state, an unordered collection of 2J2J qubits may be identified by the number of up states, which gives a total of 2J+12J+1 states. Now, an unordered collection of qubits is in other words, a fully symmetric tensor product of qubits, which themselves may be written as α|↑⟩+β|↓⟩\alpha|\uparrow\rangle+\beta|\downarrow\rangle for α,β∈C\alpha,\beta\in\mathbb C. Now, consider a state given by |ψ⟩=∑+JM=−JψM|J,M⟩|\psi\rangle = \sum_{M=-J}^{+J}\psi_M|J,M\rangle where ψM∈C\psi_M\in\mathbb C. Now, let's formally replace all instances of |↑⟩|\uparrow\rangle by zz and |↓⟩|\downarrow\rangle by 11. Since, |J,M⟩|J,M\rangle is basically a state with with k=J+Mk=J+M up qubits, we have ∑2Jk=0ψk−Jzk=∏2Jj=1(αjz+βj)\sum_{k=0}^{2J}\psi_{k-J}z^k = \prod_{j=1}^{2J}(\alpha_jz+\beta_j) This is precisely the Fundamental Theorem of Algebra. Q.E.D.                                                         *** This one requires a bit of terminology. The Möbius group is the group of transformations on the extended complex plane (i.e. ∞\infty is included) which have the following form z↦az+bcz+d,ad−bc=1z\mapsto\frac{az+b}{cz+d}, \quad ad-bc=1 The group SO+(1,3)SO^+(1,3) is the connected group of linear transformations on R4\mathbb R^4 such that for any two points (t1,x1,y1,z1)(t_1,x_1,y_1,z_1) and (t2,x2,y2,z2)(t_2,x_2,y_2,z_2), the 'Lorentz dot' product, −t1t2+x1x2+y1y2+z1z2-t_1t_2+x_1x_2+y_1y_2+z_1z_2 is preserved. The two groups above are isomorphic. And there is a very physical way of demonstrating this. Firstly, we note that all possible Möbius transformations may be generated using translations, rotations, dilations, and inversions (not the same as Steiner's circular inversion but not unrelated either), all of which send circlines to circlines (a portmanteau word describing a curve that is either a circle or a line — it's convenient to regard lines as circles passing through the point at infinity). In fact, the Möbius group is precisely the group of transformations which send circlines to circlines and are orientation preserving (the handedness of the oriented boundary of a simply connected region is preserved — if you walk along boundary in the prescribed direction, the interior of the simply connected region is on the same side before and after the transformation). In fact, if I may add a point here, the problem with Steiner's circular inversion is that it reverses orientation — Möbius inversion is therefore the circular inversion followed by a reflection about the real line (conjugation) to restore orientation. Secondly, we can use the stereographic projection (illustrated below) to transform the extended complex plane into a Riemann sphere. Since the stereographic projection sends circlines on the complex plane to circles on the sphere, the Möbius transformations may again be characterised as orientation-preserving maps which send circles on the Riemann sphere to circles. Now, we'll remember from special relativity, that  SO+(1,3)SO^+(1,3) is precisely the group of Lorentz transformations with time reversals excluded (a past-pointing timelike four-vector cannot be transformed into a future-pointing timelike four-vector and vice versa). Imagine you are on a spaceship floating in the middle of nowhere, with a star-encrusted celestial sphere cocooning you with its dispassionate grandeur. What happens when you give yourself a Lorentz boost (go off in a direction of your choice at a constant velocity) or change the orientation of the viewing gallery? How does the celestial sphere transform? The celestial sphere is basically the set of all light rays (null lines) coming from the past. As long as two distinct events lie on the same null line, they represent the same point on the celestial sphere. To obtain a physical spherical surface we simply need to choose a spacelike slice of spacetime with a constant time tt (in our frame). To be specific, let's choose the slice t=−1t=-1. The null lines, by definition, are lines along which we have −t2+x2+y2+z2=0-t^2+x^2+y^2+z^2 = 0 So, if we let the bundle of null lines intersect this spacelike slice, we get a sphere of radius √x2+y2+z2=1\sqrt{x^2+y^2+z^2} = 1. Note that in a different frame, we'd make a corresponding choice t′=−1t'=-1 and get a different sphere. Since, the time component of the null four-vectors generating the sphere are fixed at t=−1t=-1, a circle on the sphere is given by null four-vectors whose Lorentz dot product with a fixed null four-vector is constant (to clarify, because the time term of the Lorentz dot product is fixed, the usual dot product between the spacelike projections of the corresponding null four-vectors, which actually corresponds to the angle, is constant). This means that under a Lorentz transformation, circles on the celestial sphere transform into circles. Therefore, the Möbius group and SO+(1,3)SO^+(1,3) are isomorphic. Q.E.D. Advanced Quite a significant fraction of Ed Witten's work, especially the work he did with Simon Donaldson and Nathan Seiberg, involves importing deep physical insights coming from supersymmetry and gauge theory into mathematical subdisciplines such as Morse theory and topology. I'll briefly describe the strategy Witten employed to prove and strengthen what are known as Morse inequalities using ideas commonplace in supersymmetry. The details may be found in his phenomenal 1982 paper that stunned the world of differential geometry, http://Supersymmetry%20and%20Morse%20Theory. So, what's Morse theory all about? The idea is that we can glean information about the topology of a 'compact Riemannian manifold wthout boundary' (which is a fancy name for a blob (or finitely many blobs) with no sharp edges/protrusions or self-intersections over which integration may be defined in a sensible way) by looking at the critical points of a 'height' function defined over the manifold. Critical points are of course those points on the manifold where the first derivatives vanish. To each critical point we may assign a Morse index pp, which is the number of negative eigenvalues of the matrix of second derivatives at the said point. Intuitively, this is the number of independent directions along which the Morse function is decreasing. Let's look at a concrete example. Consider a torus in its familiar doughnut embedding in space. Prop it up vertically on a tabletop, and the height function may be taken to be the literal height above the tabletop. Here's a diagram. We see that there are four critical points (we restrict ourselves to only nondegenerate cases — we don't want things to be unnecessarily complicated by the presence of entire rings of critical points which is what we get when the torus is placed horizontally on the table-top). The critical point at the very bottom has Morse index 00 as the height function increases along both the independent directions. The critical point at the very top has Morse index 22 as the height function decreases along both the independent directions.  The two critical points on the inner rim of the torus have Morse index 11,  as they are saddle points and the height function decreases along one independent directions and increases along the other. The pp-th Morse number MpM_p is defined to be the number of critical points with Morse index pp. So, in this case the Morse numbers (M0,M1,M2)(M_0,M_1,M_2) are (1,2,1)(1,2,1). Note that it doesn't make sense for pp to be greater than the dimension of the manifold as the dimension is the total number of independent directions there are. Also note that the notion of Morse numbers is sensitive to the choice of the height function — had we chosen to put a kink in our doughnut before propping it up on the table, as pictured below, the Morse numbers would be (2,4,2)(2,4,2). However the important thing is that no matter how we stretch and squeeze it, the lower bounds shall always be (1,2,1)(1,2,1). This is basically the content of the weak Morse inequality Mp≥BpM_p\ge B_p where BpB_p, the pp-th Betti number, is a topological invariant of the manifold. Informally put, BpB_p is the number of ways a 'compact' pp-dimensional submanifold without boundary may be obstructed from being continuosly deformed into another such compact pp-dimensional submanifold without boundary. For instance, a compact 00-dimensional submanifold without boundary is just a point on the manifold, and it is obstructed from being continuously deformed into other points by the fact that the other point might be residing on a different path-connected component. So, B0B_0 is just the number of path-connected components. Likewise, a compact 11-dimensional submanifold without boundary is just a loop. The obstructions here are 'holes' that extend in such a way that loops wrapping 'around' them can't be pulled free (or alternatively continuously deformed into points). So, B1B_1 is the number of such 'holes' and so on. Let's revisit the torus. Since there is only one path-connected component we have B0=1B_0=1. Since loops fail to to be continuously deformable to points in two distinct ways (pictured below), we have B1=2B_1=2. And finally, there is only one obstruction to the continuous deformation of a closed surface into a point, namely, the torus itself. So, B2=1B_2=1. As would be expected, we do have Mp≥BpM_p\ge B_p for all three values of pp in case of the torus. This is all pretty vague so far, but owing to the work of the differential geometers Georges de Rham and William V. D. Hodge, amongst others, this can be dealt with in a very precise way. I'll be skipping most of the details here (you might want to look up http://en.wikipedia.org/wiki/De_Rham_cohomology and http://en.wikipedia.org/wiki/Hodge_theory to fill in the gaps) but shall quote the result we'll directly use. The Laplacian we're familiar with in harmonic analysis of functions may be generalised as the Laplace-de Rham operator, denoted by Δ:=dd†+d†d\Delta:=dd^\dagger+d^\dagger d. I will postpone the discussion regarding what the operator dd and d†d^\dagger mean for now. The Laplace-de Rham operator acts on objects called differential forms or pp-forms (pp is said to be the grade of the differential form), which are things you can integrate over a compact pp-dimensional submanifold to get a real number (in particular, a 00-form is just a scalar function). The pp-forms ω\omega which satisfy Δω=0\Delta\omega=0 are said to be harmonic, and it turns out that the harmonic pp-forms form a vector space with dimension BpB_p. This is almost immediate in the case p=0p=0 when Δ\Delta reduces to the usual Laplacian. Now the functions annihilated by the Laplacian are basically the most boring functions possible on the manifold in question (it's ironic that studying these 'boring functions' can confront us with some of the deepest connections in mathematics), and the most boring function on a blob without boundary is just the constant function. However, the manifold may be consisting of many blobs (the number of such blobs is of course B0B_0) and the harmonic function can be a different constant on different blobs. So, the space of such functions form a vector space with dimension B0B_0. As for the case p>0p>0, the intuition may be derived from the behaviour of electromagnetic fields in vacuum, which happen to be harmonic 22-forms (provided the manifold has no boundary). So, for the general case, we would have to visualise some sort of a generalised electromagnetic field, whose governing equations continue to be linear (so that the solutions form a vector space). Now, even though the region permeated by the fields is empty, we might influence it by placing specimens of a similarly generalised notion of a current in the 'holes'. Since, an entire basis of solutions may be generated by placing a nonzero 'current' in exactly one of the 'holes', we see that the dimension of the vector space of solutions is the same as the number of 'holes'. Finally, we come to Witten's coup d'etat. The idea is to interpret the operator H:=ehtΔe−htH:=e^{ht}\Delta e^{-ht}, where hh is the height function, as a Hamiltonian with the space of differential forms being the Hilbert space of quantum states. The motivation for this choice of the Hamiltonian comes from the study of supersymmetry. To understand what supersymmetry has to do with it, we need to look at the  how the operators dd and d†d^\dagger are defined.  The operator dd, referred to as the exterior derivative, is a linear map from a pp-form ω\omega to a (p+1)(p+1)-form dωd\omega such that if we have a (p+1)(p+1)-dimensional (sub)manifold M\mathcal M with a pp-dimensional boundary ∂M\partial\mathcal M, then the following generalisation of the Fundamental Theorem of Calculus holds ∫Mdω=∫∂Mω\int_\mathcal M d\omega = \int_{\partial\mathcal M} \omega The  operator d†d^\dagger is simply the adjoint of dd, in the sense of normed vector spaces. It is possible to define an inner product between differential forms of the same grade. Given such an inner product ⟨⋅,⋅⟩\langle\cdot,\cdot\rangle, we can define d†d^\dagger as follows: ⟨d†α,β⟩=⟨α,dβ⟩\langle d^\dagger\alpha,\beta\rangle = \langle \alpha,d\beta\rangle where α\alpha is a (p+1)(p+1)-form while β\beta is a pp-form. We note that for the above definition to make sense, d†d^\dagger has to be a map from the space of (p+1)(p+1)-forms to that of pp-forms, the opposite of what happens in the case of dd. So, here's the deal. Witten interprets all differential forms of even grade to be bosonic and those of odd grade to be fermionic. Since, the operators dd and d†d^\dagger carry a form of even grade to a form of odd grade and vice versa, they are basically analogous to supersymmetric transformations. It must be remembered that these operators are not self-adjoint (as bona fide quantum mechanical operators have to be to ensure real eigenvalues), but rather are the adjoints of each other. This can however be easily remedied by taking the following linear combinations. Q1=d+d†Q_1=d+d^\dagger Q2=i(d−d†)Q_2=i(d-d^\dagger) [Alright, this has been taking more time than I realised it would and I have a flight to catch. So, I figured that since I won't be having access to Internet for over a week, and since there are quite a few examples here already, which I'll be really disheartened to delete in case someone else covers them, I'll go ahead and publish this anyway. I promise to finish it up as soon as I return to civilisation. Happy New Year, y'all.]