Wikipedia : Plateau's problem | Monge-Kantorovitch transportation problem
DMOZ: Calculus of Variations
Vocabulary :
For an objective function of several variables, a "stationary point"
is a set of values of those variables where all partial derivatives
of the function vanish.
As illustrated below, a local extremum must be
a stationary point but the converse need not hold.
Because of the inclusive meaning
assigned by default to the simplest mathematical terms
(which is the exact opposite of the exclusive meaning often
attributed to simple words in everyday language)
most mathematicians consider "stationary point" and "saddlepoint"
to be synonymous:
At a saddlepoint, the relevant quantity may
rise in some directions and fall in others
but it's not required to do so...
There might very well be an extremum there!
Some authors reserve the term "saddlepoint"
to nonextremal stationary points.
We prefer to call those proper
saddlepoints
(thus following normal mathematical nomenclature).
Without the general methods presented in the rest of this page, every optimization problem appears to call for a brilliant insight. Here are a few classical such solutions, which ought to be remembered not only for the sheer beauty of the arguments leading to them but also for the simplicity in which the final results can be described. Arguably, ad hoc methods are best suited for problems whose solutions are simple (like Dido's problem).
Once guessed, such solutions may have beautiful justifications. In general, however, correctly guessing the solution of a complicated optimization problem is all but impossible and calculus is required for both the discovery and the justification of the solutions. When calculus itself becomes difficult to apply, the latest trend is to use numerical methods instead of symbolic manipulations (this should only be a last resort, though).
The solution of our first optimization problem depends on what can be construed as Euclid's postulated "optimization", namely:
Postulate : In a Euclidean plane (or Euclidean space) the shortest path between two points is a straight line.
Heron's problem
(least length and laws of reflection):
In the plane, we consider a straight line and two points A and B
on the same side of the line.
The basic question is to find which point M
on the line will yield the shortest two-legged path from A to M
and M to B.
The key observation is that any such path has the same length as a path from A to M and from M to the point B', symmetric of B with respect to the line. Conversely, any path from A to B' must cross the line at a point M.
If the path from A to B' is shortest, so is the matching path from A to M to B. By postulate, the shortest path from A to B' is a straight line and the solution M is simply found as the intersection of two straight lines.
Heron thus showed, nearly 2000 years ago, that the main law of mirror optics (the angle of reflection is equal to the angle of incidence) can be deduced from the appealing postulate that light travels along the shortest available route. A great intellectual feat for that time, or for any time!
Fermat's principle
(least time and laws of refraction):
Postulating instead (more generally) that light travels along the fastest
route, the assumption that its speed is different on either side
of the line yields the basic law of refraction (Snell's law).
The Torricelli point minimizes the sum of the distances to three given points. (This is a two-dimensional specialization of Plateau's laws.)
Wikipedia :
Fermat's theorem on stationary points
|
Hudde's rules (1657)
Video : Reasons for studying tensors
by Pavel Grinfeld (2014).
For a point x in the interior of the domain of f and a small enough value of e, both points x-e and x+e are in the allowed domain and yield values of f which are on both sides of f (x) if we just assume that f' is nonzero. Therefore, there can be an extremum at point x only when f' (x) vanishes.
On the other hand, there's no such requirement for a point x at the border of the allowed domain, because small displacements of only one sign are allowed.
Away from the border, a saddlepoint x (let's use that general term to indicate that f' (x) vanishes) will be the location of a minimum (resp. a maximum ) when f'' (x) is positive (resp. negative). If it's zero, further analysis is needed to determine whether x is the location of an extremum or not. (It's indeed an extremum if the first nonzero derivative is of even order.)
Extreme Value Theorem (real-valued continuous function on a compact set)
Regiomontanus (1436-1476) proposed to discuss how a line segment is viewed at an angle q from a variable viewpoint. That angle is maximum (and equal to 180°) from a point inside the segment. It is minimum (and equal to 0°) for a viewpoint on the same line as the segment but outside of it. The angle is arbitrarily small from viewpoints at large distances.
Regiomontanus considers a line perpendicular to the line of the segment and crossing it at a point outside the segment. He asks when the viewing angle is maximum from points on that line.
Wikipedia : Regiomontanus' angle maximization problem
A necessary (but not sufficient) condition for a smooth function of two variables to be extremal ( "minimized" or "maximized" ) on an open domain is to be at a saddlepoint where both partial derivatives vanish. In this case that means:
9x2-9y = 0 and 9y2-9x = 0
So, extrema of z can only occur when (x,y) is either (0,0) or (1,1).
To see whether a local extremum actually occurs, we must examine the second-order behavior of the function about each such saddlepoint (in the rare case where the second-order variation vanishes, further analysis would be required).
Well, if the second-order partial derivatives are L, M and N, then the second-order variation at point (x+dx, y+dy) is the quantity
½ [ L(dx)2 + 2M(dxdy) + N (dy)2 ]
We recognize the bracket as a quadratic expression whose sign remains the same (regardless of the ratio dy/dx) if and only if its (reduced) discriminant (M2-LN) is negative. If it's positive, the point in question is not an extremum.
In our example, we have L = 18x, M = -9 and N = 18y. Therefore:
M2 - LN = 81 (1 - 4xy)
At (0,0) this quantity is positive (+81). Thus, there's no extremum there. On the other hand, at the point (1,1) this quantity is negative (-243) so the point (1,1) does correspond to the only local extremum of z.
Is this a maximum or a minimum? Well, just look at the sign of L (which is always the same as the sign of N for an extremum). If it's positive, surrounding points yield higher values and, therefore, you've got a minimum. If it's negative you've got a maximum. Here, L = 18, so a minimum is achieved at (1,1).
To summarize: z has only one local extremum; it's a minimum of -3, reached at x=1 and y=1. Does this mean we have an absolute minimum? Certainly not! (When x and y are negative, a large enough magnitude of either will make z fall below any preset threshold.)
We're seeking saddlepoints (stationary points) of a scalar objective function M of n variables x1 , ... , xn which we view as components of a vector x.
M ( x1 , ... , xn ) = M ( x )
If those n variables are independent, then a saddlepoint is obtained only at a point where the differential form dM vanishes. This is to say:
0 = dM = grad M . dx
As this relation must hold for any infinitesimal vectorial displacement dx, such absolute saddlepoints of M are thus characterized by:
grad M = 0
That vectorial relation is shorthand for n separate scalar equations:
Instead of a set of independent variables (as discussed above) we may have to deal with several variables tied to each other by several constraints. The method of choice was originally introduced by Lagrange (before 1788) as an elegant way to deal with mechanical constraints:
For example, the variables may be subject to a single constraint (e.g., the cartesian equation of a surface on which a point-mass is free to move).
E ( x1 , ... , xn ) = constant
While an unconstrained saddlepoint of M was obtained when dM = 0. A constrained saddlepoint is obtained when dM is proportional to dE.
More generally, when k functions E1 ... Ek are given to be constant, a constrained saddlepoint of M is achieved when the differential form dM is a linear combination of the differentials of E1 ... Ek.
dM = l1 dE1 + l2 dE1 + ... + lk dEk
In other words, there are k constants li (each is known as the Lagrange multiplier associated with the corresponding constraint) such that the following function has an unconstrained (or absolute) saddlepoint.
M - ( l1 E1 + l2 E1 + ... + lk Ek )
[画像: Come back later, we're still working on this one... ]
Video: Lecture 2 | Modern Physics: Statistical Mechanics by Leonard Susskind (2009年04月06日)
The volume of a cone is one third of the product of its base area by its height. So, by imposing the cone's volume, we're actually imposing the height of the cone and looking for an optimal apex somewhere in a fixed plane parallel to the base...
Historically, the first problem of the type described below was the brachistochrone problem (find the shape of the curve of fastest descent) which had been considered by Galileo in 1638 and solved by Johann Bernoulli in 1696 (Bernoulli withheld his solution to turn the problem into a public challenge, which was quickly met by several prestigious mathematicians). The subject was investigated by Euler in 1744 and by Lagrange in 1760. It was named calculus of variations by Euler in 1766 and made fully rigorous by Weierstrass before 1890.
Let L be a smooth enough function of 2n+1 variables:
L = L ( q 1 , q 2 ... q n , v1 , v2 ... vn , t )
We assume that the first n variables (q) are actually functions of the last one (t) and also that the subsequent variables (v) are simply their respective derivatives with respect to t:
This makes L a function of t alone and we may consider the following integral S (called the "action" of L ) for prescribed configurations at both extremities. In this context, a configuration is defined as a complete set of values for the q-variables only (irrespective of what the v-variables may be) so, what we are really considering now are prescribed fixed values of all the 2n quantities q i (a) and q i (b).
The fundamental problem of the calculus of variations is to find what local conditions make S stationary, for optimal functions q i . Namely:
Leonhard Euler (1707-1783) Joseph-Louis Lagrange (1736-1813) Euler-Lagrange equationsProof : From a set of optimal functions q i and arbitrary (sufficiently smooth) functions h i which vanish at both extremities a and b, let's define the following family of functions, depending on one parameter e :
Those yield a value S(e) of the action which must be stationary at e = 0 (since the functions q i are optimal). Thus, the derivative of S(e) along e does vanish at e = 0, which is to say:
Since every h i vanishes at both extremities a and b , we may integrate by parts the second term of the square bracket to obtain:
As h i is arbitrary, the square bracket must vanish everywhere, for every i. (If this wasn't so, the whole equality would be violated for a choice of h i vanishing wherever the square bracket doesn't have a prescribed sign). QED
The above is most commonly used as the basis for variational mechanics and related aspects of theoretical physics based on a principle of least action. However, it's also the correct answer to more practical concerns:
As a pilot, I've always been interested in writing a proper piece of flight-planning software to optimize the plane's path with regard to time, fuel, or any combination thereof. [...]
I've always felt the professionally-provided solutions were crude approximations that do not take into account the full range of possibilities, especially over long distance, as one crosses varying jet streams at different altitudes, etc. [...] What is the correct mathematical approach?
Well, just express carefully the local cost function (L) as a function of the position of the aicraft and of the related derivatives (to a good approximation, the latter are only useful to compute horizontal speed). Predicted meteorological conditions (changing with time throughout space) can be used for best planning.
The Euler-Lagrange equations then tell the pilot what to do at all times.
Calculus of Variations (30:03) by Dr. Peyam (2018年02月16日).
A slight modification of the above proof yields a straight derivation of one of the greatest statements of mathematical physics. Let's just do it...
Suppose that, for specific functions hi , a symmetry exists which leaves L unchanged, (to first-order in e about 0) when q i is replaced by
Q i = q i + e h i
Formally, this leads to a situation similar to the previous one, since we still know that the derivative of S(e) vanishes at e = 0 (albeit for very different reasons). However, the h functions need not vanish at the extremities a and b. So, an extra "integrated term" appears in the following expression of the derivative of S(e) which results from our integration by parts:
Now, as the integral still vanishes (because the previously established Euler-Lagrange equations make every square bracket vanish) the extra term must be zero as well. This means that the following quantity doesn't change:
Conserved Noether Charge
Emmy Noether (1882-1935)
Lagrangian & Field Theory (1:40:31)
by Leonard Susskind (2008年05月05日).
Noether's Theorem and The Symmetries of Reality (13:01)
by Matt O'Dowd (2018年05月16日).
The most significant genius: Emmy Noether (10:23)
by Don Lincoln (Fermilab, 2018年07月13日).
On a 2D surface M (u,v) the infinitesimal distance corresponding to a displacement du,dv is given by the first fundamental quadratic form:
F1 (du,dv) = (dM)2 = E (du)2 + 2 F du dv + G (dv)2
The problem of minimizing the length of the path between two points on that surface is a standard exemple of the calculus of variations. Let's introduce notations compatible with the above:
q1 = u
q2 = v
v1 = du/dt
v2 = dv/dt
L = [ F1 (v1 , v2 )
]½
=
[ E v12 + 2 F v1 v2
+ G v22
]½
The two Euler-Lagrange equation for this Lagrangian are (i = 1,2) :
The first of those (i = 1, qi = u) translates into:
Since dX/dt = v1 ¶X/¶u + v2 ¶X/¶v the right-hand side is equal to:
We are now presenting the brachistochrone problem as a simple application of the calculus of variation. Historically, the relation is reversed, as this problem actually helped define the need for the latter, which was formalized many years after the brachistochrone problem was solved.
In June 1696, Johann Bernoulli (1667-1748) challenged the readers of Acta Eruditorum with his famous brachistochrone problem:
Along what curve would a point-mass fall from one prescribed point to another lower one (not directly underneath) in the least possible time?
Johann Bernoulli's own solution was published in the same journal a year later, along with 4 of the 5 solutions sent by famous contributors (the solution of Guillaume de l'Hôpital was only published in 1988).
We shall use a coordinate system where the origin is the starting point. We choose to orient the y-axis downward so that y is positive for all target points. Let's call u the slope of the trajectory dy/dx. In a gravitational field g the conservation of mechanical energy for a mass m dropped from the origin at zero speed tells us that:
m g y = ½ m [ (dx/dt)2 + (dy/dt)2 ]
Therefore, ( 2 g y ) ½ dt = ( 1 + u2 ) ½ dx
To minimize the descent time, we seek to minimize the path integral of dt or, equivalently, to minimize the integral of (2g)½ dt = L dx
L dx = L (y,u,x) dx = ( 1 + u2 )½ / y½ dx
To minimize this integral, the relevant Euler-Lagrange equation must hold:
With the above expression of L, this translates into:
- ( 1 + u2 )1/2 / 2y3/2 = d/dx [ u ( 1 + u2 )-1/2 / y1/2 ]
Recalling that u = dy/dx, we reduce this algebraically:
- ( 1 + u2 )-1/2 / 2y3/2 = (du/dx) ( 1 + u2 )-3/2 / y1/2
- ( 1 + u2 ) = 2 y du/dx = 2 u y du/dy
That last equality holds only when dy/dx doesn't vanish, which need not always be true. We put v = u2 and separate the variables to integrate:
d {Log y} = dy/y = - dv / (1+v) = - d { Log (1+v) }
Therefore, y / 2a = 1/(1+v) which is to say v = 2a / y-1 for some a.
This characterizes a cycloid corresponding to a wheel of radius a.
[画像: Come back later, we're still working on this one... ]
Video :
The Brachistochrone
by Grant Sanderson, with Steven Strogatz.
Wikipedia :
Brachistochrone curve
They purchased a site, which was named 'Bull's Hide' after the bargain
by which they should get as much land as they could enclose with a bull's hide.
Aeneid by
Virgil (70-19 BC)
The land so acquired by queen Dido, founder Carthage, is interpreted to be whatever area could be bounded by the seashore on one side and a rope of fixed length on the other. (The preliminary step was to cut the hide into very thin strips to make a long rope out of it.) The problem is to maximize that area, assuming the shore line to be perfectly straight.
To prove that the solution is a half-circle, one considers the two straight lines drawn from an arbitrary point on the optimal boundary to either extremity of the rope (on the shoreline). If we assume that both parts of the rope are rigidly attached to those two lines, there's a constant area between those lines and their respective ropes. What remains is the area of the triangle formed by two sides of fixed length and part of the shore line. This area is greatest when the two sides are perpendicular!
By the second theorem of Thales, a planar curve where an arbitrary point always sees the extremities at a right-angle is a semicircle. QED
In 1846, the great Swiss geometer Jakob Steiner (1796-1863) showed that the above solution of Dido's problem implies the familiar 2-dimensional isoperimetric inequality (i.e., the planar curve of given perimeter which encloses the greatest area is a circle). Three steps:
No such simple geometric solutions are available to justify the higher-dimensional counterparts of the isoperimetric inequality, discussed next.
The surface area S enclosed by a closed planar curve of given perimeter P is largest in the case of a circle. This ancient statement (proved geometrically by Steiner, in 1846) can be expressed by the following relation, known as the isoperimetric inequality:
S ≤ P 2 / 4p
The three-dimensional equivalent of the isoperimetric inequality says that the closed surface of area S which encloses the largest volume V is a sphere.
V 2 ≤ S 3 / 36 p
The above can be generalized to n dimensions: No hypersurface of hyperarea S encloses a larger hypervolume V than the hypersphere. This makes the relations given in the oldest Numericana article yield:
V n-1 ≤ G ( 1 + n/2 ) [ S / nÖp ] n
Action, Minkowski Addition
& Two Proofs of the Isoperimetric Inequality (2010) by Jeremy Boother.
Wikipedia :
Isoperimetric inequality
That constant vanishes when both sides of the soap film are at the same pressure, in which case the film is a surface of least area within the given border. However, the non-vanishing case was also considered by Joseph Plateau; it arises when soap film encloses pockets of air...
Besides the familiar spherical soap bubble, the simplest case is a soap bubble merged within a large planar film. It takes on the form of two spherical caps which form an outer angle of 120° (inner angle of 60°) with the planar part of the film whwere they meet. The thickness 2h and the equatorial diameter d of the lenticular bubble are proportional to the radius of curvature of the spherical parts:
The Plateau rules (1873) state that the solutions are smooth surfaces of constant mean curvature with only two possible types of inner singularities:
Because a regular tetrahedron is formed by every other vertex of a cube, two vertices of a regular tetrahedron are seen from its center at the same angular separation as between (1,1,1) and (1,-1,-1). The cosine of that angle is -1/3. Chemists call it the tetrahedral angle :
q = acos ( -1/3 ) = 1.910633236249... = 109.47122063449...°
Jean Taylor (1944-) published a proof of Plateau's laws in 1976.
By finding that the catenoid is the only minimal surface of revolution (besides planes orthogonal to the axis) Euler solved Plateau's problem (well before it was so named) when the given border consists of a pair of two sufficiently close coaxial circles.
When the parallel planes of the coaxial circles are too far apart, no catenoid can join them and the solution of Plateau's problem just consists of separate planar surfaces.
Wikipedia : Plateau's Laws | Joseph Plateau (1801-1883) | Jean Taylor (1944-)
Such surfaces are technically called complete embedded minimal surfaces. Before 1982, only four examples were known: [画像: Helicoid ]
David Hoffman (a 1973 Stanford Ph.D.) undertook the task at the University of Massachusetts at Amherst.
Loosely speaking, Costa's surface features two complementary pairs of "tunnels" through the equatorial plane which connect the inside of some "catenoidal" northern half and the outside of its southern half, or vice-versa.
Costa-Hoffman-Meeks Surface (Image courtesy of Matthias Weber)
Applied to helicoids, the idea yields yet another family of complete minimal embedded surfaces where tunnels provide shortcuts between slices of space which are otherwise connected only by circling around the helicoid's axis.
[画像: Come back later, we're still working on this one... ]
Minimal Surfaces and
their Classification (pdf 4.44 MB) by William H. Meeks, III.
Scherk's surface (1834)
|
Heinrich Scherk (1798-1885)
Bonnet family
of a minimal surface.
Proof. (HINT: This puzzle appears among optimization problems.)
The three vertices of a triangle ABC can be connected using just its two shortest sides. If the angle between those two sides is 120° or larger, this is the most economical way to do so.
Otherwise, we may define a fourth point T such that the three angles ATB, BTC and CTA are all equal to 120°. Then, the three llines TA, TB and TC form the most economical way to connect the three dots A, B and C.
The point T is called the Torricelli point of ABC. The results for the first and the second of the above cases are unified by defining T to be the vertex opposite the longest side in the first case.
Likewise, four points in a plane are best connected by introducing 0, 1 or 2 new Torricelli points.
Nobody questioned the celebrated Honeycomb conjecture before 1993, when Phelan and Weaire showed its 3D counterpart (Kelvin's conjecture) to be false! The issue was settled in 1999 with a proper proof by Thomas C. Hales : Thus, the 2D Honeycomb conjecture is now a theorem! Here it is:
Honeycomb Theorem In any partition of the plane into tiles of unit area, each tile cannot have a lesser perimeter than the regular hexagon of unit area. (Loosely speaking, the regular honeycomb tiling is the most economical one.)
The first proof by Thomas C. Hales (1999) was surprisingly intricate. It was crucial to get rid of the convexity restriction which earlier authors had reluctantly imposed. (The isoperimetric inequality implies that the boundaries between optimal shapes are either circular arcs or straight lines. Both sides of such a boundary are convex only in the latter case.) The reason curved boundaries are ruled out is not at all obvious; it is false in the 3D case.
With a vertical axis, such a curve has a vertical plane of symmetry, the vertical coordinate is unchanged by rotation of 2p/3 (1/3 of a turn) and its sign changes for half of that (p/3 rotation).
A surface containing that line and having the same symmetry will be described by an equation of the following type, in cylindrical coordinates:
z = f ( r , sin 3q ) [ f being odd for either argument ]
Let's find the functions f which make the mean curvature constant:
[画像: Come back later, we're still working on this one... ]
Mean curvature (Numericana) | Monkey saddle (Wikipedia) | Monkey saddle (MathWorld)
William Thomson, Lord Kelvin (1824-1907) Kelvin's problem is to partition space into unit-volume cells with least surface area. Simon Lhuilier (1750-1840) called this one of the most difficult problems in geometry.
In the straight polyhedron, the angle q between a square and an adjoining hexagon is slightly less than 60°. An easy way to obtain the exact value is to consider the cube formed by the six square faces. Seen from the center, the middles of the hexagons are in the directions of the corners of that cube. Therefore:
q = acos ( 1/Ö3 ) = 54.73561...°
By Plateau's rules, an optimally curved hexagonal surface always meets the plane of an adjoining tetragonal face at an angle of 60°. Therefore, the curved surface at the middle of an edge is inclined 5.26439° with respect to the plane of the original flat hexagon.
On the other hand, an edge between two hexagonal faces curves inward. (It's the outward-curving edge of a tetragon in an adjacent cell.)
Although the equal-volume foams discussed below are more economical than this one, Kelvin's foam remains best among all monohedral foams (i.e., foams whose cells are all congruent to each other).
On the division of space with minimum partitional area
Sir William Thomson (Lord Kelvin).
Acta Mathematica, 11 (1887) pp. 121-134.
When a foam isn't monohedral, it need not be equal-pressure. Actually, none of the foams below are. In all of them, different cells may be assigned different "pressures" so a face's mean curvature is the difference between the pressures of the cells it separates (not necessarily zero).
[画像: Basic pattern of the Weaire-Phelan foam (c) 2008 by Ruggero Gabbrielli ]
In 1993, Denis Weaire (1942-) and his student Robert Phelan disproved Kelvin's conjecture with a structure more economical than Kelvin's foam: The A15 Weaire-Phelan structure is one example of a Frank-Kasper foam. It consists of two distinct types of cells, with pentagonal or hexagonal faces. One cell is a squashed dodecahedron, the other is a tetradecahedron. That basic structure occurs naturally in crystals of b tungsten.
[画像: Basic pattern of the p42a foam (c) 2008 by Ruggero Gabbrielli ]In a draft (2008年10月25日) of a paper published on 2009年06月02日, Ruggero Gabbrielli presents a new type of unit-cell foam (featuring some tetragonal faces) whose cost (5.303) is intermediate between the Weaire-Phelan foam (5.288) and Kelvin's foam (5.306). He calls this structure "P42a". The repeating pattern consists of 14 curved cells of 4 different types (ten tetradecahedra and four tridecahedra). It has an average of 96/7 = 13.714285... faces per cell.
The above pictures for the A15 and P42a foams were obtained by Ruggero Gabbrielli using the GAVROG 3dt visualization software. Gabbrielli has also posted several related interactive 3D visualizations under Java. The 1993 optimization on A15 and Gabbrielli's recent research both used Ken Brakke's Surface Evolver (with specific code provided by John M. Sullivan, in the latter case at least).
At this writing, the A15 foam described by Weaire and Phelan
in
1993 is still conjectured to be the answer to Kelvin's problem.
It is about 0.3% less costly than Kelvin's original foam.
The Weaire-Phelan foam was used as the basis for the design of the celebrated Water-cube of the 2008 Olympics (Beijing National Aquatics Center) which is the largest structure covered with ETFE film.
The number of faces in a minimal foam (1992)
by Robert B. Kusner
Comparing the Weaire-Phelan
foam to Kelvin's foam (1996)
by Rob Kusner and John M. Sullivan
A new counterexample to
Kelvin's conjecture on minimal surfaces
by Ruggero Gabbrielli
Comparing the Kelvin,
Gabbrielli and Weaire-Phelan structures (Java)
The
Weaire-Phelan Structure (and the Water-Cube) by
John C. Baez (2008年08月21日).
J.J. Thomson is credited with the discovery of the electron in 1897 (to be fair, Jean Perrin had discovered it two years earlier).
This was the first subatomic particle ever discovered. For quite a while, nobody knew exactly how neutral atoms could be made from those negative electrons and some unknown positive stuff. Thomson himself envisioned a naive model which became known as the Plum-pudding model (1904) whereby mutually-repulsive electrons would sit at the surface of a spherical blob of positive stuff.
In 1909, the celebrated gold-foil experiment was conducted by Geiger and Marsden under the supervision of Ernest Rutherford (1871-1937). The observed violent bouncing of some of the alpha-particle projectiles suggested a totally different model (Rutherford, 1911) where most of the atom's mass resides in a tiny positively-charged nucleus around which electrons are maintained in orbit by electric attraction.
Classically, such accelerated charges would radiate energy away and the orbiting electrons would quickly spiral into the nucleus... It was a major achievement of early quantum mechanics to explain why they don't do so (Bohr, 1913).
Back in 1904, J.J. Thomson wondered how N like charges would be distributed on the surface of a sphere at equilibrium (minimizing potential energy, at least locally). He set it as a purely mathematical puzzle, mostly for the Coulomb potential (1/r) but possibly for other potentials as well.
Until 2010, the only known analytical solutions were the obvious configurations for 1, 2, 3, 4, 6 and 12 charges (forming a Platonic solid in the last three cases). It was proved only recently (2010) that the triangular dipyramid is the only solution for 5 charges.
Surprisingly enough, the Platonic solids for N = 8 or 20 are not solutions (better distributions have been found numerically). For example, for N=8, the cube is improved upon by a square antiprism obtained from that cube by rotating two opposing faces 45° from each other (about their common axis). That shape happens to be optimal for the Coulomb potential.
Searching potential energy surfaces by simulated annealing by Luc T. Willie, Nature 324, 46-48 (London, 1986年11月06日).
Interestingly, the electric dipole moment (EDM) of some numerical solutions differ substantially from zero for the following number of charges:
11, 13, 19, 21, 25, 26, 31, 33, 35, 43, 47, 49, 52, 53, 54, 55 ... (A268487)
Thanks to Maximilian Hasler for calling my attention to this.
Thomson problem
|
J.J. Thomson (1856-1940; Nobel 1906)
Smale's problems (18 open questions, 1998)
|
Steve Smale
(Fields medalist, b. 1930)
The 5-Electron case of Thomson's Problem
by Richard Evan Schwartz (Brown, 2010年02月09日).
A new take on the old
Thomson problem by Travis Hoppe (2017年05月04日).
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Gaspard Monge (1746-1818) | Leonid_Kantorovich (1912-1986; Nobel Prize 1975) | Linear programming
