(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 15677, 500]*) (*NotebookOutlinePosition[ 21665, 638]*) (* CellTagsIndexPosition[ 20237, 607]*) (*WindowFrame->Normal*) Notebook[{ Cell["Divergence and Curl and Their Geometric Interpretations", "Title"], Cell[CellGroupData[{ Cell["Scalar Potentials and Their Gradient and Laplacian Fields", "Subtitle"], Cell["\<\ Setting up the combined potential due to three point potentials as \ an example to visualize gradients, divergences, and curls\ \>", "Section"], Cell[BoxData[ \(Simple\ \((2 D)\)\ \ 1\/r\ \(potential : \ \ \((\ gravity, \ electrostatic\ potential\ have\ this\ dependency)\)\)\)], "Text", FontFamily->"Helvetica", CellTags->"mmtag:13:1/r_potentials"], Cell[BoxData[ \(potential[x_\ , \ y_, \ xo_\ , \ yo_] := \ \(-1\)/ Sqrt[\((x - xo)\)^2\ + \ \((y - yo)\)^2]\)], "Input"], Cell["A field source located a distance 1 south of the origin", "Text"], Cell[BoxData[ \(HoleSouth[x_, \ y_]\ := \ potential[x, y, \ Cos[3 Pi/2], \ Sin[3\ Pi/2]]\)], "Input"], Cell["Sources located distance 1 at 30\[Degree] and 150\[Degree]:", "Text"], Cell[BoxData[ \(HoleNorthWest[x_\ , \ y_]\ := \ potential[x, y, \ Cos[Pi/6], \ Sin[\ Pi/6]]\)], "Input"], Cell[BoxData[ \(HoleNorthEast[x_\ , \ y_]\ := \ potential[x, y, \ Cos[\ 5\ Pi/6], \ Sin[5\ Pi/6]]\)], "Input"], Cell["\<\ Function that returns the two dimensional (x,y) gradient field of \ any function declared a function of two arguments:\ \>", "Text"], Cell[BoxData[ \(gradfield[ scalarfunction_]\ := \ {D[scalarfunction[x, y], x] // Simplify, \ D[scalarfunction[x, y], y] // Simplify}\)], "Input", CellTags->"mmtag:13:gradient__example_function"], Cell["Generalizing the function to any arguments:", "Text"], Cell[BoxData[ \(gradfield[scalarfunction_, \ x_\ , \ y_]\ := \ {D[scalarfunction[x, y], x] // Simplify, \ D[scalarfunction[x, y], y] // Simplify}\)], "Input"], Cell["The sum of three potentials:", "Text"], Cell[BoxData[ \(ThreeHolePotential[x_, \ y_]\ := \ HoleSouth[x, y]\ + \ HoleNorthWest[x, y]\ + \ HoleNorthEast[x, y]\)], "Input"], Cell["f(x,y) visualization of the scalar potential:", "Text"], Cell[BoxData[ \(Plot3D[ ThreeHolePotential[x, y], {x, \(-2\), 2}, {y, \(-2\), 2}]\)], "Input", CellTags->"mmtag:13:Plot3D[]__example_1/r_potentials"], Cell["Contour visualization of the three-hole potential", "Text"], Cell[BoxData[ \(ContourPlot[ThreeHolePotential[x, y], {x, \(-2\), 2}, {y, \(-2\), 2}, PlotPoints \[Rule] 40, ColorFunction \[Rule] \((Hue[1 - #*0.66] &)\)]\)], "Input", CellTags->"mmtag:13:ContourPlot[]__example_1/r_potentials"], Cell["Gradient field of three-hole potential", "Text"], Cell[BoxData[ \(gradthreehole\ = \ gradfield[ThreeHolePotential]\)], "Input"], Cell[BoxData[ \(<< Graphics`PlotField`\)], "Input"], Cell[BoxData[ \(PlotVectorField[gradthreehole, {x, \(-2\), 2}, {y, \(-2\), 2}, ScaleFactor \[Rule] 0.2, ColorFunction \[Rule] \((Hue[1 - #*0.66] &)\), PlotPoints \[Rule] 21]\)], "Input", CellTags->"mmtag:13:PlotVectorField[]__example_gradient_of_1/r_potentials"],\ Cell["\<\ Function that takes a two-dimensional vector function of (x,y) as \ an argument and returns its divergence\ \>", "Text"], Cell[BoxData[ \(divergence[{xcomp_\ , \ ycomp_}]\ := \ Simplify[D[xcomp, x]\ + \ D[ycomp, y]]\)], "Input", CellTags->"mmtag:13:divergence__example_function"], Cell[BoxData[ \(divgradthreehole\ = \ divergence[gradfield[ThreeHolePotential]] // Simplify\)], "Input"], Cell[BoxData[ \(Plotting\ the\ divergence\ of\ the\ gradient\ \((\[Del]\(\(\[CenterDot]\ \)\((\[Del]\ f)\)\)\ is\ the\ ``Laplacian''\ \[Del]\^2\ f, \ sometimes\ indicated\ with\ symbol\ \[CapitalDelta]f)\)\)], "Text", FontFamily->"Helvetica", CellTags-> "mmtag:13:laplacian__example_divegence_of_gradient_of_1/r_potentials"], Cell[BoxData[ \(Plot3D[divgradthreehole, {x, \(-2\), 2}, {y, \(-2\), 2}, PlotPoints -> 60]\)], "Input", CellTags->"mmtag:13:Plot3D[]__example_divegence_of_1/r_potentials"] }, Closed]], Cell[CellGroupData[{ Cell["Coordinate Transformations in the Vector Analysis Package", "Subtitle"], Cell[TextData[{ "It is no surprise that many of these differential operations already exist \ in ", StyleBox["Mathematica", FontSlant->"Italic"], " packages." }], "Text"], Cell[BoxData[ \(<< Calculus`VectorAnalysis`\)], "Input", CellTags->"mmtag:13:VectorAnalysis_package"], Cell["Coordinate Systems", "Section"], Cell["Converting between coordinate systems", "Subsection", CellTags->"mmtag:13:coordinate_systems__converting_between"], Cell["\<\ The spherical coordinates expressed in terms of the cartesian x,y,z\ \ \>", "Text"], Cell[BoxData[ \(CoordinatesFromCartesian[{x, y, z}, Spherical[r, theta, phi]]\)], "Input", CellTags->"mmtag:13:CoordinatesFromCartesian[]"], Cell["\<\ The cartesian coordinates expressed in terms of the spherical r \ \[Theta] \[Phi]\ \>", "Text"], Cell[BoxData[ \(CoordinatesToCartesian[{r, theta, phi}, Spherical[r, theta, phi]]\)], "Input", CellTags->"mmtag:13:CoordinatesToCartesian[]"], Cell["\<\ The equation of a line through the origin in spherical \ coodinates\ \>", "Text"], Cell[BoxData[ \(Simplify[ CoordinatesFromCartesian[{a\ t, \ b\ t, \ c\ t}, Spherical[r, theta, phi]], t\ > \ 0]\)], "Input"], Cell[CellGroupData[{ Cell["\<\ An example of calculating the positions of cities in cartesian and \ spherical coordinates.\ \>", "Subsection"], Cell[BoxData[ \(<< Miscellaneous`CityData`\)], "Input", CellTags->"mmtag:13:MiscellaneousCityData_package"], Cell["\<\ Boston is located at latitude 42\[Degree] 21' 30\" N and longitude \ -71\[Degree],-3',-37\" W\ \>", "Text", CellTags->"mmtag:13:latitude_and_longitude"], Cell[BoxData[ \(boston\ = \ CityData["\", CityPosition]\)], "Input", CellTags->"mmtag:13:CityData[]"], Cell[BoxData[ \(paris\ = \ CityData["\", CityPosition]\)], "Input"], Cell[BoxData[ \(SphericalCoordinatesofCity[ cityname_String]\ := \ \[IndentingNewLine]{\[IndentingNewLine]6378.1\ \ , \[IndentingNewLine]\(\(2\ Pi\)\/360\) ToDegrees[\(CityData[cityname, CityPosition]\)[\([1]\)]], \[IndentingNewLine]\ \[IndentingNewLine]\(\(2\ Pi\)\/360\) ToDegrees[\ \(CityData[cityname, CityPosition]\)[\([2]\)]]\[IndentingNewLine]}\)], "Input", CellTags->"mmtag:13:SphericalCoordinatesofCity[]__example_function"], Cell[BoxData[ \(SphericalCoordinatesofCity["\"]\)], "Input"], Cell[BoxData[ \(CartesianCoordinatesofCity[cityname_String]\ := \ CoordinatesToCartesian[SphericalCoordinatesofCity[cityname], Spherical[r, theta, phi]]\)], "Input", CellTags->{ "mmtag:13:CartesianCoordinatesofCity[]__example_function", "mmtag:13:distances_from_Boston_to_Paris"}], Cell[BoxData[ \(CartesianCoordinatesofCity["\"]\)], "Input"], Cell[BoxData[ \(MinimumTunnel[city1_String, city2_String]\ := \[IndentingNewLine]Norm[ CartesianCoordinatesofCity[city1] - 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1)\) \(r\& \[Rule] \ \)\)\/\((\(||\)\(r\& \[Rule] \)\(||\))\)\^\(2 + n\)\)], "Text", FontFamily->"Helvetica"], Cell[BoxData[ \(Div[GradSimplePot[x, y, z, 1], Cartesian[x, y, z]] // Simplify\)], "Input"], Cell[BoxData[ \(This\ makes\ ``sense''\ because\ the\ amount\ of\ stuff\ flowing\ into\ \ a\ sphere\ is\ like\ the\ gradient*4 \[Pi]\ r\^2\ which\ is\ independent\ of\ \ the\ size\ of\ the\ sphere\ for\ n = 1\)], "Text", FontFamily->"Helvetica"] }, Closed]], Cell[CellGroupData[{ Cell["A Visualization Example of the Curl", "Subtitle"], Cell[TextData[{ "There is a very useful free software tool for solving minimal surface (and \ many other) variational problems called ", ButtonBox["Surface Evolver", ButtonData:>{ URL[ "http://www.susqu.edu/facstaff/b/brakke/evolver/"], None}, ButtonStyle->"Hyperlink"], " by Ken Brakke. To use Surface Evolver to greatest possible advantage, a \ user should be adept at using results from vector analysis. ", StyleBox["Mathematica's ", FontSlant->"Italic"], "Vector Analysis package is very helpful aid for developing powerful \ Evolver codes. The following example is extracted from the ", ButtonBox["Surface Evolver manual", ButtonData:>{ URL[ "http://www.susqu.edu/facstaff/b/brakke/evolver/manual220.pdf"], None}, ButtonStyle->"Hyperlink"], "." }], "Section"], Cell[BoxData[ \(LeavingKansas[x_, \ y_, \ z_\ , \ n_] := \ z\^n\/\(\((x^2\ + \ y^2)\) \((x^2\ + \ y^2\ + \ z^2)\)\^\(n\/2\)\)\ \ {y, \(-x\), \ 0}\)], "Input"], Cell[BoxData[ \(LeavingKansas[x, y, z, 3]\)], "Input"], Cell[BoxData[ \(<< Graphics`PlotField3D`\)], "Input"], Cell["\<\ Visualize the vector field for n=3, note that the function will be \ singular near the z-axis\ \>", "Subsection"], Cell[BoxData[ \(PlotVectorField3D[ LeavingKansas[x, y, z, 3], {x, \(-1\), 1}, {y, \(-1\), 1}, {z, \(- .5\), .5}, VectorHeads \[Rule] True, ColorFunction \[Rule] \((\((Hue[#* .66])\) &)\), PlotPoints \[Rule] 15, ScaleFactor \[Rule] 0.5]\)], "Input", CellTags->"mmtag:13:PlotVectorField3D[]"], Cell["\<\ We could make the function better behaved along the z-axis by brute \ force:\ \>", "Subsection"], Cell[BoxData[ \(LeavingKansasNicely[x_, \ y_, \ z_\ , \ n_]\ := \ Module[{CindRadsq\ = \ x^2\ + \ y^2}, \[IndentingNewLine]CindRadsq\ = \ If[CindRadsq\ \[LessEqual] \ 10\^\(-4\), 10\^\(-4\), \ CindRadsq, \ CindRadsq]; \[IndentingNewLine]\ z\^n\/\(CindRadsq \((CindRadsq\ + \ z^2)\)\^\(n\/2\)\)\ {y, \(-x\), \ \ 0}]\)], "Input", CellTags->"mmtag:13:plots__dealing_with_singularities"], Cell[BoxData[ \(PlotVectorField3D[ LeavingKansasNicely[x, y, z, 3], {x, \(-1\), 1}, {y, \(-1\), 1}, {z, \(- .5\), .5}, VectorHeads \[Rule] True, ColorFunction \[Rule] \((\((Hue[#* .66])\) &)\), PlotPoints \[Rule] 15, ScaleFactor \[Rule] 0.5]\)], "Input"], Cell["\<\ Or simply by avoiding the axis altogether and using the symmetry of \ the field\ \>", "Subsection"], Cell[BoxData[ \(PlotVectorField3D[ LeavingKansas[x, y, z, 3], {x, .01, 1}, {y, .01, 1}, {z, .01, .5}, VectorHeads \[Rule] True, ColorFunction \[Rule] \((\((Hue[#* .66])\) &)\), PlotPoints \[Rule] 15, ScaleFactor \[Rule] 0.5]\)], "Input"], Cell["\<\ Calculate the curl of the function using the VectorAnalysis \ package--note that the coordinate system is specified as cartesian. For the particular case of n=3:\ \>", "Section"], Cell[BoxData[ \(Curl[LeavingKansas[x, y, z, 3], Cartesian[x, y, z]] // Simplify\)], "Input", CellTags->"mmtag:13:Curl[]"], Cell["Define a new vector function for the curl for general n", "Subsubtitle"], Cell[BoxData[ \(Glenda[x_, y_, z_, n_] := Simplify[Curl[LeavingKansas[x, y, z, n], Cartesian[x, y, z]]]\)], "Input"], Cell["\<\ Demonstrate the assertion that the curl has a fairly simple form \ and is sphericaly symmetric for n=1\ \>", "Subsubtitle"], Cell[BoxData[ \(Glenda[x, y, z, n]\)], "Input"], Cell[BoxData[ \(Glenda[x, y, z, 1]\)], "Input"], Cell[BoxData[ \(The\ above\ is\ a\ vector\ field\ that\ points\ radially\ from\ the\ \ origin, \ with\ a\ magnitude\ that\ falls\ off\ like\ 1/ r\^2\)], "Subsubsection"], Cell["Visualize the curl for n=3", "Subsubtitle"], Cell[BoxData[ \(PlotVectorField3D[ Evaluate[Glenda[x, y, z, 3]], {x, 0, .5}, {y, 0, .5}, {z, 0.1, .5}, VectorHeads \[Rule] True, ColorFunction \[Rule] \((\((Hue[#* .66])\) &)\), PlotPoints \[Rule] 7]\)], "Input"], Cell["\<\ Demonstrate that the divergence of the curl vanishes for the above \ function independent of n\ \>", "Subsubtitle", CellTags->"mmtag:13:divergence_of_curl__example"], Cell[BoxData[ \(DivCurl = Div[Glenda[x, y, z, n], Cartesian[x, y, z]]\)], "Input"], Cell[BoxData[ \(Simplify[DivCurl]\)], "Input"] }, Closed]] }, FrontEndVersion->"5.2 for Macintosh", ScreenRectangle->{{4, 1280}, {0, 832}}, ScreenStyleEnvironment->"Presentation", CellGrouping->Manual, WindowSize->{1057, 724}, WindowMargins->{{Automatic, 394}, {Automatic, 101}}, WindowTitle->"Lecture 13 MIT 3.016 (Fall 2006) \[Copyright] W. 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