(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 4.0, MathReader 4.0, or any compatible application. 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). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. 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[ 134376, 3117]*) (*NotebookOutlinePosition[ 170318, 4349]*) (* CellTagsIndexPosition[ 170241, 4343]*) (*WindowFrame->Normal*) Notebook[{ Cell[BoxData[ \(Remove["\"]\)], "Input"], Cell[BoxData[ \(Clear["\"]\)], "Input"], Cell[CellGroupData[{ Cell[" The Three-Body Problem", "Title"], Cell[CellGroupData[{ Cell[TextData[StyleBox["Inroduction:", FontColor->GrayLevel[1]]], "Section", CellFrame->True, CellDingbat->None, Background->GrayLevel[0]], Cell[TextData[{ "This notebook will explore how three bodies behave when they interact \ gravitationally with each other. Since it has been proved that you can't \ solve the general three-body problem analytically, we will have mathematica \ help us do it numerically. There are no gerneral proofs. I will show \ concrete examples of instances that various quantities are conserved \ throughout certain time intervals of the motion. \n\nThe vector notation will \ be as follows: bold letters will represent vectors while scalars will be \ represented by just plain text. The unit vectors will be represented by bold \ i,j,k.\n\n", StyleBox["i", FontFamily->"Courier", FontWeight->"Bold"], " [unit vector in the x direction]\n", StyleBox["j ", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" [unit vector in the y direction]\n", FontVariations->{"CompatibilityType"->0}], StyleBox["k ", FontFamily->"Courier", FontWeight->"Bold", FontVariations->{"CompatibilityType"->0}], StyleBox[" [unit vector in the z direction]\n\nThe magnitude of a vector ", FontVariations->{"CompatibilityType"->0}], StyleBox["P", FontFamily->"Courier", FontWeight->"Bold", FontVariations->{"CompatibilityType"->0}], StyleBox[" will be represented by ||", FontVariations->{"CompatibilityType"->0}], StyleBox["P", FontFamily->"Courier", FontWeight->"Bold", FontVariations->{"CompatibilityType"->0}], StyleBox["||.\nIf ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ FormBox[ RowBox[{ StyleBox[ FormBox[ RowBox[{ StyleBox["P", FontWeight->"Bold"], "=", RowBox[{ SubscriptBox[ StyleBox["P", FontSlant->"Plain"], "x"], StyleBox["i", FontWeight->"Bold", FontSlant->"Plain"]}]}], "TraditionalForm"], FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], StyleBox["+", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], RowBox[{ StyleBox[\(P\_y\), FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], StyleBox["j", FontWeight->"Bold", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]}], StyleBox["+", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], RowBox[{ StyleBox[\(P\_z\), FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], StyleBox["k", FontWeight->"Bold", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]}]}], TraditionalForm]]], " then ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"||", StyleBox["P", FontWeight->"Bold", FontSlant->"Plain"], "||"}], "=", \(\@\(\((P\_x)\)\^2 + \((P\_y)\)\^2 + \((P\_z)\)\^2\)\)}], TraditionalForm]]], "\n", Cell[BoxData[""], "Input"], "\n", Cell[BoxData[""], "Input"] }], "Text", Background->RGBColor[1, 1, 0.890196]] }, Closed]], Cell[CellGroupData[{ Cell["(1) The equations of motion:", "Section", CellFrame->True, CellDingbat->"\[FilledDownTriangle]", Background->RGBColor[0.0862745, 0.133333, 0.776471]], Cell[TextData[{ "The first thing we need to do is write down the equations of motion for \ our three body system. We have three masses and the only forces on them are \ from their gravitational \ninteraction. Recall the force law for gravity is \ ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[ RowBox[{ FormBox[ RowBox[{"||", SubscriptBox[ StyleBox["F", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[\(g(i, j)\), FontSlant->"Plain"]]}], "TraditionalForm"], "||"}], FontSize->13], StyleBox["=", FontSize->13], StyleBox[ FractionBox[ RowBox[{ StyleBox["G", FontSlant->"Plain"], "*", SubscriptBox[ StyleBox["M", FontSlant->"Plain"], "i"], "*", SubscriptBox[ StyleBox["M", FontSlant->"Plain"], "j"]}], RowBox[{"||", SubscriptBox[ StyleBox["r", FontWeight->"Bold", FontSlant->"Plain"], \(i, j\)], SuperscriptBox["||", StyleBox["2", FontSlant->"Italic"]]}]], FontSize->14, FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]}], TraditionalForm]]], " where G is the gravitational proportionality constant, ", Cell[BoxData[ \(TraditionalForm\`M\_i\)]], " is the mass of object i, ", Cell[BoxData[ \(TraditionalForm\`M\_j\)]], " is the mass of object j, and ", Cell[BoxData[ \(TraditionalForm\`r\_\(i, j\)\)]], " is the distance between the two masses. We know what G, ", Cell[BoxData[ \(TraditionalForm\`M\_i\)]], ", and ", Cell[BoxData[ \(TraditionalForm\`M\_j\)]], " are, so we need to find ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["r", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[\(i, j\), FontSlant->"Italic"]], TraditionalForm]]], ", their seperation vector (a vector that goes from ", Cell[BoxData[ \(TraditionalForm\`m\_i\ to\ m\_j\)]], "). We will define the serperation vector as follows: the position vector \ of M", Cell[BoxData[ \(TraditionalForm\`\_i\)]], " added to the seperation vector will give us the position vector of ", Cell[BoxData[ \(TraditionalForm\`M\_j\)]], ".\n\nSo ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "i"], TraditionalForm]]], "+ ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["r", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[\(i, j\), FontSlant->"Italic"]], TraditionalForm]]], StyleBox[" ", FontWeight->"Bold"], "= ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "j"], TraditionalForm]]], " or ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["r", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[\(i, j\), FontSlant->"Italic"]], TraditionalForm]]], " = ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], StyleBox["j", FontSlant->"Italic"]], TraditionalForm]]], "- ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "i"], TraditionalForm]]], ". \n\nThe direction of the force is pointed at the other body so knowings \ this and the magnitude along with ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["r", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[\(i, j\), FontSlant->"Italic"]], TraditionalForm]]], " we can get the vector force law.\n", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ RowBox[{ StyleBox[ FormBox[ SubscriptBox[ StyleBox["F", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[\(i, j\), FontSlant->"Plain"]], "TraditionalForm"], FontSize->13], StyleBox["=", FontSize->13], StyleBox[ FractionBox[ RowBox[{ StyleBox["G", FontSlant->"Plain"], StyleBox["*", FontSlant->"Italic"], \(M\_i\), "*", SubscriptBox[ StyleBox["M", FontSlant->"Plain"], "j"]}], RowBox[{"||", SubscriptBox[ StyleBox["r", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[\(i, j\), FontSlant->"Italic"]], 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TraditionalForm]]], " (this is the force on particle i due to particle j)\n\n", Cell[BoxData[ FormBox[ RowBox[{ FormBox[ SubscriptBox[ StyleBox["F", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[\(i, j\), FontSlant->"Plain"]], "TraditionalForm"], "=", StyleBox[ FractionBox[ RowBox[{"G", "*", \(M\_i\), "*", \(M\_j\), "*", SubscriptBox[ StyleBox["r", FontWeight->"Bold"], \(i, j\)]}], RowBox[{"||", SubscriptBox[ StyleBox["r", FontWeight->"Bold"], \(i, j\)], \( || \^3\)}]], FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]}], TraditionalForm]], FontFamily->"Courier", FontSize->16, FontVariations->{"CompatibilityType"->0}], "\n\nand in component form, having ", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "i"], TraditionalForm]]], "= ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(x\_i\), StyleBox["i", FontFamily->"Courier", FontWeight->"Bold", FontSlant->"Plain"]}], StyleBox[" ", FontFamily->"Courier", FontWeight->"Bold"], StyleBox["+", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" ", FontFamily->"Courier", FontWeight->"Bold"], RowBox[{\(y\_i\), StyleBox["j", FontFamily->"Courier", FontWeight->"Bold", FontSlant->"Plain"]}], StyleBox[" ", FontFamily->"Courier", FontWeight->"Bold", FontSlant->"Plain"], StyleBox["+", FontFamily->"Courier", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[" ", FontFamily->"Courier", FontWeight->"Bold", FontSlant->"Plain"], RowBox[{\(z\_i\), StyleBox["k", FontWeight->"Bold", FontSlant->"Plain"]}]}], TraditionalForm]]], " and ", Cell[BoxData[ FormBox[ RowBox[{ SubscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "j"], "=", RowBox[{ RowBox[{\(x\_j\), StyleBox["i", FontFamily->"Courier", FontWeight->"Bold", FontSlant->"Plain"]}], StyleBox[" ", FontFamily->"Courier", FontWeight->"Bold"], StyleBox["+", 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"CompatibilityType"->0, "RotationAngle"->0}], RowBox[{ RowBox[{ RowBox[{ StyleBox["(", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], RowBox[{ SubscriptBox[ StyleBox["x", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], "j"], StyleBox["-", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], SubscriptBox[ StyleBox["x", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], "i"]}], StyleBox[")", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]}], StyleBox["i", FontWeight->"Bold", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]}], StyleBox["+", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], RowBox[{ RowBox[{ StyleBox["(", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], RowBox[{ SubscriptBox[ StyleBox["y", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], "j"], StyleBox["-", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], SubscriptBox[ StyleBox["y", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], "i"]}], StyleBox[")", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]}], StyleBox["j", FontWeight->"Bold", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]}], StyleBox["+", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], RowBox[{ RowBox[{ StyleBox["(", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], RowBox[{ SubscriptBox[ StyleBox["z", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], "j"], StyleBox["-", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], SubscriptBox[ StyleBox["z", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], "i"]}], StyleBox[")", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]}], StyleBox["k", FontWeight->"Bold", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]}]}], StyleBox[")", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"-> 0}]}]}], \(\((\((x\_j - x\_i)\)\^2 + \((y\_j - \ y\_i)\)\^2 + \((z\_j - z\_i)\)\^2)\)\^\(3/2\)\)], FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}]}], TraditionalForm]], FontFamily->"Courier", FontSize->16, FontVariations->{"CompatibilityType"->0}], "\n\n\nNewton's 2nd law tells us that ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\)]], StyleBox["F=", FontWeight->"Bold"], "m", StyleBox["a ", FontWeight->"Bold"], "(The sum of all the forces acting on the particle is equal to mass times \ the particles acceleration). So what we have to do is sum up all the forces \ that are acting on particle i to find its acceleration, from which we can get \ its positon and velocity at any other time we want as long as we have its \ initial conditions. \n\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\[Sum]\+\(\(j = 1\)\+\(j \[NotEqual] i\)\)\%n\), SubscriptBox[ StyleBox["F", FontWeight->"Bold", FontSlant->"Plain"], \(g(i, j)\)]}], "=", " ", RowBox[{\(m\_i\), SubscriptBox[ StyleBox["a", FontWeight->"Bold", FontSlant->"Plain"], "i"]}]}], TraditionalForm]], FontSize->15], " this is our equations for the i'th particle (notice j\ \[NotEqual]i because the particle doesnt gravitationally act with itself)\n\n\ but since this is a vector equation, it is actualy 3 seperate equations (one \ for each component)\n\n", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[ RowBox[{\(\[Sum]\+\(\(j = 1\)\+\(j \[NotEqual] i\)\)\%n\), SubscriptBox[ RowBox[{"(", SubscriptBox[ StyleBox["F", FontWeight->"Bold", FontSlant->"Plain"], \(g(i, j)\)], ")"}], "x"]}], FontSize->16], "=", " ", RowBox[{\(m\_i\), RowBox[{ SubscriptBox[ StyleBox["x", FontSlant->"Italic"], "i"], StyleBox["''", FontSlant->"Italic"]}], StyleBox[\((t)\), FontSlant->"Italic"]}]}], TraditionalForm]], FontSize->15], "\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\[Sum]\+\(\(j = 1\)\+\(j \[NotEqual] i\)\)\%n\), SubscriptBox[ RowBox[{"(", SubscriptBox[ StyleBox["F", FontWeight->"Bold", FontSlant->"Plain"], \(g(i, j)\)], ")"}], "y"]}], "=", " ", RowBox[{\(m\_i\), RowBox[{ SubscriptBox[ StyleBox["y", FontSlant->"Italic"], "i"], StyleBox["''", FontSlant->"Italic"]}], StyleBox[\((t)\), FontSlant->"Italic"]}]}], TraditionalForm]], FontSize->15], "\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\[Sum]\+\(\(j = 1\)\+\(j \[NotEqual] i\)\)\%n\), SubscriptBox[ RowBox[{"(", SubscriptBox[ StyleBox["F", FontWeight->"Bold", FontSlant->"Plain"], \(g(i, j)\)], ")"}], "z"]}], "=", " ", \(\(m\_i\) z\_i'' \((t)\)\)}], TraditionalForm]], FontSize->15], "\n\nAll we have to do is write down all the quations for all the \ components for all n particles and each of their initial conditons and have \ mathematica solve them. \n\nIf we just started typing out the equations by \ hand it would be very tedious (especially when you add more particles to your \ system). You will have to write out 3*n differential equations each having \ n-1 terms. So instead of doing that, lets start off by making some functions \ that will make writing down the equations easier. First, make a fucntion \ R[i,j] which will represent the distance between particle i and j." }], "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(\(\(R[i_, j_] := Sqrt[\((x\_j[t] - x\_i[t])\)^2 + \((y\_j[t] - y\_i[t])\)^2 + \((z\_j[t] - z\_i[t])\)^2];\)\(\n\)\)\)], "Input"], Cell["\<\ Now we make another function, r[e,i,j]. This funtion will be the e \ component (e being x, y, or z) of the seperation vector between particles i \ and j.\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(r[e_, i_, j_] := \(\(e\_j[t]\)\(-\)\(e\_i[t]\)\(\n\)\)\)], "Input"], Cell["Write down the number of particles of your system, n:", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(\(n = 3;\)\)], "Input"], Cell["\<\ Now we write down the differential equations of motion with the \ help of a function called Do. Do helps us save a lot of time because it \ repeats an operation that would have taken a long time to keep writing out.\ \ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell["the x components:", "Text"], Cell[BoxData[ \(Do[de\_\(x, i\) = m\_i*\(\(x\_i'\)'\)[t] \[Equal] Sum[G*m\_i*m\_j*r[x, i, j]/R[i, j]^3, {j, 1, i - 1}] + Sum[G*m\_i*m\_j*r[x, i, j]/R[i, j]^3, {j, i + 1, n}], {i, 1, n}]\)], "Input"], Cell["the y components:", "Text"], Cell[BoxData[ \(Do[de\_\(y, i\) = m\_i*\(\(y\_i'\)'\)[t] \[Equal] Sum[G*m\_i*m\_j*r[y, i, j]/R[i, j]^3, {j, 1, i - 1}] + Sum[G*m\_i*m\_j*r[y, i, j]/R[i, j]^3, {j, i + 1, n}], {i, 1, n}]\)], "Input"], Cell["the z components:", "Text"], Cell[BoxData[ \(Do[de\_\(z, i\) = m\_i*\(\(z\_i'\)'\)[t] \[Equal] Sum[G*m\_i*m\_j*r[z, i, j]/R[i, j]^3, {j, 1, i - 1}] + Sum[G*m\_i*m\_j*r[z, i, j]/R[i, j]^3, {j, i + 1, n}], {i, 1, n}]\)], "Input"], Cell[TextData[{ "Now we need to put the values in for our initial conditions of the sytem. \ We need to specify the intial position vector of particle ", Cell[BoxData[ \(TraditionalForm\`m\_i\)]], ", ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ FormBox[ RowBox[{ FormBox[ SubscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "i"], "TraditionalForm"], "=", \((x\_i\)}], "TraditionalForm"], ",", \(y\_i\), ",", \(z\_i\)}], ")"}], TraditionalForm]]], " and its initial velocity vector ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[ FractionBox[ SubscriptBox[ StyleBox[ RowBox[{ StyleBox["d", FontSlant->"Plain"], StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"]}]], "i"], "dt"], FontSize->14], "="}], TraditionalForm]]], " ", Cell[BoxData[ \(TraditionalForm\`\((x\_i', \ y\_i', \ z\_i')\)\)]], ":" }], "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(bc\_1 = x\_1[0] \[Equal] 7; \ bc\_2 = y\_1[0] \[Equal] 5; \ bc\_3 = z\_1[0] \[Equal] 2; \ bc\_4 = \(x\_1'\)[0] \[Equal] 6; \ bc\_5 = \(y\_1'\)[0] \[Equal] 5; \ bc\_6 = \(z\_1'\)[0] \[Equal] 1;\)], "Input"], Cell[BoxData[ \(bc\_7 = x\_2[0] \[Equal] 4; \ bc\_8 = y\_2[0] \[Equal] 2; \ bc\_9 = z\_2[0] \[Equal] \(-4\); \ bc\_10 = \(x\_2'\)[0] \[Equal] \(-6\); \ bc\_11 = \(y\_2'\)[0] \[Equal] \(-5\); \ bc\_12 = \(z\_2'\)[0] \[Equal] 2;\)], "Input"], Cell[BoxData[ \(bc\_13 = x\_3[0] \[Equal] 0; \ bc\_14 = y\_3[0] \[Equal] \(-2\); \ bc\_15 = z\_3[0] \[Equal] \(-2\); \ bc\_16 = \(x\_3'\)[0] \[Equal] 0; \ bc\_17 = \(y\_3'\)[0] \[Equal] 0; \ bc\_18 = \(z\_3'\)[0] \[Equal] 0;\)], "Input"], Cell["\<\ Now we must define the constants. For this problem we need to \ define the gravitational constant G and the masses of particle 1, 2, and 3:\ \ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(G = .6; \ m\_1 = 1000; \ m\_2 = 1000; \ m\_3 = 10;\)], "Input"], Cell["\<\ Finally, we will have mathematica solve the differential equations \ we have set up from newtons laws of motion:\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(soln = NDSolve[{Table[de\_\(x, i\), {i, 1, n}], Table[de\_\(y, i\), {i, 1, n}], Table[de\_\(z, i\), {i, 1, n}], Table[bc\_i, {i, 1, 6*n}]}, Flatten[Table[{x\_i[t], y\_i[t], z\_i[t]}, {i, 1, n}]], {t, 0, 40}, MaxSteps -> Infinity, WorkingPrecision -> 21]\)], "Input"], Cell[BoxData[""], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["\<\ (2) Define the various position and velocity functions we will \ need:\ \>", "Section", CellFrame->True, CellDingbat->"\[FilledDownTriangle]", Background->RGBColor[0.203922, 0.282353, 0.776471]], Cell["\<\ Now that mathematica has solved the differential equations, we want \ to make position and velocity functions in order to do some computations. We start by defining the component functions for the position vectors for \ each mass:\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(Do[x\_i[t_] = x\_i[t] /. soln[\([1]\)]; \ y\_i[t_] = y\_i[t] /. soln[\([1]\)]; \ z\_i[t_] = z\_i[t] /. soln[\([1]\)];, {i, 1, n}]\)], "Input"], Cell[TextData[{ "Here we define our velocity component functions (the velocity of the \ particles) as ", Cell[BoxData[ \(TraditionalForm\`Dx\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`Dy\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`Dz\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`Dx\_2\)]], "...:" }], "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(Do[Dx\_i[t_] = D[x\_i[t], t]; \ Dy\_i[t_] = D[y\_i[t], t]; \ Dz\_i[t_] = D[z\_i[t], t];, {i, 1, n}]\)], "Input"], Cell["Now we can define our positon and velocity vector funtions:", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(Do[r\_i[t_] = {x\_i[t], y\_i[t], z\_i[t]}, {i, 1, n}]\)], "Input"], Cell[BoxData[ \(Do[Dr\_i[t_] = D[r\_i[t], t], {i, 1, n}]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["(3) Properties of the system:", "Section", CellFrame->True, CellDingbat->"\[FilledDownTriangle]", Background->RGBColor[0.27451, 0.462745, 0.776471]], Cell[TextData[{ "We now have our position and velocity funtions for all times in the \ interval t: [", Cell[BoxData[ FormBox[ RowBox[{ FormBox[\(t\_start\), "TraditionalForm"], ",", \(t\_end\)}], TraditionalForm]]], "], we can begin examining various properties of our system. We can now \ directly compute all dynamical quantities of interest so we will check to see \ if energy, momentum, and angular momentum are conserved for this system \ during a chosen interval of time.\n\nThe system's kinetic energy is: ", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(i = 1\)\%n\( 1\/2\) m\_i\), "||", SubscriptBox[ StyleBox["v", FontWeight->"Bold"], "i"], \( || \^2\)}], TraditionalForm]]] }], "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(\(ke[t_] = Sum[\((1/2)\)*m\_i*\((Dr\_i[t] . Dr\_i[t])\), {i, 1, n}];\)\)], "Input"], Cell[TextData[{ "The system's potential energy will be: ", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(i = 1\)\%\(n - 1\)\), RowBox[{\(\[Sum]\+\(j = i + 1\)\%n\), FractionBox[\(\(-G\)*m\_i*m\_j\), RowBox[{"||", RowBox[{ SubscriptBox[ StyleBox["R", FontWeight->"Bold"], "j"], "-", SubscriptBox[ StyleBox["R", FontWeight->"Bold"], "i"]}], "||"}]]}]}], TraditionalForm]]], " (", ButtonBox["why?", ButtonData:>"potential", ButtonStyle->"Hyperlink"], ")" }], "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ RowBox[{ RowBox[{\(pe[t_]\), "=", RowBox[{"Sum", "[", RowBox[{ RowBox[{"Sum", "[", RowBox[{ RowBox[{\(-G\), "*", SubscriptBox["m", StyleBox["i", FontSize->11]], "*", \(m\_j/R[i, j]\)}], ",", \({j, i + 1, n}\)}], "]"}], ",", \({i, 1, n - 1}\)}], "]"}]}], ";"}]], "Input"], Cell[BoxData[""], "Input"], Cell["\<\ The system's total energy will be the sum of the kenetic and the \ potential, En=ke+pe. Now plot it and notice that it is constant throughout \ time, as it should be for interactions with conservative forces like gravity.\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(\(En[t_] = ke[t] + pe[t];\)\)], "Input"], Cell[BoxData[ \(Plot[En[t], {t, 0, 5}, PlotRange \[Rule] {En[0] - 1, En[0] + 1}]\)], "Input"], Cell["\<\ Now we will look at how the kinetic and potential vary with each \ other to give a constant total energy. We have defined the potential energy \ to be zero when it is at infinity (an infinite distance away from the other particle). The kinetic \ energy is the red, the potential the blue, and the total energy is black. \ \ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(Plot[{pe[t]/100000, ke[t]/100000, En[t]/100000}, {t, 0, 40}, PlotRange -> {\(-4\), 3}, PlotLabel \[Rule] Energy, GridLines \[Rule] None, PlotPoints \[Rule] 25, \ AspectRatio \[Rule] 1, \ PlotStyle \[Rule] {RGBColor[0, 0, 1], RGBColor[1, 0, 0], \ RGBColor[0, 0, 0]}, ImageSize \[Rule] {500, 500}]\)], "Input"], Cell[TextData[{ " So we can see from this plot that when the potential energy is at its \ most negative, the kinetic energy\nis at its maximum to equal the constant E.\ \n", Cell[BoxData[ \(\[IndentingNewLine]\)], "Input"], "\n\n\nThe total momentum of a system with no external forces should be \ conserved. Taking the system to be the three masses so there is no external \ forces, the total momentum will be: \n", StyleBox["P = ", FontWeight->"Bold"], Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(i = 1\)\%n\), RowBox[{\(m\_i\), "*", SubscriptBox[ StyleBox["v", FontWeight->"Bold"], "i"]}]}], TraditionalForm]]] }], "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(\(P[t_] = Sum[{m\_i*\(x\_i'\)[t], m\_i*\(y\_i'\)[t], m\_i*\(z\_i'\)[t]}, {i, 1, n}];\)\)], "Input"], Cell[BoxData[ \(Table[P[t], {t, 1, 10, 1}]\)], "Input"], Cell[TextData[{ "So we see from this table that the momentum of the table is indeed \ constant (besides a little calculation error).\n\n\n\nNow define the angular \ momentum of the system: ", StyleBox["L ", FontWeight->"Bold"], "= ", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(i = 1\)\%n\), RowBox[{ SubscriptBox[ StyleBox["r", FontWeight->"Bold"], "i"], "\[Times]", \(m\_i\), SubscriptBox[ StyleBox["v", FontWeight->"Bold"], "i"]}]}], TraditionalForm]]] }], "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(\(L[t_] = Sum[Cross[r\_i[t], m\_i*Dr\_i[t]], {i, 1, n}];\)\)], "Input"], Cell["\<\ Here is a table showing the angular momentum of the system L[t] is \ a conserved quantity of the motion. \ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(\(\(Table[L[t], {t, 0, 10, 1}]\)\(\[IndentingNewLine]\)\)\)], "Input"], Cell[BoxData[ \(\[IndentingNewLine]\)], "Input"], Cell[BoxData[ \(Plot[\((L[t] - L[t + Random[Real, {0, 5}]])\) . \((L[t + 1] - L[t + Random[Integer, {6, 10}]])\) + .1, {t, 0, 10}, PlotRange \[Rule] {\(- .1\), .4}]\)], "Input"], Cell[TextData[{ "And again, we see like you have learned in class, angular momentum is \ conserved for central forces with no external torques acting on the system.\n\ \nWe now want to look at the center of mass for the system. Since we have \ seen that the total momentum of the system is conserved the center of mass of \ the system should be moving with a constant velocity. The center of mass is \ defined as:\n\n", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["R", FontWeight->"Bold", FontSlant->"Plain"], "cm"], TraditionalForm]]], "= ", Cell[BoxData[ FormBox[ StyleBox[ FractionBox[ RowBox[{\(\[Sum]\+\(i = 1\)\%n\), RowBox[{\(m\_i\), "*", SubscriptBox[ StyleBox["r", FontWeight->"Bold"], "i"]}]}], \(\[Sum]\+\(i = 1\)\%n m\_i\)], FontSize->14], TraditionalForm]]] }], "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(\(Cm[ t_] = {Sum[m\_i*x\_i[t], {i, 1, n}], Sum[m\_i*y\_i[t], {i, 1, n}], Sum[m\_i*z\_i[t], {i, 1, n}]}/Sum[m\_i, {i, 1, n}];\)\)], "Input"], Cell[BoxData[ \(\(DCm[t_] = D[Cm[t], t];\)\)], "Input"], Cell[BoxData[ \(\(DDCm[t_] = D[DCm[t], t];\)\)], "Input"], Cell[BoxData[ \(ParametricPlot3D[DDCm[t], {t, 0, 40}]\)], "Input"], Cell["\<\ Above is a plot of the acceleration of the center of mass, which \ should be zero. This is due to round off error and step size of our \ numerical calculations.\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[""], "Input"], Cell[BoxData[""], "Input"], Cell[BoxData[""], "Input"], Cell[TextData[{ "Since we have defined the center of mass, we are in the position to \ directly show another theorem in mechanics. \n\n", StyleBox["L = ", FontWeight->"Bold"], Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["L", FontWeight->"Bold", FontSlant->"Plain"], "cm"], TraditionalForm]]], "+ ", Cell[BoxData[ FormBox[ StyleBox[\(L'\), FontWeight->"Bold", FontSlant->"Plain"], TraditionalForm]]], "\n\nThe angular momentum about a point is equal to the angular momentum \ about the center of mass (", StyleBox["L'", FontWeight->"Bold"], ") plus the angular momentum of the center of mass about that point (", Cell[BoxData[ FormBox[ SubscriptBox[ StyleBox["L", FontWeight->"Bold", FontSlant->"Plain"], "cm"], TraditionalForm]]], ")." }], "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(\(Lcm[t_] = Sum[Cross[r\_i[t] - Cm[t], m\_i*\((Dr\_i[t] - DCm[t])\)], {i, 1, n}];\)\)], "Input"], Cell[BoxData[ \(\(Locm[t_] = Cross[Cm[t], \((Sum[m\_i, {i, 1, n}])\)*DCm[t]];\)\)], "Input"], Cell[BoxData[ \(Lcm[1] + Locm[1]\)], "Input"], Cell[BoxData[""], "Input"], Cell[TextData[{ "**The potential energy of a system of masses can be found in this way:\nIt \ is the sum of the potential energy of all unique groups of masses. How many \ unique groups are there? Think of it as a how many unique lines you can \ connect a group of n points with. It will be 1 + 2 + 3 + .. + n-1. An easy \ way to see this is to think of two lines of n points sepereated by some \ distance. point 1 can connect with 2,3,4,..,n (so a total of n-1 ways) then \ move to point 2, u notice it has already been connected to point 1, so it can \ be connected to 3,4,5,...,n (n-2 points) and follow this sequence to point n \ (who cant connect with anyone) and you get n-1 + n-2 + n-3 + ... +3 + 2 + 1\n\ ", Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica %%AspectRatio: 1 MathPictureStart /Mabs { Mgmatrix idtransform Mtmatrix dtransform } bind def /Mabsadd { Mabs 3 -1 roll add 3 1 roll add exch } bind def %% Graphics %%IncludeResource: font Courier %%IncludeFont: Courier /Courier findfont 10 scalefont setfont % Scaling calculations -0.214286 0.238095 0.0238095 0.952381 [ [ 0 0 0 0 ] [ 1 1 0 0 ] ] MathScale % Start of Graphics 1 setlinecap 1 setlinejoin newpath 0 0 m 1 0 L 1 1 L 0 1 L closepath clip newpath 0 g gsave .02381 .97619 -68.6562 -8.90625 Mabsadd m 1 1 Mabs scale 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o`000?ooo`00000ooooo00<0003oool00000@Oooo`030000oooooooo04Soool000Soool01P000?oo o`0000000?ooo`0003ooool2000000?oool0003oool0?oooo`030000oooo0000047oool00`000?oo ooooo`18oooo0008oooo0`000003oooo0000oooo03koool010000?ooo`000000043oool00`000?oo o`000012oooo00<0003oooooool0B?ooo`001oooo`030000oooo000000800010oooo0P000003oooo 0000oooo03ooool00`000?ooo`000011oooo00<0003oooooool0BOooo`001oooo`@00011oooo0`00 047oool20000@_ooo`030000oooooooo04Woool000Ooool40000@?ooo`@00011oooo0P0004;oool0 0`000?ooo`000012oooo0P0000Goool000?oool010000?ooo`000?ooo`<0000moooo00@0003oool0 003oool30000?Oooo`040000oooo0000oooo0P000003oooo0000oooo03coool010000?ooo`000?oo o`<0000moooo00<0003oool000001?ooo`030000oooooooo00;oool000Coool01P000?ooo`000000 0?ooo`0003koool00`000?ooooooo`02000000?oool0003oool0??ooo`030000oooooooo00<0000o oooo00@0003oool0003oool20000?_ooo`040000oooooooooooo0P0000Goool000;oool500000_oo o`030000oooooooo03[oool90000>oooo`P0000moooo1@0000;oool00`000?ooooooo`0joooo1@00 0003oooo0000000000Goool000Coool00`000?ooooooo`11oooo00<0003oooooool0@Oooo`030000 oooooooo04;oool00`000?ooooooo`11oooo00<0003oooooool02?ooo`00ooooob7oool00?oooolQ oooo003ooooo8Oooo`00ooooob7oool00?oooolQoooo0000\ \>"], ImageRangeCache->{{{0, 287}, {287, 0}} -> {0, 0, 0, 0}}], "\n[(note): there arent 2 serperate collecations of masses, they are the \ same, it is easier to show how they can be connected this way.]\n\nFor each \ collection of points the potential energy is ", Cell[BoxData[ \(TraditionalForm\`U\_\(i, j\)\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\(\(-G\)*m\_i*m\_j\)\/\(\(||\)\(r\_i - \ r\_j\)\(||\)\)\)]], ", and looking at our graphic we notice it should turn out that the total \ is:\n", Cell[BoxData[ \(TraditionalForm\`U\_tot\)]], "= ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(j = \ 2\)\%n\(\(-G\)*m\_1*m\_j\)\/\(\(||\)\(r\_1 - r\_j\)\(||\)\)\)]], "+ ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(j = \ 3\)\%n\(\(-G\)*m\_2*m\_j\)\/\(\(||\)\(r\_2 - r\_j\)\(||\)\)\)]], "+...+", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(j = n\)\%n\(\(-G\)*m\_\((n - 1)\)*m\_j\)\/\ \(\(||\)\(r\_\((n - 1)\) - r\_j\)\(||\)\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\[Sum]\+\(i = 1\)\%\(n - 1\)\(\[Sum]\+\(j = i + \ 1\)\%n\(\(-G\)*m\_i*m\_j\)\/\(\(||\)\(r\_i - r\_j\)\(||\)\)\)\)]] }], "Text", CellFrame->True, Background->GrayLevel[0.900008], CellTags->"potential"] }, Closed]], Cell[CellGroupData[{ Cell["(4) The motion of the bodies:", "Section", CellFrame->True, CellDingbat->"\[FilledDownTriangle]", Background->RGBColor[0.462745, 0.658824, 0.878431]], Cell["\<\ Now that we have done all the calculations, we want to see what the \ motion of these bodies actually looks like when we plot them. So now we will \ look at our results in various ways: First we will look at each different mass in our original coordinate system \ from various views: Mass 1:\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[{x\_1[t], y\_1[t], z\_1[t]}, {t, 1, 40}, ImageSize \[Rule] {200, 200}]\)], "Input"], Cell["\<\ Now we will look at it from abolve (we are now viewing the x-y \ plane)\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot[{x\_1[t], y\_1[t]}, {t, 0, 40}]\)], "Input"], Cell["Mass 2:", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[{x\_2[t], y\_2[t], z\_2[t]}, {t, 0, 40}, ImageSize \[Rule] {200, 200}]\)], "Input"], Cell[BoxData[ \(ParametricPlot[{x\_2[t], y\_2[t]}, {t, 0, 40}]\)], "Input"], Cell["Mass 3:", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[{x\_3[t], y\_3[t], z\_3[t]}, {t, 0, 40}]\)], "Input"], Cell[BoxData[ \(ParametricPlot[{x\_3[t], y\_3[t]}, {t, 0, 40}]\)], "Input"], Cell["\<\ Now here is the motion of all three of them together on the same \ plot (in our origional coordinate system):\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[{{x\_1[t], y\_1[t], z\_1[t]}, {x\_2[t], y\_2[t], z\_2[t]}, {x\_3[t], y\_3[t], z\_3[t]}}, {t, 0, 40}, PlotRange -> {{40, \(-30\)}, {40, \(-30\)}, {\(-50\), 60}}]\)], "Input"], Cell["Overhead view:", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot[{{x\_1[t], y\_1[t]}, {x\_2[t], y\_2[t]}, {x\_3[t], y\_3[t]}}, {t, 0, 20}, PlotRange -> {{17, \(-5\)}, {20, \(-10\)}}]\)], "Input"], Cell["\<\ Here is a plot of the motion from the center of mass frame of \ reference:\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[{{x\_1[t] - \((Cm[t] . {1, 0, 0})\), y\_1[t] - \((Cm[t] . {0, 1, 0})\), z\_1[t] - \((Cm[t] . {0, 0, 1})\)}, {x\_2[ t] - \((Cm[t] . {1, 0, 0})\), y\_2[t] - \((Cm[t] . {0, 1, 0})\), z\_2[t] - \((Cm[t] . {0, 0, 1})\)}, {x\_3[ t] - \((Cm[t] . {1, 0, 0})\), y\_3[t] - \((Cm[t] . {0, 1, 0})\), z\_3[t] - \((Cm[t] . {0, 0, 1})\)}}, {t, 0, 40}, PlotRange \[Rule] {{\(-25\), 10}, {\(-20\), 10}, {\(-10\), 10}}]\)], "Input"], Cell["Overhead view of center of mass frame:", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot[{{x\_1[t] - \((Cm[t] . {1, 0, 0})\), y\_1[t] - \((Cm[t] . {0, 1, 0})\)}, {x\_2[ t] - \((Cm[t] . {1, 0, 0})\), y\_2[t] - \((Cm[t] . {0, 1, 0})\)}, {x\_3[ t] - \((Cm[t] . {1, 0, 0})\), y\_3[t] - \((Cm[t] . {0, 1, 0})\)}}, {t, 0, 10}, PlotRange \[Rule] {{\(-10\), 10}, {\(-10\), 10}}]\)], "Input"], Cell["\<\ Here is a graph of the motion of the center of mass motion. It \ should be straight line:\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[Cm[t], {t, 0, 40}, PlotRange \[Rule] {{\(-20\), 20}, {\(-20\), 20}, {\(-20\), 20}}]\)], "Input"], Cell["Overhead view (x-y):", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot[{Cm[t] . {1, 0, 0}, Cm[t] . {0, 1, 0}}, {t, 0, 20}, PlotRange \[Rule] {{\(-4\), 4}, {\(-4\), 4}}]\)], "Input"], Cell["y-z view:", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot[{Cm[t] . {0, 1, 0}, Cm[t] . {0, 0, 1}}, {t, 0, 20}, PlotRange \[Rule] {{\(-4\), 4}, {\(-4\), 4}}]\)], "Input"], Cell["\<\ How does it look from one mass to another? Here is how the motion of Mass 2 looks from Mass 1:\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[r\_2[t] - r\_1[t], {t, 0, 40}]\)], "Input"], Cell["Here is how the motion of mass 3 looks from mass 1:", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[r\_3[t] - r\_1[t], {t, 0, 40}]\)], "Input"], Cell["Here is how the whole situation looks like from mass 1", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[{{x\_2[t] - x\_1[t], y\_2[t] - y\_1[t], z\_2[t] - z\_1[t]}, {x\_3[t] - x\_1[t], y\_3[t] - y\_1[t], z\_3[t] - z\_1[t]}}, {t, 0, 40}]\)], "Input"], Cell["Here is how mass 3 looks like from mass 2:", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[r\_3[t] - r\_2[t], {t, 0, 40}]\)], "Input"], Cell[BoxData[ \(ParametricPlot3D[r\_1[t] - r\_2[t], {t, 0, 40}]\)], "Input"], Cell["How the center of mass motion looks from mass 1", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[Cm[t] - r\_1[t], {t, 0, 40}]\)], "Input"], Cell["How the center of mass motion looks from mass 2:", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[Cm[t] - r\_2[t], {t, 0, 40}]\)], "Input"], Cell["How the center of mass motion looks from mass 3:", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(ParametricPlot3D[Cm[t] - r\_3[t], {t, 0, 40}]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["(5) Animations of the system:", "Section", CellFrame->True, CellDingbat->"\[FilledDownTriangle]", Background->RGBColor[0.768627, 0.847059, 1]], Cell["\<\ Now lets see what it would like like if we animated it: As viewed from the initial reference frame:\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(state[a_, b_, c_, d_, e_, f_, g_, h_, i_]\ := \ Show[{Graphics3D[Text["\<*\>", {a, b, c}]], Graphics3D[Text["\<*\>", {d, e, f}]], Graphics3D[Point[{g, h, i}]]}, DisplayFunction \[Rule] $DisplayFunction, AspectRatio \[Rule] 1, BoxStyle \[Rule] Dashing[{0.01, .03, .01, .03}], PlotRange -> {{30, \(-20\)}, {30, \(-20\)}, {\(-40\), 50}}]\)], "Input"], Cell[BoxData[ \(Table[ state[x\_1[t], y\_1[t], z\_1[t], x\_2[t], y\_2[t], z\_2[t], x\_3[t], y\_3[t], z\_3[t]], {t, 0, 40, .5}]\)], "Input"], Cell["As viewed from the center of mass reference frame", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(state[q_, w_, e_, r_, t_, y_, u_, i_, o_]\ := \ Show[{Graphics3D[Text["\<*\>", {q, w, e}]], Graphics3D[Text["\<*\>", {r, t, y}]], Graphics3D[Text["\<*\>", {u, i, o}]]}, DisplayFunction \[Rule] $DisplayFunction, PlotRange \[Rule] {{\(-25\), 25}, {\(-25\), 25}, {\(-25\), 25}}, AspectRatio \[Rule] 1]\)], "Input"], Cell[BoxData[ \(Table[ state[x\_1[t] - Cm[t] . {1, 0, 0}, y\_1[t] - Cm[t] . {0, 1, 0}, z\_1[t] - Cm[t] . {0, 0, 1}, x\_2[t] - Cm[t] . {1, 0, 0}, y\_2[t] - Cm[t] . {0, 1, 0}, z\_2[t] - Cm[t] . {0, 0, 1}, x\_3[t] - Cm[t] . {1, 0, 0}, y\_3[t] - Cm[t] . {0, 1, 0}, z\_3[t] - Cm[t] . {0, 0, 1}], {t, 0, 40, .5}]\)], "Input"], Cell["As viewed from another mass", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(state[x_, y_, z_]\ := \ Show[Graphics3D[ Polygon[{{x, y, z}, {x + 1, y, z}, {x + 1, y + 1, z}, {x + 1, y + 1, z + 1}, {x, y + 1, z + 1}, {x, y, z + 1}, {x, y, z}}]], DisplayFunction \[Rule] $DisplayFunction, AspectRatio \[Rule] 1, PlotRange \[Rule] {{\(-35\), 40}, {\(-30\), 40}, {\(-30\), 20}}]\)], "Input"], Cell[BoxData[ \(Table[ state[x\_2[t] - x\_1[t], y\_2[t] - y\_1[t], z\_2[t] - z\_1[t]], {t, 0, 40, .5}]\)], "Input"], Cell["\<\ How the energy oscillates w/ time. The kinetic energy is on top \ and the potential is on the bottom. Remember that the more negative the \ potential energy is the closer it is to the other object. (PE=0 an infinite distance away.)\ \>", "Text", Background->RGBColor[1, 1, 0.890196]], Cell[BoxData[ \(state[q_, w_, e_, r_]\ := \ Show[{Graphics[Circle[{q, w}, .1]], Graphics[Circle[{e, r}, .1]], Graphics[ Line[{{0, \((pe[0] + ke[0])\)/10000}, {t, \((pe[t] + ke[t])\)/ 10000}}]]}, DisplayFunction \[Rule] $DisplayFunction, PlotRange \[Rule] {{0, 40}, {\(-29\), 23}}, AspectRatio \[Rule] 1, \ Axes \[Rule] True, \ PlotLabel \[Rule] "\"]\)], "Input"], Cell[BoxData[ \(Table[ state[t, pe[t]/10000, t, ke[t]/10000], {t, 0, 40, .5}]\)], "Input"] }, Closed]], Cell["by Bill Ruane", "Section", CellDingbat->None, TextAlignment->Right] }, Open ]] }, FrontEndVersion->"4.0 for X", ScreenRectangle->{{0, 1024}, {0, 768}}, WindowSize->{1016, 695}, WindowMargins->{{0, Automatic}, {Automatic, 0}}, StyleDefinitions -> Notebook[{ Cell[CellGroupData[{ Cell["Style Definitions", "Subtitle"], Cell["\<\ Modify the definitions below to change the default appearance of \ all cells in a given style. 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LineSpacing->{1, 11}, LanguageCategory->"NaturalLanguage", CounterIncrements->"Title", CounterAssignments->{{"Section", 0}, {"Equation", 0}, {"Figure", 0}, { "Subtitle", 0}, {"Subsubtitle", 0}}, FontFamily->"Helvetica", FontSize->36, FontWeight->"Bold"], Cell[StyleData["Title", "Presentation"], CellMargins->{{24, 10}, {20, 40}}, LineSpacing->{1, 0}, FontSize->44], Cell[StyleData["Title", "Condensed"], CellMargins->{{8, 10}, {4, 8}}, FontSize->20], Cell[StyleData["Title", "Printout"], CellMargins->{{2, 10}, {12, 30}}, FontSize->24] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subtitle"], CellMargins->{{12, Inherited}, {20, 15}}, CellGroupingRules->{"TitleGrouping", 10}, PageBreakBelow->False, DefaultNewInlineCellStyle->"None", InputAutoReplacements->{"TeX"->StyleBox[ RowBox[ {"T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "LaTeX"->StyleBox[ RowBox[ {"L", StyleBox[ AdjustmentBox[ "A", BoxMargins -> {{-0.36, -0.1}, {0, -0}}, BoxBaselineShift -> -0.2], FontSize -> Smaller], "T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "mma"->"Mathematica", "Mma"->"Mathematica", "MMA"->"Mathematica"}, LanguageCategory->"NaturalLanguage", CounterIncrements->"Subtitle", CounterAssignments->{{"Section", 0}, {"Equation", 0}, {"Figure", 0}, { "Subsubtitle", 0}}, FontFamily->"Helvetica", FontSize->24], Cell[StyleData["Subtitle", "Presentation"], CellMargins->{{24, 10}, {20, 20}}, LineSpacing->{1, 0}, FontSize->36], Cell[StyleData["Subtitle", "Condensed"], CellMargins->{{8, 10}, {4, 4}}, FontSize->14], Cell[StyleData["Subtitle", "Printout"], CellMargins->{{2, 10}, {12, 8}}, FontSize->18] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsubtitle"], CellMargins->{{12, Inherited}, {20, 15}}, CellGroupingRules->{"TitleGrouping", 20}, PageBreakBelow->False, DefaultNewInlineCellStyle->"None", InputAutoReplacements->{"TeX"->StyleBox[ RowBox[ {"T", AdjustmentBox[ "E", BoxMargins 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Cell[StyleData["Subsection", "Presentation"], CellMargins->{{36, 10}, {11, 32}}, LineSpacing->{1, 0}, FontSize->22], Cell[StyleData["Subsection", "Condensed"], CellMargins->{{16, Inherited}, {6, 12}}, FontSize->12], Cell[StyleData["Subsection", "Printout"], CellMargins->{{9, 0}, {7, 22}}, FontSize->12] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsubsection"], CellDingbat->"\[FilledSmallSquare]", CellMargins->{{22, Inherited}, {8, 18}}, CellGroupingRules->{"SectionGrouping", 50}, PageBreakBelow->False, DefaultNewInlineCellStyle->"None", InputAutoReplacements->{"TeX"->StyleBox[ RowBox[ {"T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "LaTeX"->StyleBox[ RowBox[ {"L", StyleBox[ AdjustmentBox[ "A", BoxMargins -> {{-0.36, -0.1}, {0, -0}}, BoxBaselineShift -> -0.2], FontSize -> Smaller], "T", AdjustmentBox[ "E", BoxMargins -> {{-0.075, -0.085}, {0, 0}}, BoxBaselineShift -> 0.5], "X"}]], "mma"->"Mathematica", "Mma"->"Mathematica", 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BoxBaselineShift -> 0.5], "X"}]], "mma"->"Mathematica", "Mma"->"Mathematica", "MMA"->"Mathematica"}, Hyphenation->True, LineSpacing->{1, 3}, CounterIncrements->"Text"], Cell[StyleData["Text", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}, FontSize->16], Cell[StyleData["Text", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, LineSpacing->{1, 1}, FontSize->11], Cell[StyleData["Text", "Printout"], CellMargins->{{2, 2}, {6, 6}}, TextJustification->0.5, FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SmallText"], CellMargins->{{12, 10}, {6, 6}}, DefaultNewInlineCellStyle->"None", Hyphenation->True, LineSpacing->{1, 3}, LanguageCategory->"NaturalLanguage", CounterIncrements->"SmallText", FontFamily->"Helvetica", FontSize->9], Cell[StyleData["SmallText", "Presentation"], CellMargins->{{24, 10}, {8, 8}}, LineSpacing->{1, 5}, FontSize->12], Cell[StyleData["SmallText", "Condensed"], CellMargins->{{8, 10}, {5, 5}}, LineSpacing->{1, 2}, FontSize->9], Cell[StyleData["SmallText", "Printout"], CellMargins->{{2, 2}, {5, 5}}, TextJustification->0.5, FontSize->7] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Input/Output", "Section"], Cell["\<\ The cells in this section define styles used for input and output \ to the kernel. Be careful when modifying, renaming, or removing these \ styles, because the front end associates special meanings with these style \ names. Some attributes for these styles are actually set in FormatType Styles \ (in the last section of this stylesheet). \ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["Input"], CellMargins->{{45, 10}, {5, 7}}, Evaluatable->True, CellGroupingRules->"InputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, DefaultFormatType->DefaultInputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, AutoItalicWords->{}, LanguageCategory->"Formula", FormatType->InputForm, ShowStringCharacters->True, NumberMarks->True, LinebreakAdjustments->{0.85, 2, 10, 0, 1}, CounterIncrements->"Input", FontWeight->"Bold", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False, "StrikeThrough"->False, "Masked"->False, "CompatibilityType"->0, "RotationAngle"->0}], Cell[StyleData["Input", "Presentation"], CellMargins->{{72, Inherited}, {8, 10}}, LineSpacing->{1, 0}, FontSize->16], Cell[StyleData["Input", "Condensed"], CellMargins->{{40, 10}, {2, 3}}, FontSize->11], Cell[StyleData["Input", "Printout"], CellMargins->{{39, 0}, {4, 6}}, LinebreakAdjustments->{0.85, 2, 10, 1, 1}, FontSize->9] }, Closed]], Cell[StyleData["InputOnly"], Evaluatable->True, CellGroupingRules->"InputGrouping", CellHorizontalScrolling->True, DefaultFormatType->DefaultInputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, AutoItalicWords->{}, LanguageCategory->"Formula", FormatType->InputForm, ShowStringCharacters->True, NumberMarks->True, LinebreakAdjustments->{0.85, 2, 10, 0, 1}, CounterIncrements->"Input", StyleMenuListing->None, FontWeight->"Bold"], Cell[CellGroupData[{ Cell[StyleData["Output"], CellMargins->{{47, 10}, {7, 5}}, CellEditDuplicate->True, CellGroupingRules->"OutputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, DefaultFormatType->DefaultOutputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, AutoItalicWords->{}, LanguageCategory->"Formula", FormatType->InputForm, CounterIncrements->"Output"], Cell[StyleData["Output", "Presentation"], CellMargins->{{72, Inherited}, {10, 8}}, LineSpacing->{1, 0}, FontSize->16], Cell[StyleData["Output", "Condensed"], CellMargins->{{41, Inherited}, {3, 2}}, FontSize->11], Cell[StyleData["Output", "Printout"], CellMargins->{{39, 0}, {6, 4}}, FontSize->9] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Message"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"OutputGrouping", PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, DefaultFormatType->DefaultOutputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, AutoItalicWords->{}, FormatType->InputForm, CounterIncrements->"Message", StyleMenuListing->None, FontSize->11, FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Message", "Presentation"], CellMargins->{{72, Inherited}, {Inherited, Inherited}}, LineSpacing->{1, 0}, FontSize->16], Cell[StyleData["Message", "Condensed"], CellMargins->{{41, Inherited}, {Inherited, Inherited}}, FontSize->11], Cell[StyleData["Message", "Printout"], CellMargins->{{39, Inherited}, {Inherited, Inherited}}, FontSize->7, FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Print"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"OutputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, DefaultFormatType->DefaultOutputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, AutoItalicWords->{}, FormatType->InputForm, CounterIncrements->"Print", StyleMenuListing->None], Cell[StyleData["Print", "Presentation"], CellMargins->{{72, Inherited}, {Inherited, Inherited}}, LineSpacing->{1, 0}, FontSize->16], Cell[StyleData["Print", "Condensed"], CellMargins->{{41, Inherited}, {Inherited, Inherited}}, FontSize->11], Cell[StyleData["Print", "Printout"], CellMargins->{{39, Inherited}, {Inherited, Inherited}}, FontSize->8] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Graphics"], CellMargins->{{4, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"GraphicsGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, DefaultFormatType->DefaultOutputFormatType, LanguageCategory->None, FormatType->InputForm, CounterIncrements->"Graphics", ImageMargins->{{43, Inherited}, {Inherited, 0}}, StyleMenuListing->None, FontFamily->"Courier", FontSize->10], Cell[StyleData["Graphics", "Presentation"], ImageMargins->{{62, Inherited}, {Inherited, 0}}], Cell[StyleData["Graphics", "Condensed"], ImageMargins->{{38, Inherited}, {Inherited, 0}}, Magnification->0.6], Cell[StyleData["Graphics", 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FontSlant->"Plain"], Cell[StyleData["TI"], StyleMenuListing->None, FontFamily->"Times", FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["TB"], StyleMenuListing->None, FontFamily->"Times", FontWeight->"Bold", FontSlant->"Plain"], Cell[StyleData["TBI"], StyleMenuListing->None, FontFamily->"Times", FontWeight->"Bold", FontSlant->"Italic"], Cell[StyleData["MR"], HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, StyleMenuListing->None, FontFamily->"Courier", FontWeight->"Plain", FontSlant->"Plain"], Cell[StyleData["MO"], HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, StyleMenuListing->None, FontFamily->"Courier", FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["MB"], HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, StyleMenuListing->None, FontFamily->"Courier", FontWeight->"Bold", FontSlant->"Plain"], Cell[StyleData["MBO"], HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, StyleMenuListing->None, FontFamily->"Courier", FontWeight->"Bold", FontSlant->"Italic"], Cell[StyleData["SR"], StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Plain", FontSlant->"Plain"], Cell[StyleData["SO"], StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SB"], StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Bold", FontSlant->"Plain"], Cell[StyleData["SBO"], StyleMenuListing->None, FontFamily->"Helvetica", FontWeight->"Bold", FontSlant->"Italic"], Cell[CellGroupData[{ Cell[StyleData["SO10"], StyleMenuListing->None, FontFamily->"Helvetica", FontSize->10, FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SO10", "Printout"], StyleMenuListing->None, FontFamily->"Helvetica", FontSize->7, FontWeight->"Plain", FontSlant->"Italic"], Cell[StyleData["SO10", "EnhancedPrintout"], StyleMenuListing->None, FontFamily->"Futura", FontSize->7, FontWeight->"Plain", FontSlant->"Italic"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Formulas and Programming", "Section"], Cell[CellGroupData[{ Cell[StyleData["InlineFormula"], CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, LanguageCategory->"Formula", ScriptLevel->1, SingleLetterItalics->True], Cell[StyleData["InlineFormula", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}, FontSize->16], Cell[StyleData["InlineFormula", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, LineSpacing->{1, 1}, FontSize->11], Cell[StyleData["InlineFormula", "Printout"], CellMargins->{{2, 0}, {6, 6}}, FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["DisplayFormula"], CellMargins->{{42, Inherited}, {Inherited, Inherited}}, CellHorizontalScrolling->True, DefaultFormatType->DefaultInputFormatType, HyphenationOptions->{"HyphenationCharacter"->"\[Continuation]"}, LanguageCategory->"Formula", ScriptLevel->0, SingleLetterItalics->True, UnderoverscriptBoxOptions->{LimitsPositioning->True}], Cell[StyleData["DisplayFormula", "Presentation"], LineSpacing->{1, 5}, FontSize->16], Cell[StyleData["DisplayFormula", "Condensed"], LineSpacing->{1, 1}, FontSize->11], Cell[StyleData["DisplayFormula", "Printout"], FontSize->10] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Program"], CellFrame->{{0, 0}, {0.5, 0.5}}, CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, Hyphenation->False, LanguageCategory->"Formula", ScriptLevel->1, FontFamily->"Courier"], Cell[StyleData["Program", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}, FontSize->16], Cell[StyleData["Program", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, LineSpacing->{1, 1}, FontSize->11], Cell[StyleData["Program", "Printout"], CellMargins->{{2, 0}, {6, 6}}, FontSize->9] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Hyperlink Styles", "Section"], Cell["\<\ The cells below define styles useful for making hypertext \ ButtonBoxes. The \"Hyperlink\" style is for links within the same Notebook, \ or between Notebooks.\ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["Hyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2]}]&), Active->True, ButtonNote->ButtonData}], Cell[StyleData["Hyperlink", "Presentation"], FontSize->16], Cell[StyleData["Hyperlink", "Condensed"], FontSize->11], Cell[StyleData["Hyperlink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell["\<\ The following styles are for linking automatically to the on-line \ help system.\ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["MainBookLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "MainBook", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["MainBookLink", "Presentation"], FontSize->16], Cell[StyleData["MainBookLink", "Condensed"], FontSize->11], Cell[StyleData["MainBookLink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["AddOnsLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Courier", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "AddOns", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["AddOnsLink", "Presentation"], FontSize->16], Cell[StyleData["AddOnsLink", "Condensed"], FontSize->11], Cell[StyleData["AddOnsLink", "Printout"], FontSize->10, FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["RefGuideLink"], 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